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Abstract argument games via modal logic

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Abstract

Inspired by some logical considerations, the paper proposes a novel perspective on the use of two-players zero-sum games in abstract argumentation. The paper first introduces a second-order modal logic, within which all main Dung-style semantics are shown to be formalizable, and then studies the model checking game of this logic. The model checking game is then used to provide a systematic game theoretic proof procedure to test membership with respect to all those semantics formalizable in the logic. The paper discusses this idea in detail and illustrates it by providing a game for the so-called skeptical preferred and skeptical semi-stable semantics.

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Notes

  1. See Prakken and Vreeswijk (2002), Bench Capon and Dunne (2007) and Baroni and Giacomin (2009) for comprehensive overviews.

  2. See Prakken and Vreeswijk (2002) and Modgil and Caminada (2009) for extensive overviews.

  3. To be precise, all is needed is the one free variable binary fragment of monadic second-order logic, which corresponds essentially to a second-order modal logic.

  4. The reader is referred to Grädel (2002) and van Benthem (2005) for accessible introductions.

  5. See also Grossi (2011b) for an introductory exposition.

  6. Note that, although often assumed in researches on abstract argumentation, we do not impose a finiteness constraint on \(A\).

  7. It might be worth noticing that this is a generalization of the sort of labeling functions studied in argumentation theory (cf. Baroni and Giacomin 2009).

  8. This logic is well-studied and well-behaved: it has a simple strongly complete axiomatics, a P-complete model checking problem and an EXPTIME-complete satisfiability problem (cf. (Blackburn et al. 2001, Ch. 7)).

  9. In the ensuing sections, we will often drop the reference to \(\mathbf{P}\) in \(\mathcal{L }_\mathbf{\mathsf{{U}} }(\mathbf{P})\), writing just \(\mathcal{L }_\mathbf{\mathsf{{U}} }\).

  10. Like for \(\mathcal{L }_\mathbf{\mathsf{{U}} }(\mathbf{P})\), we will normally drop the reference to \(\mathbf{P}\) in \(\mathcal{L }^2_\mathbf{\mathsf{{U}} }(\mathbf{P})\).

  11. For a presentation of the standard translation of modal logic we refer the reader to (Blackburn et al. (2001), Ch. 2.4).

  12. For a good presentation of the \(\mu \)-calculus, the interested reader is referred to Venema (2008). The \(\mu \)-calculus will be of relevance later also in Remarks 4 and 5.

  13. Recall that the smallest fixpoint of the characteristic function coincides with the minimal complete extension (Dung 1995).

  14. For a good introduction to model checking games the reader is referred to Grädel (2002).

  15. Equivalently, positions could be pairs formula-model.

  16. A formula is in positive normal form when negations occur only in front of propositional atoms. It must be clear that any formula of \(\mathcal{L }^2_\mathbf{\mathsf{{U}} }\) can be easily translated into an equivalent one in positive normal form.

  17. We will normally use the more compact notation \({\varepsilon ^{2}_\mathrm{U}}(\mathcal{A })@(\varphi , a)\) when all propositional variables in \(\varphi \) occur under the scope of a quantifier and, therefore, \(\mathcal{V }\) does not play any role at the instantiation of the game.

  18. Notice that although \(\textit{SkPrf}\) is technically not in positive normal form (cf. Definition 5), the tree is built by manipulating its formulation in positive normal form—which, notice, would be a rather long formula—according to the rules in Table 2. Intuitively, every time we encounter a negation followed by an operator, we apply the rules for the dual of that operator.

    Fig. 3
    figure 3

    Partial game tree of \({\varepsilon ^{2}_\mathrm{U}}(\mathcal{A })@(\textit{SkPrf}, d, \mathcal{V })\) where \(\mathcal{A }\) is the graph on the right-hand side of Fig. 2. Turn-taking is given through the \(\exists \) (\(\exists \)ve) and \(\forall \) (\(\forall \)dam) labels. Positions that are underlined denote the last position in a play. Recall the convention that the player having to play at a final position in a play is the one who loses (Table 3)

  19. Intuitively, a sequence denotes a dialogue in the argument game. If the sequence contains \(n\) elements, then the dialogue consists of \(n-1\) moves or replies.

  20. For instance, the path labeled by ‘\(\mathbf{*}\)’ in the model checking game for \(\textit{SkPrf}\) (Fig. 3) is not to be found in \(\mathcal{G }^{\textit{SkPrf}}@d\). The path is ruled out by the rules given earlier in Fig. 4, which make it impossible to make clearly counterproductive moves. In the above example, once \(\mathcal{O }\) has chosen \(\left\{ e \right\} \), it is clearly hopeless for \(\mathcal{P }\) to try to select an attacker of \(\left\{ e \right\} \) from whithn \(\left\{ e \right\} \) since \(e\) is not reflexive. Game \({\varepsilon ^{2}_\mathrm{U}}(\mathcal{A })@(\textit{SkPrf}, d)\) allows such move, which necessarily leads to \(\mathcal{P }\)’s defeat, while \(\mathcal{G }^{\textit{SkPrf}}@d\)) does not.

  21. Note that this sort of position occurs in the model checking game for skeptical preferred at the first position where \(\mathcal{P }\) moves.

  22. Note that this is precisely the position where the model checking game for skeptical preferred starts.

  23. For a set \(X\), the expression \(X \cup \left\{ a \mid a {\leftarrow }X \right\} \) is commonly referred to as its range (Caminada et al. 2011). Notice that the definition of the skeptical semi-stable semantics would actually require \(\mathcal{O }\) to show that \(X\) is not a complete extension with a maximal range. However, it must be clear that requiring him to show \(X\) is not an admissible set with maximal range achieves the same result.

  24. To the best of our knowledge, there is only one notable exception, namely (Dung and Thang 2009), where a game is studied which allows proponent and opponent to select sets of arguments. However, the game is then shown to be equivalent to existing ones, and the extra structure available in the game is not used to obtain new adequate games for extensions not yet characterized game theoretically.

  25. We wish to stress that the idea that the logical structure of the definitions of Dung-style semantics give key information for the development of algorithms for the membership problem with respect to those semantics is by no means new. The same intuition underpins, for instance, the algorithm for deciding skeptical preferred acceptance proposed in Cayrol et al. (2003). However, in the argumentation theory literature, this intuition has never been systematically developed in a unified framework (for instance, the aforementioned Cayrol et al. 2003 does not provide a formal proof of the soundness and completeness of the proposed algorithm) and it has never been mathematized in the form of a game.

  26. Cf. Grädel (2002).

  27. Figure 7 gives a pictorial representation of the values of the injection with respect to the input dialogues compatible with the structure in Fig. 4.

    Fig. 7
    figure 7

    The values of the injection \(g\) from the possible positions of \(\mathcal{G }^{\textit{SkPrf}}\) (Fig. 4) to those in the model checking game for \(\textit{SkPrf}\)

  28. Cf. the definition of the function \(\mathtt{move }\) given in Definition 7.

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Acknowledgments

This work was partly supported by the Nederlandse Organisatie voor Wetenschappelijk Onderzoek under the NWO VENI grant 639.021.816.

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Correspondence to Davide Grossi.

Appendix

Appendix

Proof

Proof of Theorem 1 The proof is by a standard argumentFootnote 26 and proceeds by induction on the syntax of \(\varphi \).

The base case is straightforward. For the induction step, the case of Boolean connectives is straightforward. As to the modal case, we cover \( \langle \mathbf{\mathsf{{U}} } \rangle \). [\(\varphi = \langle \mathbf{\mathsf{{U}} } \rangle \psi \)]. From left to right. Assume \((\varphi , a, \mathcal{V }) \in \textit{Win}_{\exists }({\varepsilon ^{2}_\mathrm{U}}(\mathcal{A }))\). It is \(\exists \)’s turn to move. It follows that there exists a position \((\psi , b, \mathcal{V })\) s.t. it is a winning position for \(\exists \). By IH we conclude that \((\mathcal{A }, \mathcal{V }), b \models \psi \) and hence \((\mathcal{A }, \mathcal{V }), a \models \langle \mathbf{\mathsf{{U}} } \rangle \psi \). From right to left. Assume \((\mathcal{A }, \mathcal{V }), a \models \varphi \). It follows that there exists \(b\) s.t. \((\mathcal{A }, \mathcal{V }), b \models \psi \). By induction hypothesis we have that \((\psi , b, \mathcal{V }) \in \textit{Win}_{\exists }({\varepsilon ^{2}_\mathrm{U}}(\mathcal{A }))\). But it is \(\exists \)’s turn to move, hence we conclude \((\varphi , a, \mathcal{V }) \in \textit{Win}_{\exists }({\varepsilon ^{2}_\mathrm{U}}(\mathcal{A }))\).

As to the propositional quantifiers: [\(\varphi = \exists p. \psi (p)\)] Left to right. Assume \((\exists p.\psi (p), a, \mathcal{V }) \in \textit{Win}_{\exists }({\varepsilon ^{2}_\mathrm{U}}(\mathcal{A }))\). As, according to Definition 5 it is \(\exists \text{ ve}\)’s turn to play, by Definition 6 it follows that there exists \(X \subseteq A\) s.t. \((\psi , a, \mathcal{V }_{p := X}) \in \textit{Win}_{\exists }(\mathcal{A })\), from which, by IH, we conclude that \((\mathcal{A }, \mathcal{V }_{p := X}), a \models \psi \) and, by Definition 4, that \((\mathcal{A }, \mathcal{V }), a \models \exists p. \psi \). Right to left. Assume \((\mathcal{A }, \mathcal{V }), a \models \exists p. \psi \). This means, by Definition 4, that there exists \(X \subseteq A\) s.t. \((\mathcal{A }, \mathcal{V }_{p := X}), a \models \psi \). By IH it follows that \((\psi , a, \mathcal{V }_{p := X}) \in \textit{Win}_{\exists }({\varepsilon ^{2}_\mathrm{U}}(\mathcal{A }))\) for some \(X\), and from the fact that it is \(\exists \text{ ve}\)’s turn to play, by Definition 6 we can thus conclude that \((\exists p.\psi (p), a, \mathcal{V }) \in \textit{Win}_{\exists }({\varepsilon ^{2}_\mathrm{U}}(\exists p.\psi (p), \mathcal{A }))\). [\(\varphi = \forall p. \psi (p)\)] Left to right. Assume \((\forall p.\psi (p), a, \mathcal{V }) \in \textit{Win}_{\exists }({\varepsilon ^{2}_\mathrm{U}}(\mathcal{A }))\). As, according to Definition 5 it is \(\forall \text{ dam}\)’s turn to play, by Definition 6 it follows that for all \(X \subseteq A\) we have that \((\psi , a, \mathcal{V }_{p := X}) \in \textit{Win}_{\exists }({\varepsilon ^{2}_\mathrm{U}}(\mathcal{A }))\). From this, by IH we obtain that for all \(X \subseteq A\) \((\mathcal{A }, \mathcal{V }_{p := X}), a \models \psi \) and by Definition 4 we can conclude that \((\mathcal{A }, \mathcal{V }), a \models \forall p. \psi \). Right to left. Assume \((\mathcal{A }, \mathcal{V }), a \models \forall p. \psi \). This means, by Definition 4, that for all \(X \subseteq A\) s.t. \((\mathcal{A }, \mathcal{V }_{p := X}), a \models \psi \). By IH it follows that \((\psi , a, \mathcal{V }_{p := X}) \in \textit{Win}_{\exists }({\varepsilon ^{2}_\mathrm{U}}(\mathcal{A }))\) for all \(X\), and from the fact that it is \(\forall \text{ dam}\)’s turn to play, by Definition 6 we can conclude that \((\forall p.\psi (p), a, \mathcal{V }) \in \textit{Win}_{\exists }({\varepsilon ^{2}_\mathrm{U}}(\mathcal{A }))\). \(\square \)

Proof

Sketch of proof of Theorem 2 The claim follows directly from Theorem 1 once we have built, for every graph \(\mathcal{A }\) and argument \(a\), an injection \(g\) from the set \(D\) of possible dialogues in \(\mathcal{G }_\mathcal{A }^\textit{SkPrf}@ a\) to the positions in \({\varepsilon ^{2}_\mathrm{U}}(\mathcal{A })@(\textit{SkPrf}, a)\), such that, for a dialogue \(\mathbf{s}\), \(\mathcal{P }\) in \(\mathcal{G }_\mathcal{A }^\textit{SkPrf}@ a\) has a winning strategy from \(\mathbf{s}\) if and only if \(g(\mathbf{s})\) is a winning position for \(\exists \text{ ve}\) in \({\varepsilon ^{2}_\mathrm{U}}(\mathcal{A })@(\textit{SkPrf}, a)\). This injection is obtainable directly from the semantics of \({\mathsf{K }^{2}_{\mathbf{\mathsf{{U}} }}}\) and is defined by case declaration as follows:Footnote 27

  • If \(\mathbf{s}= (a)\) then \(g(\mathbf{s}) = (\forall p. {\textit{Prf}}(p) \rightarrow p, \mathcal{V }, a)\)

  • If \(\mathbf{s}= (a,X)\) then \(g(\mathbf{s}) = (\lnot {\textit{Prf}}(p), \mathcal{V }_{p:= X}, a)\)

  • If \(\mathbf{s}= (a,X,Y)\) then \(g(\mathbf{s}) = ({\textit{Adm}}(q), \mathcal{V }_{q:= Y}, a)\)

  • If \(\mathbf{s}= (a,X,b)\) then \(g(\mathbf{s}) = \left\{ \begin{array}{ll} (p, \mathcal{V }_{p:= X}, b)&{{\textsc {if}}}\, b \in X \\ ( \square \lnot p, \mathcal{V }_{p:= X}, b)&{\textsc {otherwise}}. \end{array} \right.\)

  • If \(\mathbf{s}= (a,X,b,c)\) then \(g(\mathbf{s}) = (\lnot p, \mathcal{V }_{p:= X}, c)\)

  • If \(\mathbf{s}= (a,X,Y,b)\) then \(g(\mathbf{s}) = \left\{ \begin{array}{ll} ( q, \mathcal{V }_{q:= Y}, b)&{{\textsc {if}}}\, b \in Y \\ ( \square \lnot q, \mathcal{V }_{q:= Y}, b)&{\textsc {otherwise}}. \end{array} \right.\)

  • If \(\mathbf{s}= (a,X,Y,b,c)\) then \(g(\mathbf{s}) = (\lnot p, \mathcal{V }_{p:= Y}, c)\)

Function \(g\) is clearly an injection.Footnote 28 Now for the equivalence we have to prove that, for each dialogue \(\mathbf{s}\), \(\mathcal{P }\) has a winning strategy from \(\mathbf{s}\) if and only if \(\exists \text{ ve}\) has a winning strategy from \(g(\mathbf{s})\). Now, \(\mathbf{s}\) is either (i) terminal or (ii) not. (i) If it is terminal we have, for \(a \not \in X\), one of the following: \(\mathbf{s}= (a,X)\) or \(\mathbf{s}= (a,X,b)\) or \(\mathbf{s}= (a,X,b,c)\) or \(\mathbf{s}= (a,X,Y)\) or \(\mathbf{s}= (a,X,Y, b)\) or \(\mathbf{s}= (a,X,Y,b,c)\). For each of these cases, by the semantics of \({\mathsf{K }^{2}_{\mathbf{\mathsf{{U}} }}}\) and Theorem 1 it can be shown that \(g(\mathbf{s})\) is a winning position for \(\exists \text{ ve}\). We give two cases as an illustration. (1) If \(\mathbf{s}= (a,X)\), that means that neither \(Y \supset X\) s.t. \(a \not \in Y\), nor \(b \rightarrow X\) exist. Hence \(\lnot {\textit{Prf}}(p)\) cannot be true at \(a\) and \((\lnot {\textit{Prf}}(p), \mathcal{V }_{p:= X}, a)\) is a winning position for \(\forall \text{ dam}\). The same reasoning applies in the other direction. (2) If \(\mathbf{s}= (a, X, Y, b)\) then either \(b \in Y\), and hence \(Y\) is not conflict-free, or \(b \not \in Y\), and hence \(Y {\leftarrow }b\) but there exists no \(c \in Y\) s.t. \(b {\leftarrow }c\). In both cases the corresponding positions \((p, \mathcal{V }_{p:= Y}, b)\) and \(( \square \lnot q, \mathcal{V }_{q:= Y}, b)\) are clearly winning for \(\exists \text{ ve}\). The same reasoning applies in the other direction. The other cases are similar. (ii) If \(\mathbf{s}\) is not terminal then it is one of these, for \(a \not \in X\): \(\mathbf{s}= (a)\) or \(\mathbf{s}= (a,X)\) or \(\mathbf{s}= (a, X, b)\) or \(\mathbf{s}= (a, X, Y)\) or \(\mathbf{s}= (a, X, Y, b)\). In all these cases a similar argument as above applies. \(\square \)

Proof

Sketch of proof of Theorem 3 First we have to build an injection \(g\) based on the move function for \(\mathcal{G }_{\mathcal{A }@a}^{\textit{SkSStb}}\) as defined in Definition 8 (and depicted in Fig. 6). We can then prove that \(\mathcal{P }\) in \(\mathcal{G }_{\mathcal{A }}^{{SkSStb}}@a\) has a winning strategy from \(\mathbf{s}\) if and only if \(g(\mathbf{s})\) is a winning position for \(\exists \text{ ve}\) in \({\varepsilon ^{2}_\mathrm{U}}(\mathcal{A })@(\textit{SkSStb}, a)\). The proof is fully analogous to the one of Theorem 2. \(\square \)

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Grossi, D. Abstract argument games via modal logic. Synthese 190 (Suppl 1), 5–29 (2013). https://doi.org/10.1007/s11229-012-0237-1

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