Argumentative Characterisations of Non-monotonic Inference in Preferred Subtheories: Stable Equals Preferred

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1 Argumentative Characterisations of Non-monotonic Inference in Preferred Subtheories: Stable Equals Preferred Sanjay Modgil November 17, 2017 Abstract A number of argumentation formalisms provide dialectical characterisations of non-monotonic inference in Brewka s Preferred Subtheories. Given a totally ordered set B of classical well formed formulae, these formalisms establish a correspondence between the preferred subtheories of B, and the stable extensions of the argument framework instantiated by arguments constructed from B, and defeats that denote successful attacks amongst these arguments, where the success of an attack is decided by an argument preference relation defined by the ordering over B. In this research note I show that the preferred extensions of the instantiated framework coincide with the stable extensions. 1 Characterising Preferred Subtheories Inference in the ASPIC + Framework for Structured Argumentation Brewka s Preferred Subtheories [1] models the use of an ordering over a classical propositional or first order theory Γ, in order to resolve inconsistencies. It has therefore been used to both formalise default reasoning and belief revision [2]. In what follows, we as usual have that x < y iff x y and y x, and x y iff x y and y x. Definition 1 A default theory is a tuple (Γ, ) where is a total ordering over the set Γ of first order classical wff. (Γ, ) can be represented as the stratification (Γ 1,..., Γ n ), where Γ = n i=1 Γ i and α Γ i, β Γ j, i < j iff α > β. α, β Γ i iff α β (i.e., α β, β α). A preferred subtheory is a set Σ = Σ 1... Σ n such that for i = 1... n, Σ 1... Σ i is a maximal (under set inclusion) consistent subset of Γ 1,..., Γ i. Intuitively, a preferred subtheory is obtained by taking a maximal under set inclusion consistent subset of Γ 1, extending this with a maximal consistent subset of Γ 2, extending this with a maximal consistent subset of Γ 3, and so on. 1

2 Example 1 Let (Γ, ) = ({c, a, b, a b, c a b}, a b c, a a b, b a b, a b, b a, c a b a). Then (Γ, ) can be represented as the stratification: Γ 1 Γ 2 = {c} = { a b} Γ 3 = {a, b} Γ 4 = {c a b} Then there are two preferred subtheories: Σ = {c, a b, a} and Σ = {c, a b, b}. We recall the reconstruction of preferred subtheories as an instance of the ASPIC + framework [4]. Definition 2 Let (L,, R) 1 be the argumentation system, where: L is the full first order classical language; the contrary function denoting that two formulae are in conflict, is defined so as to encode standard negation: α β iff α is of the form β or β is of the form α; R is the set of strict inference rules {φ 1,..., φ n φ φ 1,..., φ n c φ} encoding entailment in classical logic ( c is the classical consequence relation). Let D = (Γ, ) be a default theory. Let K be the ASPIC + knowledge base consisting only of ordinary premises K = K p = Γ (the set of axiom premises K n is empty) 2, and the total ordering over K. Then the structured argumentation framework (SAF) (A, C, ) corresponding to D is defined as follows: A = {({φ 1,..., φ n }, φ) n i=1 φ i K, n i=1 φ i, φ 1,..., φ n φ R}. Given an argument X = ({φ 1,..., φ n }, φ) A we write conc(x) to denote the conclusion φ of X and prem(x) to denote the premises {φ 1,..., φ n } of X. (X, Y ) C iff X, Y A and conc(x) = φ, φ prem(x). We say X attacks Y on φ, equivalently, X attacks Y on the argument Y = ({φ}, φ). X Y iff α prem(x), β prem(y ): α < β 3. Then X defeats Y ((X, Y ) ) iff for some Y, X attacks Y on Y and X Y. The admissible, complete, grounded, preferred and stable extensions of the Dung framework (A, ) are then defined in the usual way [3], and designated to be the extensions of the SAF (A, C, ). 1 We do not include in the definition of an argumentation system the naming function n, as in [4], since this is used to assign names only to defeasible rules, and defeasible rules are not required in the ASPIC + reconstruction of Preferred Subtheories. 2 Recall that an ASPIC + knowledge base [4] consists of fallible ordinary premises that can be challenged by counter-arguments, and infallible axiom premises that cannot be challenged. 3 This is the well known Elitist principle used in defining strict preferences over arguments based on an ordering over their constituent premises. 2

3 Example 2 [Example 1 continued] The SAF (A, C, ) corresponding to the default theory (Γ, ) in Example 1 is: A = {(, φ) 2 Γ, c φ} C = 1. {(X, Y ) conc(x) = c, c prem(y )} 2. {(X, Y ) conc(x) = ( a b), a b prem(y )} 3. {(X, Y ) conc(x) = a, a prem(y )} 4. {(X, Y ) conc(x) = b, b prem(y )} 5. {(X, Y ) conc(x) = (c a b), c a b prem(y )} X ({c}, c) whenever prem(x) { a b, a, b, c a b} X ({ a b} a b) whenever prem(x) {a, b, c a b} X ({a}, a) whenever prem(x) {c a b} X ({b}, b) whenever prem(x) {c a b} X A: X ({c a b}, c a b) There are two stable extensions: E1 = {X prem(x) {c, a b, a}} E2 = {X prem(x) {c, a b, b}} The following theorem recapitulates Theorem 34 in [4], and uses the notation premises(e) = X E prem(x) to denote the union of premises in arguments in a set of arguments E, and the notation args(σ) = {X A prem(x) Σ} to denote the arguments whose premises are contained in a set Σ of classical wff. Theorem 3 Let (A, C, ) be a structured argumentation framework corresponding to D. Then: If E is a stable extension of (A, C, ) then premises(e) is a preferred subtheory of D; If Σ is a preferred subtheory of D then args(σ) is a stable extension of (A, C, ) Given the above correspondence, we then have that: α is classically entailed from a preferred subtheory (ps) of D iff α is the conclusion of an argument in a stable extension of the SAF corresponding to D. This result relies on the fact that the SAF corresponding to D satisfies the closure under strict rules postulate [4]. That is to say, if {X 1,..., X n } E, where E is a complete extension (stable extensions are complete), and conc(x 1 ),..., conc(x n ) φ R (recall that the rules in R are strict and encode classical entailment), then X E such that conc(e) = φ. We then have the following argumentative characterisations of non-monotonic inference in Preferred Subtheories: 3

4 φ is a sceptical (credulous) ps-inference iff φ is entailed by all (respectively at least one) ps, iff φ is a sceptical (credulous) argumentation defined inference (i.e., φ is the conclusion of an argument in all, respectively at least one, stable extension(s)). 2 Preferred Equals Stable We now show that given a SAF corresponding to a default theory D, the preferred and stable extensions coincide. It is well known that any stable extension of a Dung AF is preferred 4. Hence it suffices to show that if E is a preferred extension of a SAF corresponding to a default theory D, then E is a stable extension. The following lemma is used in the proof of this result. Lemma 4 Let E be a complete extension of a SAF (A, C, ) corresponding to a default theory D. Then X A such that prem(x) premises(e), X E. PROOF. We first show that E {X} is conflict free. Suppose 1) Y E, Y X on α. Since prem(x) premises(e), Z E s.t. α prem(z). Hence Y Z on α, contradicting E is conflict free. Suppose 2) Y E, X Y on α. By assumption of E being complete, W E, W X. But then (reasoning as in case 1) this implies W Z, Z E, contradicting E is conflict free. Suppose 3) X X. Then this implies (reasoning as in the previous cases) that X Z, Z E, and we are in case 2). It remains to show that X is acceptable w.r.t. E. Suppose Y A, Y X on α. Then reasoning as above, Z E, Y Z on α. Since E is a complete extension, W E s.t. W Y. Hence X is acceptable w.r.t. E. QED Proposition 5 Let (A, C, ) be a structured argumentation framework corresponding to D = (Γ, ). If E is a preferred extension of (A, C, ) then E is a stable extension of (A, C, ). PROOF. Let E be a preferred extension of (A, C, ). Given Proposition 3, it suffices to show that Σ = premises(e) is a preferred subtheory. Let Σ = Σ 1,..., Σ n where for i = 1,..., n, Σ i = premises(e) Γ i. Suppose for contradiction that Σ is not a preferred subtheory. Hence, without loss of generality one can assume that: and for j = 1... i 1, Σ 1... Σ i 1 is a maximal (under set inclusion) consistent subset of Γ 1,..., Γ i 1 (1) i 1 j=1 Σ j Σ i is not maximally consistent (2) 4 Suppose E is stable, X E and Y X. If Y E then this contradicts E is conflict free. If Y / E, then Z E, Z Y, and so X is acceptable w.r.t. E. Moreover, if X / E, then X cannot be acceptable w.r.t. E since Y E s.t. Y X. Hence if X acceptable w.r.t. E then Z E, Z Y, contradicting E is conflict free. Hence E is complete and must be maximal complete (i.e., preferred) since any E E cannot be conflict free by virtue of all arguments outside of E being defeated by an argument in E. 4

5 Let Λ 1,..., Λ m be the maximal subsets of Γ i \ Σ i s.t. for k = 1... m: i Σ j Λ k. (3) j=1 Now, choose any Λ {Λ 1,..., Λ m }, and let E = {Y prem(y ) (Λ premises(e)), Y / E} (4) We show that E E is admissible, contradicting E is a preferred extension. We first show that E E is conflict free. i) Suppose X E, Y E, X Y on α. It must be that α Λ else X defeats some Z E on α, contradicting E is conflict free. By Eq. 3, i j=1 Σ j α. Hence, it must be that β prem(x) s.t. β Σ j, j > i. But then X ({α}, α), contradicting X Y. ii) Suppose X E, Y E, Y X on α. Since E is admissible, Z E, Z Y. But then this leads to a contradiction as shown above. iii) Suppose X E, Y E, X Y on α Λ. By Eq. 3, i j=1 Σ j Λ α. Hence it must be that β prem(x) s.t. β Σ j, j > i. But then X ({α}, α), contradicting X Y. iv) Suppose X E, Y E, X Y on α premises(e). Then Z E, X Z and we are in case ii). We now show that all arguments in E are acceptable w.r.t. E E. We consider the following three mutually exclusive cases. 1. Suppose Y E is defeated on α by some argument X that makes use of a premise in Γ j, j > i. Then it cannot be that α Λ as X ({α}, α), contradicting X Y. Hence α premises(e) and so Z E, X Z on α. By assumption of E being admissible, W E s.t. W X. Hence Y is acceptable w.r.t. E E. It remains to consider cases where Y E is defeated on α by some argument X, all of whose premises are in Γ j, j i: 2. Suppose prem(x) premises(e) Λ. But then by Lemma 4 and Eq. 4, X E E, contradicting E E is conflict free. 3. Hence prem(x) includes at least one premise β in i j=1 Γ j ( i j=1 Σ j Λ). Suppose β Γ j Σ j for some j = 1... i 1. By Eq.1: j k=1 Σ k and j k=1 Σ k {β}. Hence j k=1 Σ k β. Hence Z such that prem(z) j k=1 Σ k, conc(z) = β, and Z ({β}, β) (since it cannot be that Z ({β}, β) given that β Γ j Σ j ). Hence Z X. Moreover, Z E by Lemma 4. Hence Y is acceptable w.r.t. 5

6 E E. Suppose β Γ i (Σ i Λ). Since i j=1 Σ j Λ is maximally consistent (Eq. 3), i j=1 Σ j Λ β. By Lemma 4 and Eq. 4, Z E E, prem(z) i j=1 Σ j Λ, conc(z) = β, and Z ({β}, β) (since it cannot be that Z ({β}, β)). Hence Z X, and Y is acceptable w.r.t. E E. QED References [1] G. Brewka. Preferred subtheories: An extended logical framework for default reasoning. In International Joint Conference on Artificial Intelligence, pages , [2] G. Brewka. Nonmonotonic Reasoning: Logical Foundations of Commonsense.. Cambridge University Press, [3] P. M. Dung. On the acceptability of arguments and its fundamental role in nonmonotonic reasoning, logic programming and n-person games. Artificial Intelligence, 77: , [4] S. Modgil and H. Prakken. A general account of argumentation with preferences. Artificial Intelligence, 195: ,