Rationality and maximal consistent sets for a fragment of ASPIC + without undercut

Jesse Heyninck, Christian Straßer
2020 Argument & Computation  
Structured argumentation formalisms, such as ASPIC + , offer a formal model of defeasible reasoning. Usually such formalisms are highly parametrized and modular in order to provide a unifying framework in which different forms of reasoning can be expressed. This generality comes at the price that, in their most general form, formalisms such as ASPIC + do not satisfy important rationality postulates, such as non-interference. Similarly, links to other forms of knowledge representation, such as
more » ... asoning with maximal consistent sets of rules, are insufficiently studied for ASPIC + although such links have been established for other, less complex forms of structured argumentation where defeasible rules are absent. Clearly, for a formal model of defeasible reasoning it is important to understand for which range of parameters the formalism (a) displays a behavior that adheres to common standards of consistency, logical closure and logical relevance and (b) can be adequately described in terms of other well-known forms of knowledge representation. In this paper we answer this question positively for a fragment of ASPIC + without the attack form undercut by showing that it satisfies all standard rationality postulates of structured argumentation under stable and preferred semantics and is adequate for reasoning with maximal consistent sets of defeasible rules. The study is general in that we do not impose any other requirements on the strict rules than to be contrapositable and propositional and in that we also consider priorities among defeasible rules, as long as they are ordered by a total preorder and lifted by weakest link. In this way we generalize previous similar results for other structured argumentation frameworks and so shed further light on the close relations between assumption-based argumentation and ASPIC + . This is in contradiction to the consistency of since D(a) ∈℘( ) and r ∈ . Lemma 3. Where ∈ MCS(D), Arg( ) ∈ stab(AS) and Arg( ) ∈ stab (AS). Proof. Let A = Arg( ), where ∈ MCS(D). By Lemma 2, A is conflict-free. Consider now some b ∈ Arg(AS) \ A. Thus, D(b) \ = ∅. Notice that since ∈ MCS(D), ⊥(D(b) ∪ ) and hence, by Fact 18, there is a r ∈ D(b) for which head(r ). Thus, there is a ∈℘( ) such that there is a proof from with conclusion head(r ). The corresponding argument a ∈ A with D(a) ⊆ (see Fact 10) attacks b. The next fact follows since any defeater/attacker of a sub-argument of some argument a also defeats/attacks a. Fact 19 (Sub-Argument Closure). Where A is a complete extension of AS, a ∈ A and b is a subargument of a, b ∈ A. Fact 20. Where A ⊆ Arg(AS) is closed under sub-arguments and r ∈ D[A], there is an argument is inconsistent. Thus, there is a finite ∈℘( ) and an r ∈ for which head(r). Since ∪ {r} ⊆ D[A] and by the sub-argument closure of A (Fact 19) and by Fact 20, for each r ∈ ∪ {r} there is a a r ∈ A with top-rule r . Let r = D(a r ). also ∈℘(D). By Lemma 15, is consistent. By Fact 19, A is closed under sub-arguments. Assume for a contradiction that is not R-consistent. Thus by Fact 23, there is a ∈℘(D) such that is R-inconsistent with . Thus, by Lemma 14, there is a consistent b ∈ Arg( ∪ ) such that (i) b defeats an a ∈ A with top-rule r and (ii) Sub(b) ⊆ A ∪ Arg( ). Since A is preferred, there is a c ∈ A that defeats b and so R[D(c)] R [D(b)]. By the conflictfreeness of A, the attack must be in some Since D(c) ∈℘( ) and ⊥(D(c) ∪ ) this contradicts the fact that is R-inconsistent with . So, is R-consistent. Suppose it is not maximal R-consistent. Thus, by Lemma 10, there is a maximal R-consistent ∈℘(D) for which ⊃ . Since by Lemma 12, Arg( ) is stable and Arg( ) ⊃ Arg( ) this contradicts that Arg( ) is preferred. Now we are in a position to prove Theorem 9: Theorem 9. For any AS = (L, S, D, K, , ) where S satisfies S1 and S2, Arg( ) | ∈ MCS R (D) = pref(AS) = stab(AS) = pref (AS) = stab (AS).
doi:10.3233/aac-200903 fatcat:np34hizvt5debbzahelteyxr7y