Splitting Argumentation Frameworks with Collective Attacks and Supports

Anna Rapberger; Giovanni Buraglio; Lydia Bl\"umel; Matti Berthold

arxiv: 2604.28112 · v1 · submitted 2026-04-30 · 💻 cs.AI · cs.LO

Splitting Argumentation Frameworks with Collective Attacks and Supports

Matti Berthold , Lydia Bl\"umel , Giovanni Buraglio , Anna Rapberger This is my paper

Pith reviewed 2026-05-07 06:04 UTC · model grok-4.3

classification 💻 cs.AI cs.LO

keywords argumentation frameworkscollective attackscollective supportsbipolar argumentationsplitting techniquessemantics preservationset-based argumentation

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The pith

Argumentation frameworks with collective attacks and supports admit splitting schemata that preserve standard semantics.

A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.

The paper develops splitting techniques for bipolar set-based argumentation frameworks that incorporate both collective attacks and collective supports between arguments. It defines schemata for splitting along attack relations, along support relations, and along both at once, then proves that the sub-frameworks produce the same outcomes as the full framework under common semantics. A sympathetic reader would care because the decompositions make it possible to analyze large argumentation systems by studying smaller independent pieces while keeping the logical conclusions unchanged.

Core claim

Bipolar set-based argumentation frameworks generalize frameworks with collective attacks and bipolar frameworks by incorporating both collective attacks and supports. We consider splits over collective attacks, splits over collective supports, as well as splits over both collective attacks and supports. We establish suitable splitting schemata and prove their correctness for the most common argumentation semantics.

What carries the argument

Splitting schemata that divide the arguments and the attack and support relations into sub-frameworks while ensuring the semantics of the original framework are recovered exactly from the semantics of the parts.

If this is right

Splits over collective attacks extend the correctness results already known for frameworks with only collective attacks.
Splits over collective supports allow the positive dependencies to be handled separately from the negative ones.
Combined splits over both attacks and supports provide a single method that works for frameworks containing mixed relations.
The correctness proofs cover the standard semantics including admissibility, preferred, and stable interpretations.

Where Pith is reading between the lines

These are editorial extensions of the paper, not claims the author makes directly.

The techniques could support modular implementations of argumentation solvers by allowing independent computation on sub-frameworks.
Similar decomposition ideas might apply to related formalisms in non-monotonic reasoning where positive and negative links appear together.
Empirical tests on concrete debate examples could measure whether the splits reduce computation time in practice.

Load-bearing premise

The splitting operations capture all interactions between the split parts exactly, so that the semantics remain identical without needing extra conditions on the framework or the semantics chosen.

What would settle it

A specific bipolar set-based argumentation framework together with a chosen split and a standard semantics such as stable semantics, where the set of accepted arguments differs between the original framework and the combination of the split parts.

read the original abstract

This work proposes novel splitting techniques for argumentation formalisms that incorporate supports between defeasible elements. We base our studies on bipolar set-based argumentation frameworks (BSAFs) which generalize argumentation frameworks with collective attacks (SETAFs), as well as bipolar argumentation frameworks (BAFs), by incorporating both collective attacks and supports. Notably, BSAFs establish a crucial link to structured argumentation as they naturally capture general (potentially non-flat) assumption-based argumentation. The increase in expressiveness calls for diverse forms of splitting. We consider splits over collective attacks (thereby generalizing the recently proposed splitting techniques for SETAFs), splits over collective supports, as well as splits over both collective attacks and supports. We establish suitable splitting schemata and prove their correctness for the most common argumentation semantics.

Editorial analysis

A structured set of objections, weighed in public.

Desk editor's note, referee report, simulated authors' rebuttal, and a circularity audit. Tearing a paper down is the easy half of reading it; the pith above is the substance, this is the friction.

Desk Editor's Note private letter to a colleague

The paper lifts SETAF splitting to BSAFs with collective supports and joint attack-support splits, but the joint case on non-flat frameworks needs explicit verification to back the correctness claim.

read the letter

The main takeaway is that this work generalizes the recent SETAF splitting techniques to bipolar set-based argumentation frameworks. It defines three kinds of splits—over collective attacks, over collective supports, and over both at once—and states that it proves the schemata preserve the standard semantics (stable, preferred, complete, grounded) on BSAFs. BSAFs are positioned as a bridge to non-flat assumption-based argumentation, which is the main motivation for adding supports and collective attacks together.

Referee Report

1 major / 2 minor

Summary. The manuscript introduces novel splitting schemata for bipolar set-based argumentation frameworks (BSAFs), which support collective attacks and supports, generalizing SETAF splitting techniques. It considers three types of splits (attacks only, supports only, and combined) and claims to prove their correctness for common argumentation semantics including stable, preferred, complete, and grounded.

Significance. If the proofs hold without hidden restrictions, this work is significant for enabling modular analysis in expressive argumentation frameworks that capture non-flat assumption-based argumentation. The extension to combined attack-support splits addresses a clear gap left by prior SETAF and BAF results.

major comments (1)

[§4.3 and Theorem 5] §4.3 and Theorem 5 (Joint splitting): the reconstruction step for cases where both a collective support and a collective attack cross the cut does not explicitly verify preservation of acceptance conditions in non-flat BSAFs; the proof sketch assumes the interaction is handled by the general definitions but provides no separate case analysis or counter-example check, which is load-bearing for the abstract's claim of applicability to arbitrary BSAFs.

minor comments (2)

[Definition 3] Notation in Definition 3: the symbol for the cut set is overloaded with the support relation; a distinct symbol would improve readability.
[Figure 3] Figure 3: the reconstruction diagram lacks labels on the recomputed joint acceptance conditions, making it harder to follow the proof of Theorem 5.

Simulated Author's Rebuttal

1 responses · 0 unresolved

We thank the referee for their careful reading and constructive feedback on our manuscript. We address the major comment below and commit to strengthening the relevant proof for greater explicitness and rigor.

read point-by-point responses

Referee: [§4.3 and Theorem 5] §4.3 and Theorem 5 (Joint splitting): the reconstruction step for cases where both a collective support and a collective attack cross the cut does not explicitly verify preservation of acceptance conditions in non-flat BSAFs; the proof sketch assumes the interaction is handled by the general definitions but provides no separate case analysis or counter-example check, which is load-bearing for the abstract's claim of applicability to arbitrary BSAFs.

Authors: We appreciate the referee highlighting this aspect of the joint splitting proof. While the general definitions of BSAF acceptance conditions and the splitting schemata are formulated to apply to arbitrary frameworks (including non-flat ones that capture assumption-based argumentation), we agree that the proof sketch for Theorem 5 does not include an explicit separate case analysis for the reconstruction step when both a collective attack and a collective support cross the cut. To address this and ensure the claim of applicability to arbitrary BSAFs is fully substantiated, we will revise Section 4.3 and the proof of Theorem 5 to add a dedicated case analysis. This will verify step-by-step the preservation of acceptance conditions for the stable, preferred, complete, and grounded semantics in non-flat BSAFs, confirming no hidden restrictions arise from the interaction. We will also consider including a brief illustrative example of such a crossing case to aid readability. revision: yes

Circularity Check

0 steps flagged

No circularity: proofs of splitting schemata are independent derivations

full rationale

The paper introduces novel splitting schemata for BSAFs (generalizing SETAFs and BAFs) and states that it establishes them with proofs of correctness for standard semantics. No equations or definitions reduce by construction to fitted parameters, self-referential inputs, or load-bearing self-citations. The central claims rest on explicit generalization and stated proofs rather than renaming known results or smuggling ansatzes. The work is self-contained as a theoretical contribution; external benchmarks (prior SETAF splitting) are cited for context but not used to force the new joint attack/support cases. This matches the default expectation for proof-based papers in argumentation theory.

Axiom & Free-Parameter Ledger

0 free parameters · 2 axioms · 0 invented entities

Review based solely on abstract; no explicit free parameters, invented entities, or non-standard axioms are mentioned. The work relies on standard definitions of argumentation semantics and framework generalizations.

axioms (2)

domain assumption Standard argumentation semantics (stable, preferred, etc.) are well-defined and applicable to BSAFs
The correctness proofs are stated to hold for the most common semantics, presupposing their standard definitions carry over to the generalized frameworks.
domain assumption BSAFs correctly generalize SETAFs and BAFs while capturing non-flat assumption-based argumentation
This link is presented as a notable property motivating the need for new splitting forms.

pith-pipeline@v0.9.0 · 5431 in / 1303 out tokens · 70046 ms · 2026-05-07T06:04:58.687020+00:00 · methodology

discussion (0)

Reference graph

Works this paper leans on

66 extracted references · 66 canonical work pages

[1]

Atkinson, K.; Baroni, P.; Giacomin, M.; Hunter, A.; Prakken, H.; Reed, C.; Simari, G

On bipolarity in argumentation frameworks.Interna- tional Journal of Intelligent Systems23:1–32. Atkinson, K.; Baroni, P.; Giacomin, M.; Hunter, A.; Prakken, H.; Reed, C.; Simari, G. R.; Thimm, M.; and Vil- lata, S. 2017. Towards artificial argumentation.AI Magazine 38(3):25–36. Baroni, P., and Giacomin, M. 2007. On principle-based evaluation of extension...

work page 2017
[2]

InCorrect Reasoning - Essays on Logic-Based AI in Honour of Vladimir Lifschitz, volume 7265 ofLNCS, 57–71

Parameterized splitting: A simple modification-based approach. InCorrect Reasoning - Essays on Logic-Based AI in Honour of Vladimir Lifschitz, volume 7265 ofLNCS, 57–71. Springer. Baumann, R.; Brewka, G.; and Ulbricht, M. 2022. Shed- ding new light on the foundations of abstract argumentation: Modularization and weak admissibility.Artificial Intelli- genc...

work page 2022
[3]

Beiser, A.; Hecher, M.; Unalan, K.; and Woltran, S

Springer. Beiser, A.; Hecher, M.; Unalan, K.; and Woltran, S. 2024. Bypassing the ASP bottleneck: Hybrid grounding by split- ting and rewriting. InProceedings of (IJCAI-24), 3250–

work page 2024
[4]

10 Ben-Eliyahu-Zohary, R

ijcai.org. 10 Ben-Eliyahu-Zohary, R. 2025. Splitting a disjunctive logic program.Theory and Practice of Logic Programming 25(4):507–521. Bengel, L., and Thimm, M. 2025. Initial models and serial- isability in abstract dialectical frameworks. InProceedings of (IJCAI-25), 4365–4373. ijcai.org. Berthold, M.; Bl ¨umel, L.; and Rapberger, A. 2025. On strong an...

work page 2025
[5]

Boella, G.; Gabbay, D

ijcai.org. Boella, G.; Gabbay, D. M.; van der Torre, L. W. N.; and Villata, S. 2010. Support in abstract argumentation. In Baroni, P.; Cerutti, F.; aspects in assumption-based argu- menno Giacomin, M.; and Simari, G. R., eds.,Proceedings of (COMMA-10), Frontiers in Artificial Intelligence and Ap- plications, 111–122. IOS Press. Brewka, G.; Ellmauthaler, S...

work page 2010
[6]

Cohen, A.; Parsons, S.; Sklar, E

A survey of different approaches to support in ar- gumentation systems.The Knowledge Engineering Review 29(5):513–550. Cohen, A.; Parsons, S.; Sklar, E. I.; and McBurney, P. 2018. A characterization of types of support between structured ar- guments and their relationship with support in abstract argu- mentation.International Journal of Approximate Reason...

work page 2018
[7]

In Zhou, Z., ed.,Pro- ceedings of (IJCAI-21), 4392–4399

Argumentative XAI: A survey. In Zhou, Z., ed.,Pro- ceedings of (IJCAI-21), 4392–4399. ijcai.org. Dung, P. M. 1995. On the acceptability of arguments and its fundamental role in nonmonotonic reasoning, logic programming and n-person games.Artificial Intelligence 77(2):321–358. Dvoˇr´ak, W.; K¨onig, M.; Ulbricht, M.; and Woltran, S. 2024. Principles and the...

work page 1995
[8]

Giacomin, M.; Baroni, P.; and Cerutti, F

Expressiveness of SETAFs and support-free ADFs under 3-valued semantics.Journal of Applied Non-Classical Logics33(3-4):298–327. Giacomin, M.; Baroni, P.; and Cerutti, F. 2021. Towards a general theory of decomposability in abstract argumenta- tion. In Baroni, P.; Benzm ¨uller, C.; and W ´ang, Y . N., eds., Proceedings of (CLAR-21), Lecture Notes in Comput...

work page 2021
[9]

Lehtonen, T.; Rapberger, A.; Toni, F.; Ulbricht, M.; and Wallner, J

AAAI Press. Lehtonen, T.; Rapberger, A.; Toni, F.; Ulbricht, M.; and Wallner, J. P. 2024. Instantiations and computational aspects of non-flat assumption-based argumentation. InProceedings of (IJCAI-24). Leofante, F.; Ayoobi, H.; Dejl, A.; Freedman, G.; Gorur, D.; Jiang, J.; Paulino-Passos, G.; Rago, A.; Rapberger, A.; Russo, F.; Yin, X.; Zhang, D.; and T...

work page 2024
[10]

Modgil, S., and Prakken, H

IOS Press. Modgil, S., and Prakken, H. 2013. A general account 11 of argumentation with preferences.Artificial Intelligence 195:361–397. Modgil, S. 2009. Reasoning about preferences in argumen- tation frameworks.Artificial Intelligence173(9-10):901– 934. Nielsen, S. H., and Parsons, S. 2006. A generalization of Dung’s abstract framework for argumentation:...

work page 2013
[11]

ThenE∈cf(F)iffE∈ cf(F ′)

LetE⊆Abe a closed set. ThenE∈cf(F)iffE∈ cf(F ′)

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[12]

ThenEis closed inFiffEis closed inF ′ Proof.1

LetE⊆A. ThenEis closed inFiffEis closed inF ′ Proof.1. Suppose firstEis not conflict-free inF, then there exists(T, h)∈RwithT⊆E, h∈E. SinceEis closed, cl(T)⊆E. So either(T, h)or(cl(T), h)∈R ′ is an attack ofEon itself inF ′, soEnot conflict-free inF ′. For the other direction suppose now,Eis not conflict- free inF ′, then there exists(T ′, h)∈R ′, such th...

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[13]

Proposition 20.LetF= (A, R, S)be a BSAF ,(T, h)∈ Ran attack and letF ′ = (A, R ′, S)withR ′ = (R\ {(T, h)})∪ {(cl(T), h)}

Since argument set and support relation coincide between FandF ′ and closedness is defined solely over the support relation, the closed sets ofFandF ′ coincide. Proposition 20.LetF= (A, R, S)be a BSAF ,(T, h)∈ Ran attack and letF ′ = (A, R ′, S)withR ′ = (R\ {(T, h)})∪ {(cl(T), h)}. Thenσ(F) =σ(F ′)for all σ∈ {stb,adm,com,pref,grd}. Proof.(adm)(⊆) First, ...

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[14]

IfE 1 =cl F1 (E1)andE 2 =cl F ⋆ 2 (E2), thenE= (E 1 ∪ E2)\ {∗ 0}=cl F (E)

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[15]

IfE=cl F (E)for someE⊆A, thenE∩A 1 =E 1 = clF1 (E1)andE∩A 2 =E 2 =cl F ⋆ 2 (E2). Proof.1. SupposeE 1 =cl F1 (E1)andE 2 =cl F ⋆ 2 (E2). We show thatE= (E 1 ∪E 2)\ {∗ 0}is closed. Leta∈As.t. there is(T, a)∈SwithT⊆E. Then, eitherT⊆E 1, T⊆E 2, orThas non-empty intersection with bothE 1 andE 2. We proceed by case distinction. • Case 1:T⊆E 1. Thena∈E 1 sinceE 1...

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[16]

• We show thatE 1 is closed: Leta∈A 1 with(T, a)∈ S1

SupposeE=cl F (E)for someE⊆A, and letE 1 = E∩A 1 andE 2 =E∩A 2. • We show thatE 1 is closed: Leta∈A 1 with(T, a)∈ S1. By construction,(T, a)∈S, thusa∈E. SinceS 1 andS 2 partitionsS, we obtaina∈E∩A 1 =E 1. • We show thatE 2 is closed: Leta∈A 2 with(T, a)∈ S2. Analogous to the first item, we obtaina∈E 2 since each support is either fully contained inA 1 orA...

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[17]

IfE 1 ∈adm(F 1)andE 2 ∈adm(F ⋆ 2 ), thenE 1 ∪E 2 ∈ cf(F). 13

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[18]

IfE∈cf(F), thenE 1 =E∩A 1 ∈cf(F 1)andE 2 = E∩A 2 ∈cf(F E1 2 ). Proof.1. SupposeE 1 ∈adm(F 1)andE 2 ∈adm(F ⋆ 2 ). Let E=E 1 ∪E 2. We showE∈cf(F). Let(T, h)∈Rwith T⊆E. Either(T, h)∈R 1,(T, h)∈R 2, or(T, h)∈R 3. We proceed by case distinction. • Case 1:(T, h)∈R 1. Thenh /∈EsinceE 1 ∈ adm(F1). • Case 2:(T, h)∈R 2. It holds thatT⊆A E1 2 , h∈A E1 2 . In this ca...

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[19]

We show thatE 1 =E∩A 1 ∈cf(F 1) andE 2 =E∩A E1 2 ∈cf(F E1 2 )

SupposeE∈cf(F). We show thatE 1 =E∩A 1 ∈cf(F 1) andE 2 =E∩A E1 2 ∈cf(F E1 2 ). •E 1 ∈cf(F 1)sinceR 1 ⊆R. • We show thatE 2 ∈cf(F E1 2 ). Let(T, h)∈R E1 2 with T⊆E 2. By definition ofR E1 2 , one of the following applies: –(T, h)∈R 2 andT⊆A E1 2 , h∈A E1 2 ; (Case 1) –(T, h) = (T ′ \A 1, h)for someT ′ ⊆Asuch that (T ′, h)∈R 3,T ′ ∩(E 1)+ R3 =∅,T∩A 1 ⊆E 1, ...

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[20]

IfE 1 ∈σ(F 1)andE 2 ∈σ(F ⋆ 2 ), thenE 1 ∪E 2 ∈σ(F)

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[21]

Proof.Closure is proven using Proposition 27; conflict- freeness is proven using Lemma 54

IfE∈σ(F), thenE 1 =E∩A 1 ∈σ(F 1)andE∩A 2 ∈ σ(F ⋆ 2 ). Proof.Closure is proven using Proposition 27; conflict- freeness is proven using Lemma 54. We proceed by proving the statement for each semantics separately. Stable semantics

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[22]

We show that E=E 1 ∪E 2 ∈stb(F)

SupposeE 1 ∈stb(F 1)andE 2 ∈stb(F ⋆ 2 ). We show that E=E 1 ∪E 2 ∈stb(F). •Eis conflict-free: Note thatF ⋆ 2 =F E1 2 becauseU E1 Rc 3 = ∅. By Lemma 54 and sinceE 1 ∈stb(F 1)andE 2 ∈ stb(F E1 2 ), we obtainE∈cf(F). •Eis closed: this follows from Proposition 27. •Eattacks all remaining arguments: Leta∈A\E. We proceed by case distinction. –Case 1:a∈A 1. Then...

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[23]

It holds thatE ⊕ R =A=A 1 ∪A 2

SupposeE∈stb(F). It holds thatE ⊕ R =A=A 1 ∪A 2. • We show thatE 1 =E∩A 1 ∈stb(F 1). From Lemma 54, we obtainE∩A 1 ∈cf(F 1). Leta∈A 1. Since (F1, F2, R3)is an attack splitting ofF, it holds thata is attacked by someT⊆A 1, thus(E∩A 1)⊕ R1 =A 1 and thereforeE 1 ∈stb(F 1). 14 • We show thatE 2 =E∩A 2 ∈stb(F ⋆ 2 ). By Lemma 54, we obtainE 2 ∈cf(F E1 2 ); by P...

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[24]

Observe that∗ 0 /∈E2 since∗ 0 is self-attacking

SupposeE 1 ∈adm(F 1)andE 2 ∈adm(F ⋆ 2 ). Observe that∗ 0 /∈E2 since∗ 0 is self-attacking. We show thatE=E 1 ∪E 2 ∈adm(F). By Lemma 54, we obtainE=cl F (E); moreover,Eis closed by Proposition 27. It remains to show thatEdefends itself against each closed set. LetT c ⊆Adenote a closed attacker ofE. Then there is (T, a)∈RwithT⊆T c anda∈E. Eithera∈E 1 or a∈E ...

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[25]

We proceed by case distinction

It holds thatcl F (T) =cl F (T ′), and thereforeT ′ =T c. We proceed by case distinction. Case 2.ii.a:T ′ ∩(E 1)+ R1∪Rc 3 ̸=∅. Thenais defended byE 1 against(T ′, a), thus also byE. Case 2.ii.b:T ′ ∩(E 1)+ R1∪Rc 3 =∅. In this case, either(T ′ ∩A 2, a)∈R ⋆ 2 (via the reduct) or((T ′ ∩A 2)∪ {∗ 0}, a)∈R ⋆ 2 (via the modification). Recall thata∈E 2 and, moreo...

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[26]

We show thatE 1 =E∩A 1 andE 2 =E∩A 2 are admissible inF 1 resp.F ⋆ 2

SupposeE∈adm(F). We show thatE 1 =E∩A 1 andE 2 =E∩A 2 are admissible inF 1 resp.F ⋆ 2 . By Lemma 54,E 1 =E∈cf(F 1)andE 2 ∈cf(F E1 2 ); by Proposition 27 both sets are closed. We show thatE 1 and E2 defend themselves inF 1 andF ⋆ 2 , respectively, against each attacker. Recall that each attacker inR c 3 is closed. •E 1 defends itself inF 1 follows sinceEde...

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[27]

Then forE=E 1 ∪ E2 we have by the previous itemE∈adm(F)

LetE 1 ∈com(F 1), E2 ∈com(F ⋆ 2 ). Then forE=E 1 ∪ E2 we have by the previous itemE∈adm(F). It is left to show thatEcontains everything it defends. So let a∈Asuch that for every closed attackerDonawith some(T, a)∈R, T⊆Dthere exists an attack(S, h) fromS⊆Etoh∈D. We proceed by case distinction (a∈A 1)ThenD⊆A 1 and therefore(S, h)∈R 1, so E1 defendsa, soa∈E ...

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[28]

Then by the previous itemE 1 =E∩A 1 is admissible inF 1 and E2 =E∩A 2 is admissible inF ⋆ 2

For the other direction supposeE∈com(F). Then by the previous itemE 1 =E∩A 1 is admissible inF 1 and E2 =E∩A 2 is admissible inF ⋆ 2 . Note that this implies that∗ 0 /∈E2. It is left to show that bothE 1 andE 2 contain every argument they defend. (a) Letabe defended byE 1 inF 1, then for every closed AttackerDonathere exists an attack(S, h)∈R 1 withS⊆E 1,...

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[29]

SupposeE 1 ∈ pref(F)andE 2 ∈pref(F ⋆ 2 )butE=E 1 ∪E 2 /∈pref(F), then by the previous itemE∈adm(F), so there ex- ists a preferred supersetE ′ ⊋EofF

The first direction follows directly from the correspon- dence results for admissible semantics. SupposeE 1 ∈ pref(F)andE 2 ∈pref(F ⋆ 2 )butE=E 1 ∪E 2 /∈pref(F), then by the previous itemE∈adm(F), so there ex- ists a preferred supersetE ′ ⊋EofF. But then either E′ ∩A 1 =E ′ 1 ⊋E 1 and sinceE ′ is admissible inF we haveE 1 ∈adm(F 1)soE 1 /∈pref(F1)orE ′ 1 ...

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[30]

ThenE∩A 1 =E 1 ∈ adm(F1)andE∩A 2 =E 2 ∈adm(F ⋆ 2 )

Now supposeE∈pref(F). ThenE∩A 1 =E 1 ∈ adm(F1)andE∩A 2 =E 2 ∈adm(F ⋆ 2 ). If there existed someE ′ 2 ⊋E 2 withE ′ 2 ∈adm(F ⋆ 2 ), thenE 1 ∪E ′ 2 ∈ adm(F)andE⊊E 1 ∪E ′ 2, soEis not preferred. Con- tradiction. SoE 2 ∈pref(F ⋆ 2 ). It is left to show thatE 1 ∈pref(F 1). Suppose there exists someE † 1 ⊋E 1, E† 1 ∈adm(F 1). We will show that in this caseE 2 ∈a...

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[31]

Now supposeE 2 /∈adm(F⋆† 2 )

We have strictly less undecided links, soR ⋆† 2 \R E† 1 2 ⊆R ⋆ 2 \R E1 2 and we haveR E† 1 2 = RE1 2 ∪ {(T\A 1, h)|(T, h)∈R 3, T∩(E † 1)+ R1∪Rc 3 = ∅, T∩A 1 ⊆E † 1, T∩A 1 ⊈E 1, h∈A 2}(sinceE † 1 is conflict-free, no attacks are lost). Now supposeE 2 /∈adm(F⋆† 2 ). ThenE 2 is either not closed, not conflict-free or does not defend itself. Noth- ing changed...

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[32]

IfE 1 ∈cf(F 1)andE 2 ∈cf(F E1 2 ), thenE 1 ∪(E 2 \ {∗2})∈cf(F)

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[33]

Proof.We prove the statements separately

IfE∈cf(F), thenE 1 =E∩A 1 ∈cf(F 1)andE∩A 2 ∈ cf(F E1 2 ). Proof.We prove the statements separately

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[34]

LetE=E 1 ∪ (E2 \ ∗2)

SupposeE 1 ∈cf(F 1)andE 2 ∈cf(F E1 2 ). LetE=E 1 ∪ (E2 \ ∗2). We distinguish two cases: •T S3 (E1) =∅. In this case, we haveF E1 2 =F 2 and thusE 2 ∈cf(F 2). Therefore,E 1 ∪E 2 ∈cf(F)follows from the fact that there is no attack fromF 1 andF 2. •T S3 (E1)̸=∅. In this case, we haveR E1 2 ⊇R 2 ∪ {(∅,∗ 1)}, so thatE 2 ∈cf(F 2). Moreover,∗ 1 /∈E2 becauseE 2 i...

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[35]

We show thatE 1 =E∩A 1 ∈cf(F 1) andE 2 =E∩A 2 ∈cf(F E1 2 )

SupposeE∈cf(F). We show thatE 1 =E∩A 1 ∈cf(F 1) andE 2 =E∩A 2 ∈cf(F E1 2 ). • SinceR 1 ⊆R, it immediately follows thatE 1 =E∩ A1 ∈cf(F 1). • We now considerE 2 =E∩A 2 ∈cf(F E1 2 ). Since R2 ⊆Rand(F 1, F2, S3)is a proper support splitting, we know thatE∩A 2 ∈cf(F 2). Further, sinceE⊆A, we derive that∗ i /∈E∩A 2 for alli∈ {1,2}. Thus, since the reduct only ...

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[36]

IfE 1 ∈adm(F 1)andE 2 ∈adm(F E1 2 ), thenE= cl F (E)whereE=E 1 ∪(E 2 \ {∗2})

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[37]

IfE∈adm(F), thenE∩A 1 =E 1 =cl F1 (E1)and E2 =cl F E1 2 (E2)whereE 2 =E∩A 2. Proof.1. LetE 1 ∈adm(F 1)andE 2 ∈adm(F E1 2 ). We show thatE=E 1 ∪(E 2 \ {∗2})is closed. Let(T, h)∈SwithT⊆E. It holds that either(T, h)∈ S1,(T, h)∈S 2, or(T, h)∈S 3. We proceed by case distinction. • Case 1:(T, h)∈S 1. ThenT⊆E 1. SinceE 1 is closed inF 1 by assumption, we geth∈E ...

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[38]

• Case 3:(T, h)∈S 3

SinceE 2 is closed inF ⋆ 2 , we obtainh∈E 2 ⊆E. • Case 3:(T, h)∈S 3. It holds thath∈A 1 by definition ofS 3. Recall thatT⊆E, thusT∩A 1 ⊆E 1. Towards a contradiction, supposeh /∈E 1. We proceed by case distinction. Case 3.i:T∩A 1 =∅. In this case,T∈ D S3 (E1). By construction of the S-reduct,(T,∗ 1)∈S ⋆

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[39]

Since(∅,∗ 1)∈R ⋆ 2,E 2 is attacked by the empty set and thus not admissible inF ⋆ 2 ; contradiction to our ini- tial assumption

Since T⊆E 2 andE 2 is closed inF ⋆ 2 , we have∗ 1 ∈E 2. Since(∅,∗ 1)∈R ⋆ 2,E 2 is attacked by the empty set and thus not admissible inF ⋆ 2 ; contradiction to our ini- tial assumption. Case 3.ii:T∩A 1 ̸=∅. In this case,T∈ T S3 (E1)\ DS3 (E1). By construction of the S-reduct,(T∩ A2,∗ 2)∈S ⋆

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[40]

Again, we can de- rive a contradiction to admissibility ofE 2 inF ⋆ 2 : By construction of the S-reduct,F ⋆ 2 contains the attack ((T∩A 2)∪ {∗ 2},∗ 2), thusE 2 /∈cf(F⋆ 2 )

SinceT∩A 2 ⊆E 2 andE 2 is closed inF ⋆ 2 , we have∗ 2 ∈E 2. Again, we can de- rive a contradiction to admissibility ofE 2 inF ⋆ 2 : By construction of the S-reduct,F ⋆ 2 contains the attack ((T∩A 2)∪ {∗ 2},∗ 2), thusE 2 /∈cf(F⋆ 2 ). In each case, we haveh∈E; therefore,E=cl F (E)

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[41]

LetE 1 =E∩A 1 and E2 =E∩A 2

AssumeEis admissible inF. LetE 1 =E∩A 1 and E2 =E∩A 2. We show thatE 1 is closed inF 1 andE 2 is closed inF E1 2 . • LetE 1 =E∩A 1. We proveE 1 =cl F1 (E1). Suppose towards contradiction thatE 1 supports somea∈A 1 inF 1, buta /∈E 1. Then for someT⊆E 1 anda∈ A1 \E 1 we have(T, a)∈S 1. SinceE 1 ⊆Eand S1 ⊆S, we derive thatEsupports somea∈A\E inF. Thus,E̸=cl ...

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[42]

IfE 1 ∈σ(F 1)andE 2 ∈σ(F E1 2 ), thenE 1∪(E2\{∗2})∈ σ(F)

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[43]

Proof.Similarly to Theorem 28, we prove the statements for each semantics

IfE∈σ(F), thenE 1 =E∩A 1 ∈σ(F 1)and (E∩A 2 ∈ σ(F E1 2 )or(E∩A 2)∪ {∗ 2} ∈σ(F E1 2 )). Proof.Similarly to Theorem 28, we prove the statements for each semantics. Admissible Semantics. 18

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[44]

We show E=E 1 ∪E 2 \ {∗2} ∈adm(F)

SupposeE 1 ∈adm(F 1)andE 2 ∈adm(F E1 2 ). We show E=E 1 ∪E 2 \ {∗2} ∈adm(F). •Eis closed inF: this follows from Lemma 56. •Eis conflict-free: this follows from Lemma 55. •Edefends itself against each closed set: Consider a set Tc ⊆Awhich is closed inF, i.e.,T c =cl F (Tc), and attacksEinF. Then there isT⊆T c,h∈E, such that(T, h)∈R. It holds that either(T,...

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[45]

We show thatE 1 =E∩A 1 ∈ adm(F1)andE 2 =E∩A 2 ∈adm(F E1 2 )

SupposeE∈adm(F). We show thatE 1 =E∩A 1 ∈ adm(F1)andE 2 =E∩A 2 ∈adm(F E1 2 ). • LetE 1 =E∩A 1.E 1 is conflict-free by Lemma 55 and closed by Lemma 56. It remains to prove thatE 1 defends itself. LetT c be a closed attacker ofE1. Then there isT⊆T c, h∈E 1, such that(T, h)∈R 1. Observe thatcl F (T)⊆ A1 sinceT⊆A 1 and there is no support(B, a)∈S withB⊆A 1 an...

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[46]

We show that E=E 1 ∪(E 2 \ {∗2})∈stb(F)

SupposeE 1 ∈stb(F 1)andE 2 ∈stb(F E1 2 ). We show that E=E 1 ∪(E 2 \ {∗2})∈stb(F). 19 •Eis conflict-free: By Lemma 55 and sinceE 1 ∈ stb(F1)andE 2 ∈stb(F E1 2 ), we obtainE∈cf(F). •Eis closed because of Lemma 56 andstb(F)⊆ adm(F)for every BSAFF. •Eattacks all remaining arguments: Leta∈A\E. We proceed by case distinction. –Case 1:a∈A 1. Thena∈(E 1)+ R1 sin...

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[47]

It holds thatE ⊕ R =A=A 1 ∪A 2

SupposeE∈stb(F). It holds thatE ⊕ R =A=A 1 ∪A 2. • We show thatE 1 =E∩A 1 ∈stb(F 1). From Lemma 55, we obtainE∩A 1 ∈cf(F 1).E 1 is closed inF 1 because of Lemma 56. Finally, we show thatE 1 at- tacks all argumentsa∈A 1 \E 1. Leta∈A 1. Since (F1, F2, S3)is a support splitting ofF, it holds thatR is partitioned intoR 1 andR 2. Thus,ais attacked by someT⊆A 1...

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[48]

We show E=E 1 ∪E 2 \ {∗2} ∈com(F)

SupposeE 1 ∈com(F 1)andE 2 ∈com(F E1 2 ). We show E=E 1 ∪E 2 \ {∗2} ∈com(F). • The premissesE 1 ∈com(F 1)andE 2 ∈com(F E1 2 ) directly entailE 1 ∈adm(F 1)andE 2 ∈adm(F E1 2 ), henceE∈adm(F)as shown above. • It remains to be shown thatEdefends itself inF. –Case 1:T S3 (E1) =∅. ThenF E1 2 =F 2, and for any (T, h)∈S 3:h∈E∨T⊊E. Lets assumeEdefends a∈A\EinF. I...

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[49]

We show thatE 1 =E∩A 1 ∈ com(F1)and (E∩A 2 ∈com(F E1 2 )or(E∩A 2)∪ {∗2} ∈ com(F E1 2 ))

SupposeE∈com(F). We show thatE 1 =E∩A 1 ∈ com(F1)and (E∩A 2 ∈com(F E1 2 )or(E∩A 2)∪ {∗2} ∈ com(F E1 2 )). • Utilizing the proof for admissible semantics, we di- rectly concludeE 1 =E∩A 1 ∈adm(F 1)andE 2 = E∩A 2 ∈adm(F E1 2 )or(E∩A 2)∪{∗ 2} ∈adm(F E1 2 )). • There are no attacks fromA 2 towardsA 1, neither are there supports fromA 1 towardsA 2. Hence any a...

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[50]

Contradiction toE ′ 2 ∈adm(F E1 2 )

Since(T ′ ∩A 2,∗ 2)∈S E1 2 , this implies E′ 2 is not closed. Contradiction toE ′ 2 ∈adm(F E1 2 ). Case 2.i.b:h∈A 2. In this case,T⊆E ′ 2 and we have a contradiction to the conflict-freeness ofE ′ 2. Case 2.ii:∗ 2 /∈E2. ThenE ′ 2 ⊋E ′ 2, contradiction toE 2 ∈ pref(F2). 21 C Omitted Proofs of Section 5 We recall the notation from Definition 48. Throughout ...

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[51]

IfE 1 ∈adm(F 1)andE 2 ∈adm(F ⊛ 2 ), thenE=E 1 ∪ (E2 \ {∗2})∈cf(F)

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[52]

IfE∈adm(F), thenE 1 =E∩A 1 ∈cf(F 1)andE∩ A2 ∈cf(F ⊛ 2 ). Proof.1. LetE 1 ∈adm(F 1)andE 2 ∈adm(F ⊛ 2 ). We show thatE=E 1 ∪(E 2 \ {∗ 2})∈cf(F). Let(T, h)∈R s.t.T⊆E. Then either(T, h)∈R 1,(T, h)∈R 2, or (T, h)∈R 3. • Case 1:(T, h)∈R 1. Thenh /∈EsinceE 1 ∈ adm(F1). • Case 2:(T, h)∈R 2. It holds thatT⊆A 2, h∈A 2. –Case 2i:cl(T)⊆A 2. Then(cl(T), h)∈R ⊛ 2 . –Ca...

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[53]

We show thatE 1 =E∩A 1 ∈ cf(F1)andE 2 =E∩A 2 ∈cf(F ⊛ 2 )

SupposeE∈adm(F). We show thatE 1 =E∩A 1 ∈ cf(F1)andE 2 =E∩A 2 ∈cf(F ⊛ 2 ). •E 1 ∈cf(F 1)sinceR 1 ⊆R. • We show thatE 2 ∈cf(F ⊛ 2 ). Let(T, h)∈R ⊛ 2 with T⊆E 2. By definition ofE 2, we haveh∈A 2. Thus, by definition ofR ⊛ 2 , one of the following applies for (T, h): –(T, h)∈ ˆR2 andT⊆A 2, h∈A 2; (Case 1) –(T, h) = (T ′ \A 1, h)for someT ′ ⊆Asuch that (T ′,...

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[54]

IfE 1 ∈adm(F 1)andE 2 ∈adm(F ⊛ 2 ), thenE=cl F (E) whereE=E 1 ∪(E 2 \ {∗2})

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[55]

IfE∈adm(F), thenE∩A 1 =E 1 =cl F1 (E1)and E2 =cl F ⊛ 2 (E2)whereE 2 =E∩A 2. Proof.We recall:F 1 = (A 1, R1, S1)andF ⊛ 2 = (A⊛ 2 , R⊛ 2 , S⊛ 2 )withA ⊛ 2 =A 2 ∪ {∗ 0,∗ 1,∗ 2}andR ⊛ 2 ,S ⊛ 2 as defined by the respective modifications for support and attack splitting. 1.E 1 ∈adm(F 1)andE 2 ∈adm(F ⊛ 2 ). We show thatE= E1 ∪(E 2 \ {∗2})is closed inF. The proof...

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[56]

We show thatE∩A 1 =E 1 = clF1 (E1)andE∩A 2 =E 2 =cl F ⊛ 2 (E2)

LetE∈adm(F). We show thatE∩A 1 =E 1 = clF1 (E1)andE∩A 2 =E 2 =cl F ⊛ 2 (E2). • LetE 1 =E∩A 1 and let(T, h)∈S 1 withT⊆E 1. SinceT⊆EandS 1 ⊆S, we obtainh∈E 1. There- fore,E 1 =cl F1 (E1). • LetE 2 =E∩A 2 and let(T, h)∈S ⊛ 2 withT⊆ E2. By construction of the S-reduct, either(T, h)∈ S2 orh=∗ 1 orh=∗ 2. Note that splitting attacks (closing links inR 3, constru...

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[57]

IfE 1 ∈σ(F 1)andE 2 ∈σ(F ⊛ 2 ), thenE 1 ∪(E 2 \ {∗2})∈ σ(F)

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[58]

Proof.We prove the theorem for all considered semantics

IfE∈σ(F), thenE 1 =E∩A 1 ∈σ(F 1)andE∩A 2 ∈ σ(F ⊛ 2 )or(E∩A 2)∪ {∗ 2} ∈σ(F ⊛ 2 ). Proof.We prove the theorem for all considered semantics. Admissible Semantics

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[59]

We show thatE=E 1 ∪E 2 ∈adm(F)

SupposeE 1 ∈adm(F 1)andE 2 ∈adm(F ⊛ 2 ). We show thatE=E 1 ∪E 2 ∈adm(F). By Lemma 57,Eis conflict-free inF; by Lemma 58,Eis closed inF. It remains to prove thatEdefends itself inF. LetT c denote a closed attacker ofEinF. That is, there isT⊆T c with(T, h)∈Randh∈E. Wlog, we can assumecl F (T) =T c. As usual, we have three cases to consider:(T, h)∈R 1, (T, h...

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[60]

We show thatE∩A 1 ∈adm(F 1) andE∩A 2 ∈adm(F ⊛ 2 )

SupposeE∈adm(F). We show thatE∩A 1 ∈adm(F 1) andE∩A 2 ∈adm(F ⊛ 2 ). By Lemma 58 we know that both sets are closed; moreover, by Lemma 57 they are also conflict-free. It remains to prove that the sets defend itself. • The case forE 1 =E∩A 1 holds since for all(T, h)∈ Rwithh∈E 1 it holds that(T, h)∈R 1. Moreover, cl F (T) =cl F1 (T)since no positive links g...

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[61]

We show thatE=E 1 ∪(E 2 \ {∗2})∈com(F)

SupposeE 1 ∈com(F 1)andE 2 ∈com(F ⊛ 2 ). We show thatE=E 1 ∪(E 2 \ {∗2})∈com(F). The premissesE 1 ∈com(F 1)andE 2 ∈com(F ⊛ 2 )di- rectly entailE 1 ∈adm(F 1)andE 2 ∈adm(F ⊛ 2 ), hence E∈adm(F)as shown above. It remains to be shown that Edefends itself inF. Assume toward contradiction that there exists an argu- mentd∈A\Ethat is defended byEinF. We proceed b...

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[62]

We show thatE 1 =E∩A 1 ∈ com(F1)and (E∩A 2 ∈com(F ⊛ 2 )or(E∩A 2)∪ {∗ 2} ∈ com(F ⊛ 2 ))

SupposeE∈com(F). We show thatE 1 =E∩A 1 ∈ com(F1)and (E∩A 2 ∈com(F ⊛ 2 )or(E∩A 2)∪ {∗ 2} ∈ com(F ⊛ 2 )). • Utilizing the proof for admissible semantics, we di- rectly concludeE 1 =E∩A 1 ∈adm(F 1)andE 2 = E∩A 2 ∈adm(F ⊛ 2 )orE 2 = (E∩A 2)∪ {∗ 2} ∈ adm(F ⊛ 2 ). • There are no attacks fromA 2 towardsA 1, neither are there supports fromA 1 towardsA 2. Hence a...

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[63]

We show that E=E 1 ∪(E 2 \ {∗2})∈stb(F)

SupposeE 1 ∈stb(F 1)andE 2 ∈stb(F ⊛ 2 ). We show that E=E 1 ∪(E 2 \ {∗2})∈stb(F). • E is conflict-free: By Lemma 57 andE 1 ∈adm(F 1) andE 2 ∈adm(F ⊛ 2 ), we obtainE∈cf(F). • E is closed because of Lemma 58 and the fact that stb(F)⊆adm(F)for every BSAFF. •Eattacks every argumenta∈A\E. Leta∈A\E. We proceed by case distinction. –Case 1:a∈A 1. Thena∈(E 1)+ R1...

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[64]

• We first show thatE 1 =E∩A 1 ∈stb(F 1)

SupposeE∈stb(F), that isE ⊕ R =E∪E + R =A. • We first show thatE 1 =E∩A 1 ∈stb(F 1). From pre- vious lemmata we know thatE 1 is closed and conflict- free. It remains to prove thatE 1 attacks everya∈ A1 \E 1. Since there is no attack fromF 2 toF 1 by def- inition of splitting, we know thata∈(E) + R1 by some T⊆E∩A 1, i.e.a∈(E∩A 1)+ R1. • We show thatE∩A 2 ∈...

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[65]

Further, fromE∈cf(F), it follows thatT c ∩(E 1)+ R1∪Rc 3 = ∅

SinceEis closed inFby hypothesis, we know thatT c ⊆E. Further, fromE∈cf(F), it follows thatT c ∩(E 1)+ R1∪Rc 3 = ∅. Moreover,T c ∩A 1 ⊆E∩A 1 becauseT c ⊆ E. Hence, by definition ofR-reduct, there is a(T c \ A1, a)∈R ⊛ 2 , and consequently,a∈E + R⊛ 2 . In both cases we finda∈E + R⊛ 2 . • Case 2.a /∈A 2. Thusa∈ {∗ 1}. By definition ofS- reduct we know that∗...

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[66]

Contradiction toE ′ 2 ∈adm(F ⊛ 2 )

Since(T ′ ∩A 2,∗ 2)∈S ⊛ 2 , this implies E′ 2 is not closed. Contradiction toE ′ 2 ∈adm(F ⊛ 2 ). Case 2.i.b:h∈A 2. In this case,T⊆E ′ 2 and we have a contradiction to the conflict-freeness ofE ′ 2 inF ⊛ 2 . Case 2.ii:∗ 2 /∈E2. ThenE ′ 2 =E 2 \ {∗2}and therefore E′ 2 ⊋E 2, contradiction toE 2 ∈pref(F ⊛ 2 ). Grounded Semantics.SupposeE 1 ∈grd(F 1)andE 2 ∈ g...

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[1] [1]

Atkinson, K.; Baroni, P.; Giacomin, M.; Hunter, A.; Prakken, H.; Reed, C.; Simari, G

On bipolarity in argumentation frameworks.Interna- tional Journal of Intelligent Systems23:1–32. Atkinson, K.; Baroni, P.; Giacomin, M.; Hunter, A.; Prakken, H.; Reed, C.; Simari, G. R.; Thimm, M.; and Vil- lata, S. 2017. Towards artificial argumentation.AI Magazine 38(3):25–36. Baroni, P., and Giacomin, M. 2007. On principle-based evaluation of extension...

work page 2017

[2] [2]

InCorrect Reasoning - Essays on Logic-Based AI in Honour of Vladimir Lifschitz, volume 7265 ofLNCS, 57–71

Parameterized splitting: A simple modification-based approach. InCorrect Reasoning - Essays on Logic-Based AI in Honour of Vladimir Lifschitz, volume 7265 ofLNCS, 57–71. Springer. Baumann, R.; Brewka, G.; and Ulbricht, M. 2022. Shed- ding new light on the foundations of abstract argumentation: Modularization and weak admissibility.Artificial Intelli- genc...

work page 2022

[3] [3]

Beiser, A.; Hecher, M.; Unalan, K.; and Woltran, S

Springer. Beiser, A.; Hecher, M.; Unalan, K.; and Woltran, S. 2024. Bypassing the ASP bottleneck: Hybrid grounding by split- ting and rewriting. InProceedings of (IJCAI-24), 3250–

work page 2024

[4] [4]

10 Ben-Eliyahu-Zohary, R

ijcai.org. 10 Ben-Eliyahu-Zohary, R. 2025. Splitting a disjunctive logic program.Theory and Practice of Logic Programming 25(4):507–521. Bengel, L., and Thimm, M. 2025. Initial models and serial- isability in abstract dialectical frameworks. InProceedings of (IJCAI-25), 4365–4373. ijcai.org. Berthold, M.; Bl ¨umel, L.; and Rapberger, A. 2025. On strong an...

work page 2025

[5] [5]

Boella, G.; Gabbay, D

ijcai.org. Boella, G.; Gabbay, D. M.; van der Torre, L. W. N.; and Villata, S. 2010. Support in abstract argumentation. In Baroni, P.; Cerutti, F.; aspects in assumption-based argu- menno Giacomin, M.; and Simari, G. R., eds.,Proceedings of (COMMA-10), Frontiers in Artificial Intelligence and Ap- plications, 111–122. IOS Press. Brewka, G.; Ellmauthaler, S...

work page 2010

[6] [6]

Cohen, A.; Parsons, S.; Sklar, E

A survey of different approaches to support in ar- gumentation systems.The Knowledge Engineering Review 29(5):513–550. Cohen, A.; Parsons, S.; Sklar, E. I.; and McBurney, P. 2018. A characterization of types of support between structured ar- guments and their relationship with support in abstract argu- mentation.International Journal of Approximate Reason...

work page 2018

[7] [7]

In Zhou, Z., ed.,Pro- ceedings of (IJCAI-21), 4392–4399

Argumentative XAI: A survey. In Zhou, Z., ed.,Pro- ceedings of (IJCAI-21), 4392–4399. ijcai.org. Dung, P. M. 1995. On the acceptability of arguments and its fundamental role in nonmonotonic reasoning, logic programming and n-person games.Artificial Intelligence 77(2):321–358. Dvoˇr´ak, W.; K¨onig, M.; Ulbricht, M.; and Woltran, S. 2024. Principles and the...

work page 1995

[8] [8]

Giacomin, M.; Baroni, P.; and Cerutti, F

Expressiveness of SETAFs and support-free ADFs under 3-valued semantics.Journal of Applied Non-Classical Logics33(3-4):298–327. Giacomin, M.; Baroni, P.; and Cerutti, F. 2021. Towards a general theory of decomposability in abstract argumenta- tion. In Baroni, P.; Benzm ¨uller, C.; and W ´ang, Y . N., eds., Proceedings of (CLAR-21), Lecture Notes in Comput...

work page 2021

[9] [9]

Lehtonen, T.; Rapberger, A.; Toni, F.; Ulbricht, M.; and Wallner, J

AAAI Press. Lehtonen, T.; Rapberger, A.; Toni, F.; Ulbricht, M.; and Wallner, J. P. 2024. Instantiations and computational aspects of non-flat assumption-based argumentation. InProceedings of (IJCAI-24). Leofante, F.; Ayoobi, H.; Dejl, A.; Freedman, G.; Gorur, D.; Jiang, J.; Paulino-Passos, G.; Rago, A.; Rapberger, A.; Russo, F.; Yin, X.; Zhang, D.; and T...

work page 2024

[10] [10]

Modgil, S., and Prakken, H

IOS Press. Modgil, S., and Prakken, H. 2013. A general account 11 of argumentation with preferences.Artificial Intelligence 195:361–397. Modgil, S. 2009. Reasoning about preferences in argumen- tation frameworks.Artificial Intelligence173(9-10):901– 934. Nielsen, S. H., and Parsons, S. 2006. A generalization of Dung’s abstract framework for argumentation:...

work page 2013

[11] [11]

ThenE∈cf(F)iffE∈ cf(F ′)

LetE⊆Abe a closed set. ThenE∈cf(F)iffE∈ cf(F ′)

work page

[12] [12]

ThenEis closed inFiffEis closed inF ′ Proof.1

LetE⊆A. ThenEis closed inFiffEis closed inF ′ Proof.1. Suppose firstEis not conflict-free inF, then there exists(T, h)∈RwithT⊆E, h∈E. SinceEis closed, cl(T)⊆E. So either(T, h)or(cl(T), h)∈R ′ is an attack ofEon itself inF ′, soEnot conflict-free inF ′. For the other direction suppose now,Eis not conflict- free inF ′, then there exists(T ′, h)∈R ′, such th...

work page

[13] [13]

Proposition 20.LetF= (A, R, S)be a BSAF ,(T, h)∈ Ran attack and letF ′ = (A, R ′, S)withR ′ = (R\ {(T, h)})∪ {(cl(T), h)}

Since argument set and support relation coincide between FandF ′ and closedness is defined solely over the support relation, the closed sets ofFandF ′ coincide. Proposition 20.LetF= (A, R, S)be a BSAF ,(T, h)∈ Ran attack and letF ′ = (A, R ′, S)withR ′ = (R\ {(T, h)})∪ {(cl(T), h)}. Thenσ(F) =σ(F ′)for all σ∈ {stb,adm,com,pref,grd}. Proof.(adm)(⊆) First, ...

work page

[14] [14]

IfE 1 =cl F1 (E1)andE 2 =cl F ⋆ 2 (E2), thenE= (E 1 ∪ E2)\ {∗ 0}=cl F (E)

work page

[15] [15]

IfE=cl F (E)for someE⊆A, thenE∩A 1 =E 1 = clF1 (E1)andE∩A 2 =E 2 =cl F ⋆ 2 (E2). Proof.1. SupposeE 1 =cl F1 (E1)andE 2 =cl F ⋆ 2 (E2). We show thatE= (E 1 ∪E 2)\ {∗ 0}is closed. Leta∈As.t. there is(T, a)∈SwithT⊆E. Then, eitherT⊆E 1, T⊆E 2, orThas non-empty intersection with bothE 1 andE 2. We proceed by case distinction. • Case 1:T⊆E 1. Thena∈E 1 sinceE 1...

work page

[16] [16]

• We show thatE 1 is closed: Leta∈A 1 with(T, a)∈ S1

SupposeE=cl F (E)for someE⊆A, and letE 1 = E∩A 1 andE 2 =E∩A 2. • We show thatE 1 is closed: Leta∈A 1 with(T, a)∈ S1. By construction,(T, a)∈S, thusa∈E. SinceS 1 andS 2 partitionsS, we obtaina∈E∩A 1 =E 1. • We show thatE 2 is closed: Leta∈A 2 with(T, a)∈ S2. Analogous to the first item, we obtaina∈E 2 since each support is either fully contained inA 1 orA...

work page

[17] [17]

IfE 1 ∈adm(F 1)andE 2 ∈adm(F ⋆ 2 ), thenE 1 ∪E 2 ∈ cf(F). 13

work page

[18] [18]

IfE∈cf(F), thenE 1 =E∩A 1 ∈cf(F 1)andE 2 = E∩A 2 ∈cf(F E1 2 ). Proof.1. SupposeE 1 ∈adm(F 1)andE 2 ∈adm(F ⋆ 2 ). Let E=E 1 ∪E 2. We showE∈cf(F). Let(T, h)∈Rwith T⊆E. Either(T, h)∈R 1,(T, h)∈R 2, or(T, h)∈R 3. We proceed by case distinction. • Case 1:(T, h)∈R 1. Thenh /∈EsinceE 1 ∈ adm(F1). • Case 2:(T, h)∈R 2. It holds thatT⊆A E1 2 , h∈A E1 2 . In this ca...

work page

[19] [19]

We show thatE 1 =E∩A 1 ∈cf(F 1) andE 2 =E∩A E1 2 ∈cf(F E1 2 )

SupposeE∈cf(F). We show thatE 1 =E∩A 1 ∈cf(F 1) andE 2 =E∩A E1 2 ∈cf(F E1 2 ). •E 1 ∈cf(F 1)sinceR 1 ⊆R. • We show thatE 2 ∈cf(F E1 2 ). Let(T, h)∈R E1 2 with T⊆E 2. By definition ofR E1 2 , one of the following applies: –(T, h)∈R 2 andT⊆A E1 2 , h∈A E1 2 ; (Case 1) –(T, h) = (T ′ \A 1, h)for someT ′ ⊆Asuch that (T ′, h)∈R 3,T ′ ∩(E 1)+ R3 =∅,T∩A 1 ⊆E 1, ...

work page

[20] [20]

IfE 1 ∈σ(F 1)andE 2 ∈σ(F ⋆ 2 ), thenE 1 ∪E 2 ∈σ(F)

work page

[21] [21]

Proof.Closure is proven using Proposition 27; conflict- freeness is proven using Lemma 54

IfE∈σ(F), thenE 1 =E∩A 1 ∈σ(F 1)andE∩A 2 ∈ σ(F ⋆ 2 ). Proof.Closure is proven using Proposition 27; conflict- freeness is proven using Lemma 54. We proceed by proving the statement for each semantics separately. Stable semantics

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[22] [22]

We show that E=E 1 ∪E 2 ∈stb(F)

SupposeE 1 ∈stb(F 1)andE 2 ∈stb(F ⋆ 2 ). We show that E=E 1 ∪E 2 ∈stb(F). •Eis conflict-free: Note thatF ⋆ 2 =F E1 2 becauseU E1 Rc 3 = ∅. By Lemma 54 and sinceE 1 ∈stb(F 1)andE 2 ∈ stb(F E1 2 ), we obtainE∈cf(F). •Eis closed: this follows from Proposition 27. •Eattacks all remaining arguments: Leta∈A\E. We proceed by case distinction. –Case 1:a∈A 1. Then...

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[23] [23]

It holds thatE ⊕ R =A=A 1 ∪A 2

SupposeE∈stb(F). It holds thatE ⊕ R =A=A 1 ∪A 2. • We show thatE 1 =E∩A 1 ∈stb(F 1). From Lemma 54, we obtainE∩A 1 ∈cf(F 1). Leta∈A 1. Since (F1, F2, R3)is an attack splitting ofF, it holds thata is attacked by someT⊆A 1, thus(E∩A 1)⊕ R1 =A 1 and thereforeE 1 ∈stb(F 1). 14 • We show thatE 2 =E∩A 2 ∈stb(F ⋆ 2 ). By Lemma 54, we obtainE 2 ∈cf(F E1 2 ); by P...

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[24] [24]

Observe that∗ 0 /∈E2 since∗ 0 is self-attacking

SupposeE 1 ∈adm(F 1)andE 2 ∈adm(F ⋆ 2 ). Observe that∗ 0 /∈E2 since∗ 0 is self-attacking. We show thatE=E 1 ∪E 2 ∈adm(F). By Lemma 54, we obtainE=cl F (E); moreover,Eis closed by Proposition 27. It remains to show thatEdefends itself against each closed set. LetT c ⊆Adenote a closed attacker ofE. Then there is (T, a)∈RwithT⊆T c anda∈E. Eithera∈E 1 or a∈E ...

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[25] [25]

We proceed by case distinction

It holds thatcl F (T) =cl F (T ′), and thereforeT ′ =T c. We proceed by case distinction. Case 2.ii.a:T ′ ∩(E 1)+ R1∪Rc 3 ̸=∅. Thenais defended byE 1 against(T ′, a), thus also byE. Case 2.ii.b:T ′ ∩(E 1)+ R1∪Rc 3 =∅. In this case, either(T ′ ∩A 2, a)∈R ⋆ 2 (via the reduct) or((T ′ ∩A 2)∪ {∗ 0}, a)∈R ⋆ 2 (via the modification). Recall thata∈E 2 and, moreo...

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[26] [26]

We show thatE 1 =E∩A 1 andE 2 =E∩A 2 are admissible inF 1 resp.F ⋆ 2

SupposeE∈adm(F). We show thatE 1 =E∩A 1 andE 2 =E∩A 2 are admissible inF 1 resp.F ⋆ 2 . By Lemma 54,E 1 =E∈cf(F 1)andE 2 ∈cf(F E1 2 ); by Proposition 27 both sets are closed. We show thatE 1 and E2 defend themselves inF 1 andF ⋆ 2 , respectively, against each attacker. Recall that each attacker inR c 3 is closed. •E 1 defends itself inF 1 follows sinceEde...

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[27] [27]

Then forE=E 1 ∪ E2 we have by the previous itemE∈adm(F)

LetE 1 ∈com(F 1), E2 ∈com(F ⋆ 2 ). Then forE=E 1 ∪ E2 we have by the previous itemE∈adm(F). It is left to show thatEcontains everything it defends. So let a∈Asuch that for every closed attackerDonawith some(T, a)∈R, T⊆Dthere exists an attack(S, h) fromS⊆Etoh∈D. We proceed by case distinction (a∈A 1)ThenD⊆A 1 and therefore(S, h)∈R 1, so E1 defendsa, soa∈E ...

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[28] [28]

Then by the previous itemE 1 =E∩A 1 is admissible inF 1 and E2 =E∩A 2 is admissible inF ⋆ 2

For the other direction supposeE∈com(F). Then by the previous itemE 1 =E∩A 1 is admissible inF 1 and E2 =E∩A 2 is admissible inF ⋆ 2 . Note that this implies that∗ 0 /∈E2. It is left to show that bothE 1 andE 2 contain every argument they defend. (a) Letabe defended byE 1 inF 1, then for every closed AttackerDonathere exists an attack(S, h)∈R 1 withS⊆E 1,...

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[29] [29]

SupposeE 1 ∈ pref(F)andE 2 ∈pref(F ⋆ 2 )butE=E 1 ∪E 2 /∈pref(F), then by the previous itemE∈adm(F), so there ex- ists a preferred supersetE ′ ⊋EofF

The first direction follows directly from the correspon- dence results for admissible semantics. SupposeE 1 ∈ pref(F)andE 2 ∈pref(F ⋆ 2 )butE=E 1 ∪E 2 /∈pref(F), then by the previous itemE∈adm(F), so there ex- ists a preferred supersetE ′ ⊋EofF. But then either E′ ∩A 1 =E ′ 1 ⊋E 1 and sinceE ′ is admissible inF we haveE 1 ∈adm(F 1)soE 1 /∈pref(F1)orE ′ 1 ...

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[30] [30]

ThenE∩A 1 =E 1 ∈ adm(F1)andE∩A 2 =E 2 ∈adm(F ⋆ 2 )

Now supposeE∈pref(F). ThenE∩A 1 =E 1 ∈ adm(F1)andE∩A 2 =E 2 ∈adm(F ⋆ 2 ). If there existed someE ′ 2 ⊋E 2 withE ′ 2 ∈adm(F ⋆ 2 ), thenE 1 ∪E ′ 2 ∈ adm(F)andE⊊E 1 ∪E ′ 2, soEis not preferred. Con- tradiction. SoE 2 ∈pref(F ⋆ 2 ). It is left to show thatE 1 ∈pref(F 1). Suppose there exists someE † 1 ⊋E 1, E† 1 ∈adm(F 1). We will show that in this caseE 2 ∈a...

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[31] [31]

Now supposeE 2 /∈adm(F⋆† 2 )

We have strictly less undecided links, soR ⋆† 2 \R E† 1 2 ⊆R ⋆ 2 \R E1 2 and we haveR E† 1 2 = RE1 2 ∪ {(T\A 1, h)|(T, h)∈R 3, T∩(E † 1)+ R1∪Rc 3 = ∅, T∩A 1 ⊆E † 1, T∩A 1 ⊈E 1, h∈A 2}(sinceE † 1 is conflict-free, no attacks are lost). Now supposeE 2 /∈adm(F⋆† 2 ). ThenE 2 is either not closed, not conflict-free or does not defend itself. Noth- ing changed...

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[32] [32]

IfE 1 ∈cf(F 1)andE 2 ∈cf(F E1 2 ), thenE 1 ∪(E 2 \ {∗2})∈cf(F)

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[33] [33]

Proof.We prove the statements separately

IfE∈cf(F), thenE 1 =E∩A 1 ∈cf(F 1)andE∩A 2 ∈ cf(F E1 2 ). Proof.We prove the statements separately

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[34] [34]

LetE=E 1 ∪ (E2 \ ∗2)

SupposeE 1 ∈cf(F 1)andE 2 ∈cf(F E1 2 ). LetE=E 1 ∪ (E2 \ ∗2). We distinguish two cases: •T S3 (E1) =∅. In this case, we haveF E1 2 =F 2 and thusE 2 ∈cf(F 2). Therefore,E 1 ∪E 2 ∈cf(F)follows from the fact that there is no attack fromF 1 andF 2. •T S3 (E1)̸=∅. In this case, we haveR E1 2 ⊇R 2 ∪ {(∅,∗ 1)}, so thatE 2 ∈cf(F 2). Moreover,∗ 1 /∈E2 becauseE 2 i...

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[35] [35]

We show thatE 1 =E∩A 1 ∈cf(F 1) andE 2 =E∩A 2 ∈cf(F E1 2 )

SupposeE∈cf(F). We show thatE 1 =E∩A 1 ∈cf(F 1) andE 2 =E∩A 2 ∈cf(F E1 2 ). • SinceR 1 ⊆R, it immediately follows thatE 1 =E∩ A1 ∈cf(F 1). • We now considerE 2 =E∩A 2 ∈cf(F E1 2 ). Since R2 ⊆Rand(F 1, F2, S3)is a proper support splitting, we know thatE∩A 2 ∈cf(F 2). Further, sinceE⊆A, we derive that∗ i /∈E∩A 2 for alli∈ {1,2}. Thus, since the reduct only ...

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[36] [36]

IfE 1 ∈adm(F 1)andE 2 ∈adm(F E1 2 ), thenE= cl F (E)whereE=E 1 ∪(E 2 \ {∗2})

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[37] [37]

IfE∈adm(F), thenE∩A 1 =E 1 =cl F1 (E1)and E2 =cl F E1 2 (E2)whereE 2 =E∩A 2. Proof.1. LetE 1 ∈adm(F 1)andE 2 ∈adm(F E1 2 ). We show thatE=E 1 ∪(E 2 \ {∗2})is closed. Let(T, h)∈SwithT⊆E. It holds that either(T, h)∈ S1,(T, h)∈S 2, or(T, h)∈S 3. We proceed by case distinction. • Case 1:(T, h)∈S 1. ThenT⊆E 1. SinceE 1 is closed inF 1 by assumption, we geth∈E ...

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[38] [38]

• Case 3:(T, h)∈S 3

SinceE 2 is closed inF ⋆ 2 , we obtainh∈E 2 ⊆E. • Case 3:(T, h)∈S 3. It holds thath∈A 1 by definition ofS 3. Recall thatT⊆E, thusT∩A 1 ⊆E 1. Towards a contradiction, supposeh /∈E 1. We proceed by case distinction. Case 3.i:T∩A 1 =∅. In this case,T∈ D S3 (E1). By construction of the S-reduct,(T,∗ 1)∈S ⋆

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[39] [39]

Since(∅,∗ 1)∈R ⋆ 2,E 2 is attacked by the empty set and thus not admissible inF ⋆ 2 ; contradiction to our ini- tial assumption

Since T⊆E 2 andE 2 is closed inF ⋆ 2 , we have∗ 1 ∈E 2. Since(∅,∗ 1)∈R ⋆ 2,E 2 is attacked by the empty set and thus not admissible inF ⋆ 2 ; contradiction to our ini- tial assumption. Case 3.ii:T∩A 1 ̸=∅. In this case,T∈ T S3 (E1)\ DS3 (E1). By construction of the S-reduct,(T∩ A2,∗ 2)∈S ⋆

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[40] [40]

Again, we can de- rive a contradiction to admissibility ofE 2 inF ⋆ 2 : By construction of the S-reduct,F ⋆ 2 contains the attack ((T∩A 2)∪ {∗ 2},∗ 2), thusE 2 /∈cf(F⋆ 2 )

SinceT∩A 2 ⊆E 2 andE 2 is closed inF ⋆ 2 , we have∗ 2 ∈E 2. Again, we can de- rive a contradiction to admissibility ofE 2 inF ⋆ 2 : By construction of the S-reduct,F ⋆ 2 contains the attack ((T∩A 2)∪ {∗ 2},∗ 2), thusE 2 /∈cf(F⋆ 2 ). In each case, we haveh∈E; therefore,E=cl F (E)

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[41] [41]

LetE 1 =E∩A 1 and E2 =E∩A 2

AssumeEis admissible inF. LetE 1 =E∩A 1 and E2 =E∩A 2. We show thatE 1 is closed inF 1 andE 2 is closed inF E1 2 . • LetE 1 =E∩A 1. We proveE 1 =cl F1 (E1). Suppose towards contradiction thatE 1 supports somea∈A 1 inF 1, buta /∈E 1. Then for someT⊆E 1 anda∈ A1 \E 1 we have(T, a)∈S 1. SinceE 1 ⊆Eand S1 ⊆S, we derive thatEsupports somea∈A\E inF. Thus,E̸=cl ...

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[42] [42]

IfE 1 ∈σ(F 1)andE 2 ∈σ(F E1 2 ), thenE 1∪(E2\{∗2})∈ σ(F)

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[43] [43]

Proof.Similarly to Theorem 28, we prove the statements for each semantics

IfE∈σ(F), thenE 1 =E∩A 1 ∈σ(F 1)and (E∩A 2 ∈ σ(F E1 2 )or(E∩A 2)∪ {∗ 2} ∈σ(F E1 2 )). Proof.Similarly to Theorem 28, we prove the statements for each semantics. Admissible Semantics. 18

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[44] [44]

We show E=E 1 ∪E 2 \ {∗2} ∈adm(F)

SupposeE 1 ∈adm(F 1)andE 2 ∈adm(F E1 2 ). We show E=E 1 ∪E 2 \ {∗2} ∈adm(F). •Eis closed inF: this follows from Lemma 56. •Eis conflict-free: this follows from Lemma 55. •Edefends itself against each closed set: Consider a set Tc ⊆Awhich is closed inF, i.e.,T c =cl F (Tc), and attacksEinF. Then there isT⊆T c,h∈E, such that(T, h)∈R. It holds that either(T,...

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[45] [45]

We show thatE 1 =E∩A 1 ∈ adm(F1)andE 2 =E∩A 2 ∈adm(F E1 2 )

SupposeE∈adm(F). We show thatE 1 =E∩A 1 ∈ adm(F1)andE 2 =E∩A 2 ∈adm(F E1 2 ). • LetE 1 =E∩A 1.E 1 is conflict-free by Lemma 55 and closed by Lemma 56. It remains to prove thatE 1 defends itself. LetT c be a closed attacker ofE1. Then there isT⊆T c, h∈E 1, such that(T, h)∈R 1. Observe thatcl F (T)⊆ A1 sinceT⊆A 1 and there is no support(B, a)∈S withB⊆A 1 an...

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[46] [46]

We show that E=E 1 ∪(E 2 \ {∗2})∈stb(F)

SupposeE 1 ∈stb(F 1)andE 2 ∈stb(F E1 2 ). We show that E=E 1 ∪(E 2 \ {∗2})∈stb(F). 19 •Eis conflict-free: By Lemma 55 and sinceE 1 ∈ stb(F1)andE 2 ∈stb(F E1 2 ), we obtainE∈cf(F). •Eis closed because of Lemma 56 andstb(F)⊆ adm(F)for every BSAFF. •Eattacks all remaining arguments: Leta∈A\E. We proceed by case distinction. –Case 1:a∈A 1. Thena∈(E 1)+ R1 sin...

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[47] [47]

It holds thatE ⊕ R =A=A 1 ∪A 2

SupposeE∈stb(F). It holds thatE ⊕ R =A=A 1 ∪A 2. • We show thatE 1 =E∩A 1 ∈stb(F 1). From Lemma 55, we obtainE∩A 1 ∈cf(F 1).E 1 is closed inF 1 because of Lemma 56. Finally, we show thatE 1 at- tacks all argumentsa∈A 1 \E 1. Leta∈A 1. Since (F1, F2, S3)is a support splitting ofF, it holds thatR is partitioned intoR 1 andR 2. Thus,ais attacked by someT⊆A 1...

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[48] [48]

We show E=E 1 ∪E 2 \ {∗2} ∈com(F)

SupposeE 1 ∈com(F 1)andE 2 ∈com(F E1 2 ). We show E=E 1 ∪E 2 \ {∗2} ∈com(F). • The premissesE 1 ∈com(F 1)andE 2 ∈com(F E1 2 ) directly entailE 1 ∈adm(F 1)andE 2 ∈adm(F E1 2 ), henceE∈adm(F)as shown above. • It remains to be shown thatEdefends itself inF. –Case 1:T S3 (E1) =∅. ThenF E1 2 =F 2, and for any (T, h)∈S 3:h∈E∨T⊊E. Lets assumeEdefends a∈A\EinF. I...

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[49] [49]

We show thatE 1 =E∩A 1 ∈ com(F1)and (E∩A 2 ∈com(F E1 2 )or(E∩A 2)∪ {∗2} ∈ com(F E1 2 ))

SupposeE∈com(F). We show thatE 1 =E∩A 1 ∈ com(F1)and (E∩A 2 ∈com(F E1 2 )or(E∩A 2)∪ {∗2} ∈ com(F E1 2 )). • Utilizing the proof for admissible semantics, we di- rectly concludeE 1 =E∩A 1 ∈adm(F 1)andE 2 = E∩A 2 ∈adm(F E1 2 )or(E∩A 2)∪{∗ 2} ∈adm(F E1 2 )). • There are no attacks fromA 2 towardsA 1, neither are there supports fromA 1 towardsA 2. Hence any a...

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[50] [50]

Contradiction toE ′ 2 ∈adm(F E1 2 )

Since(T ′ ∩A 2,∗ 2)∈S E1 2 , this implies E′ 2 is not closed. Contradiction toE ′ 2 ∈adm(F E1 2 ). Case 2.i.b:h∈A 2. In this case,T⊆E ′ 2 and we have a contradiction to the conflict-freeness ofE ′ 2. Case 2.ii:∗ 2 /∈E2. ThenE ′ 2 ⊋E ′ 2, contradiction toE 2 ∈ pref(F2). 21 C Omitted Proofs of Section 5 We recall the notation from Definition 48. Throughout ...

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[51] [51]

IfE 1 ∈adm(F 1)andE 2 ∈adm(F ⊛ 2 ), thenE=E 1 ∪ (E2 \ {∗2})∈cf(F)

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[52] [52]

IfE∈adm(F), thenE 1 =E∩A 1 ∈cf(F 1)andE∩ A2 ∈cf(F ⊛ 2 ). Proof.1. LetE 1 ∈adm(F 1)andE 2 ∈adm(F ⊛ 2 ). We show thatE=E 1 ∪(E 2 \ {∗ 2})∈cf(F). Let(T, h)∈R s.t.T⊆E. Then either(T, h)∈R 1,(T, h)∈R 2, or (T, h)∈R 3. • Case 1:(T, h)∈R 1. Thenh /∈EsinceE 1 ∈ adm(F1). • Case 2:(T, h)∈R 2. It holds thatT⊆A 2, h∈A 2. –Case 2i:cl(T)⊆A 2. Then(cl(T), h)∈R ⊛ 2 . –Ca...

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[53] [53]

We show thatE 1 =E∩A 1 ∈ cf(F1)andE 2 =E∩A 2 ∈cf(F ⊛ 2 )

SupposeE∈adm(F). We show thatE 1 =E∩A 1 ∈ cf(F1)andE 2 =E∩A 2 ∈cf(F ⊛ 2 ). •E 1 ∈cf(F 1)sinceR 1 ⊆R. • We show thatE 2 ∈cf(F ⊛ 2 ). Let(T, h)∈R ⊛ 2 with T⊆E 2. By definition ofE 2, we haveh∈A 2. Thus, by definition ofR ⊛ 2 , one of the following applies for (T, h): –(T, h)∈ ˆR2 andT⊆A 2, h∈A 2; (Case 1) –(T, h) = (T ′ \A 1, h)for someT ′ ⊆Asuch that (T ′,...

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[54] [54]

IfE 1 ∈adm(F 1)andE 2 ∈adm(F ⊛ 2 ), thenE=cl F (E) whereE=E 1 ∪(E 2 \ {∗2})

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[55] [55]

IfE∈adm(F), thenE∩A 1 =E 1 =cl F1 (E1)and E2 =cl F ⊛ 2 (E2)whereE 2 =E∩A 2. Proof.We recall:F 1 = (A 1, R1, S1)andF ⊛ 2 = (A⊛ 2 , R⊛ 2 , S⊛ 2 )withA ⊛ 2 =A 2 ∪ {∗ 0,∗ 1,∗ 2}andR ⊛ 2 ,S ⊛ 2 as defined by the respective modifications for support and attack splitting. 1.E 1 ∈adm(F 1)andE 2 ∈adm(F ⊛ 2 ). We show thatE= E1 ∪(E 2 \ {∗2})is closed inF. The proof...

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[56] [56]

We show thatE∩A 1 =E 1 = clF1 (E1)andE∩A 2 =E 2 =cl F ⊛ 2 (E2)

LetE∈adm(F). We show thatE∩A 1 =E 1 = clF1 (E1)andE∩A 2 =E 2 =cl F ⊛ 2 (E2). • LetE 1 =E∩A 1 and let(T, h)∈S 1 withT⊆E 1. SinceT⊆EandS 1 ⊆S, we obtainh∈E 1. There- fore,E 1 =cl F1 (E1). • LetE 2 =E∩A 2 and let(T, h)∈S ⊛ 2 withT⊆ E2. By construction of the S-reduct, either(T, h)∈ S2 orh=∗ 1 orh=∗ 2. Note that splitting attacks (closing links inR 3, constru...

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[57] [57]

IfE 1 ∈σ(F 1)andE 2 ∈σ(F ⊛ 2 ), thenE 1 ∪(E 2 \ {∗2})∈ σ(F)

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[58] [58]

Proof.We prove the theorem for all considered semantics

IfE∈σ(F), thenE 1 =E∩A 1 ∈σ(F 1)andE∩A 2 ∈ σ(F ⊛ 2 )or(E∩A 2)∪ {∗ 2} ∈σ(F ⊛ 2 ). Proof.We prove the theorem for all considered semantics. Admissible Semantics

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[59] [59]

We show thatE=E 1 ∪E 2 ∈adm(F)

SupposeE 1 ∈adm(F 1)andE 2 ∈adm(F ⊛ 2 ). We show thatE=E 1 ∪E 2 ∈adm(F). By Lemma 57,Eis conflict-free inF; by Lemma 58,Eis closed inF. It remains to prove thatEdefends itself inF. LetT c denote a closed attacker ofEinF. That is, there isT⊆T c with(T, h)∈Randh∈E. Wlog, we can assumecl F (T) =T c. As usual, we have three cases to consider:(T, h)∈R 1, (T, h...

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[60] [60]

We show thatE∩A 1 ∈adm(F 1) andE∩A 2 ∈adm(F ⊛ 2 )

SupposeE∈adm(F). We show thatE∩A 1 ∈adm(F 1) andE∩A 2 ∈adm(F ⊛ 2 ). By Lemma 58 we know that both sets are closed; moreover, by Lemma 57 they are also conflict-free. It remains to prove that the sets defend itself. • The case forE 1 =E∩A 1 holds since for all(T, h)∈ Rwithh∈E 1 it holds that(T, h)∈R 1. Moreover, cl F (T) =cl F1 (T)since no positive links g...

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[61] [61]

We show thatE=E 1 ∪(E 2 \ {∗2})∈com(F)

SupposeE 1 ∈com(F 1)andE 2 ∈com(F ⊛ 2 ). We show thatE=E 1 ∪(E 2 \ {∗2})∈com(F). The premissesE 1 ∈com(F 1)andE 2 ∈com(F ⊛ 2 )di- rectly entailE 1 ∈adm(F 1)andE 2 ∈adm(F ⊛ 2 ), hence E∈adm(F)as shown above. It remains to be shown that Edefends itself inF. Assume toward contradiction that there exists an argu- mentd∈A\Ethat is defended byEinF. We proceed b...

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[62] [62]

We show thatE 1 =E∩A 1 ∈ com(F1)and (E∩A 2 ∈com(F ⊛ 2 )or(E∩A 2)∪ {∗ 2} ∈ com(F ⊛ 2 ))

SupposeE∈com(F). We show thatE 1 =E∩A 1 ∈ com(F1)and (E∩A 2 ∈com(F ⊛ 2 )or(E∩A 2)∪ {∗ 2} ∈ com(F ⊛ 2 )). • Utilizing the proof for admissible semantics, we di- rectly concludeE 1 =E∩A 1 ∈adm(F 1)andE 2 = E∩A 2 ∈adm(F ⊛ 2 )orE 2 = (E∩A 2)∪ {∗ 2} ∈ adm(F ⊛ 2 ). • There are no attacks fromA 2 towardsA 1, neither are there supports fromA 1 towardsA 2. Hence a...

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[63] [63]

We show that E=E 1 ∪(E 2 \ {∗2})∈stb(F)

SupposeE 1 ∈stb(F 1)andE 2 ∈stb(F ⊛ 2 ). We show that E=E 1 ∪(E 2 \ {∗2})∈stb(F). • E is conflict-free: By Lemma 57 andE 1 ∈adm(F 1) andE 2 ∈adm(F ⊛ 2 ), we obtainE∈cf(F). • E is closed because of Lemma 58 and the fact that stb(F)⊆adm(F)for every BSAFF. •Eattacks every argumenta∈A\E. Leta∈A\E. We proceed by case distinction. –Case 1:a∈A 1. Thena∈(E 1)+ R1...

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[64] [64]

• We first show thatE 1 =E∩A 1 ∈stb(F 1)

SupposeE∈stb(F), that isE ⊕ R =E∪E + R =A. • We first show thatE 1 =E∩A 1 ∈stb(F 1). From pre- vious lemmata we know thatE 1 is closed and conflict- free. It remains to prove thatE 1 attacks everya∈ A1 \E 1. Since there is no attack fromF 2 toF 1 by def- inition of splitting, we know thata∈(E) + R1 by some T⊆E∩A 1, i.e.a∈(E∩A 1)+ R1. • We show thatE∩A 2 ∈...

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[65] [65]

Further, fromE∈cf(F), it follows thatT c ∩(E 1)+ R1∪Rc 3 = ∅

SinceEis closed inFby hypothesis, we know thatT c ⊆E. Further, fromE∈cf(F), it follows thatT c ∩(E 1)+ R1∪Rc 3 = ∅. Moreover,T c ∩A 1 ⊆E∩A 1 becauseT c ⊆ E. Hence, by definition ofR-reduct, there is a(T c \ A1, a)∈R ⊛ 2 , and consequently,a∈E + R⊛ 2 . In both cases we finda∈E + R⊛ 2 . • Case 2.a /∈A 2. Thusa∈ {∗ 1}. By definition ofS- reduct we know that∗...

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[66] [66]

Contradiction toE ′ 2 ∈adm(F ⊛ 2 )

Since(T ′ ∩A 2,∗ 2)∈S ⊛ 2 , this implies E′ 2 is not closed. Contradiction toE ′ 2 ∈adm(F ⊛ 2 ). Case 2.i.b:h∈A 2. In this case,T⊆E ′ 2 and we have a contradiction to the conflict-freeness ofE ′ 2 inF ⊛ 2 . Case 2.ii:∗ 2 /∈E2. ThenE ′ 2 =E 2 \ {∗2}and therefore E′ 2 ⊋E 2, contradiction toE 2 ∈pref(F ⊛ 2 ). Grounded Semantics.SupposeE 1 ∈grd(F 1)andE 2 ∈ g...

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