Satisfying Rationality Postulates Of Structure Argumentation Through Deductive Support Technical Report

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--- Page 1 --- April 2026 Satisfying Rationality Postulates of Structured Argumentation Through Deductive Support – Technical Report Marcos CRAMER a and Tom FRIESE b,1 asecunet Security Networks AG bTU Dresden ORCiD ID: Marcos Cramer https://orcid.org/0000-0002-9461-1245 Abstract. ASPIC-style structured argumentation frameworks provide a formal ba- sis for reasoning in artificial intelligence by combining internal argument struc- ture with abstract argumentation semantics.

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A key challenge in these frame- works is ensuring compliance with five critical rationality postulates: closure, di- rect consistency, indirect consistency, non-interference, and crash-resistance. Re- cent approaches, including ASPIC⊖and Deductive ASPIC−, have made signifi- cant progress but fall short of meeting all postulates simultaneously under a cred- ulous semantics (e.g.

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preferred) in the presence of undercuts. This paper intro- duces Deductive ASPIC⊖, a novel framework that integrates gen-rebuttals from ASPIC⊖with the Joint Support Bipolar Argumentation Frameworks (JSBAFs) of Deductive ASPIC−, incorporating preferences.

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We show that Deductive ASPIC⊖ satisfies all five rationality postulates under a version of preferred semantics. This work opens new avenues for further research on robust and logically sound struc- tured argumentation systems.

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Keywords. Argumentation, Logics for Knowledge Representation, Nonmonotonic Reasoning, Preferences 1.

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Introduction Formal argumentation has become a fruitful field of research within AI [1]. Dung [2] introduced argumentation frameworks (AF), directed graphs where nodes represent ar- guments and edges represent attacks.

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Acceptance of arguments is decided by applying argumentation semantics to AFs. Two prominent semantics are grounded semantics and preferred semantics.

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In structured argumentation, a logical language builds the basis for the internal structure of arguments, which in turn gives rise to the attack relation. The result is an argumentation framework, to which argumentation semantics can be applied.

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We consider ASPIC-style frameworks for structured argumentation, in which ar- guments are built inductively from strict and defeasible inference rules [3,4]. Attacks in ASPIC can target conclusions of arguments (rebuttals) or applications of defeasible rules (undercuts).

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Preferences over the defeasible rules are used to decide which attacks 1Corresponding Author: Tom Friese, tom.friese@tu-dresden.de. arXiv:2604.21515v1 [cs.AI] 23 Apr 2026 --- Page 2 --- April 2026 result in defeats.

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Arguments and defeats are then interpreted as abstract argumentation frameworks and the acceptance of arguments is decided by argumentation semantics. There are variants of ASPIC, such as ASPIC+ and ASPIC−, that differ in some de- tails, e.g.

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the distinction between restricted and unrestricted rebuttals [4,5]. Restricted rebuttals can only target conclusions derived from defeasible rules, which can lead to counter-intuitive results when considering argumentation in a dialectical context, as dis- cussed by Caminada et al.

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[5]. On the other hand, unrestricted rebuttals can target con- clusions derived from either defeasible or strict rules, provided the attacked argument is defeasible.

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An even more generalized approach, gen-rebuttals, can target multiple sub- arguments simultaneously, providing a powerful mechanism for reasoning [6]. To ensure logical soundness and coherence, a robust argumentation framework should satisfy key rationality postulates.

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The postulates of closure, direct consistency and indirect consistency, which were introduced by Caminada and Amgoud [7], en- sure that the accepted arguments do not conflict with one another and that whenever the antecedents of a strict rule are accepted, its conclusion is also accepted. On the other hand, the postulates of non-interference and crash-resistance, introduced by Cam- inada et al.

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[8], ensure that, when combining two argumentation frameworks that do not overlap in the knowledge they model, neither of them influences the other. Meeting all five postulates under various semantics has proven challenging.

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Heyn- inck and Straßer’s ASPIC⊖[6] introduced gen-rebuttals and demonstrated compliance with all five postulates by ignoring undercuts and focusing on grounded semantics, which only has singular solutions. Cramer and Bhadra’s Deductive ASPIC−[9] addressed these issues using Joint Support Bipolar Argumentation Frameworks (JSBAFs), which track the application of strict rules to enforce closure.

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However, an error in one of their proofs means the closure postulate can be violated by using strict rules without antecedents. This work introduces Deductive ASPIC⊖, a novel framework combining the strengths of ASPIC⊖and Deductive ASPIC−.

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Our approach leverages gen-rebuttals and preference-enhanced JSBAFs to provide a unified solution that satisfies all five rational- ity postulates under a version of preferred semantics for JSBAFs and while considering preferences. What’s more, our preferred semantics aligns with that of Dung’s abstract argumentation framework.

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To our knowledge, this is the first ASPIC-style framework for structured argumentation that satisfies all five above-mentioned rationality postulates in a credulous semantics like preferred while considering both rebuttals and undercuts and while avoiding the downsides of restricted rebuttal. The rest of this paper is structured as follows: Section 2 outlines related versions of ASPIC.

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Section 3 explains key motivations for our semantics. Section 4 defines JSBAFs and their preferred semantics.

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Section 5 introduces Deductive ASPIC⊖and presents the results that preferred semantics of Deductive ASPIC⊖satisfies all five rationality postu- lates. Section 6 introduces our version of grounded semantics and contains proofs of the existence and uniqueness of grounded labelings.

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Section 7 gives an outlook on avenues for future work and Section 8 concludes this paper. 2.

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Related Work Modgil and Prakken [10] have shown that ASPIC+ (which uses restricted rebuttals and incorporates preferences) satisfies the rationality postulates of closure, direct consistency --- Page 3 --- April 2026 and indirect consistency if the strict rules are closed under transposition. Heyninck and Straßer [11] showed that ASPIC+ without undercuts and with a total preference relation additionally satisfies non-interference.

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In [5], the same three rationality postulates were studied for ASPIC−, which uses unrestricted rebuttals. The postulates are satisifed only under grounded semantics and the limiting assumptions of a total preference order and closure under transpositions.

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Cramer and Bhadra [9] propose Deductive ASPIC−which, just like Deductive AS- PIC⊖of the current paper, is based on Joint Support Bipolar Argumentation Frameworks (JSBAFs), that track not only attacks but also the support relation between antecedents and conclusions of strict rules. In their approach, JSBAFs are flattened into classical argumentation frameworks using auxiliary arguments.

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They claim their approach satis- fies closure, direct consistency, and indirect consistency under classical semantics, but their proof for Lemma 3.17 fails to account for strict rules with no antecedents. The current paper aims to address this issue and to also satisfy the rationality postulates of non-interference and crash-resistance, while incorporating preferences.

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Wu and Podlaszewski [12] defined ASPIC Lite, where the postulates for non- interference, crash resistance, closure and consistency are satisfied by deleting inconsis- tent arguments. The approach is not extended to preferences and they present a coun- terexample to the closure postulate with preferences lifted through the last-link principle.

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Heyninck and Straßer [6] introduce the notion of gen-rebuttals (which we also use in the current paper) as a generalization of the unrestricted rebut, giving rise to ASPIC⊖. Their version of ASPIC satisfies closure, consistency and non-interference un- der grounded semantics and when lifting preferences with the weakest-link principle.

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However, their approach does not consider attacks resulting from undercuts. To the best of our knowledge, there does not yet exist an ASPIC-style formalism that satisfies all five rationality postulates introduced in the introduction in a credulous semantics like preferred semantics while incorporating preferences and undercuts and avoiding the limitations of restricted rebuttals.

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The current paper aims at filling this gap. 3.

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Conceptual Motivation This paper introduces Deductive ASPIC⊖, a new ASPIC-style framework for structured argumentation. Like in other such frameworks, arguments are constructed from strict and defeasible rules.

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Strict rules encode deductive (i.e. exceptionless) reasoning patterns, whereas defeasible rules encode reasoning patterns that allow for exceptions.

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Like in other ASPIC-style frameworks, the acceptability of arguments in Deductive ASPIC⊖is determined based on the relation between arguments, with the difference that not only the attacks between arguments are considered, but also an additional deductive support, taken from [9]. A deductive support from argument set S to argument b encodes the information that b was constructed from S using a strict rule, so that the conclusion of b is a deductive consequence of the conclusions of the arguments in S.

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Note that there is a difference between an un-supported argument and an argument supported by the empty set. When the empty set supports an argument b, the conclusion of b deductively follows from the empty set of premises, i.e.

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it is a logical tautology. The principle of closure under strict rules says that if certain formulas φ1,...,φn are accepted and there is a strict rule φ1,...,φn →ψ, then also ψ should be accepted.

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In --- Page 4 --- April 2026 Deductive ASPIC⊖, if such a strict rule exists and there are arguments for the conclusions φ1,...,φn, they will deductively support an argument for the conclusion ψ. So all that is needed for a semantics to satisfy closure in Deductive ASPIC⊖is ensuring acceptance of a supported argument whenever all supporting arguments are accepted.

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As in abstract argumentation, attacks in our approach can cause rejection of argu- ments. But there is another reason to reject an argument, namely by supporting an argu- ment that for some other reason needs to be rejected.

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This relates to the principle of con- traposition and its generalization, the principle of transposition, in classical logic: When φ1,...,φn logically entail ψ, then φ1,...,φi−1,¬ψ,φi+1,...,φn logically entail ¬φi. We generalize this mechanism for rejecting arguments through supports to a mechanism that applies not only to arguments that are clearly rejected, but also to arguments with an un- decided acceptance status.

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This mechanism for propagating the non-acceptance of argu- ments through supports is the only function that the supports play in our argumentation semantics, and its enough to ensure that the closure principle is satisfied. Note that this mechanism for propagating the non-acceptance of arguments works in the opposite direction of the support direction, due to its connection to the principles of contraposition and transposition.

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Furthermore, this mechanism is meant to propagate the non-acceptance of arguments towards further arguments, but it is not meant to bring about the non-acceptance of arguments without some initiating attack. 4.

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Joint Support Bipolar Argumentation Frameworks 4.1. Syntax The core of our approach are Joint Support Bipolar Argumentation Framework (JSBAF) from Cramer and Bhadra [9], extended with preferences between arguments: Definition 1.

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A JSBAF is a tuple J = (A ,→,⇒,⪯), where: • A is a set of arguments. • →⊆A ×A is a set of attacks.

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• ⇒⊆2A ×A is a set of supports. • ⪯⊆A ×A is a total preference ordering.2 For (a,b) ∈→we say that a attacks b.

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If no a attacks b, we call b unattacked. For (S,b) ∈⇒we say that the set of arguments S supports the argument b.

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We will use JSBAFs to model arguments and their relations in Deductive ASPIC⊖(cf. Section 5.1).

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A support (S,b) will model the case that the argument b was constructed with a strict rule that derives the conclusion of b from the conclusions of the arguments in S. We define a support chain as a non-empty set of supports {(S0,b0),...,(Sn,bn)} ⊆⇒s.t.

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for each 0 ≤i ≤n −1 we have bi ∈Si+1. We recursively define an argument a as strict iff there is (S,a) ∈⇒and we either have S = /0 or for all b ∈S, b is strict.

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We denote the set of all strict arguments of a JSBAF J by STRJ . Strict arguments will model logical proofs of a tautology in Deductive ASPIC⊖.

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The preference ordering between arguments in a JSBAF will correspond to the preference ordering between arguments in an Argumentation System (cf. Section 5.1).

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2We use a ≺b as an abbreviation for a ⪯b and b ̸⪯a. --- Page 5 --- April 2026 In Deductive ASPIC⊖, an argument encodes one particular way of reaching a con- clusion from a finite set of premises.

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Strict arguments are special in that they are deriva- tions of tautologies, making them immune against counterarguments. To adequately cap- ture arguments and their relations as constructible in Deductive ASPIC⊖, we assume the following restrictions on JSBAFs: • For all support chains {(S0,b0),..., (Sn,bn)}, bn ̸∈S0.

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• For S,S′ ⊆A and b ∈A , (S,b),(S′,b) ∈⇒implies S = S′. • |S| < ∞for all (S,b) ∈⇒.

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• If a ∈A is strict, then it is unattacked. • If a,b ∈A are both strict, then a ⪯b and b ⪯a.

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• If a,b ∈A s.t. b is strict and a is not strict, then a ≺b.

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4.2. Semantics We use a labeling-based approach for our semantics.

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For a JSBAF J = (A ,→,⇒,⪯), a labeling L is a mapping L : A 7→{IN,OUT,UNDEC} with the usual meaning of labels: IN denotes accepted arguments, OUT denotes rejected arguments and UNDEC denotes arguments whose status is undecided. The sets in(L), out(L) and undec(L) denote all arguments labeled IN, OUT and UNDEC resprectively.

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Our semantics are based on legal labelings. We define whether an argument is legally IN, legally OUT or legally UNDEC given the labels of other arguments that are con- nected to it through attacks or supports.

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The key ideas for our definitions of legal label- ings are as follows: First, the preferred semantics for our version of ASPIC will be based on our preferred semantics for JSBAFs (cf. Section 5.1).

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Our ASPIC semantics needs to satisfy closure under strict rules (cf. Section 5.2) and we enforce this already on the level of our semantics for JSBAFs.

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To do this, we utilize the support relation ⇒: Because a support (S,b) models the application of a strict rule, satisfying closure under strict rules means that an argument supported only by accepted arguments also needs to be accepted, i.e. that for any labeling L and every support (S,b), S ⊆in(L) implies b ∈in(L).

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Secondly, our preferred semantics for JSBAFs should be an extension of the preferred semantics for AFs. More precisely, in the absence of supports, preferred semantics of JSBAFs and AFs should coincide.

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Thirdly, it is helpful to think of a labeling as being constructed it- eratively, where arguments labeled UNDEC have the potential to be labeled either OUT or IN, while the labels IN and OUT remain fixed. Lastly, preferences are only relevant when choosing between arguments which jointly support another one.

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For attacks, we will instead consider preferences on the level of our ASPIC formalism, where we delete attacks from weaker arguments to stronger ones, before applying our JSBAF semantics. Based on these key notions, we first motivate our definition of legally OUT: As in AFs, an argument is legally OUT if it is attacked by an accepted argument.

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However, in a JSBAF an argument can also be legally OUT due to a support: Suppose we have a sup- port (S,b) with a ∈S and a labeling in which b is rejected, while all arguments in S\{a} are accepted. Accepting all arguments in S violates closure, thus we need to reject a.

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However, we never want to reject an argument because of accepting a weaker one, so we also need to ensure that the other arguments in S are at least as strong as a. The consider- ations so far could motivate the following definition: An argument a is legally OUT iff it is attacked by an accepted argument or if it is contained in a set S that supports a rejected argument with all arguments in S \{a} being accepted and being at least as strong as a.

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--- Page 6 --- April 2026 However, this definition would cause problems in the case of an infinite support chain C = {(S0,b0),(S1,b1),...}, where each Si contains only a single argument, while every bi is unattacked.3 It would be consistent with the proposed definition to label all bi in this infinite chain OUT, although there is no attack originating from an accepted argument which contributes to this rejection of the bi. As explained in Section 3, this is not in- tended.

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Thus, in order to avoid anomalies caused by infinite support chains, we define an argument a as legally OUT iff it is attacked by an accepted argument, or if there is a finite support chain C = {(S0,b0),...,(Sn,bn)} where a ∈S0, bn is attacked by an accepted argument, each bi is rejected, S0 \ {a} as well as every set Si+1 \ {bi} contains only ac- cepted arguments, and S0\{a} contains no arguments weaker than a. Note that C may be part of a (possibly infinite) support chain C ′ = {(S0,b0),...,(Sn,bn),(Sn+1,bn+1),...} but the finite support chain C is required to reject a.

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Now for our definition of legally IN: As in AFs, for an argument a to be legally IN, all its attackers must be rejected. However, we also need to consider the supports (S,c) for which we have a ∈S.

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If c is accepted, than this support can never constitute a reason to label a OUT or UNDEC. However, if c is not accepted, then we need to ensure that at least one argument in S is also not accepted, in order to satisfy the closure postulate.

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So if c is not accepted and all elements of S \{a} are accepted, this constitutes a reason for a not to be legally IN. Similarly to the case of legally OUT, we never want to reject an argument because of accepting a weaker argument, so we need to add the additional constraint that none of the arguments in S \ {a} is weaker than a.

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These considerations motivate the following first approach for a definition: Given some labeling L, an argument a is legally IN w.r.t. L iff all its attackers are OUT and if there is no support (S,c) ∈⇒ with a ∈S, a ⪯b for all b ∈S \ {a}, c /∈in(L) and S \ {a} ⊆in(L).

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For the case that the supported argument c is labeled UNDEC, this approach works well, but for the case that c is labeled OUT, we have one last problematic case to consider. Suppose we have a labeling L and a support (S,c) with a ∈S, L(c) = OUT, S\in(L) = {a,b}, a ̸= b, a ⪯b, b ⪯a and L(a) = L(b) = UNDEC, i.e.

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the supported argument c is rejected, there are precisely two arguments in S which are not accepted and these two arguments are of equal strength and labeled UNDEC. In line with the above approach, a would be legally IN.

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However, since b is labeled UNDEC, it still has the potential to be labeled IN. But changing the label of b to IN means a is not legally IN anymore.

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In essence, considering one of the arguments a or b as legally IN in such a case deprives the other one of their potential to be labeled IN. Therefore, if the supported argument is labeled OUT while there are no OUT-labeled arguments in the supporting set, we require not one but at least two arguments in the supporting set to be labeled UNDEC in order for any other argument in the supporting set to be considered legally IN.

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We capture all these ideas in the following definition: Definition 2. For any labeling L, the argument a is: 1.

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legally IN w.r.t. L iff for all attackers b of a we have L(b) = OUT and for all supports (S,c) with a ∈S and a ⪯b for all b ∈S\{a}, one of a−c holds: (a) L(c) = IN, or (b) L(c) = UNDEC and there is b ∈S\{a} with L(b) ∈{OUT,UNDEC}, or 3Strict inference rules will be based on the consequence relation of an underlying logical language (cf.

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see Section 5.1), thus such a chain could correspond to a statement like φ ⊨φ ⊨.... --- Page 7 --- April 2026 (c) L(c) = OUT and there is either b ∈S \ {a} with L(b) = OUT or there are b1,b2 ∈S\{a} with b1 ̸= b2 and L(b1) = L(b2) = UNDEC.

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2. legally OUT w.r.t.

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L iff there exists (b,a) ∈→with L(b) = IN or a support chain C = {(S0,b0),...,(Sn,bn)} ⊆⇒s.t. all of a−e hold: (a) a ∈S0 and S0 \{a} ⊆in(L) (b) (c,bn) ∈→for some c with L(c) = IN (c) for 0 ≤i ≤n, bi ∈out(L) (d) for 0 < i ≤n, Si \{bi−1} ⊆in(L) (e) a ⪯d for all d ∈S\{a} 3.

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legally UNDEC w.r.t. L iff a is neither legally IN nor legally OUT w.r.t.

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L Admissible and preferred labelings are based on legal labelings, the idea that strict arguments should always be labeled IN and the standard notion from labeling-based argumentation semantics that preferred labelings are subset maximal: Definition 3. Let J be a JSBAF and L a labeling of J .

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L is an admissible labeling iff: • a ∈in(L) implies a is legally IN w.r.t. L.

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• a ∈out(L) iff a is legally OUT w.r.t. L.

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• If a is a strict argument, then L(a) = IN. L is a preferred labeling of J iff L is a maximal (w.r.t.

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sub-set inclusion of in(L)) admis- sible labeling. We denote the set of all admissible labelings of J by adm(J ) and the set of all preferred labelings of J by pr(J ).

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Below, we give an example JSBAF. Here, nodes represent arguments, single arrows represent attacks between arguments and double arrows represent supports.

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Example 1. JSBAF J1 a b b c d e J1 has one attack, from b to b, the supports, (/0,a), ({a},b), ({c,d,e},b) and (/0,d), and the strict arguments a, b and d.

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Assume the following preference ordering between the arguments of J1, by which strict arguments are preferred to non-strict arguments, but there are no other preferences: ⪯= {a,b,b,c,d,e}2 \ {a,b,d}×{b,c,e}
. Let us first consider the admissible labelings of J1: Since a,b and d are strict ar- guments, they are labeled IN in every admissible labeling.

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The argument a is unattacked and only supports the accepted argument b, therefore a is legally IN. The argument b is unattacked and doesn’t support any argument, therefore b is also legally IN.

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The argu- ment d is unattacked and for the support ({c,d,e},b), we have d ̸⪯c and d ̸⪯e. Thus we don’t have to consider this support when checking the legality of the label of d, meaning d is also legally IN.

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Lastly, because b is attacked by b, it is legally OUT in every admissi- ble labeling and thus needs to be labeled OUT. Labeling c and d UNDEC results in a first admissible labeling L1 with in(L1) = {a,b,d}, out(L1) = {b} and undec(L1) = {c,e}.

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--- Page 8 --- April 2026 Now for the remaining admissible labelings: In order to label c or e OUT, they need to be legally OUT. Both c and e are unattacked – therefore they cannot be legally OUT due to an attack – but they are contained in the support chain  ({c,d,e},b) .

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Note that c ⪯d, c ⪯e, e ⪯c and e ⪯d. Thus, for this support chain, c is legally OUT if e is labeled IN and e is legally OUT if c is labeled IN.

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On the other hand, because d is always labeled IN in an admissible labeling, if e is labeled OUT, then c is legally IN and if c is labeled OUT, then e is legally IN. This results in the remaining two admissible labelings L2 and L3 with in(L2) = {a,b,c,d}, out(L2) = {b,e} and undec(L2) = /0, while in(L3) = {a,b,d,e}, out(L3) = {b,c} and undec(L3) = /0.

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L1 is not preferred, because in(L1) ⊊in(L2). Since neither of in(L2) and in(L3) is a subset of the other, both of them are preferred labelings, i.e.

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pr(J1) = {L2,L3}. 4.3.

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Existence of admissible and preferred labelings In this section, we show that there always exists at least one admissilbe and at least one preferred labeling for every JSBAF, even in cases with an infinite amount of arguments. 4.3.1.

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Existence of admissible labelings We first show that for every JSBAF, there exists at least one admissible labeling. To this end, we define a family of labelings which labels precisely the strict arguments of a JSBAF as IN, labels all arguments OUT that are legally OUT as a result from accepting the strict arguments and labels all the remaining arguments UNDEC.

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We call this the strict including minimal labeling (SIM): Definition 4. For any JSBAF J = (A ,→,⇒,⪯), the labeling SIMJ is defined as: 1.

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in(SIMJ ) = STRJ 2. O0 =  a ∈A | (b,a) ∈→,b ∈in(SIMJ ) On+1 = On ∪  a ∈A | (S,b) ∈⇒,a ∈S,b ∈On, S\{a} ⊆in(SIMJ ) out(SIMJ ) = S i≥0 Oi 3.

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undec(SIMJ ) = A \ in(SIMJ )∪out(SIMJ )
We now prove that SIMJ is an admissible labeling. To make the proof more acces- sible, we have divided it into several parts.

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We show (in this order) that SIMJ is a label- ing, that the accepted arguments are legally IN, that the rejected arguments are exactly those which are legally OUT and, finally, that SIMJ is an admissible labeling. Proposition 1.

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Let J = (A ,→,⇒,⪯) be a JSBAF. Then SIMJ is a labeling.

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Proof. From the definition of undec(SIMJ ), it is clear that every argument gets as- signed some label and that undec(SIMJ ) ∩in(SIMJ ) = /0 as well as undec(SIMJ ) ∩ out(SIMJ ) = /0.

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Thus, we have left to show that in(SIMJ )∩out(SIMJ ) = /0 also holds. For this, we will prove by induction over n ∈N that in(SIMJ )∩On = /0.

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Induction start n = 0: By construction of SIMJ , if a ∈in(SIMJ ), then a is strict and therefore unattacked in J . If a ∈O0, then there exists (b,a) ∈→, which contradicts that a is unattacked.

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We conclude that in(SIMJ )∩O0 = /0 holds. --- Page 9 --- April 2026 Inductin step n →n + 1: We concentrate on O′ = On+1 \ On.

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Suppose there is a ∈in(SIMJ )∩O′. Because a ∈in(SIMJ ), we know that a is strict.

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By a ∈O′, we know there is (S,b) ∈Supp with a ∈S, b ∈On and S\{a} ⊆in(SIMJ ) = STRJ . This means that all arguments in S are strict.

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Now b must also be strict, therefore b ∈in(SIMJ ) by construction of SIMJ . But then b ∈in(SIMJ )∩On, contradicting the induction hy- pothesis in(SIMJ ) ∩On = /0.

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We conclude that in(SIMJ ) ∩On+1 = /0 holds, therefore in(SIMJ )∩out(SIMJ ) = /0 as required. Proposition 2.

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Let J = (A ,→,⇒,⪯) be a JSBAF and a ∈A an argument. If SIMJ (a) = IN, then a is legally IN w.r.t.

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SIMJ . Proof.

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By construction of in(SIMJ ), a is strict and therefore unattacked. Now consider some support (S,c) with a ∈S.

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By the restrictions for JSBAFs mentioned in Section 4.1, if a ⪯b for all b ∈S\{a}, then all arguments in S must be strict. From this we can infer that c is also a strict argument and therefore c ∈in(SIMJ ) by construction of SIMJ .

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Thus a is legally IN w.r.t. SIMJ as required.

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Proposition 3. Let J = (A ,→,⇒,⪯) be a JSBAF and a ∈A an argument.

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We have SIMJ (a) = OUT iff a is legally OUT w.r.t. SIMJ .

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Proof. →: SIMJ (a) = OUT implies there is On s.t.

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a ∈On. We show by induction over n ∈N that, if a ∈On, then a is legally OUT w.r.t.

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SIMJ . Induction start n = 0: By construction of SIMJ , a ∈O0 implies there is b ∈ in(SIMJ ) s.t.

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(b,a) ∈→. This means a is legally OUT, as required.

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Induction step n →n + 1: We focus on O′ = On+1 \ On. By construction of SIMJ we know there is (S,b) ∈⇒with a ∈S, b ∈On and S \ {a} ⊆in(SIMJ ) = STRJ .

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Note that, by the restrictions on JSBAFs mentioned in Section 4.1, S \ {a} ⊆STRJ implies a ≺d for all d ∈S \ {a}. By the induction hypothesis, b is legally OUT w.r.t.

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SIMJ . Suppose first that b is legally OUT because of an attack (c,b) ∈→.

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Then {(S,b)} is a support chain s.t. a ∈S, (c,b) ∈→with c ∈in(SIMJ ), b ∈out(SIMJ ), S \ {a} ⊆in(SIMJ ) and a ⪯d for all d ∈S \ {a}.

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Thus a is legally OUT w.r.t. SIMJ .

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Next, suppose that b is legally OUT because of a support chain {(S0,b0),...,(Sn,bn)} which satisfies the conditions of item two of Definition 2. Then we construct the support chain {(S,b),(S0,b0),...,(Sn,bn)} which satisfies those same conditions w.r.t.

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a ∈S. By Definition 2, a is now legally OUT w.r.t.

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SIMJ , as required. We conclude that a ∈out(SIMJ ) implies a is legally OUT w.r.t.

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SIMJ . ←: Suppose first that a is legally OUT w.r.t.

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SIMJ because of an attack (b,a) ∈→ with b ∈in(SIMJ ). By construction of SIMJ , we infer that b must be a strict argument, thus a ∈O0 ⊆out(SIMJ ).

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Next, suppose that a is legally OUT w.r.t. SIMJ due to a support chain {(S0,b0),...,(Sn,bn)}.

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Then in particular a ∈S0, b0 ∈out(SIMJ ) and S0 \ {a} ⊆in(SIMJ ). By construction of SIMJ , there needs to be some m ∈N s.t.

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b0 ∈Om. Now a ∈Om+1 ⊆out(SIMJ ) as required.

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Lemma 1. Let J = (A ,→,⇒,⪯) be a JSBAF.

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Then SIMJ is an admissible labeling. Proof.

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By Proposition 1 we know that SIMJ is a labeling. By Proposition 2 we know that all a ∈in(SIMJ ) are legally IN w.r.t.

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SIMJ . By Proposition 3 we know that a ∈ out(SIMJ ) iff a is legally OUT w.r.t.

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SIMJ . Finally, STRJ ⊆in(SIMJ ) is trivially satisfied by construction of SIMJ .

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--- Page 10 --- April 2026 Corollary 1. Let J = (A ,→,⇒,⪯) be a JSBAF.

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Then we have adm(J ) ̸= /0. 4.3.2.

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Existence of preferred labelings To prove that every JSBAF has at least one preferred labeling, we have left to show that there always exists a maximal (w.r.t. sub-set inclusion) admissible labeling.

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For JSBAFs with an infinite amount of arguments, it is not immediately clear that this is always the case. Our proof will rely on the Lemma of Zorn.4 We begin by defining the following partial order on admissible labelings: Definition 5.

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Let J = (A ,→,⇒,⪯) be a JSBAF. We define the relation ≤L ⊆ adm(J )×adm(J ) as follows: L1 ≤L L2 iff in(L1) ⊆in(L2) and out(L1) ⊆out(L2).

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It is easy to verify that ≤L is a partial order and we omit a proof. Based on this relation, we now define the supremum w.r.t.

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a chain of admissible labelings: Definition 6. Let J = (A ,→,⇒,⪯) be a JSBAF.

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Furthermore, let K ⊆adm(J ) be a chain of admissible labelings w.r.t. ≤L.

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We define the supremum of K , denoted by LK , as follows: • in(LK ) = S L∈K in(L) • out(LK ) = S L∈K out(L) • undec(LK ) = A \ in(LK )∪out(LK )
We now show that this labeling LK is an admissible labeling. To make it more accessible, we have split the proof in several parts and show (in this order) that LK is a labeling, that the accepted arguments are legally IN, that the rejected arguments are exactly those which are legally OUT and, finally, that LK is an admissible labeling.

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Proposition 4. Let J = (A ,→,⇒,⪯) be a JSBAF.

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Furthermore, let K ⊆adm(J ) be a chain of admissible labelings w.r.t. ≤L and let LK be the supremum of K .

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Then LK is a labeling. Proof.

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It is clear from the definition of LK that every argument gets assigned some label and that undec(LK ) ∩in(LK ) = /0 as well as undec(LK ) ∩out(LK ) = /0. Thus, we have left to show that in(LK )∩out(LK ) = /0 also holds.

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Towards a contradiction, let a ∈in(LK ) ∩out(LK ). By construction of LK , this means there are L1,L2 ∈K with a ∈in(L1) and a ∈out(L2).

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Because K was a chain, we know that either L1 ≤L L2 or L2 ≤L L1 holds. First, lets assume L1 ≤L L2: Then a ∈in(L1) ⊆in(L2).

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Now a is labeled IN and OUT by L2, contradicting that L2 is a labeling. Next, assume L2 ≤L L1: Then a ∈out(L2) ⊆ out(L1).

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Now a is labeled OUT and IN by L1, contradicting that L1 is a labeling. We conclude that in(LK )∩out(LK ) = /0 as required.

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Proposition 5. Let J = (A ,→,⇒,⪯) be a JSBAF.

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Furthermore, let K ⊆adm(J ) be a chain of admissible labelings w.r.t. ≤L and let LK be the supremum of K .

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If a ∈in(LK ), then a is legally IN w.r.t. LK .

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4Zorn’s Lemma states that, if X is a partially ordered set s.t. every chain in X has an upper bound, then X contains a maximal element.

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--- Page 11 --- April 2026 Proof. Let a ∈in(LK ).

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We first show that any attacker of a is labeled OUT: Let L ∈K s.t. L(a) = IN.

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By construction of LK , such a labeling L must exist. Now assume that there is b ∈A s.t.

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(b,a) ∈→. Because L is an admissible labeling, we have L(b) = OUT.

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By construction of LK , we can now infer LK (b) = OUT as required. Now take some support (S,b) ∈⇒with a ∈S and a ⪯c for all c ∈S\{a}.

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We have to show that the conditions of item one of Definition 2 are satisfied. If LK (b) = IN, then this trivially holds.

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Thus we assume LK (b) ̸= IN. We first show that this implies S ̸⊆in(LK ): Towards a contradiction suppose that this is not the case.

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Let S = {a0,...,am} and let LIN = {L0,..,Lm} ⊆K s.t. ai ∈in(Li) for all 0 ≤i ≤m.

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We know that S is finite (by the restrictions mentioned in Section 4.1). Because K is a chain w.r.t.

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≤L, we know that there is Lmax ∈LIN s.t. Li ≤L Lmax for all 0 ≤i ≤m.

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Now we have in(Li) ⊆in(Lmax) for all 0 ≤i ≤m. This means S ⊆in(Lmax).

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We know that Lmax is an admissible labeling. Thus, for a ∈S in particular, we have that a is legally IN w.r.t.

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Lmax. Now we can infer that Lmax(b) = IN must hold.

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By construction of LK , this implies LK (b) = IN, contradicting our assumption LK (b) ̸= IN. Now for the remaining options of LK (b): We first consider the case LK (b) = UNDEC.

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We have argued that S ⊆in(LK ) cannot hold. This means there is at least one c ∈S with LK (c) = OUT or LK (c) = UNDEC, as required.

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Lastly, lets consider the case LK (b) = OUT: As before, we know S ̸⊆in(LK ). Let SIN = {a0,...,am} = S ∩in(LK ) and let LIN = {L0,...,Lm} ⊆K s.t.

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ai ∈Li for all 0 ≤i ≤m. Furthermore, let LOUT ∈K s.t.

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LOUT(b) = OUT. We take Lmax ∈LIN∪{LOUT} s.t.

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Li ≤L Lmax and LOUT ≤L Lmax. Then SIN ⊆in(Lmax) and b ∈out(Lmax).

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Now consider the set S′ = S\SIN: We either have |S′| = 1 or |S′| > 1. Assume first that S′ = {c}.

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Then we can infer from the admissibility of Lmax, that c ∈out(Lmax) ⊆out(LK ) must hold. Now a is legally IN w.r.t.

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LK , as required. Next, assume that |S′| > 1.

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Then there either exists an argument c ∈S′ s.t. c ∈out(Lmax) ⊆out(LK ), or there are arguments c1,c2 ∈S′ s.t.

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c1 ̸= c2 and c1,c2 ∈undec(Lmax). In the first case, we can immediately infer that a is legally IN w.r.t.

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LK . In the second case, we either have c1,c2 ∈undec(LK ), or {c1,c2}∩out(LK ) ̸= /0.

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Either way, we can again infer that a is legally IN w.r.t. LK , as required.

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Proposition 6. Let J = (A ,→,⇒,⪯) be a JSBAF.

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Furthermore, let K ⊆adm(J ) be a chain of admissible labelings w.r.t. ≤L and let LK be the supremum of K .

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Then for any a ∈A , a ∈out(LK ) iff a is legally OUT w.r.t. LK .

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Proof. →: Take a ∈out(LK ).

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By construction of LK , there exists L ∈K s.t. a ∈out(L).

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As L was an admissible labeling, we know that a is legally OUT w.r.t. L.

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Regardless of whether a is legally OUT w.r.t. L due to an attack or due to a support chain, we can use in(L) ⊆in(LK ) and out(L) ⊆out(LK ) to infer that a is legally OUT w.r.t.

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LK due to the same attacker or support chain. ←: Suppose first that a is legally OUT w.r.t.

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LK because of an attack, i.e. there is (b,a) ∈→with b ∈in(LK ).

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Let L ∈K s.t. b ∈in(L).

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As L is an admissible labeling, we can infer a ∈out(L) and by construction of LK we have a ∈out(LK ) as required. Next, suppose that a is legally OUT w.r.t.

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LK because of a support chain {(S0,b0),...,(Sn,bn)} with a ∈S0. Then in particular b0 ∈out(LK ) and for S0 \ {a} = {a0,...,am}, we have {a0,...,am} ⊆in(LK ).

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By construction of LK, we again have a set of labelings LIN = {L0,...,Lm} ⊆K s.t. ai ∈in(Li) and we have a labeling LOUT ∈K s.t.

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b ∈out(LOUT). We again take the maximal labeling Lmax ⊆LIN ∪{LOUT} for which we have Li ≤L Lmax --- Page 12 --- April 2026 for all Li ∈LIN, as well as LOUT ≤L Lmax.

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Because Lmax is admissible, we can again infer that a ∈out(Lmax), otherwise none of the ai would be legally IN w.r.t. Lmax.

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By construction of LK we now have a ∈out(LK ) as required. Proposition 7.

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Let J = (A ,→,⇒,⪯) be a JSBAF. Furthermore, let K ⊆adm(J ) be a chain of admissible labelings w.r.t.

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≤L and let LK be the supremum of K . Then LK is an admissible labeling.

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Proof. From Proposition 4 we know that LK is a labeling of J .

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By Proposition 5 we know that every a ∈in(LK ) is legally IN w.r.t. LK .

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By Proposition 6 we know that a ∈out(LK ) iff a is legally OUT w.r.t. LK .

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Finally, because all labelings L ∈K are admissible labelings, we can infer that STRJ ⊆in(L) holds. By construction of LK , we now have STRJ ⊆in(LK ).

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We conclude that LabK is an admissible labeling. With the construction of LabK , we can now argue that a maximal admissible label- ing exists even for JSBAFs with an infinite number of arguments.

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This means that for each JSBAF, there is always at least one preferred labeling: Theorem 1. Let J be a JSBAF.

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Then pr(J ) ̸= /0. Proof.

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By Corollary 1 we know that adm(J ) is non-empty. The elements of adm(J ) form a partially ordered set w.r.t.

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≤L. Now assume that K ⊆adm(J ) is a chain of admissible labelings.

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We take the supremum of K as constructed in Definition 6, i.e. LK .

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By Proposition 7, we know that LK is itself an admissible labeling. From the con- struction of LK it is clear that LK is an upper bound of K w.r.t.

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≤L. By Zorn’s Lemma we can now infer that adm(J ) contains a maximal element Lmax.

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Since Lmax is maximal w.r.t. ≤L, there is no admissible labeling L′ ∈adm(J ) s.t.

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in(Lmax) ⊂in(L′). Therefore Lmax is a preferred labeling.

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5. Deductive ASPIC⊖ 5.1.

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DeductiveASPIC⊖and JSBAFs We are assuming some underlying logical language for which we have a set of well- founded formulas F and a function Atoms : F 7→2F mapping formulas of F to the atomic formulas occurring in them.5 We say that formulas φ,ψ ∈F are syntactically disjoint, written φ||ψ, iff Atoms(φ) ∩Atoms(ψ) = /0. Similarly, we say that sets of for- mulas Γ,∆⊆F are syntactically disjoint, written Γ||∆, iff φ||ψ for all φ ∈Γ and ψ ∈∆.

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We do not make any assumptions regarding the specific syntax of the logical language we are considering, but we assume that it is closed under negation ¬ and conjunction ∧. As a shorthand notation, we use V{φ0,...,φn} to abbreviate φ0 ∧··· ∧φn and we write φ = −ψ to indicate that either φ = ¬ψ or ¬φ = ψ.

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To define the strict rules of our version of ASPIC, we make a few assumptions on the semantics of the underlying logical language: First, we assume a set of interpretations I and some model-relation ⊨M for which ¬ and ∧behave in the usual way, i.e. for any 5One can think of Propositional Logic or First Order Logic to get an idea of the logical languages we want to cover.

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--- Page 13 --- April 2026 interpretation I, we have I ⊨M φ iff not I ⊨M ¬φ, and we have I ⊨M φ ∧ψ iff I ⊨M φ and I ⊨M ψ. Secondly, we assume that for any two sets of formulas Γ and ∆with Γ||∆, if there exist interpretations I1,I2 s.t.

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I1 ⊨M φ for all φ ∈Γ and I2 ⊨M ψ for all ψ ∈∆, then there exists some interpretation I for which we have I ⊨M φ for all φ ∈Γ and I ⊨M ψ for all ψ ∈∆. We call a set of formulas Γ satisfiable, iff there exists an interpretation I s.t.

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I ⊨M φ for all φ ∈Γ. As an abbreviation, we write I ⊨M Γ to indicate I ⊨M φ for all φ ∈Γ.

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We call a set of formulas Γ tautological, iff for all I ∈I we have I ⊨M Γ. Lastly, we assume that there are no unsatisfiable or tautological atoms, i.e.

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for all φ ∈F, for all ψ ∈Atoms(φ), there exist I1, I2 s.t. I1 ⊨M ψ and I2 ⊨M ¬ψ.

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Using this model-relation, we now define a consequence relation ⊨C ⊆2F ×F as follows: Γ ⊨C ψ iff for all I ∈I , I ⊨M Γ implies I ⊨M ψ. 6 Deductive ASPIC⊖now uses such an underlying logical language as a basis.

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Definition 7. Let F be a set of well-founded formulas and let ⊨C be a consequence rela- tion associated with F.

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An Argumentation System (AS) according to Deductive ASPIC⊖ is a tuple AS = (Rs,Rd,n,≤r), where: • Rs = RAX s ∪R⊨ s is a set of strict rules consisting of axiomatic rules RAX s = {→φ | φ ∈AX} for some satisfiable set of axioms AX ⊆F and consequence-based rules R⊨ s =  φ0,...,φm−1 →φm | φ0,...,φm ∈F and {φ0,...,φm−1} ⊨C φm . • Rd is a set of defeasible rules of the form φ0,...,φm−1 ⇒φm where φi ∈F for all 0 ≤i ≤m.

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• n : Rd ⇀F is a partial function called naming function. • ≤r ⊆Rd ×Rd is a total preorder called preferences order.

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In cases where the distinction between strict and defeasible rules is not relevant, we will use ⇝and mean by that some rule r ∈Rs ∪Rd. Arguments based on AS are defined recursively: a is an argument with conclusion aC = φ iff a is of the form a : a0,...,am → φ, where a0,...,am are arguments based on AS and aC 0 ,...,aC m →φ is a strict rule, or if a is of the form a : a0,...,am ⇒φ, where a0,...,am are arguments based on AS and aC 0 ,...,aC m ⇒φ is a defeasible rule.

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We say aC 0 ,...,aC m →φ (respectively aC 0 ,...,aC m ⇒φ) is the top-rule of a, denoted TR(a). The defeasible rules of a are DR(a) = DR(a0) ∪ ···∪DR(am)∪ {TR(a)}∩Rd
and the sub-arguments of a are Sub(a) = Sub(a0)∪···∪ Sub(am) ∪{a}.

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We denote the arguments that can be constructed based on some AS by A (AS). We call a defeasible if DR(a) ̸= /0 and strict otherwise.

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We call AS inconsistent if there are strict arguments a,b ∈A (AS) s.t. aC = −bC and consistent otherwise.

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In this paper we only consider consistent AS. For a set of arguments A, we define its conclusions as AC = {aC | a ∈A}.

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We call an argument a inconsistent if there is Γ ⊆Sub(a)C s.t. Γ is unsatisfiable, otherwise we call a consistent.

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We say that a undercuts b iff there is b′ ∈Sub(b) with TR(b′) = r ∈Rd and aC = −n(r). We use the notion of gen-rebuts taken from Heyninck and Straßer [6] and say that a gen-rebuts b iff b is defeasible and there is Γ ⊆Sub(b)C s.t.

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aC = ¬VΓ. We consider the elitist weakest link principle to lift preferences between defeasible rules to preferences over arguments, denoted by the relation ⪯ewl ⊆A (AS)×A (AS).7 If arguments a and b are both strict, then we define a ⪯ewl b and b ⪯ewl a, otherwise a ⪯ewl b iff ∃ra ∈DR(a) 6It is easy to see that this consequence relation is monotone, i.e.

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Γ ⊨C φ implies Γ∪{ψ} ⊨C φ. 7We write a ≺ewl b to abbreviate a ⪯ewl b and b ̸⪯ewl a.

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--- Page 14 --- April 2026 s.t. ∀rb ∈DR(b), ra ≤r rb.

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Note that ⪯ewl is a total pre-order between arguments. We say that a defeats b if a undercuts b or if a gen-rebuts b and a ̸≺ewl b.

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Semantics of Deductive ASPIC⊖are defined via JSBAFs with the translation: Definition 8. Let AS = (Rs,Rd,n,≤r) be some AS.

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Then the JSBAF corresponding to AS is J = (A ,→,⇒,⪯) with: • A = A (AS) • (a,b) ∈→iff a defeats b. • (S,b) ∈⇒iff S = {b0,...,bm} and b is of the form b : b0,...,bm →bC • a ⪯b iff a ⪯ewl b Note that the JSBAFs we can construct based on some AS satisfy the additional constraints on JSBAFs that we introduced in Section 4 after Definition 1.

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The preferred semantics of Deductive ASPIC⊖are now based on the preferred semantics of JSBAFs: Definition 9. Let AS = (Rs,Rd,n,≤r) be some AS and J the corresponding JSBAF.

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E ⊆A (AS) is a preferred extension of AS iff E = in(L) for some L ∈pr(J ). The set of all preferred extensions of AS is pr(AS).

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The preferred conclusions of AS are Cpr(AS) = {EC | E ∈pr(AS)} Note that pr(AS) is a set of sets of arguments and Cpr(AS) is a set of sets of formulas. Below, we give an example for an AS and a translation to a JSBAF.

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For the sake of finiteness of the resulting framework, we don’t base the strict rules on the consequence relation of some underlying logical language, but rather give here a specific set of strict rules. Nevertheless, the results presented in this paper are of course also valid for argu- mentation systems with an infinite number of arguments.

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Example 2. Argumentation System AS1 = (Rs,Rd,n,≤): • Rs =  rs1 :→α; rs2 : α →¬(γ ∧δ ∧ε); rs3 :→δ; rs4 : γ,δ,ε →(γ ∧δ ∧ε) • Rd = {rd1 :⇒γ; rd2 :⇒ε} • n = /0 and rd1 ≤rd2 ≤rd1 Based on AS1 we construct the arguments a :→α, b : a →¬(γ ∧δ ∧ε), c :⇒γ, d :→δ, e :⇒ε and b : c,d,e →(γ ∧δ ∧ε).

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We have ⪯ewl = {a,b,b,c,d,e}2\ {a,b,d}× {b,c,e}
. Argument b gen-rebuts b on its conclusion, but b does not gen-rebut b since b is strict.

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Thus b defeats b. The JSBAF corresponding to AS1 is J1 of Example 1.

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In this JSBAF, the strict rules rs1, rs2, rs3 and rs4 are translated to the supports (/0,a), ({a},b), ({c,d,e},b) and (/0,d) respectively. We have pr(AS1) =  {a,b,c,d},{a,b,d,e} 5.2.

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Rationality Postulates The first three postulates we want to cover are those of closure, direct consistency and indirect consistency, which were defined by Caminada and Amgoud [7]. They state that no extension should contain contradictory arguments (consistency) and for any strict rule r, if all antecedents of r are accepted, then the conclusion of r should also be accepted (closure).

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For Deductive ASPIC⊖we define these postulates as follows: Definition 10. A Deductive ASPIC⊖semantics σ satisfies: --- Page 15 --- April 2026 • Direct consistency iff for all AS, for all E ∈σ(AS), for all φ,ψ ∈EC, φ ̸= −ψ.

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• Closure iff for all AS and for all E ∈σ(AS), EC = CLRs(EC), where for any set S, CLRs(S) is the smallest set s.t. S ⊆CLRs(S) and for all r ∈Rs, if r is of the form r : φ0,...φm →ψ and φ0,...,φm ∈CLRs(S), then ψ ∈CLRs(S).

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• Indirect consistency iff it satisfies closure and direct consistency. We can show preferred semantics of Deductive ASPIC⊖satisfies these postulates.

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Theorem 2. Preferred semantics of Deductive ASPIC⊖satisfies closure, direct consis- tency and indirect consistency.

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Proof. Let AS = (Rs,Rd,n,≤r) be some AS and let J = (A ,→,⇒,⪯) be the JSBAF corresponding to AS according to Deductive ASPIC⊖.

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We begin with the closure postulate: Let E ∈pr(AS) be some preferred extension of AS and let L ∈pr(J ) be the labeling that corresponds to E, i.e. E = in(L).

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Note that we have L ∈adm(J ). Assume that there is a strict rule r = φ0,...,φn →ψ for which we have φ0,...,φn ∈EC.

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Then there are arguments a0,...,an ∈E s.t. for each 0 ≤i ≤n, we have aC i = φi.

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Because of the strict rule r, there now also exists an argument a : a0,...,an →ψ. By construction of J from AS we know that there is a support {a0,...,an},a
.

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Because a0,...,an ∈E = in(L) and by admissibility of L, we infer that a ∈E must also hold. Therefore we have ψ ∈EC as required.

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Now for direct consistency: As before, let E ∈pr(AS) be some preferred extension of AS and let L ∈pr(J ) be the labeling that corresponds to E. Towards a contradiction, assume that there are φ,ψ ∈EC s.t.

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φ = −ψ. Then we have arguments a,b ∈E s.t.

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aC = φ and bC = ψ. Because AS is consistent, at least one of these arguments has to be defeasible.

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W.l.o.g. we assume that b is defeasible, meaning a gen-rebuts b.

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Now the argument a can be strict or not strict. If a is strict, DR(a) = /0 and we have a ̸⪯ewl b, meaning a ̸≺ewl b.

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This means a defeats b and by construction of J from AS we have (a,b) ∈→. Now a,b ∈in(L) contradicts the admissibility of L.

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On the other hand, if a is not strict, then both a and b gen-rebut each other on their respective conclusions. Furthermore, either a ̸≺ewl b or b ̸≺ewl a must hold, because otherwise we have a ⪯ewl b and b ̸⪯ewl a, while simultaneously b ⪯ewl a and a ̸⪯ewl b.

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Thus either (a,b) ∈→or (b,a) ∈→, which again means a,b ∈in(L) cannot hold by admissibility of L. Lastly, indirect consistency under preferred semantics follows directly from the fact that Deductive ASPIC⊖satisfies closure and direct consistency under preferred seman- tics.

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The remaining postulates are those of non-interference and crash-resistance. They were first defined by Caminada et al.

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[8] and cover the behavior of two distinct AS when they are combined to a single AS. We begin by introducing some additional notation.

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Given two AS, AS1 = (Rs1,Rd1,n1,≤r1) and AS2 = (Rs2,Rd2,n2,≤r2), we define a union of AS1 and AS2 as AS+ = (R+ s ,R+ d ,n+,≤+ r ), where Rs+ = RAX1 s ∪RAX2 s ∪R⊨ s , Rd+ = Rd1 ∪Rd2, n+ = n1 ∪n2 and ≤+ r ⊆R+ d × R+ d is a total pre-order s.t. ≤r1 ⊆≤+ r and ≤r2 ⊆≤+ r .

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Given any rule r ∈RAX s ∪Rd of the form r = φ0,...,φm−1 ⇝φm, we define Atoms(r) = S 0≤i≤m Atoms(φi). The atoms of the defeasible and axiomatic rules of an AS are Atoms(Rd) = S r∈Rd Atoms(r) and Atoms(RAX s ) = S r∈RAX s Atoms(r) re- spectively.

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For the name of a defeasible rule r, the atoms occurring in n(r) = φ --- Page 16 --- April 2026 are Atoms n(r)
= Atoms(φ). The atoms occurring in the naming function n are Atoms(n) = S r∈R′ Atoms n(r)
, where R′ ⊆Rd is the set of defeasible rules r for which n(r) is defined.

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Now the atoms of an AS are defined as Atoms(AS) = Atoms(Rd) ∪ Atoms(RAX s )∪Atoms(n). We say AS1 and AS2 are syntactically disjoint, written AS1||AS2, iff Atoms(AS1)||Atoms(AS2).

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Next, we define the atoms of an argument a inductively: Let a be of the form a : a0,...,an ⇝φ. If TR(a) ∈RAX s , then Atoms(a) = Atoms(φ).

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If TR(a) ∈R⊨ s , then Atoms(a) = /0∪Atoms(a0)∪···∪Atoms(an). Lastly, if TR(a) = r ∈Rd, then Atoms(a) = Atoms(a0)∪Atoms(aC 0 )
∪···∪ Atoms(an)∪Atoms(aC n )
∪Atoms(φ).

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Given some set of sets of formulas Γ ⊆2F , and a set of atoms ∆⊆Atoms(F), the restriction of Γ to ∆is defined as the set Γ|∆=  Γ′|∆| Γ′ ∈Γ,Γ′|∆= {φ ∈Γ′ | Atoms(φ) ⊆ ∆} . The nestedness of this definition is due to the fact that Γ is a set of sets of formulas (not simply a set of formulas).

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Non-interference states that, when combining AS1 with a syntactically disjoint AS2, AS1 does not influence the behavior of a semantics σ w.r.t. the atoms of AS2 and crash- resistance states that AS1 does not render the σ-consequences of AS2 irrelevant.

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Definition 11. A Deductive ASPIC⊖semantics σ satisfies: • Non-interference iff for any AS1, AS2 with AS1||AS2, Cσ(ASi) | Atoms(ASi) = Cσ(AS+)|Atoms(ASi), where i ∈{1,2} and AS+ is a union of AS1 and AS2.

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• Crash-resistance iff there is no contaminating AS1 for σ, where AS1 is contami- nating iff Atoms(AS1) ⊊Atoms(F) and for any AS2 with AS1||AS2, Cσ(AS1) = Cσ(AS+) with AS+ being a union of AS1 and AS2. Caminada et al.

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[8] showed that a logical formalism that satisfies non-interference also satisfies crash-resistance, given that the formalism in question is non-trivial. To this end, we first define non-triviality in the context of Deductive ASPIC⊖: Definition 12.

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Let F be some set of well-founded formulas and let ⊨C be a conse- quence relation associated with F. Furthermore, let Γ ⊆Atoms S φ∈F Atoms(φ)
be a non-empty set of atoms.

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We say that Deductive ASPIC⊖is non-trivial under semantics σ iff there exist AS1 = (Rs1,Rd1,n1,≤r1) and AS2 = (Rs2,Rd2,n2,≤r2), s.t. AS1||AS2, Γ = Atoms(AS1) = Atoms(AS2) and we have Cσ(AS1)|Γ ̸= Cσ(AS2)|Γ.

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Next, we show that Deductive ASPIC⊖under preferred semantics satisfies non- triviality. For this, we have adapted a proof of Wu and Podlaszewski [12]: Proposition 8.

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Deductive ASPIC⊖is non-trivial under preferred semantics. Proof.

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Let F be a set of well-founded formulas, let ⊨C be a consequence-relation as- sociated with F and let Γ ⊆ S φ∈F Atoms(φ) be a non-empty set of atoms. We define AS1 = (Rs1,Rd1,n1,≤r1) and AS2 = (Rs2,Rd2,n2,≤r2) as follows: AX1 = AX2 = /0 and n1 = n2 = /0 while for Γ = {φ0,...,φn}, Rd1 = {⇒φ0;...;⇒φn} and Rd2 = {φ0 ⇒ φ0;...;φn ⇒φn}.

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Furthermore, we define ≤r1 = Rd1 ×Rd2 and ≤r2 = Rd2 ×Rd2. We first note that for any φi ∈Γ, φi is neither unsatisfiable nor tautological.

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From this, we can infer two things: Firstly, for no φi ∈Γ is φi or ¬φi the consequence of a --- Page 17 --- April 2026 strict argument and secondly, no elements φi,φ j ∈Γ with φi ̸= φj can contradict each other. It is obvious that the first claim holds.

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To show the second claim, assume towards a contradiction that it does not hold. Then there exists interpretations I1,I2 s.t.

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I1 ⊨M φi, I2 ⊨M φj. Because φi ̸= φ j, φi and φj are syntactically disjoint.

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Therefore, there exists an interpretation I s.t. I ⊨M φi and I ⊨M φ j, but now I ⊨M φ and I ⊨M ¬φ, a contradiction.

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Using these two claims, we can infer that in AS1 there do not exist strict arguments ai with aiC = ¬φi and in AS2, there do not exist strict arguments ai with aC i = φi for 0 ≤i ≤n. We conclude: There exists Γ1 ∈Cpr(AS1)|A with {φ1,...,φn} ⊆Γ1 while there is no Γ2 ∈Cpr(AS2)|A s.t.

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{φ1,...,φn} ⊆Γ2. Therefore Cpr(AS1)|A ̸= Cpr(AS2)|A as required.

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Based on this notion of non-triviality, we now show that, if Deductive ASPIC⊖sat- isfies non-interference under preferred semantics, then it also satisfies crash resistance under preferred semantics. We adapt a proof from Wu and Podlaszewski [12] for this: Proposition 9.

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Let F be a set of well-founded formulas and let ⊨C be a consequence relation associated with F. If Deductive ASPIC⊖satisfies non-interference under pre- ferred semantics, then Deductive ASPIC⊖satisfies crash-resistance under preferred se- mantics.

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Proof. Towards a contradiction, assume that the claim does not hold.

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Then there ex- ists some AS = (Rs,Rd,n,≤r) s.t. Atoms(AS) ⊊ S φ∈F Atoms(φ) and AS is contami- nating under preferred semantics.

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Let Γ = S φ∈F Atoms(φ)
\ Atoms(AS). By Propo- sition 8, we know that Deductive ASPIC⊖is non-trivial.

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Thus there exist two fur- ther AS, AS1 = (Rs1,Rd1,n1,≤r1) and AS2 = (Rs2,Rd2,n2,≤r2), for which we have Γ = Atoms(AS1) = Atoms(AS2) and Cpr(AS1)| Γ ̸= Cpr(AS2)| Γ. Note that we have Atoms(AS1) ∩Atoms(AS) = /0 and Atoms(AS2) ∩Atoms(AS) = /0.

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Let AS12 be a union of AS1 and AS2, AS1∗be a union of AS1 and AS and let AS∗2 be a union of AS2 and AS. Because Deductive ASPIC⊖satisfies non-interference under preferred semantics by assumption, we have Cpr(AS1)|Γ = Cpr(AS1∗)|Γ and Cpr(AS2)|Γ = Cpr(AS∗2)|Γ.

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However, because AS is contaminating, we now have Cpr(AS)|Γ = Cpr(AS1∗)|Γ and Cpr(AS)|Γ = Cpr(AS∗2)|Γ. Now we have Cpr(AS1)|Γ = Cpr(AS2)|Γ, contradicting our assumption that Deductive ASPIC⊖is non-trivial.

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These proofs utilize three key insights: First, it is sufficient to concentrate on four specific “edge cases”, which can – in a simplified view – be specified as follows by reference to the argument set A ′ + = A+ \ (Ai ∪Aj): An argument a ∈Aj attacking an argument b ∈Ai. An argument a ∈A ′ + attacking an argument b ∈Ai.

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An argument a ∈Ai attacking an argument b ∈A ′ +, where b is supported only by arguments in Ai and Aj (i.e. not by arguments in A ′ +).

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An argument a ∈A ′ + attacking an argument b ∈A ′ +, where b is supported only by arguments in Ai and Aj. For better understanding, we have depicted these “edge cases” in Illustration 1 be- low.

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In this illustration, the JSBAFs resulting from ASi and AS j are depicted with clouds on the left and right, while the space between the dashed lines shows the arguments in A′ +. The arrows labeled one to four correspond to the cases described above.

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The second key insight is that, when considering these “edge cases”, we can further restrict ourselves to reduced versions of arguments w.r.t. either AS1 or AS2.

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Essentially, a --- Page 18 --- April 2026 reduced version of a w.r.t. ASi can be created by “removing” all applications of defeasible or axiomatic rules from AS j during the construction of a.

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Because reduced versions of arguments have only atoms of ASi, they are contained in Ji. Furthermore, we can show that for any preferred labeling L, if a′ is legally OUT w.r.t.

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L then a must be legally OUT w.r.t. L and if a is legally IN w.r.t.

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L then a′ must be legally IN w.r.t. L.

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Using this insight, we can make statements about arguments a ∈A+ \ (Ai ∪Aj) by considering their reduced versions a′. The last key insight is that we can utilize the support relation between arguments and our definition of legal labelings: For an attack (a,b) ∈→that is the result of a gen-rebut, we first construct a new argument a′.

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This argument a′ will only have the argument a as its direct sub-argument and the conclusion of a′ will be the negation of the conjunction of all conclusions of sub-arguments of b, which are created via axiomatic or defeasible rules. Since DR(a) = DR(a′), we also have (a′,b) ∈→.

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Accepting a would then mean a′ also needs to be accepted, whereas rejecting a′ would mean a also needs to be rejected. The conclusion of a′ has only atoms which are contained in ASi or AS j, thus for our purposes the attack originating from a′ is easier to deal with.

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Illustration 1. ASi AS j a b 1.

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2. 3.

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4. By combining all of these key insights we can show: Theorem 3.

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Preferred semantics of Deductive ASPIC⊖satisfies non-interference and crash-resistance. In the remainder of this section, we will prove Theorem 3.

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Note that from here on out, we will somewhat interweave an AS with its JSBAF counterpart. For exam- ple, we will talk about “syntactically disjoint JSBAFs J1 and J2”, by which we mean that J1 and J2 are JSBAFs that correspond to some AS1 and AS2 s.t.

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AS1 and AS2 are syntactically disjoint. Similarly, for a given JSBAF J = (A ,→,⇒,⪯), --- Page 19 --- April 2026 and an argument a ∈A , we will talk about the “sub-arguments of a”, by which we mean the arguments Sub(a) ⊆A (AS) for AS being the AS that J corresponds to.

Chunk 325

We will always assume a consistent naming between an AS and its JSBAF counter- part, i.e. for some AS1 = (Rs1,Rd1,n1,≤r1), the corresponding JSBAF will always be J1 = (A1,→1,⇒1,⪯1), while for AS2 = (Rs2,Rd2,n2,≤r2), the corresponding JSBAF will be J2 = (A2,→1,⇒2,⪯2) and for AS+ = (R+ s ,R+ d ,n+,≤+ r ) the corresponding JS- BAF will be J+ = (A+,→+,⇒+,⪯+).

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5.3. Non-Interference and Crash-Resistance for Preferred Semantics 5.3.1.

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Axiomatic and defeasible sub-arguments In order to prove Theorem 3, we begin by taking a more detailed look at the third key insight mentioned in our proof overview. Remember that, for a given attack (a,b) ∈→ which results from a gen-rebut, we wanted to leverage our definition of the support- relation between arguments and our notion of legal labelings, by constructing new argu- ments a′ and b′.

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The argument a′ will have the form a′ : a →¬VΓ, while b′ will have the form b′ : b0,...,bm →VΓ, where b0,...bm are precisely those sub-arguments of b which were created by using an axiomatic or defeasible top-rule and Γ is the set of all their conclusions. For this, we use the following definition: Definition 13.

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Let AS1 = (Rs1,Rd1,n1,≤r1) and AS2 = (Rs2,Rd2,n2,≤r2) be two AS s.t. AS1||AS2.

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Furthermore, let AS+ = (R+ s ,R+ d ,n+,≤+ r ) be the union of AS1 and AS2. The mapping ADSub : A (AS+) →2A (AS+) maps arguments to their axiomatic and defeasible sub-arguments.

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For any a ∈A (AS+) we define: ADSub(a) =  a′ ∈Sub(a) | TR(a′) ∈RAX+ s ∪Rd+ . We define the restriction of ADSub to ASi (i ∈{1,2}) as: ADSub(a)|ASi =  a′ ∈ADSub(a) | Atoms(a′C) ⊆Atoms(ASi) .

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With some abuse of notation, we will also use the mapping ADSub for sets of argu- ments. More precisely, for some S ⊆A (AS+), we define ADSub(S) = S a∈S ADSub(a).

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Next, we show that for any argument a, any conclusion of a sub-argument a′ of a can be derived from ADSub(a)C. Proposition 10.

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Let AS = (Rs,Rd,n,≤r) be some AS and let a be some argument. For every non-empty ∆⊆Sub(a)C we have ADSub(a)C ⊨C ψ for all ψ ∈∆.

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Proof. We show the claim via structured induction over the construction of the argument a.

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Let Γ = ADSub(a)C and let ∆⊆Sub(a)C be non-empty. Base case: Let a be of the form a :⇝φ.

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Then ∆= {φ}. If we have TR(a) ∈RAX s ∪Rd, then Γ = {φ}.

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Obviously {φ} ⊨C φ, therefore the claim holds. On the other hand, if TR(a) ∈R⊨ s , then we have Γ = /0 and by the way that arguments are constructed, /0 ⊨C φ.

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Thus we again have Γ ⊨C ψ for all ψ ∈∆. Induction step: Let a be of the form a : a0,...,am ⇝φ.

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For each 0 ≤i ≤m, let Γi = S a′∈ADSub(ai) a′C. Furthermore, let ∆i = ∆∩Sub(ai)C.

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Then we have ∆̸= S i∈{0,...,m} ∆i iff ∆= {φ} ∪ S i∈{0,...,m} ∆i. Similarly, we have Γ ̸= S i∈{0,...,m} Γi iff Γ = {φ} ∪ S i∈{0,...,m} Γi.

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Take an arbitrary ψ ∈∆. We have to show Γ ⊨C ψ.

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--- Page 20 --- April 2026 First, we assume that there is some 0 ≤i ≤m, s.t. ψ ∈∆i.

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By our induction hypothe- sis, we now have Γi ⊨C ψ. Since Γi ⊆Γ, we can infer Γ ⊨C ψ by monotonicity of ⊨C.

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Now assume that there is no such i ∈{0,...,m}. Then we must have ∆= {ψ} ∪ S i∈{0,...,m} ∆i, meaning ψ = φ.

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Now we can either have φ ∈Γ or φ ̸∈Γ, depending on whether or not TR(a) ∈Rd ∪RAX s (φ ∈Γ) or TR(a) ∈R⊨ s (φ ̸∈Γ). Suppose first that φ ∈Γ.

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Then trivially Γ ⊨C φ = ψ. On the other hand, if φ ̸∈Γ, then we have {aC 0 ,...,aC m} ⊨C φ by the way that arguments are constructed.

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For all 0 ≤i ≤m, we have Γi ⊨C aC i by the induction hypothesis. Because Γi ⊆Γ, we now also have Γ ⊨C aC i for each ai.

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By transitivity of ⊨C, we again infer Γ ⊨C φ = ψ, as required. In particular, we can use this proposition to infer that the conclusion of any argument is entailed by the conclusions of its axiomatic and defeasible sub-arguments: Corollary 2.

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Let AS = (Rs,Rd,n,≤r) be some AS and let a be some argument. Then ADSub(a)C ⊨C aC.

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Next, we show that, if we have a gen-rebut from an argument a to an argument b, then there exists an argument a′ with only a as a direct sub-argument, s.t. a′ gen-rebuts b on ADSub(b): Proposition 11.

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Let AS = (Rs,Rd,n,≤r) be an AS and let a,b be arguments s.t. a gen- rebuts b.

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Then there exists a′ of the form a′ : a →¬VADSub(b)C. Proof.

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Since a is gen-rebutting b by assumption, we know that aC = ¬V{φ0,...,φm} for {φ0,...,φm} ⊆Sub(b)C. We only have to show that ¬V{φ0,...,φm} ⊨C ¬VADSub(b)C.

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Towards a contradiction, assume that this is not the case. Then there exists an interpre- tation I s.t.

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I ⊨M ¬V{φ0,...,φm} and I ⊨M VADSub(b)C. By Proposition 10 we can in- fer that I ⊨M V{φ0,...,φm} also holds, contradicting I ⊨M ¬V{φ0,...,φm}.

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Therefore the argument a′ can be constructed as claimed and gen-rebuts b on ADSub(b)C. 5.3.2.

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Closure under sub-arguments of preferred labelings The next step will be to show that preferred labelings are closed under sub-arguments, which will be very helpful in our later proofs. For this, we will first introduce the notion of a propagated labeling, which can be obtained from a labeling L and an argument a by adding a to in(L) and propagating the effect of this acceptance throughout the JSBAF.

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For this propagation, we need to ensure two things: Fristly, if there is a support (S,b) where all arguments in S are accepted, then b is also accepted. Secondly, if there is a support (S,b) where b is rejected and all arguments in S except for one are accepted, this last remaining argument is rejected.

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Note that we will not refer to preferences between arguments in our definition. Instead, we will later use this method of propagation only on admissible labelings.

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We will show that under certain conditions, admissibility can be retained with this construction, even though we do not explicitly account for preferences between arguments. Definition 14.

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Let J = (A ,→,⇒,⪯) be a JSBAF, let L be a labeling of J and let a ∈A be an argument. We define the propagated labeling of L and a, denoted La, as follows: --- Page 21 --- April 2026 I0 = in(L)∪{a} Ik+1 = Ik ∪{b ∈A | ∃(S,b) ∈⇒,S ⊆Ik} in(La) = [ k≥0 Ik O0 = {b ∈A | ∃(c,b) ∈→,c ∈in(La)} Ok+1 = Ok ∪{b ∈A | ∃(S,c) ∈⇒,b ∈S, S\{b} ⊆in(La),c ∈Ok} out(La) = [ k≥0 Ok undec(La) = A \ in(La)∪out(La)
Note that, because supports (S,b) ∈⇒are based on the (transitive) entailment SC ⊨C bC, for every argument in Ik for k ≥2, there exists an “equivalent” argument in I1.

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This is easy to see with the following proof: Proposition 12. Let J = (A ,→,⇒,⪯) be a JSBAF, let L be labeling of J , let a ∈A be an argument and let La be the propagated labeling of L and a as constructed in Definition 14.

Chunk 364

If x ∈Ik \ I0, then there exists x′ ∈I1 \ I0 s.t. xC = x′C and ADSub(x) = ADSub(x′).

Chunk 365

Proof. We show the claim via structured induction over n ∈N for x ∈In \ I0.

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Induction start, n = 1: In this case the claim trivially holds. Induction step, n →n + 1: Let x be of the form x : x0,...,xm →xC.

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By the in- duction hypothesis, there is x′ i for each 0 ≤i ≤m s.t. x′ i ∈I1 \ I0, xC i = x′C i and ADSub(xi) = ADSub(x′ i).

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Let x′ i be of the form x′ i : x′ i1,...,x′ ik →x′C i . We define the sets Si as Si = {x′ i1,...,x′ ik} and the set S′ as S′ = S 0≤i≤m Si.

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By transitivity of ⊨C, we have S′C ⊨C xC, i.e. there exists the argument x′ of the form x′ : x′′ 1,...,x′′ l →xC, where {x′′ 1,...x′′ l } = S′.

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Obviously we have xC = x′C. By the induction hypothesis, we have ADSub(xi) = ADSub(x′ i) for each xi.

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From this, we can infer that ADSub(x) = ADSub(x′) also holds. By the induction hypothesis we also have x′ i ∈I1 \I0 for each x′ i.

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From this we can infer that S′ ⊆in(L)∪{a} must hold. By construction of La, we now have x′ ∈I1 \I0 as required.

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Next, we show that propagated labelings will be admissible if three conditions are met: First, we need to start with an already admissible labeling L. Secondly, for the argument a that we choose as the “starting point” for our propagation, we need L(a) ̸= OUT and for all attackers b of a we need to have L(b) = OUT.

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Lastly, we need to ensure that there does not exist a support chain C =  (S0,b0),...,(Sn,bn) starting at a ∈S0 s.t. we have Si \ {bi−1} ⊆in(L), while there exists (c,bn) ∈→with L(c) ̸= OUT.

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We will later ensure that these conditions are always satisfied when creating a propagated labeling La. To make the actual proof more accessible, we have split it into several parts.

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We begin by showing that La is a labeling: --- Page 22 --- April 2026 Proposition 13. Let J = (A ,→,⇒,⪯) be a JSBAF, let L ∈adm(J ) be an admissible labeling of J and let a ∈A be an argument.

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Furthermore, assume that L(a) ̸= OUT and that for all attackers b of a, we have L(b) = OUT. Lastly, assume that there does not exist a support chain  (S0,b0),...,(Sn,bn) ⊆⇒and an attack (c,bn) ∈→with a ∈S0, L(c) ̸= OUT, S0 \{a} ⊆in(L) and Si \{bi−1} ⊆in(L) for 0 < i ≤n.

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Then La is a labeling. Proof.

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It is easy to see that each argument gets some label and that undec(La)∩in(La) = undec(La)∩out(La) = /0. Therefore, we have left to show that in(La)∩out(La) = /0 also holds.

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Towards a contradiction, assume that there is an argument x ∈out(La) ∩in(La). We show the contradiction by induction over n ∈N for x ∈On.

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Induction start, n = 0: This means there exists some (y,x) ∈→with y ∈in(La). We make two case distinctions, based on the origins of x ∈in(La) and y ∈in(Ly): We have x ∈in(La) either because x ∈I0 or because x ∈Ik for some k ≥1.

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Similarly, we have y ∈in(La) either because y ∈I0 or because y ∈in(Ik) for some k ≥1. Assume first that x ∈I0 and y ∈I0 holds: If we have x ∈I0 because x ∈in(L), then we can use admissibility of L to infer that L(y) = OUT.

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On the other hand, if we have x ∈I0 because x = a, then we can use our assumption that all attackers of a are labeled OUT in L to infer L(y) = OUT. Either way, we have y ∈out(L)∩I0, a contradiction.

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Now assume that x ∈I0 and y ∈Ik for some k ≥1: As we have dealt with the case y ∈I0 above, we assume y ∈Ik \I0. By Proposition 12 there exists an argument y′ ∈I1 \I0 s.t.

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y′C = yC and ADSub(y′) = ADSub(y). Because y′ ∈I1 \ I0, we can infer from the construction of I1 that there exists a support (S,y′) ∈⇒with a ∈S and S \{a} ⊆in(L).

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Similar to the case of y above, we have L(y′) = OUT, because all attackers of x are labeled OUT in L and y′ attacks x. Now we have the support (S,y′) with L(y′) = OUT and S\{a} ⊆in(L).

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Because L was an admissible labeling, we can now infer that L(a) = OUT must hold, contradicting our assumption L(a) ̸= OUT. Next, let us consider the cases where x ∈Ik for some k ≥1.

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We assume x ̸∈I0 as we have dealt with this case before. We again use Proposition 12 to infer that there exists an argument x′ ∈I1 \I0 s.t.

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ADSub(x) = ADSub(x′) and xC = x′C. Next, consider the attack (y,x): If (y,x) ∈→because y undercuts x, then y also undercuts x′.

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On the other hand, if y gen-rebuts x, then we can use Proposition 11 to infer the existence of an argument ey : y →¬VADSub(xC). Because ADSub(x) = ADSub(x′), we now also infer (ey,x′) ∈→.

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Now, let us consider the origin of y ∈in(La) again: We either have y ∈I0 or y ∈Ik for some k ≥1. Assume first that y ∈I0, i.e.

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y ∈in(L) or y = a. Note that, because x′ ∈I1 \ I0, there must exist a support (Sx,x′) ∈⇒s.t.

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a ∈Sx and Sx \ {a} ⊆in(L). By our assumption on the support chains starting at a, for this support chain  (Sx,x′) ⊆⇒, we must have L(y) = OUT (if y undercuts x) or L(ey) = OUT (if y gen-rebuts x), which in turn implies L(y) = OUT by admissibility of L and construction of ey.

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Either way, we can infer L(y) = OUT, which contradicts y ∈in(L)∪{a}. Now, assume that y ∈Ik \ I0 for some k ≥1.

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We use Proposition 12 again to infer: If y undercuts x, then there exists an argument y′ ∈I1 \I0 s.t. y′C = yC and ADSub(y′) = ADSub(y) and if y gen-rebuts x, then there exists an argument ey′ ∈I1 \I0 s.t.

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ey′C = eyC and ADSub(ey′) = ADSub(ey). Note that, because y′ ∈I1 \I0 (respectively ey′ ∈I1 \I0), we can infer that there exists a support (Sy,y′) ∈⇒(respectively (Sy,ey′) ∈⇒) with a ∈Sy and Sy\{a} ⊆in(L).

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Because ADSub(y′) = ADsub(y) (respectively ADSub(ey′) = ADSub(ey)), and y′C = yC (respectively ey′C = eyC), we can again infer that (y′,x′) ∈→(respectively (ey′,x′) ∈→) holds. By the assumptions made on the support chains starting at a, we can now again infer that L(y′) = OUT (respectively L(ey′) = OUT).

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However, for the --- Page 23 --- April 2026 supporting set Sy we can now use the admissibility of L and the fact that Sy \ {a} ⊆ in(L) to infer L(a) = OUT, which contradicts our original assumption L(a) ̸= OUT. This concludes the induction start.

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Induction step, n ⇝n + 1: By the induction hypothesis we have On ∩in(La) = /0, therefore we restrict ourselves to On+1\On. Assume that x ∈On+1\On.

Chunk 400

This means there exists some support (S,c) ∈⇒, with x ∈S, S\{x} ⊆in(La) and c ∈On. By assumption, x ∈in(La) also holds.

Chunk 401

Now we can infer that there must be some k ∈N for which we have S ⊆Ik. This implies c ∈Ik+1 ⊆in(La) by construction of in(La).

Chunk 402

Now c ∈On ∩in(La), which contradicts the induction hypothesis. Next, we show that arguments accepted in La are also legally accepted: Proposition 14.

Chunk 403

Let J = (A ,→,⇒,⪯) be a JSBAF, let L ∈adm(J ) be an admissible labeling of J and let a ∈A be an argument. Furthermore, assume that L(a) ̸= OUT and that for all attackers b of a, we have L(b) = OUT.

Chunk 404

Lastly, assume that there does not exist a support chain  (S0,b0),...,(Sn,bn) ⊆⇒and an attack (c,bn) ∈→with a ∈S0, L(c) ̸= OUT, S0 \{a} ⊆in(L) and Si \{bi−1} ⊆in(L) for 0 < i ≤n. If x ∈in(La), then x is legally IN w.r.t.

Chunk 405

La. Proof.

Chunk 406

First, let us consider the attackers of x: Assume that x ∈in(La) and (y,x) ∈→. If x ∈I0, then we either have x ∈in(L) or x = a.

Chunk 407

In both cases, we know L(y) = OUT. Since in(L) ⊆in(La), we can infer that La(y) = OUT holds due to the same attack or support chain that lead to L(y) = OUT.

Chunk 408

Now assume x ∈Ik \I0 for some k ≥1. By Proposition 12, there exists x′ ∈I1 \ I0 s.t.

Chunk 409

x′C = xC and ADSub(x′) = ADSub(x). As for the proof that showed La is a labeling, we again want to point out that x′ ∈I1 \I0 implies the existence of a support (S,x′) ∈⇒with a ∈S and S\{a} ⊆in(L).

Chunk 410

Now for the attack (y,x) ∈→: If y undercuts x, then (y,x′) ∈→also holds. On the other hand, if y gen-rebuts x, then we can again use Proposition 11 to infer the existence of an argument ey : y →¬VADSub(x)C and because ADSub(x) = ADSub(x′) and DR(x) = DR(x′), we have (ey,x′) ∈→.

Chunk 411

Since x ∈I1 and by the assumptions made on support-chains starting at a, we can infer: If y undercuts x, then L(y) = OUT must thold. On the other hand, if y gen-rebuts x, then L(ey) = OUT must hold and by admissibility of L this implies L(y) = OUT.

Chunk 412

In both cases, we have L(y) = OUT and can again infer La(y) = OUT due to the same attack or support chain that lead to L(y) = OUT. Lastly, for the supports (S,b) ∈⇒with x ∈S: By construction of in(La), we cannot have S ⊆in(La) while b ̸∈in(La).

Chunk 413

Furthermore, if |S\in(La)| = 1 and La(b) = OUT, then S\in(La) ⊆out(La) by construction of out(La). From this we can infer that, if x ⪯c for all c ∈S\{x}, then one of the items 1.a to 1.c of Definition 2 holds, as required.

Chunk 414

Before we continue with the actual proof, we show a small auxiliary statement which tells us that, if we have an admissible labeling L and two arguments a and a′, where the defeasible rules and the conclusions of sub-arguments of a′ are also defeasible rules and conclusions of sub-arguments of a, then L(a′) = OUT implies L(a) = OUT: Proposition 15. Let J = (A ,→,⇒,⪯) be a JSBAF and let L ∈adm(J ) be an admis- sible labeling of J .

Chunk 415

Furthermore, let a,a′ ∈A be two arguments s.t. DR(a′) ⊆DR(a) and ADSub(a′)C ⊆ADSub(a)C.

Chunk 416

Then L(a′) = OUT implies L(a) = OUT. --- Page 24 --- April 2026 Proof.

Chunk 417

We go through the possible cases for why a′ is rejected in L: Suppose first that we have L(a′) = OUT due to an attack (b,a′) ∈→with L(b) = IN. If this attack is the result of an undercut, then we can use DR(a′) ⊆DR(a) to infer that b also undercuts a, meaning (b,a′) ∈→also holds.

Chunk 418

By admissiblity of L, this implies L(a) = OUT. Next, suppose this attack is the result of a gen-rebut.

Chunk 419

By Proposition 11, there exists an argument b′ which is of the form b′ : b →¬VADSub(a′)C. By admissibility of L, we have L(b′) = IN.

Chunk 420

Since ADSub(a′)C ⊆ADSub(a)C, we can infer that b′ gen-rebuts a. Because (b,a′) ∈→by assumption, we have b ̸≺a′.

Chunk 421

Since DR(b) = DR(b′), while DR(a′) ⊆DR(a), we can infer that b′ ̸≺a holds. Now (b′,a) ∈→and by admissibility of L, we have L(a) = OUT.

Chunk 422

Lastly, assume that L(a′) = OUT due to a support chain C =  (S0,b0),...,(Sn,bn) and an attack (c,bn) ∈→with a′ ∈S0, L(c) = IN, bi ∈out(L) for 0 ≤i ≤n, S0 \{a′} ⊆ in(L) and Si \{bi−1} ⊆in(L) for 0 < i ≤n. Let S = S0 \{a′}
∪ S 0<i≤n Si \{bi−1}
, i.e.

Chunk 423

S contains all the IN-labeled arguments of the support-chain C . First, assume that the attack (c,bn) ∈→is the result of an undercut.

Chunk 424

Then this attack must also be directed towards one of the arguments in S ∪{a}. By assumption, L(c) = IN, L is an admissible lableing and S ⊆in(L).

Chunk 425

From this we can infer that (c,a) ∈→must hold. By admissibility of L this implies L(a) = OUT.

Chunk 426

Next, assume that (c,bn) ∈→is the result of a gen- rebut. We use Proposition 11 to infer that there exists an attack (c′,bn) ∈→, where c′ is of the form c′ : c →¬VADSub(bn)C.

Chunk 427

Now we construct the argument b′ as follows: b′ : a,b′ 0,...,b′ m →V{a,b′ 0,...,b′ m}C, where {b′ 0,...,b′ m} = S. We want to point out three facts for the argument b′: Firstly, we know that {b′ 0,...,b′ m} = S ⊆in(L) holds.

Chunk 428

Secondly, from ADSub(a′)C ⊆ADSub(a)C, we can infer ADSub(bn)C ⊆ADSub(b′)C. This means c′ gen-rebuts b′.

Chunk 429

Thirdly, we have DR(bn) ⊆DR(b′). By assumption, (c,bn) ∈→is the result of a gen-rebut.

Chunk 430

This means we have c′ ̸≺bn (since DR(c) = DR(c′)), which implies c′ ̸≺b′. Now we have (c′,b′) ∈→.

Chunk 431

By admissibility of L and by construction of c′, we have L(c′) = IN. This means b′ is legally OUT w.r.t.

Chunk 432

L, and by admissibility of L we can infer L(b′) = OUT. Since {b′ 0,...,b′ m} = S ⊆in(L), this implies a ∈out(L), otherwise none of the arguments in S would be legally IN w.r.t.

Chunk 433

L. This proves the claim.

Chunk 434

Now we can continue with our actual proof of the admissibility of La. In the next step, we show that an argument is rejected in La iff it is legally rejected: Proposition 16.

Chunk 435

Let J = (A ,→,⇒,⪯) be a JSBAF, let L ∈adm(J ) be an admissible labeling of J and let a ∈A be an argument. Furthermore, assume that L(a) ̸= OUT and that for all attackers b of a, we have L(b) = OUT.

Chunk 436

Lastly, assume that there does not exist a support chain  (S0,b0),...,(Sn,bn) ⊆⇒and an attack (c,bn) ∈→with a ∈S0, L(c) ̸= OUT, S0 \{a} ⊆in(L) and Si \{bi−1} ⊆in(L) for 0 < i ≤n. We have x ∈out(La) iff x is legally OUT w.r.t.

Chunk 437

La. Proof.

Chunk 438

←: First, assume that x is legally OUT w.r.t. La due to an attack.

Chunk 439

Then x ∈O0 ⊆ out(La). Next, assume x is legally OUT w.r.t.

Chunk 440

La due to a support (S,b) with x ∈S and b ∈out(La). By construction of La, there needs to exist some n ∈N s.t.

Chunk 441

b ∈On. Now x ∈On+1 ⊆out(La) as required.

Chunk 442

→: We need to show that, if x ∈out(La), then x is legally OUT w.r.t. La.

Chunk 443

Note that, if x ∈O0, then this is trivial. We therefore assume x ̸∈O0.

Chunk 444

By the construction of Ok+1 from Ok, it is clear that, if x ∈Ok+1 \ Ok, then there exists a support chain C =  (S0,b0),...,(Sn,bn) ⊆⇒and an attack (y,bn) ∈→, s.t. x ∈S0, La(y) = IN, for --- Page 25 --- April 2026 0 ≤i ≤n we have bi ∈out(La), for 0 < i ≤n we have Si \{bi −1} ⊆in(La) and for S0 we have S0 \{x} ⊆in(La) (i.e.

Chunk 445

items 2.a to 2.d of Definition 2 all hold). However, from the construction of Ok+1 it is not clear that x ⪯d for all d ∈S\{x} holds (i.e.

Chunk 446

item 2.e of Definition 2) and in fact this is not necessarily the case. Thus, for the cases where item 2.e of Definition 2 does not hold, we will show that there is an “alternative reason” for x to be legally OUT w.r.t.

Chunk 447

La. Suppose that x ∈Ok+1 \ Ok due to the support chain C =  (S0,b0),...,(Sn,bn) and an attack (y,bn) ∈→as described above.

Chunk 448

Furthermore, assume that there is some d ∈S0 \{x} s.t. d ≺x (i.e.

Chunk 449

because of d, the support chain C is no reason for x to be con- sidered legally OUT w.r.t. La).

Chunk 450

Let S = S0 \{x}
∪ S 0<i≤n Si \{bi−1}
, i.e. S contains all arguments labeled IN in the support chain C .

Chunk 451

We now claim that, for S = {z0,...,zm}, we have ADSub(x)C ⊨C ¬VADSub(y) ∪ADSub(z0) ∪··· ∪ADSub(zm)
C. To see that this claim holds, assume towards a contradiction that it does not.

Chunk 452

Then there exists an inter- pretation I s.t. I ⊨M ADSub(x)C, I ⊨M ADSub(zi)C for each zi ∈S and I ⊨M ADSub(y)C.

Chunk 453

Let yC = ¬VΓ for some Γ ⊆Sub(bn)C. From I ⊨M ADSub(x)C and I ⊨M ADSub(zi)C for each zi ∈S, we can infer that I ⊨M VΓ holds.

Chunk 454

However, from I ⊨M ADSub(y)C, we can infer I ⊨M ¬VΓ, a contradiction. Next, we construct the arguments y′ : y,z0,...,zm →V{y,z0,...,zm}C and x′ : x0,...,xk →¬V∆, where {x0,...xk} = ADSub(x) and ∆= ADSub(y) ∪ADSub(z0) ∪ ···∪ADSub(zm)
C.

Chunk 455

Now x′ gen-rebuts y′. Note that, because d ≺x by assumption, there exists rd ∈DR(d) ⊆DR(y′) s.t.

Chunk 456

for all rx ∈DR(x) = DR(x′), rd ≤r rx. This implies y′ ⪯ewl x′, therefore x′ ̸≺ewl y′ and thus (x′,y′) ∈→.

Chunk 457

Because y ∈in(La) and z0,...,zm ⊆in(La), we can infer that y′ ∈in(La) holds by construction of La. This implies that either y′ ∈I0 or, by Proposition 12, there exists y′′ ∈I1 \I0 s.t.

Chunk 458

ADSub(y′) = ADSub(y′′) and y′C = y′′C. In the first case we have (x′,y′) ∈→and can infer L(x′) = OUT from y′ ∈I0.

Chunk 459

In the second case we have (x′,y′′) ∈→because ADSub(y′) = ADSub(y′′). Furthermore, we can use y′′ ∈I1 \ I0 to infer that there exists a support (S′′,y′′) ∈⇒with a ∈S′′ and S′′ \ {a} ⊆in(L).

Chunk 460

Now the support chain C ′′ =  (S′′,y′′) ⊆⇒is a support chain of which we know by assumption that all attackers of y′′ are labeled OUT in L. With this we can again infer L(x′) = OUT.

Chunk 461

By construction of x′, we obviously have DR(x′) ⊆DR(x) and ADSub(x′)C ⊆ ADSub(x)C. This means we can apply Proposition 15 to infer that L(x) = OUT must thold.

Chunk 462

By assumption, L is an admissible labeling. This means x is legally OUT w.r.t.

Chunk 463

L, either due to an attack or due to a support chain. Since in(L) ⊆in(La), we can now infer that x is legally OUT w.r.t.

Chunk 464

La due to the same attack or support chain. This finishes the proof.

Chunk 465

Finally, we are ready to combine the previous propositions to show that La is an admissible labeling: Proposition 17. Let J = (A ,→,⇒,⪯) be a JSBAF, let L ∈adm(J ) be an admissible labeling of J and let a ∈A be an argument.

Chunk 466

Furthermore, assume that L(a) ̸= OUT and that for all attackers b of a, we have L(b) = OUT. Lastly, assume that there does not exist a support chain  (S0,b0),...,(Sn,bn) ⊆⇒and an attack (c,bn) ∈→with a ∈S0, L(c) ̸= OUT, S0 \{a} ⊆in(L) and Si \{bi−1} ⊆in(L) for 0 < i ≤n.

Chunk 467

Then the propagated labeling La is an admissible labeling. --- Page 26 --- April 2026 Proof.

Chunk 468

By Proposition 13 we know that La is a labeling. By Proposition 14 we know that x ∈in(La) implies x is legally IN w.r.t.

Chunk 469

La. By Proposition 16 we know that x ∈out(La) iff x is legally OUT w.r.t.

Chunk 470

La. Lastly, because L ∈adm(J ), it is clear from the construction of La that STRJ ⊆in(La) holds.

Chunk 471

We conclude that La is an admissible labeling. Next, we use propagated labelings to show that preferred labelings are closed under sub-arguments.

Chunk 472

Lemma 2. Let J = (A ,→,⇒,⪯) be a JSBAF and let L ∈pr(J ) be a preferred labeling of J .

Chunk 473

Then for all a ∈in(L), we have Sub(a) ⊆in(L). Proof.

Chunk 474

Towards a contradiction, suppose that the claim does not hold. Then there is a′ ∈ Sub(a)\{a} s.t.

Chunk 475

a′ ̸∈in(L). We claim that all attackers of a′ are labeled OUT by L and that there does not exist a support chain C =  (S0,b0),...,(Sn,bn) and an attack (c,bn) ∈→ s.t.

Chunk 476

a′ ∈S0, L(c) ̸= OUT, S0 \{a′} ⊆in(L) and Si \{bi−1} ⊆in(L) for 0 < i ≤n. If this holds, then by Proposition 17 the propagated labeling La′ is an admissible labeling.

Chunk 477

Since a′ ̸∈in(L) by assumption, this implies in(L) ⊂in(La′), contradicting the assumption that L is a preferred labeling. Now for the attackers of a′ and the support chains C =  (S0,b0),...,(Sn,bn) with a′ ∈S0: From the way arguments and attacks are constructed in an AS and from the translation of an AS into a JSBAF, it is clear that any attacker of a′ is also an attacker of a.

Chunk 478

Since a ∈in(L) by assumption, we can infer that a is legally IN, meaning all attackers of a (and therefore all attackers of a′) are labeled OUT. Next, for the support chains starting at a′: Towards a contradiction, assume that there is a support chain C =  (S0,b0),...,(Sn,bn) and an attack (c,bn) ∈→s.t.

Chunk 479

a′ ∈S0, L(c) ̸= OUT, S0 \ {a′} ⊆in(L) and Si \ {bi−1} ⊆in(L) for 0 < i ≤n. Note that, if c is undercutting bn, then it is also undercutting a′ or some argument x ∈ S 0≤i≤n Si ∩in(L)
.

Chunk 480

As we have argued above, if the undercut is directed towards a′, then it is also directed towards a. Since a is legally IN but L(c) ̸= OUT, this assumption leads to a contradiction.

Chunk 481

We therefore assume that c is not undercutting a′ but some other argument x ∈ S 0≤i≤n Si ∩ in(L)
. However, if c is undercutting such an argument x, then L(c) ̸= OUT implies x ̸∈in(L) by admissibility of L.

Chunk 482

This is also a contradiction, therefore we assume that c is not undercutting bn, which means c must be gen-rebutting bn. By Proposition 11, there now exists c′ ∈A which is of the form c′ : c →¬VADSub(bn)C.

Chunk 483

This means c′ also gen-rebuts bn. For the argument c, we have c ̸≺bn because (c,bn) ∈→by assumption and because this attack results from a gen-rebut.

Chunk 484

Since DR(c) = DR(c′), we can infer that c′ ̸≺bn also holds, meaning (c′,bn) ∈→. Note that because c ∈in(L), we have c′ ∈in(L) by admissibility of L.

Chunk 485

Now, for each 0 ≤i ≤n take S′ i = Si ∩in(L) and create the set S′ = {a} ∪ S 0≤i≤n S′ i. We have S′C ⊨C VS′C, therefore there exists a support (S′,d) for an argument d with dC = VS′C.

Chunk 486

Note that for all arguments x ∈S′, we have x ∈in(L), thus by admissibility of L we also have d ∈in(L). Now consider again the attacker c′: As before, we can use Sub(a′)C ⊆Sub(a)C to infer that, because c′ is gen-rebutting bn, c′ is also gen-rebutting d.

Chunk 487

Since L(d) = IN, we can infer L(c′) = OUT by admissibility of L. Now c is legally OUT w.r.t.

Chunk 488

L and by admissibility of L we have L(c) = OUT. This contradicts our assumption L(c) ̸= OUT.

Chunk 489

--- Page 27 --- April 2026 5.3.3. Reduced versions of arguments With the results from the previous sub-section, we are now equipped to consider the second key-insight mentioned in our proof overview: Reduced verisons of arguments.

Chunk 490

We will show that – when considering preferred labelings – the labels of arguments and their reduced versions somewhat coincide. Definition 15.

Chunk 491

Let J1 = (A1,→1,⇒1,⪯1),J2 = (A2,→1,⇒2,⪯2) be two JSBAF which are syntactically disjoint. Furthermore, let J+ = (A+,→+,⇒+,⪯+) be the union of J1 and J2.

Chunk 492

Lastly, let a,a′ ∈A+ be arguments s.t. Atoms(aC) ⊆Atoms(ASi) for i ∈{1,2}.

Chunk 493

The argument a′ is a reduced version of a w.r.t. ASi iff all of the following are satisfied: • a′C = aC • Atoms(a′) ⊆Atoms(ASi) • For all r ∈RAXi s ∪Rdi, there is b ∈Sub(a) with TR(b) = r iff there is b′ ∈Sub(a′) with TR(b′) = r.

Chunk 494

• If b′ ∈Sub(a′) with b′C = φ, then there is b ∈Sub(a) with bC = φ. The following is clear from the definition of the atoms of an argument: Corollary 3.

Chunk 495

Let J1 = (A1,→1,⇒1,⪯1),J2 = (A2,→1,⇒2,⪯2) be two JSBAF which are syntactically disjoint. Furthermore, let J+ = (A+,→+,⇒+,⪯+) be a union of J1 and J2.

Chunk 496

Lastly, let a ∈A+ be an argument s.t. Atoms(aC) ⊆Atoms(ASi) (i ∈{1,2}).

Chunk 497

If a′ is the reduced version of a w.r.t. ASi, then a′ ∈Ai.

Chunk 498

Next, we introduce a handy way to describe those arguments that we can reach by traversing the support-relation as far back as possible. We will call these arguments crucial sub-arguments.

Chunk 499

Definition 16. Let AS1 = (Rs1,Rd1,n1,≤r1) and AS2 = (Rs2,Rd2,n2,≤r2) be two AS s.t.

Chunk 500

AS1||AS2 and let AS+ = (R+ s ,R+ d ,n+,≤+ r ) be a union of AS1 and AS2. Furthermore, let J1 = (A1,→1,⇒1,⪯1), J2 = (A2,→1,⇒2,⪯2) and J+ = (A+,→+,⇒+,⪯+) be the JSBAFs corresponding to AS1, AS2 and AS+ respectively.

Chunk 501

The mapping CSub : A+ → 2A+ maps arguments to their crucial sub-arguments as follows: • If TR(a) ∈RAX+ s ∪Rd+, then CSub(a) = {a}. • If TR(a) ∈R⊨ s , then we define: CSub(a) =  a′ ∈Sub(a) | TR(a′) ∈RAX+ s ∪Rd+ and C =  (S0,b0),...,(Sm,a) ⊆⇒+ is a support chain with a′ ∈S0 We define the restriction of CSub to ASi (i ∈{1,2}) as: CSub(a)|ASi =  a′ ∈CSub(a) | Atoms(a′C) ⊆Atoms(ASi) .

Chunk 502

We will somewhat abuse the above notation for sets of arguments, i.e. for S ⊆ A (AS+) we define CSub(S) = S a∈S CSub(a).

Chunk 503

--- Page 28 --- April 2026 By transitivity of ⊨C and because the support-relation between arguments corre- sponds to the application of rules from R⊨ s , the following is easy to see: Corollary 4. Let AS = (Rs,Rd,n,≤r) be an AS.

Chunk 504

For any a ∈A (AS), we have CSub(a)C ⊨C aC. We will now show that, for a consistent argument a, there always exists a reduced verison a′ of a.

Chunk 505

We start with the following auxilliary statement regarding the logical languages that we consider: Proposition 18. Let Γ = {φ0,...,φm,φ} and ∆= {ψ0,...,ψn} be syntactically disjoint sets of formulas.

Chunk 506

Furthermore, let {ψ0,...ψn} be satisfiable. If we have {φ0,...,φm,ψ0,...,ψn} ⊨C φ, then {φ0,...,φm} ⊨C φ also holds.

Chunk 507

Proof. Towards a contradiction, assume that the claim does not hold.

Chunk 508

Then there exists an interpretation I1 s.t. I1 ⊨M φi for all φi ∈{φ0,...,φm} but I1 ̸⊨M φ.

Chunk 509

This means I1 ⊨M ¬φ. By assumption, the set {ψ0,...,ψn} is satisfiable, i.e.

Chunk 510

there exists an interpretation I2 s.t. I2 ⊨M ψj for all of these ψj.

Chunk 511

Because Γ and ∆are syntactically disjoint, there now exists an interpretation I s.t. I ⊨M φi for all φi ∈{φ0,...,φm}, I ⊨M ψj for all ψ j ∈{ψ0,...,ψn} and I ⊨M ¬φ.

Chunk 512

However, because {φ0,...,φm,ψ0,...,ψn} ⊨C φ by assumption, we now also have I ⊨M φ, a contradiction to I ⊨M ¬φ. Now for the existence of reduced versions of arguments: Proposition 19.

Chunk 513

Let J1 = (A1,→1,⇒1,⪯1),J2 = (A2,→1,⇒2,⪯2) be two JSBAF which are syntactically disjoint. Furthermore, let J+ = (A+,→+,⇒+,⪯+) be a union of J1 and J2.

Chunk 514

Lastly, let a ∈A+ be a consistent argument s.t. Atoms(aC) ⊆Atoms(ASi) (i ∈{1,2}).

Chunk 515

Then there exists an argument a′ s.t. a′ is the reduced version of a w.r.t.

Chunk 516

ASi. Proof.

Chunk 517

We show the claim by structured incution over the construction of a. Induction start: Let a be of the form a :⇝φ.

Chunk 518

By assumption, Atoms(aC) ⊆Atoms(ASi). Therefore, a itself is a reduced version of a w.r.t.

Chunk 519

ASi. Induction step: Let a be of the form a : a0,...,am ⇝φ.

Chunk 520

First, we make a distinction regarding the top-rule of a: We disregard the case TR(a) ∈RAX+ s , as it was covered by the induction start. If TR(a) ∈Rd, we can infer from Atoms(aC) ⊆Atoms(ASi), that TR(a) ∈Rdi must hold.

Chunk 521

Now for each 0 ≤h ≤m, we have Atoms(aC h ) ⊆Atoms(ASi). By the induction hypothesis, there exists reduced versions a′ h (w.r.t.

Chunk 522

ASi) for each of these arguments ah. Because aC h = a′C h , we can use these arguments to construct a′ as a′ : a′ 0,...a′ m ⇒φ.

Chunk 523

It is clear that a′ satisfies the conditions of Definition 15. Now suppose that TR(a) ∈R⊨ s holds and let X = CSub(a).

Chunk 524

Note that XC ⊨C φ. We partition the set X according to the top-rules of the arguments x ∈X: Let X = Xi ∪Xj with Xi containing precisely those x for which we have TR(x) ∈RAXi s ∪Rdi and Xj con- taining precisely those x for which we have TR(x) ∈R AXj s ∪Rd j.

Chunk 525

Since a is consistent by assumption, the set XC j is satisfiable. Since Atoms(a) ⊆Atoms(ASi) by assumption, we can now use Proposition 18 to infer XC i ⊨C φ.

Chunk 526

Now we construct a set of arguments Y = {y0,...,yl} from Xi ∪Xj as follows: If x ∈Xi, then we add a reduced version of x w.r.t. ASi to Y.

Chunk 527

Such a reduced version exists by the induction hypothesis. On the other hand, if x ∈Xj, then we collect all arguments --- Page 29 --- April 2026 z ∈Sub(x) for which we have TR(z) ∈RAXi s ∪Rdi and add their reduced versions w.r.t.

Chunk 528

ASi to Y. Again, these reduced versions exist by the induction hypothesis.

Chunk 529

We now claim that the argument a′ : y0,...yl →φ exists and satisfies the conditions of Definition 15. We first argue for the existence of this argument.

Chunk 530

For this, we have to show YC ⊨C φ. Note that by construction of Y, we have XC i ⊆YC.

Chunk 531

As we have argued above, XC i ⊨C φ, thus monotonicity of ⊨C yields YC ⊨C φ as required. Next, we argue that a′ satisfies the conditions of Definition 15: We trivially have aC = a′C.

Chunk 532

By construction of Y, we only added reduced arguments w.r.t. ASi to Y, thus we have Atoms(a′) ⊆Atoms(ASi).

Chunk 533

From the construction of Y, it is clear that for all r ∈RAXi s ∪Rdi for which we have some b′ ∈Sub(a′) with TR(b′) = r, there is some b ∈Sub(a) with TR(b) = r. Next, let r ∈RAXi s ∪Rdi s.t.

Chunk 534

there is some b ∈Sub(a) with TR(b) = r. By construction of X, we can infer that b ∈Sub(x) for some x ∈Xi or some x ∈Xj must hold.

Chunk 535

In the first case, we added a reduced version of x to Y. In the second case, we collected all z ∈Sub(x) for which we have TR(z) ∈RAXi s ∪Rdi and added their reduced versions z′ to Y.

Chunk 536

In both cases, we can infer from the induction hypothesis, that there exists some b′ ∈Sub(a′) with TR(b′) = r. Lastly, because we only added reduced versions of arguments to Y and because a′C = aC, it is clear that if b′ ∈Sub(a′) with b′C = φ, then there is some b ∈Sub(a) s.t.

Chunk 537

bC = φ. This finishes the proof.

Chunk 538

Now we are ready to show the main result regarding reduced versions of arguments: For any argument a and its reduced version a′, if a is accepted in a preferred labeling, then a′ is also accepted and if a′ is rejected, then a is also rejected. Lemma 3.

Chunk 539

Let J1 = (A1,→1,⇒1,⪯1),J2 = (A2,→1,⇒2,⪯2) be two JSBAF which are syntactically disjoint. Furthermore, let J+ = (A+,→+,⇒+,⪯+) be the union of J1 and J2.

Chunk 540

Let a ∈A+ be a consistent argument s.t. Atoms(aC) ⊆Atoms(ASi) (i ∈ {1,2}) and let a′ be a reduced version of a w.r.t.

Chunk 541

ASi. If L ∈pr(J+) and a is legally IN w.r.t.

Chunk 542

L, then L(a′) = IN. If L is a labeling of J+ s.t.

Chunk 543

for all x ∈A+, x ∈out(L) iff x is legally OUT w.r.t. L, then a′ ∈out(L) implies a ∈out(L).

Chunk 544

Proof. We begin by showing that any attacker of a′ is also an attacker of a and that any support-chain C ′ =  (S′ 0,b′ 0),...,(S′ n,b′ n) with a′ ∈S′ 0 can be “recreated” with a instead of a′.

Chunk 545

First, assume that there is (b,a′) ∈→+. Then b either undercuts or gen-rebuts a′.

Chunk 546

If b undercuts a′, then we can use DR(a′) ⊆DR(a) to infer that b also undercuts a, meaning (b,a) ∈→+. If b gen-rebuts a′, then we have bC = ¬VΓ with Γ ⊆Sub(a′)C ⊆Sub(a)C.

Chunk 547

Because (b,a′) ∈→+, we have b ̸≺ewl a′, thus either b ̸⪯ewl a′ or a′ ⪯ewl b must hold. If b ̸⪯ewl a′, then for all rb ∈DR(b) there exists ra′ ∈DR(a′) ⊆DR(a), s.t.

Chunk 548

rb ̸≤+ r ra′. On the other hand, if a′ ⪯ewl b, then there exists ra′ ∈DR(a′) ⊆DR(a) s.t.

Chunk 549

for all rb ∈DR(b), ra′ ≤+ r rb. In both cases, we can infer b ̸≺ewl a and thus (b,a) ∈→+, as required.

Chunk 550

Next, assume that there is a support chain C ′ =  (S′ 0,b′ 0),...,(S′ n,b′ n) with a′ ∈ S′ 0. Let S′ 0 = {c0,...,a′,...,cm} and S0 = {c0,...,a,...,cm}.

Chunk 551

We have S′C 0 ⊨C b′C 0 and because a′C = aC, we must also have SC 0 ⊨C b′C 0 . Therefore, there exists the argument b0 : c0,...,a,...,cm →b′C 0 and the support (S0,b0).

Chunk 552

With the same reasoning, we can create a support chain C =  (S0,b0),...,(Sn,bn) with a ∈S0, b′C k = bC k and Sk = (S′ k \{b′ k−1})∪ {bk−1} for all 0 < k ≤n. Now we can use DR(a′) ⊆DR(a) to infer the following: If d ∈S′ 0 \ {a′} s.t.

Chunk 553

a′ ⪯ewl d, then there is some ra ∈DR(a′) ⊆DR(a) s.t. for all rd ∈ DR(d), we have ra ≤+ r rd.

Chunk 554

This, in turn, implies we also have a ⪯ewl d. With the same --- Page 30 --- April 2026 reasoning, for any 0 < k ≤n, if we have d ∈S′ k \{b′ k−1} s.t.

Chunk 555

b′ k−1 ⪯ewl d, then bk−1 ⪯ewl d also holds. Furthermore, because a′ is a reduced version of a by assumption, we have DR(b′ n) ⊆DR(bn) and Sub(b′ n)C ⊆Sub(bn)C.

Chunk 556

Now, let (d,b′ n) ∈→+ be an attack towards b′ n. Similar to the case (b,a′) ∈→+ above, we can infer that (d,bn) ∈→+ must also hold: If d undercuts b′ n, then it must do so on an argument x ∈CSub(b′ n) and because a′ is a reduced version of a, we can infer that d also undercuts bn, meaning (d,bn) ∈→+.

Chunk 557

On the other hand, if d gen-rebuts b′ n, then dC = ¬VΓ and d ̸≺ewl b′ n, for Γ ⊆Sub(b′ n)C ⊆ Sub(bn)C. Analogous to before, we can show that d ̸≺ewl bn also holds, meaning we again have (d,bn) ∈→+.

Chunk 558

Therefore, we can essentially “recreate” the support chains C ′ starting at a′ as support chains C which start at a. Now for the actual claim: Suppose first that L is a labeling of J+ for which we have that, x ∈out(L) iff x is legally OUT w.r.t.

Chunk 559

L. Assume that a′ is legally OUT w.r.t.

Chunk 560

L. Then this must be due to an attack or a support.

Chunk 561

If there is an attack (d,a′) ∈→+ with d ∈in(L), then we have argued above that (d,a) ∈→+ also holds, therefore a is legally OUT w.r.t. L.

Chunk 562

On the other hand, if there is a support chain C ′ =  (S′ 0,b′ 0),...,(S′ n,b′ n) s.t. a′ ∈S′ 0 is legally OUT w.r.t.

Chunk 563

L, then we have argued above that there exists a support chain C =  (S0,b0),...,(Sn,bn) with a ∈S0. Note that in the support chain C , all arguments in the supporting sets Sk are the same as in the supporting sets S′ k in the support chain C ′, except for the newly created arguments bk.

Chunk 564

More precisely, we have S′ 0 \{a′} = S0 \{a} and for each 0 < k ≤n we have S′ k \{b′ k−1} = Sk \{bk−1}. Since a′ is legally OUT w.r.t.

Chunk 565

L by assumption, we have L(b′ l) = OUT for each 0 ≤l ≤n. Now we can make an inductive argument to show that L(bl) = OUT also holds: For the induction start, we note that L(b′ n) = OUT due to an attack (c,b′ n) ∈→+ with L(c) = IN.

Chunk 566

We have argued above that this means (c,bn) ∈→+ also holds, therefore bn is legally OUT w.r.t. L and by our assumption this implies L(bn) = OUT.

Chunk 567

For the induction step, we first note that, since each b′ k ∈{b′ 0,...,b′ n−1} is labeled OUT, they must be legally OUT w.r.t. L by our assumption.

Chunk 568

From this we can infer that b′ k ⪯d for all d ∈ S′ k+1 \{b′ k} and by the way JSBAFs are constructed from their AS counterparts, we can infer b′ k ⪯ewl d. As we have argued above, this implies bk ⪯ewl d for all d ∈Sk+1 \{b′ k}, therefore bk ⪯d also holds for these arguments d.

Chunk 569

Since S′ k+1 \ {b′ k} = Sk+1 \ {bk}, we now have that each bk is legally OUT w.r.t. L and therefore L(bk) = OUT also holds.

Chunk 570

In particular, this means L(b0) ∈OUT. Since S′ 0 \ {a′} = S0 \ {a} and since a′ ⪯d for all d ∈S′ 0 \{a′}, we can now also infer that a is legally OUT w.r.t.

Chunk 571

L, as required. Now suppose that L ∈pr(J+) and that a is legally IN w.r.t.

Chunk 572

L. We will show that for the reduced argument a′, the prerequisites for Proposition 17 are satisfied.

Chunk 573

If this holds, then we can use Proposition 17 to infer that La′ is an admissible labeling. Since L is a preferred labeling, we can then infer that L(a′) = IN holds as required, since L(a′) ̸= IN would contradict the maximality (w.r.t.

Chunk 574

set-inclusion of accepted arguments) of L. Thus we only need to show that the prerequisites for Proposition 17 are satisfied.

Chunk 575

First of all, we can use the above proof to infer that L(a′) ̸= OUT must hold: If L(a′) = OUT, then by admissibility of L, a′ is legally OUT w.r.t. L.

Chunk 576

By our argumen- tation above, this implies a is legally OUT w.r.t. L, contradicting a ∈in(L).

Chunk 577

Next, for the attackers of a′: Towards a contradiction, suppose that there is an attack (d,a′) ∈→ with L(d) ̸= OUT. We have argued above that this implies (d,a) ∈→+.

Chunk 578

Now a is not legally IN w.r.t. L, a contradiction.

Chunk 579

Lastly, for the support chains starting at a′: Towards a contradiction, assume that there is a support chain C ′ =  (S′ 0,b′ 0),...,(S′ n,b′ n) and an attack (d,b′ n) ∈→+ with a′ ∈S′ 0, L(d) ̸= OUT, S′ 0 \{a′} ⊆in(L) and S′ k \{b′ k−1} ⊆in(L) --- Page 31 --- April 2026 for all 0 < k ≤n. As we have argued above, there now also exists a support chain C =  (S0,b0),...,(Sn,bn) and an attack (d,bn) with a ∈S0, L(d) ̸= OUT, S0 \{a} ⊆in(L) and Sk \{bk−1} ⊆in(L) for all 0 < k ≤n.

Chunk 580

Now none of the bk ∈{b0,...,bn} are legally IN w.r.t. L, and by admissiblity of L we can infer that none of them are actually labeled IN.

Chunk 581

For the support (S0,b0) we now have L(b0) ̸= IN but S0 ⊆in(L), which means none of the arguments in S0 are legally IN w.r.t. L.

Chunk 582

This contradicts the admissibility of L and finishes the proof. 5.3.4.

Chunk 583

Interactions between argumentation systems In this section, we will consider the first key insight mentioned in our proof overview: The four “edge cases” depicted in Illustration 1. We start with an auxiliary proposition which tells us that, if the conclusion of an argument a ∈Ai contains only atoms of AS j, then either a is inconsistent or its conclusion is a tautology.

Chunk 584

Proposition 20. Let AS1 = (Rs1,Rd1,n1,≤r1) and AS2 = (Rs2,Rd2,n2,≤r2) be two AS s.t.

Chunk 585

AS1||AS2. Furthermore, let a ∈Ai be an argument with TR(a) ∈R⊨ s .

Chunk 586

If Atoms(aC) ⊆ Atoms(ASj), then either a is inconsistent or φ is a tautology. Proof.

Chunk 587

Let a be of the form a : a0,...,am →φ. Towards a contradiction, suppose that the claim does not hold, meaning we have Atoms(φ) ⊆Atoms(ASj), a ∈Ai is consistent and φ is not a tautology.

Chunk 588

Let Γ = ADSub(a)C. By Corollary 2, we know that Γ ⊨C φ.

Chunk 589

Because a ∈Ai, we know that for each ψ ∈Γ, we have Atoms(ψ) ⊆Atoms(ASi). By AS1||AS2, we can now infer that φ and Γ must be syntactically disjoint.

Chunk 590

Since a is consistent by assumption, there is an interpretation I1 s.t. I1 ⊨M ψ for each ψ ∈Γ.

Chunk 591

Because we also assumed that φ is not a tautology, there exists an interpretation I2 s.t. I2 ⊨M −φ.

Chunk 592

Because φ and Γ are disjoint, there now exists an interpretation I s.t. I ⊨M ψ for each ψ ∈Γ, but I ⊨M −φ also holds.

Chunk 593

This contradicts Γ ⊨C φ, therefore the claim must hold. Now we are ready to consider the first case of Illustration 1.

Chunk 594

Using the proposition above, we will show that, for any syntactically disjoint AS1 and AS2 as well as their union AS+, if an argument a ∈Ai attacks an argument b ∈Aj, then either a or b is rejected by any admissible labeling of the corresponding JSBAFs. Proposition 21.

Chunk 595

Let AS1 = (Rs1,Rd1,n1,≤r1), AS2 = (Rs2,Rd2,n2,≤r2) be two AS s.t. AS1||AS2 and let AS+ be their union.

Chunk 596

Furthermore, let J1 = (A1,→1,⇒1,⪯1), J2 = (A2,→1,⇒2,⪯2) and J+ = (A+,→+,⇒+,⪯+) be the JSBAFs corresponding to AS1, AS2 and AS+. Lastly, for i, j ∈{1,2} with i ̸= j, let a ∈Ai and b ∈Aj be two arguments.

Chunk 597

If (a,b) ∈→+, then there exists c ∈Ai ∩Aj ∩A+ s.t. c is of the form c :→φ, TR(c) ∈R⊨ s and either (c,a) ∈→i or (c,b) ∈→j.

Chunk 598

Proof. By (a,b) ∈→+, we know that a is either undercutting or gen-rebutting b.

Chunk 599

We first show the undercut case. W.l.o.g.

Chunk 600

, we assume that a undercuts b on b itself. Let a and b be of the form a : a0,...,am ⇝φ and b : b0,...,bk ⇒φ with TR(b) = r ∈Rd j and −φ = nj(r).

Chunk 601

Be- cause TR(b) ∈Rd j, we have Atoms(φ) ⊆Atoms(ASj). Thus we know that TR(a) ∈R⊨ s must hold, otherwise the syntactic disjointness of AS1 and AS2 would be violated.

Chunk 602

By Proposition 20, we can now infer that either φ is a tautology or a is inconsistent. If a is inconsistent, then there is Γ ⊆Sub(a)C s.t.

Chunk 603

Γ is unsatisfiable. Now ¬VΓ is a tautology --- Page 32 --- April 2026 and there exists an argument a ∈Ai ∩Aj ∩A+ of the form a :→¬VΓ which gen-rebuts a.

Chunk 604

Since a is a strict argument, we have a ̸≺a, therefore (a,a) ∈→i, as claimed. Next, assume that a is consistent.

Chunk 605

Then φ must be a tautology. This means there exists an ar- gument b ∈Ai ∩Aj ∩A+ of the form b :→φ which undercuts b, meaning (b,b) ∈→j.

Chunk 606

This concludes the undercut case. Now for the gen-rebut case: If a gen-rebuts b, then b is defeasible and there is Γ ⊆ Sub(b)C s.t.

Chunk 607

aC = φ = ¬VΓ. We make a distinction w.r.t.

Chunk 608

TR(a). In the first case, assume that TR(a) ∈Rdi ∪RAXi s .

Chunk 609

Then for all ψ ∈Γ, we have Atoms(ψ) ⊆Atoms(ASi). Thus for all b′ ∈Sub(b) with b′C = ψ′ ∈Γ, we must have TR(b′) ∈R⊨ s , otherwise syntactic disjointness of AS1 and AS2 would be violated again.

Chunk 610

Now we can use Proposition 20 again to infer that for each of these sub-arguments of b, either b′ is inconsistent or ψ′ a tautology. If any of the b′ are inconsistent, then b is also inconsistent and we can construct the argument b ∈Ai∩Aj ∩A+ as before.

Chunk 611

We therefore assume that all b′ are consistent. But then all ψ′ ∈Γ are tautologies, which means VΓ is also a tautology.

Chunk 612

Now ¬VΓ = aC is unsatisfiable, thus a is inconsistent and we can construct the argument a ∈Ai ∩Aj ∩A+ as before. Now for the second case, i.e.

Chunk 613

TR(a) ∈R⊨ s . Let ∆a = ADSub(a)C and ∆b = ADSub(b)C be the conclusions of the axiomatic and defeasible sub-arguments of a and b respectively.

Chunk 614

By Proposition 10, we have ∆a ⊨C ¬VΓ and ∆b ⊨C VΓ. Because AS1 and AS2 are syntactically disjoint, ∆a and ∆b must also be syntactically disjoint.

Chunk 615

Now, towards a contradiction, assume that a and b are both consistent. Then in particular ∆a and ∆b are satisfiable, i.e.

Chunk 616

there exists interpretations Ia,Ib s.t. Ia ⊨M φa for all φa ∈∆a and Ib ⊨M φb for all φb ∈∆b.

Chunk 617

Because ∆a and ∆b are syntactically disjoint, there now exists an interpretation I s.t. I ⊨M φ for all φ ∈∆a ∪∆b.

Chunk 618

Because we have ∆a ⊨C ¬VΓ and ∆b ⊨C VΓ, we can now infer that I ⊨M ¬VΓ and I ⊨M VΓ must hold, a contradiction. We conclude that either a or b be need to be inconsistent and we can again construct one of the arguments a, b as before.

Chunk 619

Before we continue with our next results, we need one more definition, which will be used to adequately describe those arguments that correspond to the argument b in Il- lustration 1. More precisely, we use the term minimal A+-argument for those arguments b ∈A+ \ (Ai ∪Aj), which have crucial sub-arguments in Ai and Aj, while not having any crucial sub-arguments in A+ \(Ai ∪Aj).

Chunk 620

Definition 17. Let AS1 = (Rs1,Rd1,n1,≤r1), AS2 = (Rs2,Rd2,n2,≤r2) be two AS s.t.

Chunk 621

AS1||AS2 and let AS+ be their union. Furthermore, let a ∈A+ be an argument.

Chunk 622

We say that a is a minimal A+-argument, iff CSub(a)| ASi ̸= /0, CSub(a)| AS j ̸= /0 and CSub(a) ⊆Ai ∪Aj. Now we prove three propositions which, respectively, cover cases two, three and four of Illustration 1.

Chunk 623

For now, we only show that if these cases exist, then we can infer the existence of arguments and attacks which are wholly contained in either J1 or J2. Afterwards, we will combine all these cases to infer that a preferred labeling of Ji can be “extended” to a preferred labeling of J+ and a preferred labeling of J+ can be “reduced” to a preferred labeling of Ji.

Chunk 624

Proposition 22. Let AS1 = (Rs1,Rd1,n1,≤r1), AS2 = (Rs2,Rd2,n2,≤r2) be two AS s.t.

Chunk 625

AS1||AS2 and let AS+ be their union. Let J1 = (A1,→1,⇒1,⪯1), J2 = (A2,→1,⇒2,⪯2) --- Page 33 --- April 2026 and J+ = (A+,→+,⇒+,⪯+) be the JSBAFs corresponding to AS1, AS2 and AS+ re- spectively.

Chunk 626

Furthermore, let a,b ∈A+ be arguments s.t. a is a consistent minimal A+- argument.

Chunk 627

Lastly, let (a,b) ∈→+. If b ∈Ai, then there exists an argument a′ ∈Ai s.t.

Chunk 628

(a′,b) ∈→i and CSub(a)|ASi = CSub(a′). Proof.

Chunk 629

Throughout this proof, let Γi = CSub(a)|ASi and Γj = CSub(a)|AS j. Note that this means Γi ⊆Ai and Γ j ⊆Aj because a is a minimal A+-argument.

Chunk 630

First, assume that (a,b) ∈→+ is the result of an undercut. Then we know Atoms(aC) ⊆Atoms(ASi).

Chunk 631

Furthermore, Atoms(ΓC i ) ⊆Atoms(ASi) and Atoms(ΓC j ) ⊆Atoms(ASj), i.e. ΓC i ∪{aC} and ΓC j are syntactically disjoint sets of formulas.

Chunk 632

By assumption, we also know that ΓC i ∪{aC} as well as ΓC j are consistent sets of formulas. Now we can use Proposition 18 to infer ΓC i ⊨C aC.

Chunk 633

Thus we can construct the argument a′ ∈Ai as follows: a′ : a0,...,am → aC, where {a0,...,am} = Γi. Clearly, a′ undercuts b, therefore (a′,b) ∈→i as required.

Chunk 634

Next, assume that (a,b) ∈→+ is the result of a gen-rebut. Let ∆= {φ0,...,φm} and aC = ¬V∆.

Chunk 635

By Proposition 10, we know that ADSub(b)C ⊨C V∆. We claim that ΓC i ⊨C ¬VADSub(b)C also holds.

Chunk 636

Towards a contradiction, assume that this is not the case. Then there exists an interpretation I1 s.t.

Chunk 637

I1 ⊨M ΓC i and I1 ⊨M VADSub(b)C. By assumption a is consistent, thus there also exists an interpretation I2 s.t.

Chunk 638

I2 ⊨M ΓC j . Note that Atoms(ADSub(b)C) ∪Atoms(ΓC i ) ⊆Atoms(ASi) since b ∈Ai, while Atoms(ΓC j ) ⊆ Atoms(ASj).

Chunk 639

Now there exists an interpretation I s.t. I ⊨M ΓC i ∪ΓC j , i.e.

Chunk 640

I ⊨M ¬V∆ and I ⊨M ADSub(b)C, i.e. I ⊨M V∆.

Chunk 641

Obviously, this is a contradiction. We infer ΓC i ⊨C ¬VADSub(b)C.

Chunk 642

With this, we can construct the argument a′ ∈Ai as follows: a′ : a0,...,am → ¬VADSub(b)C, where {a0,...,am} = Γi. Because (a,b) ∈→+, we know that a ̸≺b.

Chunk 643

Thus we must have either a ̸⪯b or b ⪯a. In the first case, we can infer that for all ra ∈DR(a) ⊇DR(a′), there exists rb ∈DR(b) s.t.

Chunk 644

ra ̸≤rb. In the second case, we can infer that there exists rb ∈DR(b) s.t.

Chunk 645

for all ra ∈DR(a) ⊇DR(a′), we have rb ≤ra. In both cases we can infer a′ ̸≺b, therefore (a′,b) ∈→i as required.

Chunk 646

Proposition 23. Let AS1 = (Rs1,Rd1,n1,≤r1), AS2 = (Rs2,Rd2,n2,≤r2) be two AS s.t.

Chunk 647

AS1||AS2 and let AS+ be their union. Let J1 = (A1,→1,⇒1,⪯1), J2 = (A2,→1,⇒2,⪯2) and J+ = (A+,→+,⇒+,⪯+) be the JSBAFs corresponding to AS1, AS2 and AS+ re- spectively.

Chunk 648

Furthermore, let a,b ∈A+ be consistent arguments, s.t. a ∈Ai and b is a minimal AS+-argument (i ∈{1,2}).

Chunk 649

Lastly, let (a,b) ∈→+. Then either there exist ar- guments a′,b′ ∈Ai, s.t.

Chunk 650

CSub(a′) = CSub(a), CSub(b′) = CSub(b)|ASi and (a′,b′) ∈→i, or there exists arguments a′,b′ ∈Ai s.t. CSub(a′) = CSub(a), CSub(b′) = ADSub(b)|ASi and either (a′,b′) ∈→i or (b′,a′) ∈→i.

Chunk 651

Proof. First, assume that (a,b) ∈→+ is the result of an undercut.

Chunk 652

Then this undercut must be directed towards an argument in CSub(b)|ASi or in CSub(b)|ASj. In the second case, we have Atoms(aC) ⊆Atoms(ASj).

Chunk 653

With the same argumentation as in the proof of Proposition 21, we can show that this implies either a or b need to be inconsistent. Because this contradicts our assumptions, we infer that, if a undercuts b, then it must do so on some b′ ∈CSub(B)|ASi.

Chunk 654

Now, for CSub(b)|ASi = {b0,...,bm}, let b′ ∈Ai be of the form b′ : b0,...,bm →V{b0,...,bm}C. Obviously we have CSub(b′) = CSub(b)|ASi.

Chunk 655

Now a undercuts b′, meaning (a,b′) ∈→i as required. --- Page 34 --- April 2026 Now for the gen-rebut case: Let aC = ¬VΓ.

Chunk 656

We claim thatCSub(a)C ⊨C ¬VADSub(b)| ASi
C. Towards a contradiction, assume that this is not true.

Chunk 657

Then there exists an interpre- tation I1 s.t. I1 ⊨M CSub(a)C and I1 ⊨M VADSub(b)|ASi
C.

Chunk 658

Because b was consistent by assumption, there also exists an interpretation I2 for which we have I2 ⊨M ADSub(b)| ASj
C. Note that CSub(a)C ∪ ADSub(b)|ASi
C and ADSub(b)|ASj
C are syntactically disjoint sets of formulas.

Chunk 659

Now there exists an interpretation I for which we have both I ⊨M CSub(a)C and I ⊨M ADSub(b)|ASi
C ∪ ADSub(b)|AS j
C, i.e. I ⊨M ADSub(b)C.

Chunk 660

By Corollary 4 we can infer that I ⊨M ¬VΓ and by Proposition 10 we can infer I ⊨M VΓ. Obviously, this is a contradiction, therefore CSub(a)C ⊨C ¬VADSub(b)|ASi
C must hold.

Chunk 661

Now we construct the arguments a′,b′,b′′ ∈Ai as follows: For CSub(a) = {a0,...,am}, let a′ : a0,...,am →¬VADSub(b) | ASi
C, and for ADSub(b) | ASi = {b0,...,bn}, let b′ : b0,...,bn →V{b0,...,bn}C and b′′ : b′ →¬¬V{b0,...,bn}C. Clearly, a′ gen-rebuts b′ and b′′ gen-rebuts a′.

Chunk 662

Now, if a′ ̸≺b′, then (a′,b′) ∈→i and the claim holds. Thus we assume that a′ ≺b′.

Chunk 663

Because DR(b′) = DR(b′′), this also implies a′ ≺b′′, i.e. a′ ⪯b′′ and b′′ ̸⪯a′.

Chunk 664

From this we infer b′′ ̸≺a′, therefore (b′′,a′) ∈→i as required. Proposition 24.

Chunk 665

Let AS1 = (Rs1,Rd1,n1,≤r1), AS2 = (Rs2,Rd2,n2,≤r2) be two AS s.t. AS1||AS2 and let AS+ be their union.

Chunk 666

Let J1 = (A1,→1,⇒1,⪯1), J2 = (A2,→1,⇒2,⪯2) and J+ = (A+,→+,⇒+,⪯+) be the JSBAFs corresponding to AS1, AS2 and AS+ re- spectively. Furthermore, let a,b ∈A+ be arguments, s.t.

Chunk 667

both a and b are consistent minimal A+-arguments. Lastly, let (a,b) ∈→+.

Chunk 668

Then one of the following four cases holds: 1. There are a′,b′ ∈Ai s.t.

Chunk 669

CSub(a)|ASi = CSub(a′), CSub(b)|ASi = CSub(b′) and (a′,b′) ∈→i. 2.

Chunk 670

There are a′,b′ ∈Aj s.t. CSub(a)|ASj = CSub(a′), CSub(b)|ASj = CSub(b′) and (a′,b′) ∈→j.

Chunk 671

3. There are a′,b′ ∈Ai s.t.

Chunk 672

CSub(a)|ASi = CSub(a′), ADSub(b)|ASi = CSub(b′) and either (a′,b′) ∈→i or (b′,a′) ∈→i. 4.

Chunk 673

There are a′,b′ ∈Aj s.t. CSub(a)|AS j = CSub(a′), ADSub(b)|ASj = CSub(b′) and either (a′,b′) ∈→j or (b′,a′) ∈→j.

Chunk 674

Proof. Throughout this proof, let Γai = CSub(a)|ASi, Γa j = CSub(a)|ASj and let Γbi = CSub(b)|ASi, Γb j = CSub(b)|AS j.

Chunk 675

First, assume that (a,b) ∈→+ is the result of an undercut. Then either Atoms(aC) ⊆Atoms(ASi) or Atoms(aC) ⊆Atoms(ASj).

Chunk 676

Because CSub(b) = CSub(b)|ASi ∪CSub(b)|AS j, we know that this undercut needs to be directed towards an argument which is either in Γbi or in Γb j. Note that we have Γbi ⊆Ai and Γb j ⊆Aj.

Chunk 677

Now we can use Proposition 18 to infer that, if Atoms(aC) ⊆Atoms(ASi), then ΓC ai ⊨C aC and if Atoms(aC) ⊆Atoms(ASj), then ΓC a j ⊨C aC. In the first case, we can construct the arguments a′,b′ ∈Ai which are of the form a′ : a0,...,am →aC and b′ : b0,...,bn →V{b0,...,bn}C, where {a0,...,am} = Γai and {b0,...,bn} = Γbi.

Chunk 678

Then CSub(a)|ASi = CSub(a′), CSub(b)|ASi = CSub(b′) and (a′,b′) ∈→i, as required by item one of the claim. In the second case, we can construct the arguments a′′,b′′ ∈Aj which are of the form a′′ : a0,...,am →aC and b′′ : b0,...,bn →V{b0,...,bn}C, where --- Page 35 --- April 2026 {a0,...,am} = Γa j and {b0,...,bn} = Γbj.

Chunk 679

ThenCSub(a)|ASj =CSub(a′′),CSub(b)|AS j = CSub(b′′) and (a′′,b′′) ∈→j, as required by item two of the claim. Now for the gen-rebut case.

Chunk 680

Let ∆= {φ0,...,φm} and aC = ¬V∆. We claim that either ΓC ai ⊨C ¬VADSub(b)|ASi
C or ΓC a j ⊨C ¬VADSub(b)|ASj
C must hold.

Chunk 681

Towards a contradiction, assume that this is not the case. Then there exists two interpretations, I1,I2 s.t.

Chunk 682

I1 ⊨M ΓC ai, I1 ⊨M VADSub(b)|ASi
C and I2 ⊨M ΓC aj, I2 ⊨M VADSub(b)|ASj
C. Note that ΓC ai ∪ ADSub(b)|ASi
C and ΓC aj ∪ ADSub(b)|ASj
C are syntactically disjoint sets of formulas.

Chunk 683

Now there exists an interpretation I s.t. I ⊨M Γai ∪Γa j, i.e.

Chunk 684

I ⊨M ¬V∆and I ⊨M ADSub(b)|ASi
C ∪ ADSub(b)|ASj
C, i.e. I ⊨M V∆.

Chunk 685

Obviously, this is a contradiction. We infer that either ΓC ai ⊨C ¬VADSub(b)|ASi
C or ΓC a j ⊨C ¬VADSub(b)|ASj
C must hold.

Chunk 686

In the first case, we can construct the arguments a′,b′,b′′ ∈Ai which are of the form a′ : a0,...,am →¬V{b0,...,bn}C, b′ : b0,...,bn →V{b0,...,bn}C and b′′ : b′ → ¬¬V{b0,...,bn}C, where {a0,...,am} = Γai and {b0,...,bn} = ADSub(b)|ASi. Then a′ gen-rebuts b′ and b′′ gen-rebuts a′.

Chunk 687

Analogous to the proof of the gen-rebut case in Proposition 22, we either have (a′,b′) ∈→i (if a′ ̸≺b′), or we have (b′′,a′) ∈→i (if a′ ≺b′). Either way, item three of the claim is satisfied.

Chunk 688

In the second case, we can construct the arguments a′,b′,b′′ ∈Aj which are of the form a′ : a0,...,am →¬V{b0,...,bn}C, b′ : b0,...,bn →V{b0,...,bn}C and b′′ : b′ → ¬¬V{b0,...,bn}C, where {a0,...,am} = Γa j and {b0,...,bn} = ADSub(b)|ASj. Again, we either have (a′,b′) ∈→j or (b′′,a′) ∈→j, as required by item four of the claim.

Chunk 689

5.3.5. Non-Interference postulate Finally, we are ready to show that, for any two syntactically disjoint AS AS1, AS2 and their union AS+, for the corresponding JSBAFs J1, J2 and J+, any preferred labeling L1 of J1 can be turned into a preferred labeling L+ of J+ and any preferred labeling L+ of J+ can be reduced to a preferred labeling L1 of J1.

Chunk 690

For this, we will show three sub-results: First, we take a preferred labeling L of J1 and show that by “adding” to in(L) only the strict arguments of J2, we can turn L into an admissible labeling of J+. Next, we will show that a preferred labeling L+ of J+ can be turned into an admissible labeling L1 of J1 by simply “removing” all arguments of A+ \A1 from in(L+), out(L+) and undec(L+).

Chunk 691

Lastly, we will then show that, if there is any admissible labeling L′ 1 of J1 s.t. in(L1) ⊆in(L′ 1), we can “add back” the arguments of in(L+) that were removed in the previous step and create another admissible labeling from that.

Chunk 692

By combining the first and second of these sub-results, we will be able to show that Cσ(AS1)|Atoms(AS1) ⊆Cσ(AS+)|Atoms(AS1) holds, while combining the second and third of these sub-results will prove that Cσ(AS1)|Atoms(AS1) ⊇Cσ(AS+)|Atoms(AS1) holds. The proof for turning L1 into L+ works by first creating an admissible labeling Ladm + of J+ s.t.

Chunk 693

in(L1) ⊆in(Ladm + ). To do this, we use the following definition: Definition 18.

Chunk 694

Let AS1 = (Rs1,Rd1,n1,≤r1), AS2 = (Rs2,Rd2,n2,≤r2) be two AS s.t. AS1||AS2 and let AS+ be their union.

Chunk 695

Furthermore, let J1 = (A1,→1,⇒1,⪯1), J2 = (A2,→1,⇒2,⪯2) and J+ = (A+,→+,⇒+,⪯+) be the JSBAF’s corresponding to AS1, AS2 and AS+ respectively. Lastly, for i, j ∈{1,2} with i ̸= j, let Li ∈pr(Ji) and let SIMJj be the strict including minimal labeling of Jj.

Chunk 696

We construct the combined min- imal labeling of Li and SIMJj, denoted Lmin + , as follows: --- Page 36 --- April 2026 I0 = in(Li)∪in(SIMJj) Ik+1 = Ik ∪{a ∈A+ | ∃(S,a) ∈⇒+,S ⊆Ik} in(Lmin + ) = [ k≥0 Ik O0 = {a ∈A+ | ∃(b,a) ∈→+,b ∈in(Lmin + )} Ok+1 = Ok ∪{a ∈A+ | ∃(S,b) ∈⇒+,a ∈S, S\{a} ⊆in(Lmin + ), b ∈Ok} out(Lmin + ) = [ k≥0 Ok undec(Lmin + ) = A+ \ in(Lmin + )∪out(Lmin + )
Note that the only arguments in A+\(Ai∪Aj) that are accepted in Lmin + , are minimal A+-arguments. Furthermore, no “new” arguments in Ai and Aj are accepted by Lmin + , i.e.

Chunk 697

if a ∈in(Lmin + )∩Ai, then a ∈in(Li) and if a ∈in(Lmin + )∩Aj, then a ∈SIMJj. We now prove that Lmin + is indeed an admissible labeling.

Chunk 698

Later, we will use this result to show that any preferred labeling of Ji can be turned into a preferred labeling of J+. As usual, we have divided this proof into several parts to make it more accessible.

Chunk 699

Proposition 25. Let AS1 = (Rs1,Rd1,n1,≤r1), AS2 = (Rs2,Rd2,n2,≤r2) be two AS s.t.

Chunk 700

AS1||AS2 and let AS+ be their union. Furthermore, let J1 = (A1,→1,⇒1,⪯1), J2 = (A2,→1,⇒2,⪯2) and J+ = (A+,→+,⇒+,⪯+) be the JSBAF’s corresponding to AS1, AS2 and AS+ respectively.

Chunk 701

Lastly, for i, j ∈{1,2} with i ̸= j, for any preferred labeling Li ∈pr(Ji) and for SIMJj being the strict including minimal labeling of Jj, let Lmin + be the combined minimal labeling of Li and SIMJj. Then Lmin + is a labeling.

Chunk 702

Proof. We begin by arguing that Lmin + is a labeling.

Chunk 703

Similar to the proof of Proposition 13, the actual proof proceeds by induction over the construction of out(Lmin + ). However, since it is clear from the construction of Lmin + that we only need to ensure in(Lmin + )∩out(Lmin + ) = /0 and since the induction step of this proof is trivial, we focus here on the induction start.

Chunk 704

That is, we argue why there cannot be some a ∈in(Lmin + )∩O0: Towards a contradiction, suppose that this does not hold, i.e. there are arguments a,b ∈in(Lmin + ) s.t.

Chunk 705

(b,a) ∈→+. We first note that both a and b are consistent arguments.

Chunk 706

If they are contained in in(Li) or in in(SIMJj), then this is easy to see because these are admissible labelings. If they are contained in A+ \ (Ai ∪Aj), then we can make a simple model-theoretic argument to show the consistency: For c ∈{a,b}, let c ∈A+ \ (Ai ∪Aj).

Chunk 707

Towards a contradiction, assume that c is inconsistent, i.e. there is Γ ⊆Sub(c)C s.t.

Chunk 708

Γ is unsatisfiable. Take ADSub(c)|ASi and ADSub(c)|AS j.

Chunk 709

Note that these sets of arguments are contained in in(Li) and in(SIMJj) respectively. Because Li and SIMJj are admissible labelings, we can now infer that both ADSub(c)|ASi
C and ADSub(c)|AS j
C are satisfiable, i.e.

Chunk 710

there exists interpretations I1 ⊨M VADSub(c)|ASi
C and I2 ⊨M VADSub(c)|AS j
C. Note that these are sets of syntactically disjoint formulas.

Chunk 711

Now there exists an interpretation I s.t. --- Page 37 --- April 2026 I ⊨M ADSub(c)|ASi
C and I ⊨M ADSub(c)|ASi
C.

Chunk 712

By Proposition 10 we can now infer that Γ is satisfied by I, contradicting our assumptions. Now for the actual proof: Because we used the admissible labelings Li and SIMJj as a starting point for the construction of in(Lmin + ), we know that we cannot have a,b ∈Ai or a,b ∈Aj.

Chunk 713

By Proposition 21 (which corresponds to case one of Illustration 1), we can also infer that we cannot have a ∈Ai and b ∈Aj or a ∈Aj and b ∈Ai, as this would mean either a or b is attacked by a strict argument and therefore not accepted by Li or SIMJj respectively. This means at least one of the arguments a and b needs to be from A+ \(Ai ∪Aj).

Chunk 714

Suppose first that we have a ∈Ai and b ∈A+ \ (Ai ∪Aj) (this corresponds to case two of Illustration 1). Then a ∈in(Li) by construction of Lmin + .

Chunk 715

From Propo- sition 22, we can infer that there exists an argument b′ ∈Ai s.t. (b′,a) ∈→i and CSub(b)|ASi = CSub(b′).

Chunk 716

By construction of Lmin + , we have CSub(b′) ⊆in(Li) and by admissibility of Li, we can infer that b′ ∈in(Li) must also hold. Now (b′,a) ∈→i con- tradicts the admissibility of Li.

Chunk 717

The same argument can be made for the case a ∈Aj and b ∈A+ \(Ai ∪Aj). Next, suppose that we have a ∈A+ \ (Ai ∪Aj) and b ∈Ai (this corresponds to case three of Illustration 1).

Chunk 718

Note again that this means b ∈in(Li) by construction of Lmin + . From Proposition 23, we can infer that there exist arguments b′,a′ ∈Ai, s.t.

Chunk 719

CSub(b′) = CSub(b), CSub(a′) = CSub(a)|ASi and (b′,a′) ∈→i, or that there exists arguments b′,a′ ∈Ai s.t. CSub(b′) = CSub(b), CSub(a′) = ADSub(a)|ASi and either (b′,a′) ∈→i or (a′,b′) ∈→i.

Chunk 720

By construction of in(Lmin + ), we must have CSub(a)|ASi ⊆ in(Li). Furthermore, since Li was a preferred labeling, we can infer from Lemma 2 that CSub(b) ⊆in(Li).

Chunk 721

Similarly, we can use Lemma 2 and CSub(a)|ASi ⊆in(Li) to infer that ADSub(a)|ASi ⊆in(Li) also holds. Now we can use admissibility of Li to infer Li(a′) = Li(b′) = IN.

Chunk 722

However, now (b′,a′) ∈→i and (a′,b′) ∈→i both contradict the admissibility of Li. The same argument holds for the case that a ∈A+ \ (Ai ∪Aj) and b ∈Aj (note that SIMJj is by definition closed under sub-arguments).

Chunk 723

Lastly, let us suppose that we have a,b ∈A+ \ (Ai ∪Aj) (this corresponds to case four of Illustration 1). The argumentation in this case is similar to the one that cor- responds to case three of Illustration 1.

Chunk 724

However, now we need to use Proposition 24 in order to infer that there are arguments a′,b′ ∈Ai or a′,b′ ∈Aj for which we have Li(a′) = Li(b′) = IN or SIMJj(a′) = SIMJj(b′) = IN. Either way, we can again infer a contradiction, either for Li being admissible or for SIMJj being admissible.

Chunk 725

As all pos- sible cases lead to a contradiction, we conclude that in(Lmin + )∩O0 = /0 must hold, just as we claimed. Proposition 26.

Chunk 726

Let AS1 = (Rs1,Rd1,n1,≤r1), AS2 = (Rs2,Rd2,n2,≤r2) be two AS s.t. AS1||AS2 and let AS+ be their union.

Chunk 727

Furthermore, let J1 = (A1,→1,⇒1,⪯1), J2 = (A2,→1,⇒2,⪯2) and J+ = (A+,→+,⇒+,⪯+) be the JSBAF’s corresponding to AS1, AS2 and AS+ respectively. Lastly, for i, j ∈{1,2} with i ̸= j, for any preferred labeling Li ∈pr(Ji) and for SIMJj being the strict including minimal labeling of Jj, let Lmin + be the combined minimal labeling of Li and SIMJj.

Chunk 728

We have a ∈out(Lmin + ), iff a is legally OUT w.r.t. Lmin + .

Chunk 729

Proof. From the construction of out(Lmin + ), it is easy to see that, if a is legally OUT w.r.t.

Chunk 730

Lmin + , then a ∈out(Lmin + ). We therefore focus here on the other direction, i.e.

Chunk 731

showing --- Page 38 --- April 2026 that if a ∈out(Lmin + ), then a is legally OUT w.r.t. Lmin + .

Chunk 732

Similar to the proof of Proposi- tion 16, it is not immediately clear from the construction of out(Lmin + ), that this holds. More precisely, consider the case that we have x ∈Ok+1 \ Ok.

Chunk 733

Then there must exist some support chain C =  (S0,b0),...,(Sn,bn) ⊆⇒s.t. x ∈S0, for all 0 < k ≤n we have Sk \ {bk−1} ⊆in(Lmin + ), for S0 we have S0 \ {x} ⊆in(Lmin + ) and there exists an at- tack (y,bn) ∈→with Lmin + (y) = IN.

Chunk 734

However, it is not necessarily the case that we have x ⪯+ d for all d ∈S0 \{x}. Assume that there is d ∈S0 \{x} s.t.

Chunk 735

d ≺+ x. As in the proof of Proposition 16, we will show that this implies the existence of an “alternative reason” for why x is legally labeled OUT w.r.t.

Chunk 736

Lmin + . We begin by constructing some helper-arguments: Let S = S0 \ {x}
∪ S 0<k≤n Sk \ {bk−1}
, i.e.

Chunk 737

S contains all arguments labeled IN along the support chain C . Anal- ogous to the proof of Proposition 16, for {z0,...,zm} = S we have ADSub(x)C ⊨C ¬VADSub(y) ∪ADSub(z0) ∪··· ∪ADSub(zm)
C.

Chunk 738

Now we construct the arguments y′ : y,z0,...,zm →V{y,z0,...,zm}C and x′ : x0,...,xk →¬V∆, where {x0,...,xk} = ADSub(x) and ∆= ADSub(y)∪ADSub(z0)∪···∪ADSub(zm)
C. Clearly, x′ gen-rebuts y′.

Chunk 739

Furthermore, because we have d ≺+ x, there must exist rd ∈DR(d) ⊆DR(y′) s.t. for all rx ∈DR(x) = DR(x′), we have rd ≤+ r rx.

Chunk 740

This implies y′ ⪯+ x′, therefore x′ ̸≺+ y′ and thus (x′,y′) ∈→+. Note that by construction of Lmin + , y′ ∈in(Lmin + ) holds.

Chunk 741

Next, we make a case distinction based on the origins of x′,y′: Suppose first that we have x′,y′ ∈Ai. By construction of in(Lmin + ), we can infer that y′ ∈in(Li) must hold and by admissibility of Li, this implies x′ ∈out(Li).

Chunk 742

By construction of x′, we obviously have DR(x′) ⊆DR(x) and ADSub(x′)C ⊆ADSub(x)C. By assumption, Li is a preferred labeling, which means it is admissible.

Chunk 743

Now we can apply Proposition 15 to infer x ∈ out(Li), which implies that x is legally OUT w.r.t. Li either due to an attack or due to a support chain.

Chunk 744

Because in(Li) ⊆in(Lmin + ) we can now infer that x is legally OUT w.r.t. Lmin + as required.

Chunk 745

The case x′,y′ ∈Aj can be proven analogously. Next, assume that we have x′ ∈Aj and y′ ∈Ai (this corresponds to case one of Illustration 1).

Chunk 746

By Proposition 21, we can now infer that either x′ or y′ are inconsistent. Similar to before, we can infer from the construction of Lmin + that y′ ∈in(Li) must hold.

Chunk 747

Since Li is an admissible labeling, y′ must be consistent, therefore x′ must be inconsistent. This means there exists a strict argument x′ and an attack (x′,x) ∈→j.

Chunk 748

By Proposition 11, we can use this strict argument to construct a strict argument which is attacking x. This means x is legally OUT w.r.t.

Chunk 749

Lmin + as required. The case x′ ∈Ai and y′ ∈Aj can be proven analogously.

Chunk 750

Next, assume that we have y′ ∈A+\(Ai∪Aj), while x′ ∈Ai holds (this corresponds to case three of Illustration 1). Note that by construction of in(Lmin + ), this implies that y′ is a minimal A+-argument.

Chunk 751

Furthermore, we want to point out that we can assume both y′ and x′ are consistent: For y′ we can make a model-theoretic argument similar as in the proof of Proposition 25, while if x′ is inconsistent then we can immediately infer that x must be inconsistent as well. By Proposition 23, we can now have two cases: 1.

Chunk 752

There exists arguments x′′,y′′ ∈Ai s.t. CSub(x′′) = CSub(x′), CSub(y′′) = CSub(y′)|ASi and (x′′,y′′) ∈→i.

Chunk 753

2. There exist arguments x′′,y′′ ∈Ai s.t.

Chunk 754

CSub(x′′) = CSub(x′), CSub(y′′) = ADSub(y′)|ASi and either (x′′,y′′) ∈→i, or (y′′,x′′) ∈→i --- Page 39 --- April 2026 In the first case, we can infer from y′ ∈in(Lmin + ) that CSub(y′)| ASi ⊆in(Li) must hold. Because Li is an admissible labeling, we can now infer y′′ ∈in(Li), which im- plies x′′ ∈out(Li).

Chunk 755

We want to point out that by construction of x′ and x′′, we have ADSub(x) = ADSub(x′) = ADSub(x′′) and DR(x) = DR(x′) = DR(x′′). In particular, this means DR(x′′) ⊆DR(x) and ADSub(x′′)C ⊆ADSub(x)C.

Chunk 756

Since Li is an admissible la- beling, we can now apply Proposition 15 again to infer that Li(x) = OUT must hold. This implies that there is an attack or a support chain s.t.

Chunk 757

x is legally OUT w.r.t. Li.

Chunk 758

By in(Li) ⊆in(Lmax + ) we can now infer that x is legally OUT w.r.t. Lmin + as required.

Chunk 759

In the second case, we first use Lemma 2 (and the fact that SIMJj is by definition closed under sub-arguments) to infer that ADSub(y′)|ASi ⊆in(Li) holds. This, in turn, implies y′′ ∈in(Li).

Chunk 760

Now, if (y′′,x′′) ∈→i, we can use the argument y′′ as a reason for why x′′ is legally OUT w.r.t. Li.

Chunk 761

On the other hand, if (x′′,y′′) ∈→i, we can use the admissibility of Li to infer that x′′ must be legally OUT w.r.t. Li.

Chunk 762

Either way, we can again use the reason for why x′′ is legally OUT w.r.t. Li to construct a reason for why x is legally OUT w.r.t.

Chunk 763

Li. By in(Li) ⊆in(Lmin + ) we can infer that x is legally OUT w.r.t.

Chunk 764

Lmin + as required. The case y′ ∈A+ \(Ai ∪Aj) and x′ ∈Aj can be proven analogously.

Chunk 765

For our last few cases, suppose that we have x′ ∈A+\(Ai∪Aj). Note that x′ doesn’t necessarily need to be a minimal A+-argument.

Chunk 766

Thus, we first construct such a minimal A+-argument so that we can apply either Proposition 22 or Proposition 24: Remember that x′ is of the form x′ : x0,...,xk →¬V∆, where for each 0 ≤l ≤k, we have xl ∈ ADSub(x). This implies Atoms(xC l ) ⊆Atoms(ASi) or Atoms(xC l ) ⊆Atoms(ASj) for each of these arguments.

Chunk 767

Note that, if any of the arguments xl is inconsistent, then x is also inconsistent. This implies x is attacked by a strict argument and thus legally OUT w.r.t.

Chunk 768

Lmin + because of this strict attacker. We therefore assume that all the arguments xl are consistent.

Chunk 769

Now we can use Proposition 19 to infer that for each 0 ≤l ≤k, there exists an argument x′ l which is the reduced version of xl either w.r.t. ASi or w.r.t.

Chunk 770

AS j (depending on whether Atoms(xC l ) ⊆Atoms(ASi) or Atoms(xC l ) ⊆Atoms(ASj)). With these reduced versions, we can construct the argument x′′ which is of the form x′′ : x′ 0,...x′ k →¬V∆.

Chunk 771

Note that we have DR(x′′) ⊆DR(x′) = DR(x), which implies (x′′,y′) ∈→+. Furthermore, x′′ is a minimal A+-argument and by construction of x′ and x′′, we have ADSub(x′′)C ⊆ ADSub(x)C.

Chunk 772

Lastly, we can assume that x′′ is consistent, since otherwise x would be inconsistent and legally OUT w.r.t. Lmin + as required.

Chunk 773

Now, assume that we have y′ ∈Ai (this corresponds to case two of Illustration 1). By construction of y′, this implies y′ ∈in(Li) ⊆in(Lmin + ).

Chunk 774

Furthermore, we can infer that y′ is consistent. Because x′′ is a (consistent) minimal A+-argument, we can now use Proposition 22 to infer that there exists an argument x′′′ ∈Ai s.t.

Chunk 775

CSub(x′′)|ASi = CSub(x′′′) and (x′′′,y′) ∈→i. By admissibility of Li, we can now infer that x′′′ ∈out(Li) must hold.

Chunk 776

This implies x′′′ is legally OUT w.r.t. Li, which means there is either an attack (u,x′′′) ∈→i or a support chain C =  (S0,b0),...,(Sn,bn) giving us a reason for why x′′′ is legally OUT w.r.t.

Chunk 777

Li. Note that, by CSub(x′′)|ASi = CSub(x′′′), we have DR(x′′′) ⊆ DR(x′′) ⊆DR(x) and ADSub(x′′′)C ⊆ADSub(x′′)C ⊆ADSub(x)C.

Chunk 778

Suppose first that we have an attack (u,x′′′) ∈→i with u ∈in(Li). Then u is either undercutting or gen-rebutting x′′′.

Chunk 779

If u is undercutting x′′′, then we can use DR(x′′′) ⊆ DR(x) to infer that (u,x) ∈→+ also holds. If u is gen-rebutting x′′′, then we can again use Proposition 11 as well as ADSub(x′′′)C ⊆ADSub(x)C and DR(x′′′) ⊆DR(x) to infer that there exists u′ ∈in(Li) s.t.

Chunk 780

(u′,x) ∈→+. Either way, there exists an attacker of x --- Page 40 --- April 2026 which is labeled IN in Li.

Chunk 781

By in(Li) ⊆in(Lmin + ) we can infer that x is legally OUT w.r.t. Lmin + as required.

Chunk 782

Now suppose that x′′′ is legally OUT w.r.t. Li due to a support chain C =  (S0,b0),...,(Sn,bn) and an attack (c,bn) ∈→i with x′′′ ∈S0 and Li(c) = IN.

Chunk 783

Let S = S0 \ {x′′′} ∪ S 0<k≤n Sk \ {bk−1} be the set of all arguments labeled IN in this sup- port chain. If (c,bn) ∈→i is the result of an undercut, we can use S ⊆in(Li) to infer that c must be undercutting x′′′.

Chunk 784

By DR(x′′′) ⊆DR(x), we can infer that c is also un- dercutting x and since in(Li) ⊆in(Lmax + ), this means x is legally OUT w.r.t. Lmin + as re- quired.

Chunk 785

Now suppose that (c,bn) ∈→i is the result of a gen-rebut. We construct the argu- ment b′ ∈A+ \(Ai ∪Aj) which is of the form b′ : x,d0,...dm →V{x,d0,...,dm}C, for {d0,...,dm} = S.

Chunk 786

Furthermore, we use Proposition 11 to construct the argument c′ which is of the form c′ : c →¬VADSub(bn)C. Note that we have ADSub(bn)C ⊆ADSub(b′)C because ADSub(x′′′)C ⊆ADSub(x)C.

Chunk 787

This means c′ is also gen-rebutting b′. Because (c,bn) ∈→i is the result of a gen-rebut, we have c ̸≺i bn.

Chunk 788

Since DR(x′′′) ⊆DR(x), we can now infer that c ̸≺+ b′ also holds. Because DR(c) = DR(c′) this means (c′,b′) ∈→+.

Chunk 789

By admissibility of Li, we can infer c′ ∈in(Li) ⊆in(Lmin + ). By construction of Lmin + we now have b′ ∈O0 ⊆out(Lmin + ).

Chunk 790

We can make a simple inductive argument to show that we have x′′′ ⪯i d for all d ∈S. This means for all d ∈S, there is rx ∈DR(x′′′) s.t.

Chunk 791

for all rd ∈DR(d), rx ≤i rd. Because DR(x′′′) ⊆DR(x) and because ≤ri ⊆≤+ by construction of AS+ from ASi and AS j, we now have x ⪯+ d for all d ∈S.

Chunk 792

Since S ⊆in(Li) ⊆in(Lmin + ), we can now infer that x is legally OUT w.r.t. Lmin + due to the support chain C ′ =  (S ∪{x},b′) and the attack (c′,b′) ∈→+.

Chunk 793

The case y′ ∈Aj can be proven analogously. Lastly, suppose that we have y′ ∈A+ \ (Ai ∪Aj) (this corresponds to case four of Illustration 1).

Chunk 794

Then both x′ and y′ are minimal A+-arguments. Note that we can argue for the consistency of y′ and x′ as before: For y′ we can make a model-theoretic argument for its consistency, while inconsistency of x′ implies that x is legally OUT w.r.t.

Chunk 795

Lmin + . Now we can use Proposition 24 to infer that one of the following four cases must hold: 1.

Chunk 796

There are x′′,y′′ ∈Ai s.t. CSub(x′)|ASi = CSub(x′′), CSub(y′)|ASi = CSub(y′′) and (x′′,y′′) ∈→i.

Chunk 797

2. There are x′′,y′′ ∈Aj s.t.

Chunk 798

CSub(x′)|ASj =CSub(x′′), CSub(y′)|ASj =CSub(y′′) and (x′′,y′′) ∈→j. 3.

Chunk 799

There are x′′,y′′ ∈Ai s.t. CSub(x′)|ASi = CSub(x′′), ADSub(y′)|ASi = CSub(y′′) and either (x′′,y′′) ∈→i or (y′′,x′′) ∈→i.

Chunk 800

4. There are x′′,y′′ ∈Aj s.t.

Chunk 801

CSub(x′)|ASj = CSub(x′′), ADSub(y′)|AS j = CSub(y′′) and either (x′′,y′′) ∈→j or (y′′,x′′) ∈→j. With the same reasoning that we used to prove the cases which corresponded to cases two and three of Illustration 1, we can show that in all of the four cases described above, x′′ needs to be legally OUT w.r.t.

Chunk 802

Li or L j (depending on the specific case). Because we have DR(x′′) ⊆DR(x), ADSub(x′′)C ⊆ADSub(x)C and in(Li) ∪in(SIMJj)
⊆in(Lmin + ), we can use this to infer that x is legally OUT w.r.t.

Chunk 803

Lmin + , similar to the cases corresponding to cases two and three of Illustration 1. We conclude that regardless of the origins of x′ and y′, if x ∈out(Lmin + ), then x is legally OUT w.r.t.

Chunk 804

Lmin + . Proposition 27.

Chunk 805

Let AS1 = (Rs1,Rd1,n1,≤r1), AS2 = (Rs2,Rd2,n2,≤r2) be two AS s.t. AS1||AS2 and let AS+ be their union.

Chunk 806

Furthermore, let J1 = (A1,→1,⇒1,⪯1), J2 = --- Page 41 --- April 2026 (A2,→1,⇒2,⪯2) and J+ = (A+,→+,⇒+,⪯+) be the JSBAF’s corresponding to AS1, AS2 and AS+ respectively. Lastly, for i, j ∈{1,2} with i ̸= j, for any preferred labeling Li ∈pr(Ji) and for SIMJj being the strict including minimal labeling of Jj, let Lmin + be the combined minimal labeling of Li and SIMJj.

Chunk 807

If a ∈in(Lmin + ), then a is legally IN w.r.t. Lmin + .

Chunk 808

Proof. Let a ∈in(Lmin + ) be an accepted argument.

Chunk 809

We first consider the supports (S,b) ∈ ⇒+ where a ∈S holds: By construction of in(Lmin + ), if S ⊆in(Lmin + ), then b ∈in(Lmin + ) also holds. Furthermore, by construction of out(Lmin + ), if |S \ in(Lmin + )| = 1 while b ∈ out(Lmin + ), then S\in(Lmin + ) ∈out(Lmin + ) also holds.

Chunk 810

Therefore, the conditions of items 1.a to 1.c of Definition 2 are always satisfied and we only need to show that all attackers of a are labeled OUT in Lmin + . Note that, when we argued that Lmin + is a labeling, we have already shown that, if a ∈in(Lmin + ) and there is (b,a) ∈→+, then Lmin + (b) ̸= IN.

Chunk 811

Thus, we only have left to prove that Lmin + (b) ̸= UNDEC. Towards a contradiction, suppose that this does not hold.

Chunk 812

We proceed in a similar matter as for the proof that showed Lmin + is a labeling by considering all possible cases for the origins of a and b. First, suppose that a,b ∈Ai (or a,b ∈Aj).

Chunk 813

Then b is legally OUT w.r.t. Li (or SIMJj respectively) and because in(Li) ∪in(SIMJj) ⊆in(Lmin + ), we can infer that b is legally OUT w.r.t.

Chunk 814

Lmin + . By Proposition 26 this implies Lmin + (b) = OUT, contradicting our assumption Lmin + (b) = UNDEC.

Chunk 815

Next, suppose that a ∈Ai and b ∈Aj (or a ∈Aj and b ∈Ai), which corresponds to case one of Illustration 1. Then by Proposition 21, either a or b are legally OUT w.r.t.

Chunk 816

Lmin + , which again means either Lmin + (a) = OUT or Lmin + (b) = OUT, contradicting our assumptions. Now suppose that a ∈Ai, while b ∈A+ \(Ai ∪Aj), which corresponds to case two of Illustration 1 (we leave out the case a ∈Aj, while b ∈A+ \ (Ai ∪Aj), as it can be proven analogous).

Chunk 817

From (b,a) ∈→+, we can infer that b undercuts or gen-rebuts a. In the first case, we have Atoms(bC) ⊆Atoms(ASi).

Chunk 818

In the second case, we can use Propo- sition 11 to infer that there exists b′ ∈A+ which is of the form b′ : b →¬VADSub(a)C. Note that we have Atoms(b′C) ⊆Atoms(ASi).

Chunk 819

Now, regardless of whether the attack (b,a) was the result of an undercut or a gen-rebut, there is some (x,a) ∈→+ s.t. Atoms(x) ⊆Atoms(ASi).

Chunk 820

By Proposition 19, we can infer that there exists a reduced ver- sion of x w.r.t. ASi, i.e.

Chunk 821

some x′ ∈Ai for which we have xC = x′C and DR(x′) ⊆DR(x). Now x′ undercuts or gen-rebuts a.

Chunk 822

If x′ gen-rebuts a, then we note that (x,a) ∈→+ means x ̸≺a, which in turn implies x′ ̸≺a. Thus we can infer that (x′,a) ∈→i must hold, re- gardless of whether x′ undercuts or gen-rebuts a.

Chunk 823

Since Li was an admissible labeling and a ∈in(Li), we can infer that Li(x′) = OUT holds and because in(Li) ⊆in(Lmin + ), we must also have Lmin + (x′) = OUT by construction of Lmin + . Now we can use Lemma 3 and Propo- sition 26, to infer that Lmin + (x) = OUT also holds, contradicting our assumption that a is attacked by an argument which is labeled UNDEC.

Chunk 824

This proves case two of Illustration 1. Now, suppose that we have a ∈A+ \ (Ai ∪Aj) and b ∈Ai, which corresponds to case three of Illustration 1 (we again leave out the case a ∈A+ \(Ai ∪Aj) and b ∈Aj, as it can be proven analogous).

Chunk 825

Suppose that (b,a) ∈→+ is the result of an undercut. Then this attack must also be directed towards some x ∈CSub(a) and by construction of in(Lmin + ), we have x ∈Ai or x ∈Aj.

Chunk 826

In the first case, we can again infer that x is legally OUT w.r.t. Li by admissibility of Li and by construction of Lmin + , while in the second case either x or b must be legally OUT w.r.t.

Chunk 827

SIMJj or Li respectively. All of these options contradict our assumptions for the labels of a and b.

Chunk 828

--- Page 42 --- April 2026 Now suppose that (b,a) ∈→+ is the result of a gen-rebut. By Proposition 11, we can again infer that there must be an argument b′ ∈Ai of the form b′ : b →¬VADSub(a)C.

Chunk 829

Since DR(b) = DR(b′), it is clear that (b′,a) ∈→+. We can make a simple model- theoretic argument to show that ADSub(a)|ASj
C ∪{b′C} ⊨C ¬VADSub(a)|ASi
C.

Chunk 830

With this, we can construct the argument c ∈A+ \ (Ai ∪Aj) and the arguments a′,a′′ ∈Ai, as follows: c : aj0,...,ajm,b′ →¬VADSub(a)|ASi
C, a′ : ai0,...,ain →VADSub(a)| ASi
C and a′′ : a′ →¬¬VADSub(a)|ASi
C, where {aj0,...,ajm} = ADSub(a)|AS j and {ai0,...,ain} = ADSub(a)|ASi. Note that we have ADSub(a)|ASi ⊆in(Li) by Lemma 2 and ADSub(a)|ASj ⊆in(SIMJj) by definition of the strict including minimal labeling.

Chunk 831

Clearly, c gen-rebuts a′ and a′′ gen-rebuts c. It is also clear that either (c,a′) ∈→+ or (a′′,c) ∈→+ must hold.

Chunk 832

Suppose first that we have (c,a′) ∈→+. From ADSub(a)|ASi ⊆in(Li), we can infer a′ ∈in(Li) ⊆in(Lmin + ).

Chunk 833

Now the attack (c,a′) corresponds to case two of Illustration 1. We have argued above, that in this case, a′ ∈in(Lmin + ) implies c ∈out(Lmin + ).

Chunk 834

Since ADSub(a)| ASj = {aj0,...,ajm} ⊆in(Lmin + ), we can infer from the construction of out(Lmin + ) that we must have b′ ∈out(Lmin + ), which also implies b ∈out(Lmin + ). However, now b ∈out(Lmin + ) contradicts our assumption Lmin + (b) = UNDEC.

Chunk 835

Now suppose that we have (a′′,c) ∈→+. From ADSub(a)|ASi = {ai0,...,ain} ⊆in(Li), we can infer that a′′ ∈in(Li) ⊆in(Lmin + ) must hold.

Chunk 836

Now c is legally OUT w.r.t. Lmin + and by Proposition 26, this implies Lmin + (c) = OUT, which in turn implies that b′ and b are legally OUT.

Chunk 837

Again, we can now infer Lmin + (b) = OUT, which contradicts our assumption Lmin + (b) = UNDEC. This proves case three of Illustration 1.

Chunk 838

Finally, suppose that we have a,b ∈A+ \(Ai ∪Aj), which corresponds to case four of Illustration 1. First, let us assume that (b,a) ∈→+ is the result of an undercut.

Chunk 839

Then this attack must also be directed towards some a′ ∈CSub(a) ∈Ai ∪Aj. This corresponds to case two of Illustration 1.

Chunk 840

By the construction of in(Lmin + ), we can infer a′ ∈in(Lmin + ) from a ∈in(Lmin + ). We have argued above that this implies Lmin + (b) = OUT, contradicting the assumption Lmin + (b) = UNDEC.

Chunk 841

Now suppose that (b,a) ∈¸→+ is the result of a gen-rebut. The proof for this case proceeds analogous for that of case three of Illustration 1, which we have proven above: We construct the arguments b′ : b →¬VADSub(a)C (by Proposition 11), and with the sets ADSub(a)|ASj = {aj0,...,ajn} and ADSub(a)|ASi = {ai0,...,aim} we construct the argugment c : aj0,...,ajm,b′ →¬ ADSub(a)|ASi
C, the argument a′ : ai0,...,ain → VADSub(a)|ASi
C and the argument a′′ : a′ →¬¬VADSub(a)|ASi
C.

Chunk 842

Note that c ∈ A+ \ (Ai ∪Aj), while a′,a′′ ∈Ai. Furthermore, we have a′,a′′ ∈in(Li) ⊆in(Lmin + ) by Lemma 2 and by admissibility of Li.

Chunk 843

We again either have (c,a′) ∈→+ or (a′′,c) ∈→+. If (c,a′) ∈→+, then this corresponds to case two of Illustration 1, for which we have argued that Lmin + (c) = OUT, which – together with aj0,...,ajn ⊆in(SIMJj) ⊆in(Lmin + ) – implies that b′ and b are legally OUT, which means Lmin + (b) = OUT.

Chunk 844

On the other hand, if (a′′,c) ∈→+, then we can infer that c is legally OUT w.r.t. Lmin + , which again implies b′ and b are legally OUT, thus Lmin + (b) = OUT as required.

Chunk 845

We conclude: If a ∈in(Lmin + ), then a is legally IN w.r.t. Lmin + .

Chunk 846

Proposition 28. Let AS1 = (Rs1,Rd1,n1,≤r1), AS2 = (Rs2,Rd2,n2,≤r2) be two AS s.t.

Chunk 847

AS1||AS2 and let AS+ be their union. Furthermore, let J1 = (A1,→1,⇒1,⪯1), J2 = (A2,→1,⇒2,⪯2) and J+ = (A+,→+,⇒+,⪯+) be the JSBAF’s corresponding to AS1, --- Page 43 --- April 2026 AS2 and AS+ respectively.

Chunk 848

Lastly, for i, j ∈{1,2} with i ̸= j, for any preferred labeling Li ∈pr(Ji) and for SIMJj being the strict including minimal labeling of Jj, let Lmin + be the combined minimal labeling of Li and SIMJj. Then Lmin + is an admissible labeling of J+.

Chunk 849

Proof. By Proposition 25 we know that Lmin + is a labeling.

Chunk 850

By Proposition 26 we know that a ∈out(Lmin + ) iff a is legally OUT w.r.t. Lmin + .

Chunk 851

By Proposition 27 we know that a ∈ in(Lmin + ) implies a is legally IN w.r.t. Lmin + .

Chunk 852

Lastly, it is clear from the definition of in(Lmin + ) that STRJ+ ⊆in(Lmin + ) holds. With this, we have shown that any preferred labeling of Ji can be turned into an admissible labeling of J+.

Chunk 853

For the next step, we will show that any preferred labeling of J+ can be turned into an admissible labeling of Ji. We begin by defining the restriction of a labeling: Definition 19.

Chunk 854

Let AS1 = (Rs1,Rd1,n1,≤r1), AS2 = (Rs2,Rd2,n2,≤r2) be two AS s.t. AS1||AS2 and let AS+ be their union.

Chunk 855

Let J1 = (A1,→1,⇒1,⪯1), J2 = (A2,→1,⇒2,⪯2) and J+ = (A+,→+,⇒+,⪯+) be the JSBAFs corresponding to AS1, AS2 and AS+ re- spectively. Furthermore, let L+ be a labeling of J+.

Chunk 856

For any set of arguments A ⊆A+, we define the restriction of L+ to A , denoted L+|A , as the following labeling: in(L+| A ) = in(L+)∩A , out(L+|A ) = out(L+)∩A and undec(L+|A ) = undec(L+)∩A . Proposition 29.

Chunk 857

Let AS1 = (Rs1,Rd1,n1,≤r1), AS2 = (Rs2,Rd2,n2,≤r2) be two AS s.t. AS1||AS2 and let AS+ be their union.

Chunk 858

Let J1 = (A1,→1,⇒1,⪯1), J2 = (A2,→1,⇒2,⪯2) and J+ = (A+,→+,⇒+,⪯+) be the JSBAFs corresponding to AS1, AS2 and AS+ re- spectively. Furthermore, let L+ ∈pr(J+) be a preferred labeling of J+ and for any i ∈{1,2}, let Li = L+|Ai be the restriction of L+ to Ai.

Chunk 859

Then Li ∈adm(Ji). Proof.

Chunk 860

It is clear that Li is a labeling (i.e. that every argument receives exactly one label), that all strict arguments are accepted in Li, that a ∈in(Li) implies a is legally IN w.r.t.

Chunk 861

Li and that, if a is legally OUT w.r.t. Li, then Li(a) = OUT.

Chunk 862

Therefore, we only need to show that a ∈out(Li) implies a is legally OUT w.r.t. Li.

Chunk 863

The case is clear if L+(a) = OUT due to an attacker b ∈Ai or a support chain C = {(S0,b0),...(Sn,bn)} ⊆⇒i with a ∈S0. We therefore assume that these cases do not hold.

Chunk 864

First, suppose that L+(a) = OUT because of an attack (b,a) ∈→+, where b ∈A+ \ Ai and L+(b) = IN. We make a case distinction for the origins of b: If b ∈Aj, then we know by Proposition 21 that either a or b are legally OUT due to a strict argument.

Chunk 865

By assumption, L+(b) = IN and L+ was a preferred labeling, therefore we must have that a is legally OUT due to a strict argument a. This argument a is also present in Ai, therefore a is legally OUT w.r.t.

Chunk 866

Li as required. Next, suppose that L+(a) = OUT due to some attacker b ∈A+ \ (Ai ∪Aj) with L+(b) = IN.

Chunk 867

We make a distinction regarding for the reason of (b,a) ∈→+: If this attack results from an undercut, then we have Atoms(bC) ⊆Atoms(ASi). Let b′ be a reduced version of b w.r.t.

Chunk 868

ASi. Then b′ ∈Ai and bC = b′C, i.e.

Chunk 869

b′ undercuts a. By Lemma 3 we can infer that L+(b′) = Li(b′) = IN, which means a is legally OUT w.r.t.

Chunk 870

Li as required. Now assume (b,a) ∈→+ is the result of a gen-rebut.By Proposition 11 we infer that there exists an argument b′ ∈A+ of the form b′ : b →¬VADSub(a)C and an attack (b′,a) ∈→+.

Chunk 871

Now we take the sets {bi0,...,bim} = CSub(b′)|ASi = CSub(b)|ASi and {bj0,...,bjn} = CSub(b′)|ASj = CSub(b)|AS j. Note that Atoms(bC k ) ⊆Atoms(ASi) for --- Page 44 --- April 2026 0 ≤k ≤m and Atoms(bC l ) ⊆Atoms(ASj) for 0 ≤l ≤n, while Atoms(b′C) ⊆Atoms(ASi).

Chunk 872

Furthermore, note that {bi0,...,bim}C ∪{bj0,...,bjn}C ⊨C b′C. By Proposition 18, we can now infer that {b0,...,bm}C ⊨C ¬VADSub(a)C.

Chunk 873

Next, let b′ i0,...,b′ im ∈Ai be reduced versions of bi0,...,bim w.r.t. ASi.

Chunk 874

We construct the argument eb ∈Ai as follows: eb : b′ i0,...,b′ im →¬VADSub(a)C. It is clear that eb gen- rebuts a.

Chunk 875

Since each b′ ik is a reduced version of some bik ∈CSub(b), we have DR(b′ ik) ⊆ DR(b), therefore DR(eb) ⊆DR(b). By assumption, we have (b,a) ∈→+ as the result of a gen-rebut, therefore b ̸≺a.

Chunk 876

Now eb ̸≺a also holds and therefore (eb,a) ∈→i. Since b ∈in(L+) by assumption, we can use Lemma 2 to infer {bi0,...bim} ⊆in(L+).

Chunk 877

Now we can use Lemma 3 to infer that {b′ i0,...,b′ im} ⊆in(L+) holds. By admissibility of L+ we now have eb ∈in(L+) and by construction of Li from L+ we have eb ∈in(Li).

Chunk 878

Now (eb,a) ∈→i and Li(eb) = IN, thus a is legally OUT w.r.t. Li as required.

Chunk 879

Lastly, assume that L+(a) = OUT because of a support chain C =  (S0,b0),...,(Sn,bn) ⊆ ⇒+, with a ∈S0, (c,bn) ∈→+, L+(c) = IN, L+(bk) = OUT for 0 ≤k ≤n, S0 \ {a} ⊆ in(L+), Sk \{bk−1} ⊆in(L+) for 0 < k ≤n and a ⪯d for all d ∈S0 \{a}. We will show that this implies that there is an attack (ec,a) ∈→+ with L+(ec) = IN.

Chunk 880

As we have argued above, this implies a is legally OUT w.r.t. Li.

Chunk 881

We show the existance of such an argument ec via induction over n ∈N for n being the length of the support chain C . Induction start: n = 1, i.e.

Chunk 882

C =  (S0,b0) . Let S0 = {a,a0,...,am}.

Chunk 883

Suppose first that the attack (c,b0) ∈→+ is the result of an undercut. Then this attack must also be directed towards one of the arguments in S0.

Chunk 884

By S0 \ {a} ⊆in(L+), we can infer that c must be undercutting a. Now we have (c,a) ∈→+ and L+(c) = IN as required.

Chunk 885

Next, suppose that (c,b0) ∈→+ is the result of a gen-rebut. We first use Propo- sition 11 to infer that there exists c′ of the form c′ : c →¬VADSub(bn)C.

Chunk 886

It is easy to see that we have {c′}C ∪ S 0≤k≤m ADSub(ak)
C ⊨C ¬VADSub(a)C. With this, we construct the argument c′′ as c′′ : c0,...,cl →¬VADSub(a)C, where {c0,...,cl} = {c′}∪ S 0≤k≤m ADSub(ak).

Chunk 887

Clearly, c′′ gen-rebuts a. Furthermore, we have c′ ∈in(L+) by admissibility of L+ and – since the arguments a0,...,am are labeled IN by assumption – S 0≤k≤m ADSub(ak)
⊆in(L+) by Lemma 2.

Chunk 888

By admissibility of L+ we can now infer that L+(c′′) = IN also holds. Now, take ra ∈DR(a) s.t.

Chunk 889

ra is minimal (w.r.t. ≤+ r ) among all rules in DR(a).

Chunk 890

Be- cause r+ is a total pre-order and because a ⪯d for all d ∈S0 \ {a}, we can infer that for all al ∈{a0,...,am}, for all ral ∈DR(al), we have ra ≤+ r ral. We now claim that for all rc ∈DR(c), ra ≤rc must hold.

Chunk 891

To see this, assume towards a contradiction that there is some rc ∈DR(c) s.t. rc <+ r ra holds.

Chunk 892

By transitivity of ≤+ r , this implies rc <+ r r for all r ∈DR(b0). This means c ⪯bn.

Chunk 893

Because (c,bn) is the result of a gen-rebut, we have c ̸≺bn, i.e. either c ̸⪯bn or bn ⪯c.

Chunk 894

Obviously c ̸⪯bn contradicts c ⪯bn, therefore we assume bn ⪯c. This means there exists some rb ∈DR(bn) s.t.

Chunk 895

for all r′ c ∈DR(c), rb ≤+ r r′ c holds. Now for r′ c = rc in particular we can infer rc <+ r rb ≤+ r rc, a contradiction.

Chunk 896

With this, we have shown that for all rc ∈DR(c), ra ≤+ r rc holds. Now we can infer that for all r ∈DR(c′′), we have ra ≤+ r r.

Chunk 897

This, in turn, implies a ⪯c′′, therefore c′′ ̸≺a and (c′′,a) ∈→+ as required. Induction step: n →n+1, i.e.

Chunk 898

C =  (S0,b0),...,(Sn,bn),(Sn+1,bn+1) ⊆⇒+. Now C ′ =  (S1,b1),...,(Sn,bn),(Sn+1,bn+1) ⊆⇒+ is a support chain of length n.

Chunk 899

By the --- Page 45 --- April 2026 induction hypothesis, there now exists an attack (ec,b0) ∈→+ with L+(ec) = IN. Since  (S0,b0) is a support-chain of lengths 1 ≤n, we can now use the induction hypothesis again to infer that there is an attack (ec,a) ∈→+ for which we have L+(ec) = IN.

Chunk 900

With this, we have shown that, if a is legally OUT w.r.t. L+ due to a support chain, then there also exists an attacker of a which is labeled IN in L+.

Chunk 901

As we have argued above, this implies that a is legally OUT w.r.t. Li as required.

Chunk 902

Next, we will show that, when a labeling L+ of J+ is reduced to a labeling Li of Ji, this reduced labeling is not only admissible, but even preferred. For this, we first introduce a combined labeling similar to the combined minimal labeling of Definition 18: Definition 20.

Chunk 903

Let AS1 = (Rs1,Rd1,n1,≤r1), AS2 = (Rs2,Rd2,n2,≤r2) be two AS s.t. AS1||AS2 and let AS+ be their union.

Chunk 904

Let J1 = (A1,→1,⇒1,⪯1), J2 = (A2,→1,⇒2,⪯2) and J+ = (A+,→+,⇒+,⪯+) be the JSBAFs corresponding to AS1, AS2 and AS+ re- spectively. Furthermore, let L+ ∈pr(J+) be a preferred labeling of J+.

Chunk 905

Lastly, for any i ∈{1,2}, let Li = L+|Ai be the restriction of L+ to Ai and let L ∈pr(Ji) be a preferred labeling of Ji s.t. in(Li) ⊆in(L).

Chunk 906

We define the combined labeling of L and L+, denoted as Li+, as follows: Ii = in(L) Iold = in(L+)\Ai Inew0 =  a ∈A+ \(Ai ∪Aj) | CSub(a) ⊆Ii ∪Iold Inewk+1 =  a ∈A+ \(Ai ∪Aj) | CSub(a) ⊆Inewk ∪Inewk in(Li+) = Ii ∪Iold ∪ [ k≥0 Inewk O0 = {a ∈A+ | ∃(b,a) ∈Att+ with b ∈in(Li+)} Ok+1 =  a ∈A+ | ∃(S,b) ∈⇒+ with a ∈S, S\{a} ⊆in(Li+)and b ∈Ok ∪Ok out(Li+) = [ k≥0 Ok undec(Li+) = A+ \ in(Li+)∪out(Li+)
Intuitively, the combined labeling of Li and L+ can be obtained by labeling all arguments in either Li or L+ as IN and then iteratively accepting all arguments in A+ \(Ai ∪Aj) which need to be accepted to retain admissibility. We will now show that this construction yields an admissible labeling of J+ again.

Chunk 907

As ususal, we have divided the proof into several parts to make it more accesible: Proposition 30. Let AS1 = (Rs1,Rd1,n1,≤r1), AS2 = (Rs2,Rd2,n2,≤r2) be two AS s.t.

Chunk 908

AS1||AS2 and let AS+ be their union. Let J1 = (A1,→1,⇒1,⪯1), J2 = (A2,→1,⇒2,⪯2) --- Page 46 --- April 2026 and J+ = (A+,→+,⇒+,⪯+) be the JSBAFs corresponding to AS1, AS2 and AS+ re- spectively.

Chunk 909

Furthermore, let L+ ∈pr(J+) be a preferred labeling of J+. Lastly, for any i ∈{1,2}, let Li = L+|Ai be the restriction of L+ to Ai and let L ∈pr(Ji) be a preferred labeling of Ji s.t.

Chunk 910

in(Li) ⊆in(L). Then the combined labeling of L and L+ is a labeling.

Chunk 911

Proof. Let Li+ be the combined labeling of L and L+.

Chunk 912

Similar to the proofs of Proposi- tion 13 and Proposition 25, the actual proof proceeds via induction over the construction of out(Li+). However, it is clear that we only need to ensure in(Li+)∩out(Li+) = /0.

Chunk 913

Fur- thermore, the induction step of this proof is again trivial. We therefore only focus on the induction start, i.e.

Chunk 914

we show that there cannot be some a ∈in(Li+)∩O0. Towards a con- tradiction, suppose that this does not hold.

Chunk 915

Then there are a,b ∈in(Li+) s.t. (b,a) ∈→+.

Chunk 916

Note that both a and b are consistent arguments. If a,b ∈in(L) or a,b ∈Iold, then this is easy to see by the admissibility of L and L+.

Chunk 917

In the case that we have c ∈{a,b} with c ∈Inewk \(Ii ∪Iold) for some k ≥0, then we can make a model theoretic argument to show the consistency of c: Let {ci0,...,cik} =CSub(c)∩in(L) and let {cold0,...,coldl} = CSub(c) ∩in(L+) \ Ai. First, we construct the set Cnew = S ci∈{ci0,...,cik} ADSub(ci).

Chunk 918

Since L is a preferred labeling, we can use Lemma 2 to infer that Cnew ⊆in(L) holds. Next, for each argument cold ∈{cold0,...,coldl}, take ADSub(cold) = {bc0,...,bcp} and let ADcold = {bc′ 0,...,bc′ p} be their reduced versions w.r.t.

Chunk 919

ASi or AS j (depending on the respecitve top- rules). By Lemma 2 we can infer that ADSub(cold) ⊆in(L+) holds and by Lemma 3 this implies ADcold ⊆in(L+).

Chunk 920

Now, take Coldi = S cold∈{cold0,...,coldl } ADcold
∩Ai and Cold j = S cold∈{cold0,...,coldl } ADcold}
∩ Aj (note that reduced versions of arguments are contained in either Ai or Aj by Corollary 3). By construction of L from L+, we have in(L+) ∩Ai ⊆in(L), therefore Coldi ∪Cnew ⊆in(L).

Chunk 921

By admissibility of L, we can now infer that (Coldi ∪Cnew)C is satisfiable. Similarly, we can use the admissibility of L+ to infer that C C oldj is satisfiable.

Chunk 922

By construction, (Coldi ∪Cnew)C and C C oldj are syntactically disjoint sets of formulas, therefore there exists an interpretation I s.t. I ⊨M VCnew ∪Coldi ∪Coldj
C.

Chunk 923

It is clear that we have Cnew ∪Coldi ∪Coldj
C = ADSub(c)C. Now let Γ ⊆Sub(c)C.

Chunk 924

By Proposition 10 we can infer that Cnew ∪Coldi ∪Coldj
C ⊨C Γ holds, therefore I is a model of Γ. We conclude that each Γ ⊆Sub(c)C is satisfiable, therefore c is consistent.

Chunk 925

Now for the actual proof: Similar to the proof for Proposition 25, we will show the claim by going through the different cases for the origins of a and b. Note that for any c ∈{a,b}, if we have c ̸∈(Ai ∪Aj), we can infer from the construction of in(Li+) that CSub(c) ⊆in(L)∪in(L+) must hold.

Chunk 926

Because both L and L+ are preferred labelings, this also implies ADSub(c) ⊆in(L)∪in(L+) by Lemma 2. Now for the different cases: Because L and L+ are admissible labelings, it is clear that a,b ∈Ai, or a,b ∈Aj is a contradiction.

Chunk 927

By Proposition 21, we can also infer that a ∈Ai while b ∈Aj, and a ∈Aj while b ∈Ai both contradict the admissibility of L and L+. These two options correspond to case one of Illustration 1.

Chunk 928

Next, suppose that we have a ∈Ai and b ∈A+ \ (Ai ∪Aj), which corresponds to case two of Illustration 1 (the case a ∈Aj and b ∈A+ \ (Ai ∪Aj) can be proven analogous). Note that b is not necessarily a minimal A+-argument, as it might be the case that there is some b′ ∈CSub(b) ∩ A+ \ (Ai ∪Aj)
s.t.

Chunk 929

b′ is not a minimal A+- --- Page 47 --- April 2026 argument. However, we can construct an “alternative” version of b which is a mini- mal A+-argument: Let {bi0,...,bik} = Csub(b)∩Ai, {bj0,...,bjl} = CSub(b)∩Aj and {b+0,...,b+m} = CSub(b)∩ A+ \(Ai ∪Aj)
.

Chunk 930

For each b+ ∈{b+0,...,b+m}, let b′ + be a reduced version of b+ w.r.t. ASi or ASj (depending on TR(b+)).

Chunk 931

As we have mentioned above, we have b+ ∈in(L+), therefore we can use Lemma 3 to infer that b′ + ∈in(L+) also holds. Note that each b′ + is either in Ai or in Aj by Corollary 3.

Chunk 932

By construction of in(Li+) this also means that, if b′ + ∈Ai, then L(b′ +) = IN. Now we use bC + = b′C + , to construct a minimal A+ argument b′ which is of the form b′ : bi0,...,bik,b j0,...,bjl,b′ +0,...,b′ +m →bC.

Chunk 933

Obviously b′ undercuts or gen-rebuts a (depending on whether (b,a) ∈→+ was the result of an undercut or a gen-rebut). In the first case, we can immediately infer (b′,a) ∈→+, while in the second case we can use DR(b′) ⊆DR(b) to infer that b′ ̸≺a and therefore again (b′,a) ∈→+.

Chunk 934

Either way, we now have that b′ is a minimal A+-argument which is attacking a. By Proposition 22, this implies that there is b′′ ∈Ai s.t.

Chunk 935

(b′′,a) ∈→i and CSub(b′)|ASi = CSub(b′′). Because b ∈ in(Li+) by assumption and because L is closed under sub-arguments by Lemma 2, we can infer {bi0,...,bik} ⊆in(L).

Chunk 936

As we have argued above, we also have {b′ +0,...,b′ +m} ∩ Ai
⊆in(L+) ∩Ai ⊆in(L). Because L was a preferred labeling, we can therefore use Lemma 2 again to infer that CSub(b′′) ⊆in(L) holds.

Chunk 937

By admissibility of L we now have b′′ ∈in(L), contradicting a ∈in(Li+)∩Ai. Next, suppose that we have a ∈A+ \(Ai ∪Aj) while b ∈Ai (the case of a ∈A+ \ (Ai ∪Aj) while b ∈Aj can be proven analogously).

Chunk 938

Again, it is not necessarily the case that a is a minimal A+-argument, but we can construct an “alternative” version of a: Let {a0,...,am} = ADSub(a). As we have argued above, we know that {a0,...,am} ⊆ in(L)∪in(L+) holds.

Chunk 939

Let a′ 0,...,a′ m be reduced versions of a0,...,am (w.r.t. ASi or AS j, depending on the respective top-rules).

Chunk 940

By Lemma 3 we can infer that a′ 0,...a′ m ⊆in(L)∪ in(L+) also holds. By Corollary 3 we know that these arguments a′ 0,...a′ m are contained in Ai ∪Aj.

Chunk 941

Using Proposition 10, we can now construct the minimal A+-argument a′ as follows: a′ : a′ 0,...a′ m →aC. Note that, since we used ADSub(a) for the construction of a′, we have DR(a) = DR(a′).

Chunk 942

Now, if (b,a) ∈→+ is the result of an undercut, then we can immediately infer that (b,a′) ∈→+ also holds. On the other hand, if this attack results from a gen-rebut, then we can use Proposition 11 to construct the argument b′ of the form b′ : b →¬VADSub(a)C.

Chunk 943

Since ADSub(a)C = ADSub(a′)C, we can infer (b′,a′) ∈→+. Either way, there exists an argument eb ∈in(L) s.t.

Chunk 944

(eb,a′) ∈→+. By Proposition 23, we can now infer that one of the following two cases must hold: 1.

Chunk 945

There are a′′,eb′ ∈Ai s.t. CSub(eb′) = CSub(eb), CSub(a′′) = CSub(a′)|ASi and (eb′,a′′) ∈→i.

Chunk 946

2. There are a′′,eb′ ∈Ai s.t.

Chunk 947

CSub(a′′) = ADSub(a′)|ASi, CSub(eb′) = CSub(eb) and either (a′′,eb′) ∈→i or (eb′,a′′) ∈→i. By construction of a′′, we have CSub(a′′) = {a′ 0,...,a′ m}∩Ai
⊆in(L) and by admissi- bility of L this means a′′ ∈in(L) must also hold.

Chunk 948

Furthermore, we can use Lemma 2 and b ∈in(L) to infer that CSub(eb′) = CSub(b) ⊆in(L) must also hold. By admissibility of L, this also implies eb′ ∈in(L).

Chunk 949

Now both (a′′,eb′) ∈→i and (eb′,a′′) ∈→i contradict the admissibility of L. Lastly, suppose that we have a,b ∈A+ \(Ai ∪Aj), which corresponds to case four of Illustration 1.

Chunk 950

Analogous to the case above, we first construct minimal A+-arguments --- Page 48 --- April 2026 a′,b′ from a and b which are of the form a′ : a′ 0,...,a′ k →aC and b′ : b′ 0,...,b′ l →bC, where a′ 0,...,a′ k are reduced versions of the arguments in {a0,...,ak} = ADSub(a) and b′ 0,...,b′ l are reduced versions of the arguments in {b0,...,bl} = ADSub(b). Again, we can infer that {a′ 0,...,a′ k} ⊆in(L)∪in(L+) as well as {b′ 0,...,b′ l} ⊆in(L)∪in(L+) holds.

Chunk 951

Furthermore, we can again infer that (b′,a′) ∈→+ must hold (possibly utilizing Propo- sition 11 as well as DR(b) = DR(b′) and DR(a) = DR(a′), depending on whether or not (b,a) ∈→+ is the result of an undercut or a gen-rebut). Then we can apply Proposition 24 to infer for the arguments a′,b′ that one of the following four cases must hold: 1.

Chunk 952

There are a′′,b′′ ∈Ai s.t. CSub(a′)|ASi = CSub(a′′), CSub(b′)|ASi = CSub(b′) and (b′′,a′′) ∈→i.

Chunk 953

2. There are a′′,b′′ ∈Aj s.t.

Chunk 954

CSub(a′)|ASj = CSub(a′′), CSub(b′)|ASj = CSub(b′) and (b′′,a′′) ∈→j 3. There are a′′,b′′ ∈Ai s.t.

Chunk 955

ADSub(a′)|ASi = CSub(a′′), CSub(b′)|ASi = CSub(b′′) and either (a′′,b′′) ∈→i, or (b′′,a′′) ∈→i. 4.

Chunk 956

There are a′′,b′′ ∈Aj s.t. ADSub(a′)|AS j = CSub(a′′), CSub(b′)|ASj = CSub(b′′) and either (a′′,b′′) ∈→j, or (b′′,a′′) ∈→j.

Chunk 957

In each of these four cases we can infer a contradiction, since we either have a′′,b′′ ∈ in(L) (cases one and three) or a′′,b′′ ∈in(L+) (cases two and four). With this, we have shown that each possible case for a,b ∈in(L+) ∩O0 leads to a contradiction, as required.

Chunk 958

Proposition 31. Let AS1 = (Rs1,Rd1,n1,≤r1), AS2 = (Rs2,Rd2,n2,≤r2) be two AS s.t.

Chunk 959

AS1||AS2 and let AS+ be their union. Let J1 = (A1,→1,⇒1,⪯1), J2 = (A2,→1,⇒2,⪯2) and J+ = (A+,→+,⇒+,⪯+) be the JSBAFs corresponding to AS1, AS2 and AS+ re- spectively.

Chunk 960

Furthermore, let L+ ∈pr(J+) be a preferred labeling of J+. Lastly, for any i ∈{1,2}, let Li = L+|Ai be the restriction of L+ to Ai, let L ∈pr(Ji) be a preferred labeling of Ji s.t.

Chunk 961

in(Li) ⊆in(L) and let Li+ be the combined labeling of L and L+. We have a ∈out(Li+) iff a is legally OUT w.r.t.

Chunk 962

Li+. Proof.

Chunk 963

This proof is very similar to that of Proposition 26. We will therefore focus on the differences here.

Chunk 964

As in the proof for Proposition 26, the main task is to find an “alternative reason” for an argument a to be legally OUT in the case that a is labeled OUT in the construction of Li+ because of a support chain C =  (S0,b0),...,(Sn,bn) ⊆⇒+ and an attack (c,bn) ∈→+, where a ∈S0, for all 0 < k ≤n we have Sk \{bk−1} ⊆in(Li+), for S0 we have S0 \{a} ⊆in(Li+) and there is d ∈S0 \{a} s.t. d ≺a.

Chunk 965

The proof idea is also the same as for Proposition 26: Let S = S0\{a}∪ S 0<k≤n Sk\{bk−1}
= {d0,...,dm}, let c′ be an argument of the form c′ : c,d0,...,dm →V{c,d0,...,dm}C and for the set ADSub(a) = {a0,...,al}, let a′ be of the form a′ : a0,...,al →¬VADSub(c) ∪ADSub(d0) ∪··· ∪ ADSub(dm)
C. Note that by construction of Li+, this implies c′ ∈in(Li+), from which we can also infer CSub(c′) ⊆in(L)∪in(L+) and – by Lemma 2 – ADSub(c′) ⊆in(L)∪ in(L+).

Chunk 966

We have (a′,c′) ∈→+ and we use this as a starting point to infer that there either exists an attack or a support chain which satisfies the conditions for a to be legally OUT w.r.t. Li+.

Chunk 967

To do this, we go through the possible cases for the origins of a′ and c′. The cases for a′,c′ ∈Ai (or a′,c′ ∈Aj) and a′ ∈Ai while c′ ∈Aj (or a′ ∈Aj while c′ ∈Ai) are identical to those of Proposition 26.

Chunk 968

We therefore skip these cases and assume that we have a′ ∈Ai and c′ ∈A+ \(Ai ∪Aj), which corresponds to case three of Illustra- --- Page 49 --- April 2026 tion 1 (the case a′ ∈Aj and c′ ∈A+ \(Ai ∪Aj) can be proven analogous). Note that c′ is not necessarily a minimal A+-argument.

Chunk 969

We therefore utilize a similar technique as in the Proof of Propsotion 30 and construct an “equivalent” argument to c′ as follows: Let ADSub(c′) = {c0,...,cp} and let c′ 0,...,c′ p be the reduced versions of these arguments (w.r.t. ASi or ASj, depending on the respective top-rules).

Chunk 970

Then we construct the minimal A+-argument c′′ as follows: c′′ : c′ 0,...,c′ p →c′C. Note that by using ADSub(c′) as a start- ing point, we have DR(c′) = DR(c′′).

Chunk 971

Furthermore, since {c0,...,cp} ⊆in(L)∪in(L+), we can use Lemma 3 to infer that {c′ 0,...,c′ p} ⊆in(L)∪in(L+) also holds. By construc- tion of in(Li+), this also implies c′′ ∈in(Li+).

Chunk 972

Lastly, we want to point out that both a′ and c′′ are consistent arguments. For a′, we can assume consistency since we could oth- erwise directly construct an attack towards the original argument a, while for c′′ we can assume consistency since otherwise we would be able to construct an attack from a strict argument towards c′, which would contradict Li+(c′) = IN by construction of Li+ and Proposition 30.

Chunk 973

Now we can use Proposition 23 to infer that one of the following two cases must hold: 1. There are ea,ec ∈Ai s.t.

Chunk 974

CSub(ea) =CSub(a′), CSub(ec) =CSub(c′′)|ASi and (ea,ec) ∈ →i. 2.

Chunk 975

There are ea,ec ∈Ai s.t. CSub(ea) =CSub(a′), CSub(ec) = ADSub(c′′)|ASi and either (ea,ec) ∈→i or (ec, ea) ∈→i.

Chunk 976

Since {c′ 0,...,c′ p} ⊆in(L)∪in(L+) and since c′ 0,...,c′ p are the direct sub-arguments of c′′, we can use Lemma 2 and the admissibility of L to infer that in either of the two cases, ec ∈in(L) must hold. This implies ea ∈out(L) by admissibility of L.

Chunk 977

By construction of a′ and ea we have DR(ea) ⊆DR(a′) = DR(a) as well as ADSub(ea)C ⊆ADSub(a′)C = ADSub(a)C. Since a′ ∈Ai by assumption and since L is admissible, we can now use Proposition 15 to infer a′ ∈out(L).

Chunk 978

This means a′ is legally OUT w.r.t. L due to an attack or a support chain.

Chunk 979

Because DR(a′) = DR(a) and ADSub(a′)C = ADSub(a)C, we can use the same attacker or support chain to infer that a is legally OUT w.r.t. Li+ (possibly with the help of Proposition 11).

Chunk 980

The case that a′ ∈A+ \ (Ai ∪Aj while c′ ∈Ai (or c′ ∈Aj) is identical to that of Proposition 26 and we skip it here. This takes care of case two of Illustration 1.

Chunk 981

For the fourth and last case of Illustration 1, i.e. both a′ and c′ are arguments in A+ \(Ai ∪Aj), we first need to construct minimal A+-arguments a′′ and c′′ as follows: The argument c′′ is constructed as we did above, i.e.

Chunk 982

for ADSub(c′) = {c0,...,cp} we use reduced versions c′ 0,...,c′ p to construct c′′ as c′′ : c′ 0,...,c′ p →c′C. Analogously, for ADSub(a′) = {a0,...,al}, we use their reduced versions a′ 0,...,a′ l an construct a′′ as a′′ : a′ 0,...,a′ l →a′C.

Chunk 983

Note that we have {c0,...,cp} ⊆in(L)∪in(L+). By Lemma 3 this implies {c′ 0,...,c′ p} ⊆in(L)∪in(L+), which in turn implies c′′ ∈in(Li+) by construction of Li+.

Chunk 984

Since both a′′ and c′′ are minimal A+-arguments, we can now employ Proposi- tion 24 to infer that one of the following four cases holds: 1. There are ea,ec ∈Ai s.t.

Chunk 985

CSub(ea) = CSub(a′′)|ASi, CSub(ec) = CSub(c′′)|ASi and (ea,ec) ∈→i. 2.

Chunk 986

There are ea,ec ∈Aj s.t. CSub(ea) = CSub(a′′)|ASj, CSub(ec) = CSub(c′′)|ASj and (ea,ec) ∈→j.

Chunk 987

3. There are ea,ec ∈Ai s.t.

Chunk 988

CSub(ea) = CSub(a′′)|ASi, CSub(ec) = ADSub(c′′)|ASi and either (ea,ec) ∈→i, or (ec, ea) ∈→i. --- Page 50 --- April 2026 4.

Chunk 989

There are ea,ec ∈Aj s.t. CSub(ea) = CSub(a′′)|AS j, CSub(ec) = ADSub(c′′)|ASj and either (ea,ec) ∈→j, or (ec, ea) ∈→j.

Chunk 990

Since {c′ 0,...,c′ p} ⊆in(L) ∪in(L+), we can use Lemma 2 to infer that CSub(c′′) ⊆ in(L) ∪in(L+) and ADSub(c′′) ⊆in(L) ∪in(L+) holds, which in turn implies ec ∈in(L) (if ec ∈Ai) or ec ∈in(L+) (if ec ∈Aj) by admissibility of L and L+. In the first case, we can infer ea ∈out(L) by the admissibility of L, while in the second case we can in- fer ea ∈out(L+) by the admissibility of L+.

Chunk 991

Either way, we can use the fact that ea is legally OUT w.r.t. L or L+ and apply the same reasoning as before to infer that a is legally OUT w.r.t.

Chunk 992

Li+. Similar to our previous argumentations, this is possible because we have DR(ea) ⊆DR(a′′) ⊆DR(a′) = DR(a) as well as ADSub(ea)C ⊆ADSub(a′′)C ⊆ ADSub(a′)C = ADSub(a)C.

Chunk 993

This finishes the proof. Proposition 32.

Chunk 994

Let AS1 = (Rs1,Rd1,n1,≤r1), AS2 = (Rs2,Rd2,n2,≤r2) be two AS s.t. AS1||AS2 and let AS+ be their union.

Chunk 995

Let J1 = (A1,→1,⇒1,⪯1), J2 = (A2,→1,⇒2,⪯2) and J+ = (A+,→+,⇒+,⪯+) be the JSBAFs corresponding to AS1, AS2 and AS+ re- spectively. Furthermore, let L+ ∈pr(J+) be a preferred labeling of J+.

Chunk 996

Lastly, for any i ∈{1,2}, let Li = L+|Ai be the restriction of L+ to Ai, let L ∈pr(Ji) be a preferred labeling of Ji s.t. in(Li) ⊆in(L) and let Li+ be the combined labeling of L and L+.

Chunk 997

If a ∈in(Li+), then a is legally IN w.r.t. Li+.

Chunk 998

Proof. Let a ∈in(Li+).

Chunk 999

Similar to the proof of Proposition 27, it is clear from the con- struction of in(Li+) that, for any support (S,c) ∈⇒+ with a ∈S and a ⪯b for all b ∈S\{a}, one of the items 1.a−1.c of Definition 2 holds. Thus, we only need to ensure that all attackers of a are labeled OUT in Li+.

Chunk 1000

Let (b,a) ∈→+. Towards a contradiction, assume that Li+(b) ̸= OUT.

Chunk 1001

Note that, if Li+(b) = IN, then Li+(a) = OUT by construc- tion of Li+, which contradicts Li+(a) = IN by Proposition 30. Therefore, we assume that Li+(b) = UNDEC holds.

Chunk 1002

As usual, we go through the possible cases for the origins of a and b to derive a contradiction in each of them. Note that we can assume that both a and b are consistent, since they would otherwise be attacked by a strict argument, which again contradicts our assumptions on their labels (any strict argument x is contained in Ai and accepted by the admissible labeling L which implies x ∈in(Li+) and by construction of out(Li+) any argument y attacked by x is labeled OUT in Li+).

Chunk 1003

It is obvious that a,b ∈Ai or a,b ∈Aj contradicts the admissibility of L or L+. Furthermore, if a ∈Ai and b ∈Aj, then this corresponds to case one of Illustration 1 and we can utilize Proposition 21 to infer that either a or b are inconsistent.

Chunk 1004

Again, this inconsistency would contradict our assumptions on the labels of a and b. The case a ∈Aj and b ∈Ai can be proven analogous.

Chunk 1005

Next, suppose that we have a ∈Ai and b ∈A+ \ (Ai ∪Aj), which corresponds to case two of Illustration 1. If (b,a) ∈→+ is the result of an undercut, then this implies Atoms(bC) ⊆Atoms(ASi).

Chunk 1006

By Proposition 19 there now exists an argument b′ which is a reduced version of b w.r.t. ASi.

Chunk 1007

By Corollary 3 we have b′ ∈Ai, therefore admissibility of L yields b′ ∈out(L), which also implies b′ ∈out(Li+) by in(L) ⊆in(Li+). By Propo- sition 31 we know that x ∈out(Li+) iff x is legally OUT w.r.t.

Chunk 1008

Li+, which means we can apply Lemma 3 to infer b ∈out(Li+), contradicting our assumption b ∈undec(Li+). Now suppose that (b,a) ∈→+ is the result of a gen-rebut.

Chunk 1009

Then we can use Proposition 11 to construct b′ which is of the form b′ : b →¬VADSub(a)C. By Proposition 19 there exists an argument b′′ whcih is a reduced version of b′ w.r.t.

Chunk 1010

ASi. We have DR(b′′) ⊆DR(b′), --- Page 51 --- April 2026 thus we can infer (b′′,a) ∈→i.

Chunk 1011

Now admissibility of L implies b′′ ∈out(L). As in the undercut-case we can now use Proposition 31 in order to apply Lemma 3 and infer that b′ ∈out(Li+) must hold.

Chunk 1012

This implies b ∈out(Li+) by construction of Li+, which contra- dicts our assumption on Li+(b). The case a ∈Aj and b ∈A+ \(Ai ∪Aj) can be proven analogous.

Chunk 1013

Next, suppose that we have a ∈A+ \ (Ai ∪Aj) while b ∈Ai. This corresponds to case three of Illustration 1.

Chunk 1014

If (b,a) ∈→+ is the result of an undercut, then this attack must also be directed towards some a′ ∈CSub(a) ⊆in(L)∪in(L+). If a′ ∈Ai, then we can use the admissibility of L to infer that b ∈out(L) holds and by construction of Li+, this implies b ∈out(Li+), contradicting our assumption b ∈undec(Li+).

Chunk 1015

On the other hand, if a′ ∈A+ \ Ai, then we must have b ∈out(L+) by admissibility of L+, which also means b ∈out(L+|Ai). By Proposition 29, L+|Ai is an admissible labeling, therefore there must exist an attack or support chain in Ji, s.t.

Chunk 1016

b is labeled OUT because of this attack or support chain.. By assumption, we have in(L+|Ai) ⊆in(L) ⊆in(Li+), therefore b ∈out(Li+), contradicting our assumption b ∈undec(Li+).

Chunk 1017

Now suppose that (b,a) ∈→+ is the result of a gen-rebut. Note that a is not nec- essarily a minimal A+-argument.

Chunk 1018

Similar to the proof of Proposition 31, we there- fore again use ADSub(a) = {a0,...,am} and their reduced versions a′ 0,...,a′ m to con- struct a′ : a′ 0,...,a′ m →aC, which is a minimal A+-argument. By Lemma 2 we have ADsub(a) ⊆in(L) ∪in(L+) and by Lemma 3 this implies a′ 0,...,a′ m ⊆in(L) ∪in(L+).

Chunk 1019

By construciton of Li+ we now have a′ ∈in(Li+). Note that DR(a) = DR(a′), therefore (b,a′) ∈→+.

Chunk 1020

Now we can apply Proposition 23 to infer that one of the following cases must hold: 1. There are ea,eb ∈Ai s.t.

Chunk 1021

CSub(ea) =CSub(a′)|ASi, CSub(eb) =CSub(b) and (eb, ea) ∈ →i. 2.

Chunk 1022

There are ea,eb ∈Ai s.t. CSub(ea) = ADSub(a′)|ASi, CSub(eb) = CSub(b) and either (eb, ea) ∈→i, or (ea,eb) ∈→i.

Chunk 1023

Since a′ ∈in(Li+), we have CSub(a′) ⊆in(L) ∪in(L+) by construction of Li+ and by Lemma 2 we have ADSub(a′) ⊆in(L)∪in(L+), which implies ea ∈in(L) by admissibility of L. Now, again by admissibility of L, we have eb ∈out(L).

Chunk 1024

Note that CSub(eb) =CSub(b) implies ADSub(eb) = ADSub(b). Since eb,b ∈Ai and since L is admissible, we can now utilize Proposition 15 to infer L(b) = OUT.

Chunk 1025

By admissibility of L this means b is labeled OUT in L either due to an attack or due to a support chain. By in(L) ⊆in(Li+), we can infer that b ∈out(Li+) due to the same attacker or support chain.

Chunk 1026

This contradicts our assumption L ∈undec(Li+). The case a ∈A+ \ (Ai ∪Aj) while b ∈Aj can be proven similarly.

Chunk 1027

Finally, suppose that a,b ∈A+ \ (Ai ∪Aj) holds. Again, we start by assuming (b,a) ∈→+ is the result of an undercut.

Chunk 1028

Then this attack is directed towards some x ∈CSub(a) ⊆in(L)∪in(L+). If x ∈in(L+), then we can infer b ∈out(L+) by admissi- bility of L+, which implies b ∈out(Li+) by construction of in(Li+) from in(L+).

Chunk 1029

On the other hand, if x ∈in(L), then by Proposition 19 there exists b′, which is a reduced version of b w.r.t. ASi.

Chunk 1030

We have b′ ∈Ai by Corollary 3, therefore b′ ∈out(L) by admissibility of L. By construction of in(Li+) from in(L), this implies b′ ∈out(Li+).

Chunk 1031

As before, we can now utilize Proposition 26 to apply Lemma 3, and infer that b ∈out(Li+) also holds. This contradicts our assumption b ∈undec(Li+).

Chunk 1032

--- Page 52 --- April 2026 Lastly, assume that (b,a) ∈→+ is the result of a gen-rebut. As before, we use ADSub(a) = {a0,...,ak} and the reduced versions a′ 0,...,a′ k of these arguments to con- struct the minimal A+-argument a′ which is of the form a′ : a′ 0,...,a′ m →aC.

Chunk 1033

Similarly, we use ADSub(b) = {b0,...,bl} and the reduced versions b′ 0,...,b′ l of these arguments to construct the minimal A+-argument b′ which is of the form b′ : b′ 0,...,b′ m →bC. Because DR(a) = DR(a′) and DR(b) = DR(b′), we can infer (b′,a′) ∈→+.

Chunk 1034

Similar to before, we can also infer a′ ∈in(Li+) by Lemma 2, Lemma 3 and by construction of in(Li+). Now we can apply Proposition 24 to infer that one of the following four cases holds: 1.

Chunk 1035

There are ea,eb ∈Ai s.t. CSub(ea) = CSub(a′)|ASi, CSub(eb) = CSub(b′)|ASi and (eb, ea) ∈→i.

Chunk 1036

2. There are ea,eb ∈Aj s.t.

Chunk 1037

CSub(ea) = CSub(a′)|ASj, CSub(eb) = CSub(b′)|ASj and (eb, ea) ∈→j. 3.

Chunk 1038

There are ea,eb ∈Ai s.t. CSub(ea) = ADSub(a′)|ASi, CSub(eb) = CSub(b′)|ASi and either (eb, ea) ∈→i, or (ea,eb) ∈→i.

Chunk 1039

4. There are ea,eb ∈Aj s.t.

Chunk 1040

CSub(ea) = ADSub(a′)|AS j, CSub(eb) = CSub(b′)|AS j and either (eb, ea) ∈→j, or (ea,eb) ∈→j. Similar to before, we can infer by the construction of Li+ and by Lemma 2 that a′ ∈ in(Li+) implies ea ∈in(L) (in cases 1 and 3) or ea ∈in(L+) (in cases 2 and 4).

Chunk 1041

Either way, this again implies eb ∈out(L) (in cases 1 and 3) or eb ∈out(L+) (in cases 2 and 4) by admissibility of L and L+. Note that, by construction of eb and b′, we have DR(eb) ⊆ DR(b′) = DR(b) as well as ADSub(eb)C ⊆ADSub(b′)C = ADSub(b)C.

Chunk 1042

We only show the case of eb ∈out(L) here, as the case for eb ∈out(L+) can be proven analogously: Assume that we have eb ∈out(L) because of an attack (c,eb) ∈→i with c ∈in(L). Regardless of whether this attack is the result of an undercut or a gen-rebut, we can use DR(eb) ⊆DR(b), ADSub(eb)C ⊆ADSub(b)C, Proposition 11 and the admissibility of L to infer that there is an attack (c′,b) ∈→+ from an attacker c′ ∈in(L) ⊆in(Li+).

Chunk 1043

By construction of Li+ this implies b ∈out(Li+), contradicting our assumption b ∈undec(Li+). Lastly, suppose that eb ∈out(L) because of a support chain C =  (S0,b0),...,(Sn,bn) ⊆ ⇒i and an attack (c,bn) ∈→i with eb ∈S0 and c ∈in(L).

Chunk 1044

Similar to our proof for Proposition 26, we can now use the set S = S 0≤k≤n Sk ∩in(L) to create a new support chain C ′ =  (S ∪{b},b′) ⊆⇒+. If (c,bn) ∈→i is the result of an undercut we can infer from S ⊆in(L) that (c,eb) ∈→i holds.

Chunk 1045

By DR(eb) ⊆DR(b) we then have (c,b) ∈→+ and by c ∈in(L) ⊆in(Li+) we can infer that Li+(b) = OUT, contradict- ing our assumption Li+(b) = UNDEC. On the other hand, if (c,bn) ∈→i is the re- sult of a gen-rebut, then we can use Proposition 11 to construct an argument c′ which is of the form c′ : c →¬VADSub(bn)C.

Chunk 1046

Because DR(c′) = DR(c), DR(eb) ⊆DR(b) and ADSub(eb)C ⊆ADSub(b)C, we can infer that we also have (c′,b′) ∈→+. By con- struction of Li+, we have c′ ∈in(Li+).

Chunk 1047

By construciton of out(Li+), we now have b′ ∈O0 ⊆out(Li+) and – because S ⊆in(L) ⊆in(Li+) – b ∈O1 ⊆out(Li+). This con- tradicts our assumption b ∈undec(Li+) and finishes the proof.

Chunk 1048

Proposition 33. Let AS1 = (Rs1,Rd1,n1,≤r1), AS2 = (Rs2,Rd2,n2,≤r2) be two AS s.t.

Chunk 1049

AS1||AS2 and let AS+ be their union. Let J1 = (A1,→1,⇒1,⪯1), J2 = (A2,→1,⇒2,⪯2) --- Page 53 --- April 2026 and J+ = (A+,→+,⇒+,⪯+) be the JSBAFs corresponding to AS1, AS2 and AS+ re- spectively.

Chunk 1050

Furthermore, let L+ ∈pr(J+) be a preferred labeling of J+. Lastly, for any i ∈{1,2}, let Li = L+|Ai be the restriction of L+ to Ai, let L ∈pr(Ji) be a preferred labeling of Ji s.t.

Chunk 1051

in(Li) ⊆in(L) and let Li+ be the combined labeling of L and L+. Then Li+ ∈adm(J+).

Chunk 1052

Proof. By Proposition 30 we know that Li+ is a labeling of J+.

Chunk 1053

By Proposition 31 we know that a ∈out(Li+) iff a is legally OUT w.r.t. Li+.

Chunk 1054

By Proposition 32 we know that any a ∈in(Li+) is legally IN w.r.t. Li+.

Chunk 1055

By construction of in(Li+), we have in(L) ⊆ in(Li+), as well as in(L+) ⊆in(Li+). By admissibility of L and L+ we can infer that STRJ+ ⊆in(Li+) holds.

Chunk 1056

We conclude that Li+ is an admissible labeling of J+. With this, we are finally ready to prove Theorem 3.

Chunk 1057

We begin by proving that pre- ferred semantics of Deductive ASPIC⊖satisfies non-interference: Lemma 4. Preferred semantics of Deductive ASPIC⊖satisfies non-interference.

Chunk 1058

Proof. Let AS1 = (Rs1,Rd1,n1,≤r1) and AS2 = (Rs2,Rd2,n2,≤r2) be two AS s.t.

Chunk 1059

AS1||AS2. Furthermore, let AS+ = (R+ s ,R+ d ,n+,≤+ r ) be the union of AS1 and AS2.

Chunk 1060

We need to show that for each i ∈{1,2}, Cpr(ASi)| Atoms(ASi) = Cpr(AS+)| Atoms(ASi) holds. Let J1 = (A1,→1,⇒1,⪯1), J2 = (A2,→1,⇒2,⪯2) be the JSBAFs corresponding to AS1 and AS2 and let J+ = (A+,→+,⇒+,⪯+) be the JSBAFs corresponding to AS+.

Chunk 1061

Lastly, let j ∈{1,2} with i ̸= j. ⊆: Let Ei ∈pr(ASi) be a preferred extension of ASi and let Li ∈pr(Ji) be the pre- ferred labeling of Ji corresponding to Ei.

Chunk 1062

We first show that there is a preferred exten- sion E+ ∈pr(AS+) s.t. Ei = E+ ∩A (ASi): Let SIMj be the strict including minimal la- beling of Jj and let L+ be the combined minimal labeling of Li and SIMj.

Chunk 1063

By Proposi- tion 28 we know that L+ is an admissible labeling of J+. That means there is a preferred labeling Lpr+ ∈pr(J+) s.t.

Chunk 1064

in(Li) ⊆in(L+) ⊆in(Lpr+). Towards a contradiction, as- sume that in(Li) ⊂in(Lpr+)|A (ASi).

Chunk 1065

Take L′ i = Lpr+|A (ASi). By Proposition 29, we know that L′ i is an admissible labeling of Ji.

Chunk 1066

It is clear that in(Li) ⊆in(L′ i) holds. Now, if we have in(Li) ⊂in(Lpr+)|A (ASi), then this means in(Li) ⊂in(L′ i), which contradicts that Li is a preferred labeling.

Chunk 1067

Therefore, we must have in(Li) = in(Lpr+)|A (ASi). We conclude that for Ei ∈pr(ASi) there exists E+ ∈pr(AS+) s.t.

Chunk 1068

Ei = E+ ∩A (ASi). Now, let Γi = EC i , Γ+ = EC + be the sets of conclusions of Ei and E+.

Chunk 1069

Clearly, we have Γi|Atoms(ASi) ⊆Γ+|Atoms(ASi). Towards a contradiction, assume that we have Γi|Atoms(ASi) ⊂ Γ+|Atoms(ASi).

Chunk 1070

Let φ ∈Γ+|Atoms(ASi) s.t. φ ̸∈Γi|Atoms(ASi) and let a be the corresponding argument, i.e.

Chunk 1071

aC = φ. Then we have Atoms(aC) ⊆Atoms(ASi).

Chunk 1072

By Proposition 19, there exists a reduced version a′ of a. By Corollary 3, a′ ∈A (ASi) and by Lemma 3, we can infer that a′ ∈Ei.

Chunk 1073

Since aC = a′C, we infer φ ∈Γi|Atoms(ASi), contradicting our assumption. With this, we conclude that Cpr(ASi)|Atoms(ASi) ⊆Cpr(AS+)|Atoms(ASi) holds.

Chunk 1074

⊇: Take E+ ∈pr(AS+) and let L+ be the corresponding labeling of J+. Let Li = L+|Ai be the restriction of L+ to Ai.

Chunk 1075

By Proposition 29, we know that Li is an admissible labeling of Ji. Towards a contradiction, assume that Li is not a preferred labeling.

Chunk 1076

Then there exists L ∈pr(Ji) s.t. in(Li) ⊂in(L).

Chunk 1077

Now, let Li+ be the combined labeling of L and L+. By Proposition 33, we know that Li+ is an admissible labeling of J+.

Chunk 1078

Since in(Li) ⊂in(L), we now have in(L+) ⊂in(L′ +), contradicting that L+ is an admissible --- Page 54 --- April 2026 labeling. We conclude that for E+ ∈pr(AS+), there exists Ei ∈pr(ASi) s.t.

Chunk 1079

Ei = E+ ∩ A )(ASi). Now, take Γ+ = EC + and Γi = EC i .

Chunk 1080

Clearly, we have Γi|Atoms(ASi) ⊆Γ+|Atoms(ASi). To- wards a contradiction, assume Γi|Atoms(ASi) ⊂Γ+|Atoms(ASi).

Chunk 1081

Let φ ∈Γ+|Atoms(ASi) s.t. we have φ ̸∈Γi|Atoms(ASi) and let a be the corresponding argument, i.e.

Chunk 1082

aC = φ. By Proposi- tion 19 there exists a reduced version a′ of a and by Corollary 3 we know that a′ ∈Ai holds.

Chunk 1083

By Lemma 3 we can infer that a′ ∈Ei holds. Now φ ∈Γi|Atoms(ASi), contradicting our assumption.

Chunk 1084

We conclude that Cpr(ASi)|Atoms(ASi) ⊇Cpr(AS1 ⊎AS2)|Atoms(ASi) holds. This finishes the proof.

Chunk 1085

By using Proposition 8, Proposition 9 and Lemma 4, we can now infer that Deduc- tive ASPIC⊖satisfies crash-resistance under preferred semantics: Corollary 5. Deductive ASPIC⊖satisfies crash-resistance under preferred semantics.

Chunk 1086

6. Grounded Semantics In this section, we give a definition of a grounded semantics for JSBAFs.

Chunk 1087

We need to point out that this semantics was developed somewhat independently from the preferred semantics for JSBAFs. In particular, the notion of preferences between arguments is not considered in our grounded semantics.

Chunk 1088

We therefore begin this section by giving an overview of some definitions and results for JSBAFs without preferences. For brevity, we do not include any proofs for these preliminary results.

Chunk 1089

However, the actual proofs are very similar to those for JSBAFs with preferences, which are contained in Section 4. 6.1.

Chunk 1090

Preliminaries We define JSBAFs without preferences as follows: Definition 21. A JSBAF is a triple J = (Args,Att,Supp), with: • Args, the set of arguments • Att ⊆Args×Args, the set of attacks between arguments • Supp ⊆2Args ×Args, the set of supports between arguments We denote the set of all possible JSBAFs without preferences as J.

Chunk 1091

Note that we use the notations Args, Att and Supp when talking about JSBAFs with- out preferences, as opposed to A , →and ⇒when talking about JSBAFs with prefer- ences. We do so to avoid confusion between the two.

Chunk 1092

Furthermore, throughout this sec- tion we will denote arguments by capital letters A,B,C,... to distinguish these JSBAFs from those considered in Section 4.

Chunk 1093

In the context of JSBAFs without preferences, strict arguments and chains of sup- ports are defined analogous to the case of JSBAFs with supports (see Section 4). Given a chain of supports C = {(S0,B0),...,(Sn,Bn)} and some A ∈S0, we will sometimes say that C starts at A and ends at Bn or that C starts with (S0,B0).

Chunk 1094

If C = {(S0,B0)} we call C trivial. We use the following additional notation regarding chains of supports: --- Page 55 --- April 2026 Definition 22.

Chunk 1095

Let J = (Args,Att,Supp) be some JSBAF and let A,B ∈Args be argu- ments of J . We say there is a support-path of length n ≥1 from A to B in Supp iff there is a chain of supports C ⊆Supp which starts at A and ends at B.

Chunk 1096

The support ancestors of B in Supp, denoted SA(B), are defined as: SA(B) = {A ∈Args | there is a support-path from A to B in Supp}. The support children of A in Supp, denoted SC(A), are defined as: SC(A) = {B ∈Args | there is a support-path from A to B in Supp}.

Chunk 1097

Similar to the case of JSBAFs with preferences, we only consider JSBAFs that can be constructed on the basis of an argumentation system (while ignoring preferences be- tween arguments). To this end, we use the following definition: Definition 23.

Chunk 1098

Let J = (Args,Att,Supp) be some JSBAF. We say that J is con- structible according to Deductive ASPIC⊖iff the following conditions hold: 1.

Chunk 1099

For any A ∈Args, we have A ̸∈SA(A). 2.

Chunk 1100

Let S,S′ ⊆Args and let B ∈Args. If (S,B) ∈Supp and (S′,B) ∈Supp, then S = S′.

Chunk 1101

3. For any (S,B) ∈Supp, we have |S| < ∞.

Chunk 1102

4. If A ∈STRJ , then A is unattacked in J .

Chunk 1103

We denote by JDA⊖⊆J the set of all JSBAFs constructible according to Deductive AS- PIC⊖. As for the case of JSBAFs with supports, we use a labeling based approach for our semantics, with the possible labels identical to those we used for JSBAFs with prefer- ences (see Section 4).

Chunk 1104

We define legal labelings in the context of JSBAFs without pref- erences by utilizing the following auxiliary definition: Definition 24. Let L be a multiset of labels and l a label.

Chunk 1105

Then L ≤l iff any of the following conditions hold: 1. l = IN 2.

Chunk 1106

l = UNDEC and either UNDEC ∈L or OUT ∈L 3. l = OUT and OUT ∈L or |L(Undec)| ≥2.8 Note that with this definition, for L = /0 we have L ≤l, iff l = IN.

Chunk 1107

Based on this ordering, we now define what it means for an argument to be legally IN, OUT and UN- DEC: Definition 25. Let J = (Args,Att,Supp) be a JSBAF s.t.

Chunk 1108

J ∈JDA⊖, let A,B,C ∈Args be arguments, S ⊆Args a set of arguments and let Lab be a labeling of J . 1.

Chunk 1109

A is legally IN w.r.t. Lab, iff (a) for all attackers B of A, we have Lab(B) = OUT and (b) for every deductive support S ⇒B s.t.

Chunk 1110

A ∈S, we have that ˙ Lab(C) | C ∈ S\{A}˙ ≤Lab(B).9 2. A is legally OUT w.r.t.

Chunk 1111

Lab, iff (a) there is some attacker B of A, s.t. Lab(B) = IN or 8Here, |L(Undec)| denotes the number of UNDEC-labels in the multiset L.

Chunk 1112

9Here, ˙{ ˙} denotes a multiset. --- Page 56 --- April 2026 (b) there is a chain of supports C = {(S0,B0),...,(Sn,Bn)} ⊆Supp which starts at A and ends at some argument Bn s.t.

Chunk 1113

there exists an attack (C,Bn) ∈Att with Lab(C) = IN and for each (Si,Bi) ∈C the following conditions hold: • Bi ∈out(Lab) and • |Si ∩in(Lab)| = |Si|−1 3. A is legally UNDEC iff (a) A is not legally IN and (b) A is not legally OUT.

Chunk 1114

4. Lab is a legal labeling of J , iff every argument is legally labeled w.r.t.

Chunk 1115

Lab. For brevity, we will sometimes write a labeling Lab as the tuple in(Lab),out(Lab),undec(Lab)
.

Chunk 1116

Admissible labelings, preferred labelings and the strict including minimal labeling (SIM) of JSBAFs without preferences are defined analogous to the case of JSBAFs with pref- erences (see Section 4). The following statements can be proven similar as for the case of JSBAFs with preferences: Proposition 34.

Chunk 1117

Let J = (A ,→,⇒,⪯) be a JSBAF s.t. J ∈JDA⊖and let SIMJ be the strict including minimal labeling of J .

Chunk 1118

Then SIMJ is an admissible labeling. Proposition 35.

Chunk 1119

Let J = (Args,Att,Supp) be a JSBAF s.t. J ∈JDA⊖.

Chunk 1120

Then adm(J ) ̸= /0. 6.2.

Chunk 1121

Motivation 6.2.1. Introductory example It turns out that defining a grounded semantics for our JSBAFs in such a way that they intuitively correspond to the ideas of the standard grounded semantics of abstract argu- mentation is not that straight forward.

Chunk 1122

One property of the grounded semantics in abstract argumentation is that there is no admissible labeling Ladm for which in(Ladm) attacks the accepted arguments of the grounded labeling Lgr.10 In other words, a grounded labeling accepts only arguments that can never be rejected by an admissible labeling. In particu- lar, all preferred labelings agree on the arguments accepted in the grounded labeling.11 At the same time, the grounded semantics is still a very skeptical semantics.

Chunk 1123

Consider the following as an introductory example, in order to see how these ideas can be translated into the realm of JSBAFs. 10To see that this is the case, consider the characteristic function of abstract argumentation, defined in [2, Definition 16], which maps a set of argument S to the set of all arguments defended by S.

Chunk 1124

Towards a contradic- tion, assume that there is an admissible labeling Ladm s.t. in(Ladm) attacks in(Lgr) for Lgr being the grounded labeling.

Chunk 1125

By iterating over the characteristic function with starting point in(Ladm) until a fixpoint in(Lcmp) is reached, one can find a complete labeling Lcmp. In particular, Lcmp is conflict-free.

Chunk 1126

Since the characteristic function is monotone, we have in(Ladm) ⊆in(Lcmp). Since the grounded labeling is the smallest (w.r.t.

Chunk 1127

sub-set inclusion of accepted arguments) complete labeling, we also have in(Lgr) ⊆in(Lcmp). Now conflict-freeness of Lcmp is violated, a contradiction.

Chunk 1128

11Since every preferred labeling is a complete labeling and the grounded labeling is the smallest (w.r.t. sub-set inclusion of accepted arguments) complete labeling.

Chunk 1129

--- Page 57 --- April 2026 Example 3. JSBAF J2 A1 A1 A2 B B Suppose for a moment that we are not dealing with JSBAFs but with an abstract argumentation framework which does not contain the support ({A1,A2},B).

Chunk 1130

Then the grounded labeling of this framework would correspond to the labeling Lab1, for which we have in(Lab1) = {A2}, out(Lab1) = /0 undec(Lab1) = {A1,A1,b,B}. In particular, in the case of abstract argumentation, there could never be a reason to reject A2 – after all, this argument is unattacked.

Chunk 1131

However, in our JSBAFs we actually can have a reason to reject A2, namely the the following labeling, which we define to be Lab2: in(Lab2) = {A1,B}, out(Lab2) = {A1,A2,B}, undec(Lab2) = /0. This is because accepting B means we have to reject B, while simultaneously accepting A1 means we have to make sure that A2 is rejected.12 This tells us that in the example above, A2 should actually not be accepted in the grounded labeling for JSBAFs.

Chunk 1132

As a first (naive) attempt at defining our grounded semantics, assume that we sim- ply define a grounded labeling as the smallest (w.r.t. sub-set inclusion of accepted ar- guments) admissible labeling which accepts every argument that is legally IN.

Chunk 1133

In the case of J2, the smallest admissible labeling is the the strict including minimal la- beling, which we denote here as SIMJ2. We have in(SIMJ2) = out(SIMJ2) = /0 and undec(SIMJ2) = {A1,A1,A2,B,B}.

Chunk 1134

Lets see if SIMJ2 is already a grounded labeling according to this approach: Note that none of the arguments A1,A1,B and B are legally IN w.r.t. SIMJ2, because each of them is attacked by at least one argument which is not labeled OUT.

Chunk 1135

However, the argument A2 is legally IN w.r.t. SIMJ2, because it is unattacked and because ˙ SIMJ2(A1)˙ ≤SIMJ2(B) holds.

Chunk 1136

Thus, according to our naive approach, SIMJ2 would not actually be a grounded labeling because we would have to accept A2. On the other hand, the following labeling which we denote as Lab3, would be a grounded labeling according to this first attempt: in(Lab3) = {A2}, out(Lab3) = /0, undec(Lab3) = {A1,A1,B,B}.

Chunk 1137

The problem with this naive approach now lies in the fact that we actually have a preferred labeling in which A2 is rejected, as stated above. There- fore it seems unintuitive to consider Lab3 as a grounded labeling.

Chunk 1138

What this first approach tells us, is that when we want to check if a specific labeling Lab should be a grounded labeling, we cannot confine ourselves just to Lab. Rather, we have to take a broader approach and consider a variety of labelings.

Chunk 1139

Furthermore, when 12Remember again that supports correspond to the application of strict rules, meaning that if we accept both A1 and A2 while B is rejected, the closure postulate would be violated. --- Page 58 --- April 2026 trying to determine if a specific argument A should be accepted in the grounded labeling, it is not just the label of A and its attackers that we have to think about.

Chunk 1140

Instead, for any support (S,B) where A is contained in the supporting set S, we also have to take into account the labels of all the other arguments that are part of S as well as the label of B. 6.2.2.

Chunk 1141

Important ideas for grounded semantics Now, how should we go about finding a grounded labeling in general? In abstract ar- gumentation, one way is to start with the labeling that labels all arguments as UNDEC and then iteratively accepting arguments that can never be rejected (by iterating over the characteristic function) until a fixpoint is reached.

Chunk 1142

As we have done above, in the case of JSBAFs the most natural starting point for this procedure is SIM. Thus we could start with SIM and then go about checking arguments A one after the other to see if there exists a reason for why they could be rejected.

Chunk 1143

If we don’t find such a reason, we accept A and continue until we don’t find any more arguments that we need to accept. An obvious reason for rejection would be an attack from an argument C to A s.t.

Chunk 1144

C is not labeled OUT. However, another reason could be that there is a support (S,B) with A ∈S and an admissible labeling in which we cannot accept A because of this support.

Chunk 1145

Then this support would give us a valid reason for why A should not be contained in the grounded labeling. If none of these conditions are satisfied (i.e.

Chunk 1146

all attackers are OUT and we cannot find such a support), we are in a way forced to accept A in the grounded labeling. However, when considering a specific labeling Lab and an argument A that is part of a support (S,B), not every admissible labeling in which we cannot legally accept A should count as a valid reason to dismiss the idea of accepting A in the ground labeling.

Chunk 1147

To see this, consider this following, slightly modified example: Example 4. JSBAF J3 A1 A2 B B Let us suppose that we again start with the strict including minimal labeling of J3 and want to check if A2 should be accepted in the grounded labeling of this framework.

Chunk 1148

We have SIMJ3 = {B},{B},{A1,A2}
, meaning we start with a labeling where B is accepted (because it is strict) and B is rejected. Now A2 is not legally IN w.r.t.

Chunk 1149

SIMJ3, since the label SIMJ3(B) = OUT requires us to either reject an argument in the sup- porting set {A1,A2} or to leave both of these arguments as UNDEC. However, since A1 is a self-attacker, we can never accept it.

Chunk 1150

On the other hand, if we were to accept A2, we could still satisfy ˙ SIMJ3(A1)}˙ ≤SIMJ3(B) by labeling A1 as OUT. In fact, even though A2 is not legally IN w.r.t.

Chunk 1151

SIM in J3, there now does not seem to be a valid reason anymore to ever reject A2. Because we can never accept A1, we will always have --- Page 59 --- April 2026 an argument in the supporting set {A1,A2} which can be labeled OUT in order to satisfy the support-condition for A2.

Chunk 1152

This leads us an important observation: When we have a labeling Lab and are considering a (possibly different) labeling Lab′ in order to see if we can find a reason to not accept A, we can allow ourselves to alter Lab′. Naturally, there should be some restrictions placed on this idea.

Chunk 1153

Most importantly, we should not be allowed to change the labeling of arguments to which we have already committed ourselves. That is, we should not be allowed to change the label of argu- ments already labeled IN or OUT, but restrict our changes only to the arguments labeled UNDEC.

Chunk 1154

We capture this in the following definition: Definition 26. Let J = (Args,Att,Supp) be a JSBAF and let Lab,Lab′ be two labelings of J .

Chunk 1155

We say that Lab′ extends Lab iff in(Lab) ⊆in(Lab′) and out(Lab) ⊆out(Lab′). Another restriction we should put on the idea of altering the labelings that we con- sider is that we should not allow ourselves to “cheat” by labeling the supported argument IN (in cases where this is possible while still adhering to admissibility).

Chunk 1156

After all, for any support (S,B), accepting B means this support can never give us a reason to reject any of the arguments in the supporting set S.13 The next question now is, which labelings we should even consider when searching for a valid reason to not accept an argument in the grounded labeling. As we have stated above, the fundamental problem we have in J2 of Example 3 is that, whenever B is labeled UNDEC or OUT, we have to make sure that not both A1 and A2 are labeled IN.

Chunk 1157

Thus we need to check at least the labelings where B is UNDEC or OUT. This motivates the following definition that we need for our grounded semantics: Definition 27.

Chunk 1158

Let l,l′ ∈{IN,OUT,UNDEC} be two labels. We say that the label l′ is more informative than the label l, denoted l′ ≥p l iff l = UNDEC or l′ = l.

Chunk 1159

Note that with this definition, for every label l ∈{IN,OUT,UNDEC} we have l ≥p l and for every three labels l1,l2,l3, if l1 ≥p l2 ≥p l3, then l1 ≥p l3 also holds. To see how we will be utilizing this definition, imagine again that we are iteratively labeling arguments and we are interested in whether or not we should accept an argument A that is part of a supporting set S for an argument B, i.e.

Chunk 1160

we have a support (S,B) with A ∈S. We want to know if there exists an admissible labeling, in which the support (S,B) gives us a reason to reject A.

Chunk 1161

Therefore, when we have a current labeling Lab, we are going to check all admissible labelings Lab′, for which we have Lab′(B) ≥p Lab(B). If our current labeling Lab labels B as UNDEC, then this means we also need to check those labelings where B is labeled IN or OUT.

Chunk 1162

On the other hand, if our current labeling already labels B as OUT (IN), then we don’t need to additionally check those labelings where B is labeled UNDEC or IN (OUT). After all, when we are iteratively labeling arguments and we have already committed to labeling B as IN or OUT, then this label will not be changed anymore.

Chunk 1163

13This restriction might seem a bit harsh now. For example, consider the case that we are currently inves- tigating a labeling Lab in which Lab(B) = UNDEC while B is unattacked and doesn’t support anything, i.e.

Chunk 1164

there will never be a valid reason to reject B. Then depriving us of the option to label B as IN does not seem reasonable.

Chunk 1165

However, when iteratively accepting arguments, we can always accept B first and get back to A ∈S later. When we are then reconsidering the status of A at a later time, this support will not be a reason for the rejection of A anymore.

Chunk 1166

--- Page 60 --- April 2026 So far, we can summarize our findings as follows: A natural starting point to look for a grounded labeling is the most skeptical labeling of all – the strict including minimal labeling. When considering a particular labeling Lab and checking if an argument A should be accepted according to the grounded semantics, we not only need to check the attackers of A but also the supports (S,B) that A is a part of.

Chunk 1167

Out of these supports, we can dismiss those where Lab(B) = IN holds. We then need to check every labeling Lab′ where Lab′(B) is more informative than Lab(B).

Chunk 1168

What we need to ensure is that for all of those labelings Lab′, we can still find a labeling Lab′′ which extends Lab′ and where A can be legally accepted. However, in our extension of Lab′ we should not change the label of any arguments that are already accepted or rejected and we should not change the label of B.

Chunk 1169

With this, we are almost ready to give our most important definition for the grounded semantics. However, there is one more caveat we need to consider, namely that of JS- BAFs with an infinite amount of arguments.

Chunk 1170

Suppose we have a JSBAF that only consists of one argument which is the starting point for an infinitely long chain of supports: Example 5. JSBAF J4 A1 A2 A3 ...

Chunk 1171

Since there are no strict arguments in this JSBAF, the strict including minimal la- beling doesn’t accept any arguments. However, in the absence of any attacks to any of the Ai, there does not seem to exist a valid reason to ever reject any of these arguments.

Chunk 1172

In fact, is seems most natural to say that the grounded labeling for this particular JSBAF should just accept all the (infintely many) arguments in this chain.14 On the other hand, when applying our previous ideas, we are at an impasse: Suppose that we consider the strict including minimal labeling of this JSBAF and want to check if there is some ar- gument Ai which we need to accept. Then in particular we need to check if we can alter the strict including minimal labeling itself in such a way, that Ai is legally IN without altering the label of Ai+1.

Chunk 1173

However, as Ai is the only argument in the support for Ai+1 and since Ai+1 has the label UNDEC, we cannot make this modification. To remedy this, we will use one last idea for our grounded labelings.

Chunk 1174

Consider some Ai from J4: Since there are no attackers for this particular Ai or any A j that is a support- child of Ai, no chain of arguments starting at the support {Ai},Ai+1) can ever be a reason for us to label Ai as OUT. In this sense, the support {Ai},Ai+1) is safe for Ai and we don’t actually need to consider it when checking if Ai should be accepted in the grounded labeling.

Chunk 1175

This idea is captured in the following definition: Definition 28. Let J = (Args,Att,Supp) be a JSBAF s.t.

Chunk 1176

J ∈JDA⊖. Furthermore, let A ∈Args be an argument and (S,B) ∈Supp a support with A ∈S.

Chunk 1177

Lastly, let Lab be a labeling of J . We say that (S,B) is safe for A in Lab iff for all chains of supports C that start with (S,B), we have that for all (S′,B′) ∈C and all (D,B′) ∈Att, Lab(D) = OUT.

Chunk 1178

We denote by SSLab(A) the set of all safe supports for A in Lab, that is SSLab(A) =  (S,B) ∈Supp | A ∈S and (S,B) is safe for A in Lab . 14This is most evident when remembering that supports in our JSBAFs will ultimately correspond to the application of strict rules, which we want to base on the entailment relation of some underlying logic.

Chunk 1179

Thus a JSBAF like that would correspond to something like an infinite chain of entailments φ ⊢φ ⊢φ ⊢.... --- Page 61 --- April 2026 With this, we are finally ready to define the property that an argument needs to satisfy in order for us to be forced to accept it in the grounded labeling: Definition 29.

Chunk 1180

Let J = (Args,Att,Supp) be a JSBAF s.t. J ∈JDA⊖.

Chunk 1181

Furthermore, let A ∈Args be an argument and Lab be a labeling of J . We say A is forced IN w.r.t.

Chunk 1182

Lab iff all of the following conditions hold: 1. For any argument B ∈Args with (B,A) ∈Att, we have Lab(B) = OUT.

Chunk 1183

2. For every support (S,B) ∈Supp with A ∈S and Lab(B) ̸= IN, one of the following conditions hold: (a) For every admissible labeling Lab′ with Lab′(B) ≥p Lab(B) there exists an admissible labeling Lab′′ which extends Lab′ s.t.

Chunk 1184

A is legally IN w.r.t. Lab′′ and Lab′′(B) = Lab′(B), or (b) (S,B) is safe for A in Lab.

Chunk 1185

For any labeling Lab of J , we denote by FI(Lab) the set of all arguments of J which are forced IN w.r.t. Lab.

Chunk 1186

Based on this notion of forced IN, we now define a family of labelings that will be the basis for our grounded semantics: Definition 30. Let J = (Args,Att,Supp) be a JSBAF s.t.

Chunk 1187

J ∈JDA⊖and Lab a labeling of J . We say that Lab is a ground-complete labeling iff both of the following conditions hold: 1.

Chunk 1188

Lab ∈adm(J ) 2. If A ∈Args is forced IN w.r.t.

Chunk 1189

Lab, then Lab(A) = IN. We denote the set of all ground-complete labelings of J as grcmp(J ).

Chunk 1190

Finally, we are ready to define our grounded semantics. A grounded labeling is sim- ply a minimal (w.r.t.

Chunk 1191

set-inclusion of accepted arguments) ground-complete labeling: Definition 31. Let J = (Args,Att,Supp) be a JSBAF s.t.

Chunk 1192

J ∈JDA⊖and Lab a labeling of J . We say that Lab is a grounded labeling iff both of the following conditions hold: 1.

Chunk 1193

Lab ∈grcmp(J ) 2. There is no Lab′ ∈grcmp(J ) with in(Lab′) ⊂in(Lab).

Chunk 1194

The set of all grounded labelings of J is denoted by gr(J ). We give a detailed example of the grounded labeling in Section 6.3 and only want to state the grounded labelings of the examples above for completeness: We have gr(J2) = n/0, /0,{A1,A1,A2,B,B}
o and gr(J3) = n{B,A2},{A1,B}, /0
o , while gr(J4) = n{Ai | i ≥1}, /0, /0
o .

Chunk 1195

6.3. Grounded Construction 6.3.1.

Chunk 1196

Definition One useful property of the grounded semantics for abstract argumentation is that it only contains a single labeling. For the grounded labelings as we have defined them, it is not --- Page 62 --- April 2026 immediately clear that this is the case.

Chunk 1197

However, we will show in this section that this property is actually satisfied. We begin by turning the idea of iteratively labeling argu- ments – which we have used to explain our motivation for the definition of a grounded labeling – into an algorithmic approach for constructing a labeling for a JSBAF.

Chunk 1198

This approach will use transfinite recursion and sequences of potentially transfinite length in order to be applicable in the case of JSBAFs with an infinite amount of arguments. We will discuss in Example 6 why this approach is necessary.

Chunk 1199

Definition 32. Let J = (Args,Att,Supp) be a JSBAF s.t.

Chunk 1200

J ∈JDA⊖and let SIM be the strict including minimal labeling of J . A sequence GC = (Lab0,...,Labα) is called a grounded construction iff: • Lab0 = SIM, • FI(Labα)\in(Labα) = /0 and • for 0 ≤β < α there is some A ∈FI(Labβ)\in(Labβ) s.t.

Chunk 1201

: in(Labβ+1) =in(Labβ)∪{A}∪ [ (S,B)∈SSLabβ (A) {B}∪SC(B)
O0 β+1 ={D ∈Args | (C,D) ∈Att, C ∈in(Labβ+1)} O j+1 β+1 =O j β+1 ∪{D ∈Args | (S,B) ∈Supp, D ∈S,B ∈Oj β+1 and S\{D} ⊆in(Labβ+1)} out(Labβ+1) = [ j≥0 Oj β+1 undec(Labβ+1) = Args\ in(Labβ+1)∪out(Labβ+1)
• in(Labδ) = S α<δ in(Labα) out(Labδ) = S α<δ out(Labα) undec(Labδ) = Args\ in(Labδ)∪out(Labδ)
We call Labα the result of the grounded construction GC. Essentially, our construction works as follows: We start with the strict including minimal labeling.

Chunk 1202

In each step, we accept all arguments that we have previously ac- cepted (in(Labβ) ⊆in(Labβ+1)). We take a single argument A which is forced IN w.r.t.

Chunk 1203

to the previous labeling and add A to the accepted arguments ({A} ⊆in(Labβ+1)). Furthermore, for each support (S,B) which is safe for A w.r.t.

Chunk 1204

the previous labeling, we add each support-child of B, as well as B itself to the set of accepted arguments ( S (S,B)∈SSLabβ (A) {B}∪SC(B)
⊆in(Labβ+1)). Afterwards, we compute the effect that ac- --- Page 63 --- April 2026 cepting these arguments has on the resulting framework.

Chunk 1205

For this, we begin by rejecting all arguments that are attacked by an accepted argument (via the set O0 β+1). Then we re- ject all arguments B for which we have a chain of supports that satisfies the conditions of Definition 25 (via the sets Oj+1 β+1).

Chunk 1206

Lastly, we label all remaining arguments as UNDEC. 6.3.2.

Chunk 1207

Example To see how this approach works in practice, we give an example below. Note that this JS- BAF is supposed to contain an infinitely long chain of supports starting, with {D1},D2
as well as an infinitely long chain of attacks starting at (E1,E2).

Chunk 1208

We have indicated this by adding ... below the arguments D4 and E4.

Chunk 1209

Furthermore, suppose that for each Ei we have (Ei,H) ∈Att iff (Ei+1,H) ̸∈Att, beginning with the chain of arguments E1 to E4 that is shown. Example 6.

Chunk 1210

JSBAF J5 A B C D1 D2 D3 D4 ... E1 E2 E3 E4 ...

Chunk 1211

G F I H Now for our grounded construction. We begin with the strict including minimal la- beling of J5.

Chunk 1212

It is easy to see that this is the labeling Lab0 = {A},{B},{C,D1,...,E1,...,F,G,H,I}
. Now let us consider the arguments in this framework to see which of them is forced IN w.r.t.

Chunk 1213

Lab0. The arguments C,G and H as well as all the Ei are attacked by some argu- ment not labeled OUT, therefore they cannot be forced IN.

Chunk 1214

Thus the only arguments that we really have to consider are F,I and each of the Di. --- Page 64 --- April 2026 Lets begin with the argument F.

Chunk 1215

Here we need to check the support {F,I},G
, because Lab0(G) ̸= IN. Note that this is not a safe support, since G is attacked by E1 and E1 is not labeled OUT in Lab0.

Chunk 1216

We have to check all admissible labelings Lab′ for which we have that Lab′(G) is more informative than Lab0(G). Since Lab0(G) = UNDEC this essentially means we have to check all cases where Lab′(G) = OUT, or Lab′(G) = UNDEC.

Chunk 1217

Technically, we also have to check the cases where Lab′(G) = IN, but as we said in our motivation, in these cases the support {F,I},G
can never give us a reason to not accept F. Since we are only interested in knowing if there is a reason to not accept F, we can skip these labelings.

Chunk 1218

Note that there exists the admissible labeling Lab′ = {A,I},{B},{C,D1,...,E1,...,F,G,H}
, which is exactly like Lab0 except for the fact that we have accepted I. We have that F is not legally IN w.r.t.

Chunk 1219

Lab′, thus F is not forced IN w.r.t. Lab0.

Chunk 1220

Obviously, we can make a similar argument for the case of I. Now let us consider an argument Di in the infinite chain of supports.

Chunk 1221

For exam- ple, lets consider D3. Here we only have to check one support, namely {D3},D4
.

Chunk 1222

Note that none of the arguments Di are attacked by any argument, therefore this sup- port {D3},D4
is actually a safe support for D3 w.r.t. Lab0.

Chunk 1223

This means we have found our first argument that is legally IN w.r.t. Lab0.

Chunk 1224

Obviously, with the same ar- gumentation we actually have that all arguments Di for i ≥2 are forced IN w.r.t. Lab0.

Chunk 1225

However, for the sake of the example, lets stick with D3. Thus we choose D3 as the argument in FI(Lab0)\in(Lab0) which we want to accept in the first ac- tual step of our grounded construction.

Chunk 1226

This gives us our next labeling, Lab1 = {A,D3,D4,...},{B},{C,D1,D2,E1,...,F,G,H,I}
. Note that, because we chose D3 as the argument we want to label as IN, we also have to consider all chains of supports C which start at the safe support {D3},D4
and for each (S,B) ∈C we have to label B IN as well.

Chunk 1227

Therefore we also had to accept all arguments Di for i ≥4 in addition to D3. Now let us check again if there are any more arguments that we need to accept according to our construction.

Chunk 1228

Similar to the case of Lab0, for the arguments F and I we can easily turn Lab1 into an admissible labeling where either F or I cannot be accepted. This means neither F nor I are forced IN w.r.t.

Chunk 1229

Lab1. Next, lets consider D1.

Chunk 1230

Here we have to check two supports, namely {D1},D2
and {C,D1},B
, since both B and D2 are not labeled IN. First, we point out that, similar to the case of D3 in Lab0, we actually have that the support {D1},D2
is safe for D1 in Lab1.

Chunk 1231

Therefore, we don’t have to consider any other admissible labelings for the case of this support. However, the same does not hold for the support {C,D1},B
, since B is attacked by A and A is labeled IN by Lab1.

Chunk 1232

This means we have to check if we can satisfy item 2.a of Definition 29 in order to find out if D1 is forced IN w.r.t. Lab1.

Chunk 1233

Assume that we have some admissible labeling Lab′ for which we have that Lab′(B) ≥p Lab(B). Since Lab(B) = OUT, this means Lab′(B) = OUT must also hold.

Chunk 1234

The question is now, if we can extend Lab′ to turn it into an admissible labeling Lab′′ s.t. D1 is legally IN w.r.t.

Chunk 1235

Lab′′ and Lab′′(B) = OUT still holds. To check if D1 is legally IN, we only have to consider the supports that D1 is a part of, i.e.

Chunk 1236

{D1},D2
and {C,D1},B
. Note that the only restriction we have when considering if we can extend Lab′ to Lab′′, is that the labeling of B and the labeling of any argument which is IN or OUT cannot be changed.

Chunk 1237

However, if we have Lab′(D2) = UNDEC, then we are still allowed to extend Lab′ in such a way that Lab′′(D2) = IN holds, so long as the label- ing Lab′′ is admissible in the end and we don’t change the label of any argument la- beled IN or OUT by Lab′. Since we cannot have the case that Lab′(Di) = OUT for any --- Page 65 --- April 2026 i ≥2 (because all these arguments are unattacked and thus can never be labeled OUT in any admissible labeling), this means we can simply extend Lab′ to Lab′′ in such a way that Lab′′(Di) = IN for any i ≥2.

Chunk 1238

Then for the support {D1},D2
, we have that ˙˙ ≤Lab′′(D2) is trivially satisfied. So lets suppose that our extension of Lab′ is such that Lab′′(Di) = IN for any i ≥2.

Chunk 1239

Now for the second support that D1 is a part of, namely {C,D1},B
. Remember that we must adhere to Lab′′(B) = Lab′(B) = OUT, i.e.

Chunk 1240

we are not allowed to change the label of B itself. However, if Lab′(C) = UNDEC, then we can still change the label of C.

Chunk 1241

Note that C is a self-attacker in J5, therefore we cannot actually have the case that Lab′(C) = IN. Furthermore, it is easy to see that if Lab′(C) = UNDEC, then we cannot have Lab′(D1) = OUT.

Chunk 1242

Thus we can simply extend Lab′ to Lab′′ in such a way that Lab′′(C) = OUT while Lab′′(D1) = IN holds. Then C is legally OUT w.r.t.

Chunk 1243

Lab′′ (which is required for admissibility of Lab′′) and we have ˙ Lab′′(C)˙ ≤Lab′′(B), therefore D is now legally IN w.r.t. Lab′′.

Chunk 1244

Lastly, we have to ensure that no label of any other arguments labeled IN or OUT in Lab′ are changed by our extension to Lab′′. For this, we have to make sure that the effect of accepting D1 does not cause any interference with F,G,H,I or any of the Ei.

Chunk 1245

For Lab′, we note that we cannot have Lab′(D1) = OUT, as this would require either C to be labeled IN (so that we could reject D1 because of the support {C,D1},B
) or D2 to be labeled OUT. However, as we have pointed out above, these cases can never occur.

Chunk 1246

This means for Lab′, we actually either have Lab′(D1) = IN or Lab′(D1) = UNDEC. In the first case, the extension to Lab′′ described in the previous paragraph does not actually change the label of D1, so we don’t need to check any of the other arguments F,G,H,I or Ei because Lab′ is already admissible.

Chunk 1247

In the second case, we note that Lab′(E1) = UNDEC must hold, thus we can now simply label E1 OUT in our extension Lab′′ to achieve admissibility of Lab′′. After that, no further modifications to the labeling need to be made, since E1 being UNDEC means Lab′(G) = Lab′(H) = UNDEC as well as Lab′(Ei) = UNDEC for any i ≥2.

Chunk 1248

We conclude that, for any admissible labeling Lab′ for which we have Lab′(B) = OUT, we can find another admissible labeling Lab′′ for which we have Lab′′(B) = OUT, while D1 is legally IN w.r.t. Lab′′.

Chunk 1249

Therefore D1 is forced IN w.r.t. Lab1 and we can actually choose it as the argument for the next step of our grounded con- struction.

Chunk 1250

As before, we now also have to accept the argument D2 because it is part of the safe support {D1},D2
of D1. This gives us the following labeling: Lab2 = {A,D1,...},{B,C,E1},{E2,...,F,G,H,I}
.

Chunk 1251

Lets see if there are any more arguments in FI(Lab2)\in(Lab2) which we need to label IN according to our grounded construction. The situation for the arguments F and I still has not changed.

Chunk 1252

However, for the argument G we now actually have that all attackers of G, namely E1, are labeled OUT in Lab2. Whats more, G does not support any argument in J5, which means that G is now forced IN.

Chunk 1253

We thus get our next labeling Lab3 = {A,D1,...,G},{B,C,E1},{E2,...,F,H,I}
. After this step, we now finally have that both F and I are forced IN w.r.t.

Chunk 1254

Lab3. This is because they are unattacked and the only support that they are a part of is {F,I},G
, which we don’t have to check because the supported argument G is labeled IN.

Chunk 1255

We summarize the next two steps in our grounded construction into one and immediately state the labeling that we get from accepting both F and I: Lab2 = {A,D1,...,F,G,I},{B,C,E1},{E2,...,H}
. --- Page 66 --- April 2026 Finally, let us consider the chain of attacks starting at E1.

Chunk 1256

Because we have ac- cepted D1, we needed to reject E1, meaning E2 is now forced IN because its only attacker is OUT and because it does not support anything. This means we would have to accept E2 and label E3 as OUT as a consequence.

Chunk 1257

Then we could continue with E4 and so on and so on. Since this chain of arguments is infinite, situations like these are the reason for why we actually needed to define our grounded con- struction via transfinite recursion.

Chunk 1258

After accepting all the arguments Ei for i being a multiple of two and rejecting all Ej for j not being a multiple of two, we arrive at the labeling Labω = {A,D1,...,E2,E4,...,F,G,I},{B,C,E1,E3,...},{H}
. Finally, we only have one argument left to check, namely H.

Chunk 1259

All attackers of H are now la- beled OUT in Labω and H does not support any arguments. Therefore, H is forced IN w.r.t.

Chunk 1260

Labω and we arrive at the result of our grounded construction: Labω+1 = {A,D1,...,E2,E4,...,F,G,H,I},{B,C,E1,E3,...}, /0
. Note that we only described a specific grounded construction for this example, while there are actually infinitely many possible grounded constructions for J5.

Chunk 1261

In particular, we could have chosen to not start with D3, but with another argument in the chain of Di’s. Similarly, instead of accepting G right after the step where we accepted D1 and rejected E1, we could have also first accepted (some) arguments in the infinite chain of Ei’s before accepting F,G and I.

Chunk 1262

We will therefore show in the next section that the result of each grounded construction is identical, namely the unique grounded labeling. To see this, we will first show that each grounded construction ends with a labeling which is ground-complete and minimal (w.r.t.

Chunk 1263

set-inclusion of accepted arguments) and then we will show that there is actually only one such minimal labeling that is ground-complete for each JSBAF. This also immediately implies that there is a unique grounded labeling for each JSBAF.

Chunk 1264

6.4. Uniqueness of the grounded labeling 6.4.1.

Chunk 1265

Overview We begin by showing that the labeling produced by the algorithmic approach of Defini- tion 32 is a ground-complete labeling. This proof will mainly consist of showing that the constructed labeling is an admissible labeling.

Chunk 1266

Afterwards, we will prove the uniqueness of the grounded labeling by showing that the labeling we construct is contained in every ground-complete labeling. For the proof of admissibility we will show via transfinite induction that, for a given grounded construction GC = (Lab0,...Labα), each of the labelings Labβ are admissi- ble.

Chunk 1267

The induction start will be trivial, since the construction begins with Lab0 = SIM of which we know that it is an admissible labeling. To make the induction step more accessible, we prove three statements separately before combining them and proving the limit-case for the transfinite induction in Proposition 43.

Chunk 1268

These three statements are the following: 1. Labβ+1 is a labeling.

Chunk 1269

2. If any argument A is labeled IN by Labβ+1, then A is legally IN w.r.t.

Chunk 1270

Labβ+1. 3.

Chunk 1271

For any argument A we have Labβ+1(A) = OUT iff it is legally OUT w.r.t. Labβ+1.

Chunk 1272

--- Page 67 --- April 2026 6.4.2. Induction step We begin by proving that, if Labβ is an admissible labeling, then Labβ+1 is a labeling.

Chunk 1273

In this proof, we will use several small auxiliary statements. The first one tells us that, if A is forced IN w.r.t.

Chunk 1274

to Labβ, then for any support (S,B) where A ∈S and S\{A} ⊆in(Labβ), this support must either be safe for A or B must already be labeled IN in Labβ. This way we know that, when labeling A as IN in the step β +1, we don’t end up with a support S where all arguments are labeled IN, while the supported argument B is not IN.

Chunk 1275

Proposition 36. Let J = (Args,Att,Supp) be a JSBAF s.t.

Chunk 1276

J ∈JDA⊖and let Lab be an admissible labeling of J . Furthermore, let (S,B) ∈Supp be some support of J and A ∈S be some argument of the supporting set S.

Chunk 1277

If A ∈FI(Lab) and S\{A} ⊆in(Lab), then Lab(B) = IN or (S,B) ∈SSLab(A). Proof.

Chunk 1278

We show that if (S,B) ̸∈SSLab(A), then Lab(B) = IN holds. Towards a contra- diction, assume that this is not the case, i.e.

Chunk 1279

(S,B) ̸∈SSLab(A) and Lab(B) ̸= IN. As A ∈FI(Lab) by assumption, we know that either item 2.a or 2.b of Definition 29 must hold.

Chunk 1280

By assumption, item 2.b does not hold. Note that Lab itself is an admissible la- beling for which we have Lab(B) ≥p Lab(B).

Chunk 1281

Thus there exists an admissible labeling Lab′ ∈adm(J ) that extends Lab s.t. A is legally IN w.r.t.

Chunk 1282

Lab′ and Lab(B) = Lab′(B). As Lab′ extends Lab we know that S\{A} ⊆in(Lab′) must hold.

Chunk 1283

Now if A is legally IN w.r.t. Lab′, then Lab(B) ̸= IN is a contradiction.

Chunk 1284

The second auxiliary statement tells us that, if an argument A is forced IN w.r.t. some admissible labeling Labβ, then A cannot be labeled OUT in Labβ: Proposition 37.

Chunk 1285

Let J = (Args,Att,Supp) be a JSBAF s.t. J ∈JDA⊖and let Lab be an admissible labeling of J .

Chunk 1286

If A ∈FI(Lab), then Lab(A) ̸= OUT. Proof.

Chunk 1287

Suppose towards a contradiction that Lab(A) = OUT holds. Because Lab is an admissible labeling, we know that either of the following two cases must hold: • There is B ∈Args with (B,A) ∈Att and Lab(B) = IN, or • there is a chain of supports  (S0,B0),...,(Sn,Bn) stat starts at A s.t.

Chunk 1288

that there is an attack (C,Bn) ∈Att with Lab(C) = IN and for each 0 ≤i ≤n we have Bi ∈ out(Lab) while |Si ∩in(Lab)| = |Si|−1. The first case cannot hold because A is forced IN w.r.t.

Chunk 1289

Lab, meaning all its attackers need to be labeled OUT in Lab. Now for the second case: Note that (S0,B0) is a support with A ∈S0 s.t.

Chunk 1290

B0 ̸= IN, while Lab is itself a admissible labeling. It is obvious that the support (S0,B0) is not safe for A in Lab because of the attacker C.

Chunk 1291

Thus item 2.b of Definition 29 does not hold, which means item 2.a must hold. Therefore, there needs to be an admissible labeling Lab′ which extends Lab in such a way that A is legally IN w.r.t.

Chunk 1292

Lab′. Because Lab′ needs to extend Lab, we know that Lab′(B0) = OUT and S0\{A} ⊆ in(Lab′) must hold.

Chunk 1293

Now for the support (S0,B0) we have A ∈S0, while ˙ Lab(C) | C ∈ S0\{A}˙ ̸≤Lab(B0), therefore A is not legally IN w.r.t. Lab′.

Chunk 1294

This contradicts A being forced IN w.r.t. Lab.

Chunk 1295

We conclude that both cases lead to a contradiction, meaning we have Lab(A) ̸= OUT. --- Page 68 --- April 2026 The third auxiliary statement tells us that, if an argument A is labeled OUT in Labβ, then it cannot be in the union S (S,B)∈SSLabβ (X) {B}∪SC(B)
for X being the argument that is added in the step β +1 of the grounded construction.

Chunk 1296

Proposition 38. Let J = (Args,Att,Supp) be a JSBAF s.t.

Chunk 1297

J ∈JDA⊖. Furthermore, let A,B ∈Args be two arguments and Lab an admissible labeling of J s.t.

Chunk 1298

Lab(A) = OUT. Then A ̸∈ S (S,C)∈SSLab(B) {C}∪SC(C)
.

Chunk 1299

Proof. Towards a contradiction, suppose that the claim does not hold.

Chunk 1300

Then there exists a support (S,C) ∈SSLab(B) s.t. A = C or A ∈SC(C).

Chunk 1301

First, we assume that (S,A) ∈SSLab(B). Because A ∈out(Lab) by assumption, we know that there either is an attack (D,A) ∈Att with Lab(D) = IN or there is a chain of supports  (S0,B0),...,(Sn,Bn) which starts at A and for which there exists an at- tack (D,Bn) ∈Att with Lab(D) = IN.

Chunk 1302

In the first case,  (S,A) is a (trivial) chain of supports starting with (S,A) s.t. A is attacked by an argument D ∈in(Lab).

Chunk 1303

Thus (S,A) ∈SSLab(B) is a contradiction according to the definition of safe supports in 28. In the second case, we can construct a new chain of supports starting with (S,A), namely  (S,A),(S0,B0),...,(Sn,Bn) , s.t.

Chunk 1304

Bn is attacked by an argument D ∈in(Lab). Again, this contradicts (S,A) ∈SSLab(B).

Chunk 1305

Next, assume that A ∈SC(C) for some safe support (S,C) ∈SSLab(B). Then there exists a chain of supports C =  (S,C),...,(S′,A) that starts with (S,C).

Chunk 1306

Because Lab(A) = OUT, there either is an attack (D,A) ∈Att with Lab(D) = IN or there is a chain of supports  (S′ 0,B′ 0),...,(S′ m,B′ m) for which we have A ∈S′ 0 and an attack (D,B′ m) ∈Att with Lab(D) = IN. In the first case, C itself is a chain of supports start- ing with (S,C) which contradicts (S,C) ∈SSLab(B) and in the second case we can con- struct the chain of supports  (S,C),...,(S′,A),(S′ 0,B′ 0),...,(S′ m,B′ m) which contradicts (S,C) ∈SSLab(B).

Chunk 1307

Now for the prove that Labβ+1 is a valid labeling: Proposition 39. Let J = (Args,Att,Supp) be a JSBAF s.t.

Chunk 1308

J ∈JDA⊖. Furthermore, let GC = (Lab0,...,Labα) be a grounded construction and let Labβ be a labeling of this grounded construction s.t.

Chunk 1309

Labβ ∈adm(J ). Then Labβ+1 is a labeling.

Chunk 1310

Proof. Throughout this proof, let X be the argument in FI(Labβ) that is chosen for this step of the grounded construction, i.e.

Chunk 1311

Labβ+1 = Labβ ∪{X}∪ S (S,B)∈SSLabβ (X) {B}∪ SC(B)
. From the construction of Labβ+1 it is clear that each A ∈Args receives some label and that undec(Labβ+1)∩in(Labβ+1) = undec(Labβ+1)∩out(Labβ+1) = /0.

Chunk 1312

It remains to be shown that in(Labβ+1) ∩out(Labβ+1) = /0 also holds. For this, we will argue via induction over n ∈N that in(Labβ+1)∩On β+1 = /0.

Chunk 1313

Towards a contradiction, assume that this does not hold and let A ∈in(Labβ+1)∩On β+1. Induction start n = 0: Then there exists an argument C ∈in(Labβ+1) s.t.

Chunk 1314

(C,A) ∈Att. Because A ∈in(Labβ+1), we can have three cases: Either A ∈in(Labβ), A = X (i.e.

Chunk 1315

A was forced IN w.r.t. Labβ), or A ∈ S (S,B)∈SSLab(X) {B}∪SC(B)
.

Chunk 1316

We first argue that each --- Page 69 --- April 2026 of these cases means C ∈out(Labβ) must hold: This is clear for the first and second case. For the third case, assume first that there is a safe support (S,B) ∈SSLabβ (X) for X in Labβ s.t.

Chunk 1317

A = B. Then  (S,A) is a chain of supports that starts with (S,A).

Chunk 1318

By definition of a safe support in 28, all attackers of A are OUT in Labβ, i.e. we have Labβ(C) = OUT.

Chunk 1319

On the other hand, if there is a safe support (S,B) ∈SSLabβ (X) s.t. A ∈SC(B), then we must have a chain of supports  (S,B),...,(S′,A) .

Chunk 1320

Again, by definition of a safe support this means all attackers of A are OUT in Labβ, i.e. we have Labβ(C) = OUT.

Chunk 1321

Now to see that C cannot be in Labβ+1: C ∈Labβ+1 means we either have C ∈ in(Labβ), C = X or C ∈ S (S,B)∈SSLab(X) {B}∪SC(B)
. Because Labβ(C) = OUT, we must have Labβ(C) ̸= IN.

Chunk 1322

By Proposition 37 we can also infer that C ̸∈FI(Labβ), i.e. C ̸= X.

Chunk 1323

Lastly, by Proposition 38 we know that C ̸∈ S (S,B)∈SSLabβ (X) {B} ∪SC(B)
. Thus all the possible cases for C ∈in(Labβ+1) lead to a contradiction and we infer in(Labβ+1) ∩ O0 β+1 = /0.

Chunk 1324

Induction step n →n + 1: By IH we know in(Labβ+1) ∩On β+1 = /0, thus we con- centrate on O′ = On+1 β+1\On β+1. If A ∈O′, then there must be a support (S,B) ∈Supp with A ∈S, S\{A} ⊆in(Labβ+1) and B ∈On β+1.

Chunk 1325

Because A ∈in(Labβ+1) by as- sumption, we can now infer S ⊆in(Labβ+1). We argue over the possible cases for S ⊆in(Labβ)∪{X}∪ S (S,B)∈SSLabβ (X) {B}∪SC(B)
and show that each of them leads to a contradiction.

Chunk 1326

From here on, let U = S (S′,B′)∈SSLabβ (X) {B′}∪SC(B′)
. If S = /0, then B is a strict argument, thus B ∈Labβ ⊆in(Labβ+1), contradicting our induction hypothesis IH.

Chunk 1327

If S ⊆in(Labβ), then by admissibility of Labβ we have B ∈in(Labβ). Thus B ∈in(Labβ+1), again contradicting IH.

Chunk 1328

For the case S ⊆{X} we first note that X was forced IN in Labβ. We ignore the case Labβ(B) = IN because this trivially contradicts IH again.

Chunk 1329

Thus for Labβ(B) ̸= IN, either (S,B) was safe for X in Labβ or item 2.a of Definition 29 was satisfied. In the first case, we have B ∈U , therefore B ∈in(Labβ+1), which again contradicts IH.

Chunk 1330

In the second case, we note that Labβ itself was admissible, so there must have been an extension Lab′ of Labβ s.t. Lab′(B) ̸= IN while X was legally IN w.r.t.

Chunk 1331

Lab′. Clearly this cannot hold, thus we again have a con- tradiction.

Chunk 1332

Next, for the case that S ⊆in(Labβ)∪{X}. Then by Proposition 36 we either have Labβ(B) = IN or (S,B) ∈SSLabβ (X).

Chunk 1333

Either way, we can infer Labβ+1(B) = IN which again contradicts IH. Lastly, for the remaining cases we note that in each of them we have S∩U ̸= /0.

Chunk 1334

We show that this cannot hold. Towards a contradiction, suppose there exists D ∈S ∩U .

Chunk 1335

Let (S′,B′) ∈SSLabβ (X) be the safe support for X in Labβ for which we have D ∈{B′} ∪SC(B′). In par- ticular, this means that there must be a chain of supports  (S′,B′),...,(S′′,D) start- ing at X.

Chunk 1336

Because D ∈S, we can infer from the construction of On+1 β+1 that there ex- ists a chain of supports  (S,B),...,(S′′′,B′′′) starting at D, s.t. (C,B′′′) ∈Att and C ∈in(Labβ+1).

Chunk 1337

Now we can combine these two chains to construct a new chain of sup- ports  (S′,B′),...,(S′′,D),(S,B),...,(S′′′,B′′′) with (C,B′′′) ∈Att and Labβ+1(C) = IN. We note that, because (S′,B′) ∈SSLabβ (X), by Definition of a safe support in 28, we must have Labβ(C) = OUT.

Chunk 1338

From C ∈in(Labβ+1) we infer C ∈in(Labβ)∪{X}∪U . Clearly, --- Page 70 --- April 2026 C ∈in(Labβ)∩out(Labβ) cannot hold.

Chunk 1339

By Proposition 37 we also know that C = X and Labβ(C) = OUT cannot hold. Lastly, suppose that C ∈U .

Chunk 1340

Because Labβ(C) = OUT, we know that there must have been an attack (E,C) ∈Att with Labβ(E) = IN or yet another chain of supports  (bS, bB),...,(bS′, bB′) s.t. C ∈bS and there is E ∈in(Labβ) with (E, bB′) ∈Att.

Chunk 1341

Clearly, this means there cannot be a safe support (bS′′, bB′′) ∈SSLabβ (X) s.t. C ∈{c B′′}∪SC(bB′′) holds.

Chunk 1342

With this, we infer that S∩U = /0 must hold, therefore all cases for S ⊆in(Labβ)∪ {X} ∪U have lead to a contradiction. We conclude that in(Labβ+1) ∩On+1 β+1 = /0 must hold.

Chunk 1343

The proof for A ∈out(Labβ+) iff A is legally OUT w.r.t. Labβ+1 will be omitted because this is clear from the construction of Labβ+1.

Chunk 1344

Proposition 40. Let J = (Args,Att,Supp) be a JSBAF s.t.

Chunk 1345

J ∈JDA⊖. Furthermore, let GC = (Lab0,...,Labα) be a grounded construction and let Labβ be a labeling of this grounded construction s.t.

Chunk 1346

Labβ ∈adm(J ). Then A ∈out(Labβ+1) iff A is legally OUT w.r.t.

Chunk 1347

Labβ+1. Next, we will show another auxiliary statement concerning the arguments labeled OUT by two different labelings Lab,Lab′ s.t.

Chunk 1348

in(Lab) ⊆in(Lab′). Proposition 41.

Chunk 1349

Let J = (Args,Att,Supp) be a JSBAF s.t. J ∈JDA⊖.

Chunk 1350

Furthermore, let Lab,Lab′ be two labelings of J s.t. Lab ∈adm(J ).

Chunk 1351

Lastly, suppose that if A is legally OUT w.r.t. Lab′, then A ∈out(Lab′).

Chunk 1352

If in(Lab) ⊆in(Lab′), then we have out(Lab) ⊆ out(Lab′). Proof.

Chunk 1353

Assume in(Lab) ⊆in(lab′) and let A ∈out(Lab). Because Lab is admissible by assumption, we know that A is legally OUT w.r.t.

Chunk 1354

Lab. Thus there either is an attack (B,A) ∈Att with B ∈in(Lab), or there is a chain of supports  (S0,B0),...,(Sn,Bn) that starts at A, ends at Bn and satisfies the conditions of Definition 25.

Chunk 1355

In the first case, we have that A is legally OUT w.r.t. Lab′ and by assumption this implies A ∈out(Lab′).

Chunk 1356

For the second case we can prove via induction over n ∈N, for n being the length of the chain, that A is again legally OUT w.r.t. Lab′, which implies A ∈out(Lab′).

Chunk 1357

As this induction is trivial, we omit it here. Corollary 6.

Chunk 1358

Let J = (Args,Att,Supp) be a JSBAF s.t. J ∈JDA⊖.

Chunk 1359

Furthermore, let Lab,Lab′ ∈adm(J ). If we have in(Lab) ⊆in(Lab′), then out(Lab) ⊆out(Lab′) also holds.

Chunk 1360

Lastly, we will show that any argument that is labeled IN by Labβ+1 is also legally IN w.r.t. Labβ+1.

Chunk 1361

Proposition 42. Let J = (Args,Att,Supp) be a JSBAF s.t.

Chunk 1362

J ∈JDA⊖. Furthermore, let GC = (Lab0,...,Labα) be a grounded construction and let Labβ be a labeling of this grounded construction s.t.

Chunk 1363

Labβ ∈adm(J ). If A ∈in(Labβ+1), then A is legally IN w.r.t.

Chunk 1364

Labβ+1. Proof.

Chunk 1365

Let A ∈in(Labβ+1). Towards a contradiction, assume that A is not legally IN w.r.t.

Chunk 1366

Labβ+1. First, assume that there is an attack (B,A) ∈Att with Labβ+1(B) ̸= OUT.

Chunk 1367

By our --- Page 71 --- April 2026 grounded construction we have A ∈Labβ, A = X or A ∈∪ S (S,B)∈SSLabβ (X) {B}∪SC(B)
for X ∈FI(Labβ)\in(Labβ). In either of these cases, we know that all attackers B of A were labeled OUT in Labβ.

Chunk 1368

By in(Labβ) ⊆in(Labβ+1) and the same argumentation that was used in the proof for Proposition 41 we can infer that B is legally OUT w.r.t. Labβ+1.

Chunk 1369

By Proposition 40 we have Labβ+1(B) = OUT, contradicting the assumption Labβ+1(B) ̸= OUT. Now take some support (S,B) ∈Supp with A ∈S and assume that ˙ Labβ+1(C) | C ∈S\{A}˙ ̸≤Labβ+1(B).

Chunk 1370

We only need to consider the case where S ⊆in(Labβ+1) while B ̸∈in(Labβ+1) (we have excluded the case |S ∩in(Labβ+1)| = |S| −1 while Labβ+1(B) = OUT and S ∩out(Labβ+1) = /0 by construction of On+1 β+1 whereas the cases |S ∩in(Labβ+1)| < |S| −1 while Labβ+1(B) = OUT and |S ∩in(Labβ+1)| = |S| −1 while Labβ+1(B) = UNDEC are trivial). Let S ⊆in(Labβ+1) and suppose Labβ+1(B) ̸= IN.

Chunk 1371

Now we again consider all the possible cases for S ⊆in(Labβ)∪{X}∪ S (S′,B′)∈SSLabβ (X) {B′}∪SC(B′)
. If S = /0 then B is strict, thus B ∈in(Labβ) ⊆in(Labβ+1), contradicting our assumption.

Chunk 1372

If S ⊆in(Labβ), then B ∈in(Labβ) ⊆in(Labβ+1) by admissibility of Labβ, which again contradicts our assumption. If S ⊆{X} or S ⊆ in(Labβ) ∪{X}, then we know by Proposition 36 that B ∈in(Labβ) ⊆in(Labβ+1) which again contradicts our assumption.

Chunk 1373

For the remaining cases we note that we have S (S′,B′)∈SSLabβ (X) {B′}∪SC(B′)
∩S ̸= /0. However, this means there is (S′,B′) ∈SSLabβ (X) s.t.

Chunk 1374

D ∈{B′}∪SC(B′) and D ∈S. Now B ∈{B′}∪SC(B′) holds, meaning Labβ+1(B) = IN, which again contradicts our assumption.

Chunk 1375

As all possible cases for S have lead to a contradiction, we conclude that S ⊆ in(Labβ+1) implies B ∈in(Labβ+1), as required. 6.4.3.

Chunk 1376

Transfinite Induction Finally, we are ready to show that every labeling that we create during our grounded construction is an admissible labeling: Proposition 43. Let J = (Args,Att,Supp) be a JSBAF s.t.

Chunk 1377

J ∈JDA⊖. Furthermore, let GC = {Lab0,...,Labα} be a grounded construction.

Chunk 1378

Then for all ordinals δ, each Labδ is an admissible labeling. Proof.

Chunk 1379

We show the claim via transfinite induction. Induction start δ = 0.

Chunk 1380

Then Labδ is the strict including minimal labeling of J . By Proposition 34 we know that Labδ is an admissible labeling.

Chunk 1381

Induction step δ →δ + 1: By the induction hypothesis we know that Labδ was an admissible labeling. By Propositions 39, 40 and 42, Labδ+1 is an admissible labeling (STRJ ⊆in(Labδ+1) is trivial).

Chunk 1382

Limit step α: From here on, we denote our induction hypothesis for the transfinite induction as IHα. Note that STRJ ⊆in(Labα) is again trivially satisfied.

Chunk 1383

We first show that Labα is a labeling. By construction it is clear that all arguments receive at least one label and that in(Labα)∩undec(Labα) = out(Labα)∩undec(Labα) = /0.

Chunk 1384

Thus we have left to show in(Labα) ∩out(Labα) = /0. Towards a contradiction, assume that this does --- Page 72 --- April 2026 not hold.

Chunk 1385

Then there are ordinals γ < α and δ < α s.t. A ∈in(Labγ) and A ∈out(Labδ).

Chunk 1386

We either have γ ≤δ or δ ≤γ. In the first case, A ∈in(Labδ)∩out(Labδ) by construction of in(Labδ).

Chunk 1387

This contradicts IHα. In the second case, we note that A ∈out(Labδ) means A is legally OUT w.r.t.

Chunk 1388

Labδ. Thus there is an attack (B,A) ∈Att with B ∈in(Labδ) or a chain of supports  (S0,B0),...,(Sn,Bn) which starts at A and for which we have an attack (C,B) ∈Att with C ∈in(Labδ) while for 0 < i < n |Si ∩in(Labδ)| = |S|−1.

Chunk 1389

We know by construction of in(Labγ) that in either of those cases A ∈out(Labγ)∩in(Labγ) holds, which again contradicts IHα. Next, we show that A ∈in(Labα) implies that A is legally IN w.r.t.

Chunk 1390

Labα. Let γ < α s.t.

Chunk 1391

Labγ(A) = IN. Then A was legally IN w.r.t.

Chunk 1392

Labγ, meaning all attackers of A were labeled OUT in Labγ. By construction of Labα, we can infer that each of these attackers is labeled OUT in Labα.

Chunk 1393

Now take some support (S,B) with A ∈S. Towards a contradic- tion, suppose first that S ⊆in(Labα) while B ̸∈in(Labα).

Chunk 1394

Let S = {A,A0,...,An} and take Labγ,Labγ0,...,Labγn s.t. A ∈in(Labγ) and for each 0 < i < n we have Ai ∈in(Labγi).

Chunk 1395

Now let γmax be s.t. γ ≤γmax and γi ≤γmax for each 0 ≤i ≤n.

Chunk 1396

Then S ⊆in(Labγmax), meaning B ∈in(Labγmax) ⊆in(Labα), contradicting our assumption. Next we note that if |S∩in(Labα)| ≤|S|−2 and Labα(B) = OUT or |S∩in(Labα)| ≤|S|−1 and Labα(B) = UNDEC, then ˙ Labα(C) | C ∈S\{A}˙ ≤Labα(B) is trivially satisfied.

Chunk 1397

We therefore as- sume S\in(Labα) = {A′}, Labα(B) = OUT and for A′ we have Labα(A′) = UNDEC. By construction of Labα there must exist Labγ s.t.

Chunk 1398

Labγ(B) = OUT. Similarly, for S′ = {A0,...,An} there must exist Labγi for each 0 ≤i ≤n s.t.

Chunk 1399

Labγi(Ai) = IN. Take γmax ∈{γ,γ0,...,γn} s.t.

Chunk 1400

γ ≤γmax and γi ≤γmax for each 0 ≤i ≤n. Then S′ ⊆in(Labγmax) while B ∈out(Labγmax) by construction of Labγmax.

Chunk 1401

Since γmax < α, Labγmax is admis- sible by IHα, therefore Labγmax(A′) = Labα(A′) = OUT. Note that A′ = A cannot hold since we assumed Labα(A) = IN.

Chunk 1402

Therefore there is an argument labeled OUT in S s.t. ˙ Labα(C) | C ∈S\{A}˙ ≤Labα(B) is satisfied.

Chunk 1403

Lastly, we show A ∈out(Labα) iff A is legally OUT w.r.t. Labα.

Chunk 1404

First, assume that A ∈out(Labα). Then by construction of Labα there is γ < α s.t.

Chunk 1405

A ∈out(Labγ). Since Labγ was an admissible labeling by IHα, we again have an attack or a chain of supports s.t.

Chunk 1406

A is legally OUT w.r.t. Labγ.

Chunk 1407

By construction of Labα we have in(Labγ) ⊆in(Labα) and out(Labγ) ⊆out(Labα). Therefore we can infer that A is legally OUT w.r.t.

Chunk 1408

Labα. Now suppose that A is legally OUT w.r.t.

Chunk 1409

Labα. Assume first that there is an attack (B,A) ∈Att with Labα(B) = IN.

Chunk 1410

Then there is γ < α s.t. Labγ(B) = IN.

Chunk 1411

By IHα we know that Labγ was an admissible labeling, thus A ∈out(Labγ) ⊆out(Labα). Now assume that there is a chain of supports  (S0,B0),...,(Sn,Bn) s.t.

Chunk 1412

A ∈S0, there exists an attack (C,Bn) ∈Att with C ∈in(Labα) and for each 0 < i < n we have |Si ∩Labα| = |Si| −1 while Bi ∈out(Labα). We show the claim via induction over n ∈N for n being the length of that chain.

Chunk 1413

Let S′ = S0\{A}. Induction start n = 1: By construction of Labα there ex- ists a labeling Labγ for each X ∈{C}∪S′ s.t.

Chunk 1414

Labγ(X) = IN. Let L = {Labγ0,...,Labγm} be the set of all these labelings Labγ and take Labγmax ∈L s.t.

Chunk 1415

γi ≤γmax for each 0 ≤i ≤m. By construction of in(Labγmax) we know that {C} ∪S′ ⊆in(Labγmax).

Chunk 1416

Thus B ∈out(Labγmax) and therefore A ∈out(Labγmax), meaning A ∈out(Labα) by con- struction of Labα. Induction step n →n + 1: By IH we know that for the chain of supports  (S1,B1),...,(Sn+1,Bn+1) that starts at B0 ∈S1 and has length n, we have B0 ∈out(Labα).

Chunk 1417

By construction of Labα there is γ < α s.t. Labγ(B0) = OUT holds.

Chunk 1418

Since Labγ was an admissible labeling, we again know that there is an attack from an argument labeled IN or a chain of supports satisfying the conditions of Definition 25. --- Page 73 --- April 2026 Again, for each X ∈S′ = S0\{A}, take a labeling Labγ′ s.t.

Chunk 1419

Labγ′(X) = IN and let L = {Labγ′ 0,...,Labγ′n} be the set of all these labelings. Now take γmax ∈{γ,γ′ 0,...,γ′ n} s.t.

Chunk 1420

for 0 ≤i ≤n we have γ′ i ≤γmax and γ ≤γmax. By construction of Labγmax, we know S′ ⊆in(Labγmax) and B ∈out(Labγmax), therefore A must be legally OUT w.r.t.

Chunk 1421

Labγmax. Since Labγmax was an admissible labeling we have Labγmax(A) = Labα(A) = OUT as re- quired.

Chunk 1422

It is now easy to see that the result of a grounded construction is a ground-complete labeling: Proposition 44. Let J = (Args,Att,Supp) be a JSBAF s.t.

Chunk 1423

J ∈JDA⊖and let Lab be the result of a grounded construction GC = (Lab0,...,Labα). Then Lab is a ground- complete labeling.

Chunk 1424

Proof. By Proposition 43 we know that Lab = Labα is a admissible labeling.

Chunk 1425

By Defi- nition 32 we also know that FI(Lab)\in(Lab) = /0. We can therefore infer that for every A ∈FI(Lab), A ∈in(Lab) must hold.

Chunk 1426

Thus Lab is a ground-complete labeling according to Definition 30. At this point, we quickly state that since our grounded construction results in some ground-complete labeling, the set of all ground-complete labelings is non-empty (for this, we also note that the starting point of our grounded construction, SIM, always exists).

Chunk 1427

Therefore, there has to exist at least one grounded labeling for each JSBAF: Corollary 7. Let J = (Args,Att,Supp) be a JSBAF s.t.

Chunk 1428

J ∈JDA⊖. Then gr(J ) ̸= /0.

Chunk 1429

6.4.4. Uniqueness of the grounded labeling Before we can show the uniqueness of grounded labelings, we need one more definition that will be used in the upcoming proof: Definition 33.

Chunk 1430

Let J = (Args,Att,Supp) be a JSBAF s.t. J ∈JDA⊖.

Chunk 1431

Furthermore, let Labα be the result of a grounded construction GC and let A ∈in(Labα)\STRJ be a non-strict argument labeled IN by Labα. We define the computation step of A w.r.t.

Chunk 1432

GC, denoted STEP(A) as follows: STEP(A) = β where GC = (Lab0,...,Labβ,...,Labα) and A ∈in(Labβ) while there is no γ < β s.t. A ∈in(Labγ).

Chunk 1433

The key part of this proof is to show that during the grounded construction, every argument A which is labeled IN in some computation step β, is also forced IN w.r.t. every other ground-complete labeling.

Chunk 1434

Therefore all arguments that are labeled IN in Labα are also labeled IN in every other ground-complete labeling. Proposition 45.

Chunk 1435

Let J = (Args,Att,Supp) be a JSBAF s.t. J ∈JDA⊖.

Chunk 1436

Furthermore, let Lab be the result of some grounded construction GC = (Lab0,...,Labα) for J . Then Lab is the unique grounded labeling of J .

Chunk 1437

Proof. Let L ∈grcmp(J ) be some ground-complete labeling of J .

Chunk 1438

We first show that in(Lab) ⊆in(L) holds. Towards a contradiction, assume in(Lab) ̸⊆in(L), i.e.

Chunk 1439

there is A ∈in(Lab)\in(L). Let β = STEP(A) be the computation step of A w.r.t.

Chunk 1440

GC. W.l.o.g.

Chunk 1441

we assume that there --- Page 74 --- April 2026 does not exist an argument A′ ∈in(Lab)\in(L) s.t. STEP(A′) < β, i.e.

Chunk 1442

A is the “first” argument which was labeled IN by Lab and which is not labeled IN by L. For this, we note that in(SIM) ⊆in(Lab) and in(SIM) ⊆in(L) because both of these labelings are admissible.

Chunk 1443

This means that, although we did not define STEP for strict arguments, this is not relevant here because A ̸∈in(SIM) must hold. Lastly, we note that for each labeling Labγ s.t.

Chunk 1444

γ < β, we have in(Labγ) ⊆in(L), because there is no A′ ∈in(Lab)\in(L) with STEP(A′) < β. By Corollary 6, this means out(Labγ) ⊆out(L) also holds.

Chunk 1445

We will now show that A is forced IN w.r.t. L.

Chunk 1446

By construction of Labβ, we know that there was γ < β s.t. A ∈FI(Labγ) or there is an argument X ∈in(Labβ) s.t.

Chunk 1447

for STEP(X) = γ there is (S,B) ∈SSLabγ(X) with A = B or A ∈SC(B). First, let us consider the attackers of A and their label in L.

Chunk 1448

Because we have A ∈FI(Labγ) ∪  S (S,B)∈SSLabγ (X) {B} ∪SC(B)
 , we can infer that for all (C,A) ∈Att, Labγ(C) = OUT must hold. Because out(Labγ) ⊆out(L), we know that L(C) = OUT must also hold.

Chunk 1449

Therefore, all attackers of A are labeled OUT by L. Next, let us consider an arbitrary support (S,B) ∈Supp with A ∈S and L(B) ̸= IN.

Chunk 1450

We first note that, if (S,B) was safe for A in Labγ, then we can use Corollary 6 to infer that (S,B) is safe for A in L. Therefore, if (S,B) is not safe for A in L, then (S,B) cannot be safe for A in Labγ.

Chunk 1451

Now we assume that (S,B) is not safe for A in L and show that this means for (S,B), item 2.a of Definition 29 is satisfied. Take an arbitrary admissible labeling L′ with L′(B) ≥p L(B).

Chunk 1452

From L′(B) ≥p L(B) we know that either L(B) = UNDEC or L′(B) = L(B) must hold. Suppose first that L(B) = UNDEC.

Chunk 1453

Then by in(Labγ) ⊆in(L) and out(Labγ) ⊆out(L) we can infer that Labγ(B) = UNDEC must also hold. Therefore we have L′(B) ≥p Labγ(B).

Chunk 1454

Now assume that L′(B) = L(B) and let us consider the possible cases for Lab(B). The case that L(B) = UNDEC is analogous to above.

Chunk 1455

If L(B) = IN, then we must have either Labγ(B) = IN or Labγ(B) = UNDEC. Similarly, if Lab(B) = OUT we must have either Labγ(B) = OUT or Labγ(B) = UNDEC.

Chunk 1456

In all cases we again infer L′(B) ≥p Labγ(B). By assumption, A was forced IN w.r.t.

Chunk 1457

Labγ while (S,B) was not safe for A in Labγ. By Definition 29, we now infer that there is an admissible labeling L′′ which extends L′ s.t.

Chunk 1458

A is legally IN w.r.t. Lab′′ and Lab′′(B) = L′(B).

Chunk 1459

We have just shown that for an arbitrary support (S,B) with A ∈S and L(B) ̸= IN, if (S,B) is not safe for A in L, then for every admissible labeling L′ with L′(B) ≥p L(B), we can find an admissible labeling L′′ which extends L′ s.t. L′′(B) = L′(B) and A is legally IN w.r.t.

Chunk 1460

L′′. By Definition 29, A is now forced IN w.r.t.

Chunk 1461

L. Since L is a ground- complete labeling by assumption, we infer that A ∈in(L) must hold, a contradiction to our assumption A ∈in(Lab)\in(L).

Chunk 1462

With this, we have shown that for every L ∈grcmp(J ) and for Lab being the result of a grounded construction, we have in(Lab) ⊆in(L). In particular, this means that there is no L ∈grcmp(J ) s.t.

Chunk 1463

in(L) ⊂in(Lab), i.e. the result of our grounded construction is a grounded labeling according to Definition 31.

Chunk 1464

Lastly, to show the uniqueness of the grounded labeling, assume towards a contra- diction that there is some L ∈gr(J ) with Lab ̸= L. By Definition 31, L ∈grcmp(J ).

Chunk 1465

We have argued above that in(Lab) ⊆in(L) must hold. Because L ∈gr(J ) we can- not have in(Lab) ⊂in(L), therefore in(Lab) = in(L).

Chunk 1466

By Corollary 6 this clearly im- plies out(Lab) = out(L). Because Lab ̸= L by assumption, it must now be the case that undec(Lab) ̸= undec(L).

Chunk 1467

Obviously this contradicts the fact that both Lab and L are la- --- Page 75 --- April 2026 belings which assign a single label to every argument in J . We conclude that Lab = L must hold, i.e.

Chunk 1468

Lab is the unique grounded labeling of J . 7.

Chunk 1469

Future Work While the postulates we discuss have been established for grounded semantics, we thought it might be interesting to adapt our approach to grounded semantics as well. We can show that closure, direct and indirect consistency hold for this semantics, but the proofs for non-interference and crash-resistance are left for future work.

Chunk 1470

Note that in ASPIC-style formalisms that apply restricted rebut, like ASPIC+, ad- missibility is required – albeit not sufficient – to satisfy the closure postulate (see [13]). In Deductive ASPIC⊖, the support-relation between arguments alone can be used to sat- isfy the closure postulate.

Chunk 1471

Thus non-admissibility based semantics also have a chance to satisfy the closure postulate. It would be interesting to define JSBAF based analogues of naive-based semantics, like CF2 semantics, defined by Baroni et al.

Chunk 1472

[14], stage seman- tics, defined by Verheij [15], stage2 semantics, defined by Dvoˇr´ak and Gaggl [16,17], and SCF2 semantics, defined by Cramer and van der Torre [18,19]. 8.

Chunk 1473

Conclusion Although various versions of ASPIC have been proposed in the literature, to the best of our knowledge, so far none of them satisfy all five of the rationality postulates defined by Caminada and Amgoud [7] and Caminada et al. [8] in a credulous semantics like pre- ferred, while considering both rebuttals and undercuts and while avoiding the downsides of restricted rebuttal.

Chunk 1474

In this paper, we proposed Deductive ASPIC⊖to address this issue. Deductive ASPIC⊖combines the notion of Joint Support Bipolar Argumentation Frameworks defined by Cramer and Bhadra [9] with the notion of gen-rebuttals of Heyn- inck and Straßer [6].

Chunk 1475

We have given definitions of legal labelings and admissibility for JSBAFs (and by extension Deductive ASPIC⊖) and used them to define a preferred se- mantics for JSBAFs. This semantics intuitively correspond to the preferred semantics of abstract argumentation.

Chunk 1476

Furthermore, we have shown that with preferred semantics, De- ductive ASPIC⊖satisfies the rationality postulates of closure, direct consistency, indirect consistency, non-interference and crash-resistance. References [1] Rahwan I, Simari GR.

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Argumentation in Artificial Intelligence. 1st ed.

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Springer Publishing Company, Incorporated; 2009. [2] Dung PM.

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On the Acceptability of Arguments and its Fundamental Role in Nonmonotonic Reasoning, Logic Programming and n-Person Games. Artif Intell.

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1995;77(2):321-58. [3] Amgoud L, Bodenstaff L, Caminada M, McBurney P, Parsons S, Prakken H, et al.

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Final Review and Report on Formal Argumentation System; 2006. [4] Prakken H.

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An abstract framework for argumentation with structured arguments. Argument Comput.

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2010;1(2):93-124. Available from: https://doi.org/10.1080/19462160903564592.

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--- Page 76 --- April 2026 [5] Caminada M, Modgil S, Oren N. Preferences and Unrestricted Rebut.

Chunk 1485

In: Parsons S, Oren N, Reed C, Cerutti F, editors. Computational Models of Argument - Proceedings of COMMA 2014, Atholl Palace Hotel, Scottish Highlands, UK, September 9-12, 2014.

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IOS Press; 2014. p.

Chunk 1488

209-20. Available from: https://doi.org/10.3233/ 978-1-61499-436-7-209.

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[6] Heyninck J, Straßer C. Revisiting Unrestricted Rebut and Preferences in Structured Argumentation.

Chunk 1490

In: Sierra C, editor. Proceedings of the Twenty-Sixth International Joint Conference on Artificial Intelli- gence, IJCAI 2017, Melbourne, Australia, August 19-25, 2017.

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ijcai.org; 2017. p.

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1088-92. Available from: https://doi.org/10.24963/ijcai.2017/151.

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[7] Caminada M, Amgoud L. On the evaluation of argumentation formalisms.

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Artif Intell. 2007;171(5- 6):286-310.

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Available from: https://doi.org/10.1016/j.artint.2007.02.003. [8] Caminada MWA, Carnielli WA, Dunne PE.

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[9] Cramer M, Bhadra M. Deductive Joint Support for Rational Unrestricted Rebuttal.

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In: Prakken H, Bistarelli S, Santini F, Taticchi C, editors. Computational Models of Argument - Proceedings of COMMA 2020, Perugia, Italy, September 4-11, 2020.

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vol. 326 of Frontiers in Artificial Intelligence and Applications.

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[10] Modgil S, Prakken H. A general account of argumentation with preferences.

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Artif Intell. 2013;195:361- 97.

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Available from: https://doi.org/10.1016/j.artint.2012.10.008. [11] Heyninck J, Straßer C.

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Available from: https://doi.org/10.1093/logcom/ exu017. [13] Caminada M.

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Artif Intell. 2005;168(1-2):162-210.

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[15] Verheij B. Two approaches to dialectical argumentation: admissible sets and argumentation stages.

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[16] Dvor´ak W, Gaggl SA. Incorporating stage semantics in the scc-recursive schema for argumentation semantics.

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