Bipolar Argumentation Frameworks, Modal Logic and Semantic Paradoxes

Abstract

Bipolar Argumentation Frameworks (BAF) are a natural extension of Dung’s Argumentation Frameworks (AF) where a relation of support between arguments is added to the standard attack relation. Despite their interest, BAF present several difficulties and their semantics are quite complex. This paper provides a definition of semantic concepts for BAF in terms of fixpoints of the functions of neutrality and defense, thus preserving most of the fundamental properties of Dung’s AF. From this angle it becomes easy to show that propositional dynamic logic provides an adequate language to talk about BAF. Finally, we illustrate how this framework allows to encode the structure of the referential discourse involved in semantic paradoxes such as the Liar. It turns out that such paradoxes can be seen as BAF without a stable extension.

Carlo Proietti gratefully acknowledges funding received from the European Commission (Marie Skłodowska-Curie Individual Fellowship 2016, 748421) for his research. The authors would like to thank the anonymous reviewers of LORI’19 for many helpful suggestions.

Appendix

Proof of Theorem 1 :

• 1. It suffices to show that no $n{+}$-attack is possible. Suppose $\texttt {Adm}^{ns}_{\mathcal {A}}(X)$ and that X contains a and b such that $a (\Rightarrow ^{-1})^{*} \rightarrow \Rightarrow ^{*} b$. Then there is a c such that $a (\Rightarrow ^{-1})^{*}c$ and $c \rightarrow \Rightarrow ^{*} b$ (as in Fig. 2(c)). Therefore c carries a secondary attack towards b. But since X is admissible it defends b against c, i.e. there is $d \in X$ such that $d \rightarrow \Rightarrow ^{*} c$. But then $d \rightarrow \Rightarrow ^{*} a$, i.e. $d\in X$ attacks $a\in X$, against the assumption that $\texttt {Cfr}^{ns}_{\mathcal {A}}(X)$.

• 2. Suppose $a \in X$ and $b \Rightarrow a$. As $\texttt {Cmp}^{ns}_{\mathcal {A}}(X)$ implies $X = \mathbf {d}^{ns}_{\mathcal {A}}(X)$, it is enough to show that $b \in \mathbf {d}^{ns}_{\mathcal {A}}(X)$. Indeed, if $b\not \in \mathbf {d}^{ns}_{\mathcal {A}}(X)$ then $\exists c \rightarrow \Rightarrow ^{*}b$ and not $X \rightarrow \Rightarrow ^{*}c$. But then $c \rightarrow \Rightarrow ^{*}a$ and X does not defend a, from which we get a contradiction by the completeness of X. Therefore $b\in \mathbf {d}^{ns}_{\mathcal {A}}(X)$.

• 3. Suppose $a \in X$ and $\exists b \in X$ such that $b \rightarrow \Rightarrow ^{*}a$. Then there is a c such that $b \rightarrow c$ and $c \Rightarrow ^{*}a$. Since $a \in X$ we get, by $\Rightarrow ^{-1}$-closure, that $c\in X$, which entails that $X \not \subseteq \mathbf {n}_{\mathcal {A}}(X)$. Contradiction.

• 4. The proof exploits the equivalence $\texttt {Stb}^{ns}_{\mathcal {A}}(X)$ iff $X = \mathbf {n}^{ns}_{\mathcal {A}}(X)$ (Definition 4). It is not difficult to prove that $X = \mathbf {n}^{ns}_{\mathcal {A}}(X)$ implies $X = \mathbf {n}_{\mathcal {A}}(X) \cap [\Rightarrow ^{-1}](X)$ even without restriction to well-foundedness of $\Rightarrow$. We skip this part here.

For the other direction we need to prove that $\mathbf {n}^{ns}_{\mathcal {A}}(X) = X$. The only difficult part is $\mathbf {n}^{ns}_{\mathcal {A}}(X) \subseteq X$, the converse inclusion being almost immediate. For this it suffices to show that $a \not \in X$ implies $a \not \in \mathbf {n}_{\mathcal {A}}(X)$. Suppose $a \not \in X$. Therefore, by $X = \mathbf {n}_{\mathcal {A}}(X) \cap [\Rightarrow ^{-1}](X)$, either (a) $\exists c_{0} \in X$ such that $c_{0} \rightarrow a$, in which case $a \not \in \mathbf {n}^{ns}_{\mathcal {A}}(X)$ and we are done, or else (b) $\exists b_{0} \not \in X$ such that $b_{0} \Rightarrow a$. The same reasoning applies to $b_{0}$: either (a) $\exists c_{1} \in X$ such that $c_{1} \rightarrow b$, in which case $a \not \in \mathbf {n}^{ns}_{\mathcal {A}}(X)$ (since $c_{1} \rightarrow \Rightarrow ^{*}a$), or else (b) $\exists b_{1} \not \in X$ such that $b_{1} \Rightarrow b_{0}$. Alternative (b) can only apply a finite number of times, otherwise it would determine an infinite descending chain of supports, which is excluded by the well-foundedness of $\Rightarrow$. Therefore $a \not \in \mathbf {n}^{ns}_{\mathcal {A}}(X)$ and the inclusion is proved.

Proof of Theorem 2 :

• 1. Suppose that $\texttt {Adm}^{ds}_{\mathcal {A}}(X)$ and X contains both a and b with $a \Rightarrow ^{*}\rightarrow (\Rightarrow ^{-1})^{*} b$. Then there is a c such that $a \Rightarrow ^{*}c$ and $c \rightarrow (\Rightarrow ^{-1})^{*} b$. Therefore c carries a mediated attack towards b. Since X is admissible it defends b against c, i.e. there is $d \in X$ such that $d \rightarrow (\Rightarrow ^{-1})^{*}c$. But then $d \rightarrow (\Rightarrow ^{-1})^{*}a$, against the conflict-freeness of X.

• 2. Suppose $a \in X$ and $a \Rightarrow b$. As $\texttt {Cmp}^{ds}_{\mathcal {A}}(X)$ implies $X = \mathbf {d}_{\mathcal {A}}(X)$, it is enough to show that $b\in \mathbf {d}_{\mathcal {A}}(X)$. Suppose $b\not \in \mathbf {d}_{\mathcal {A}}(X)$; then $\exists c \rightarrow (\Rightarrow ^{-1})^{*}b$ and not $X \rightarrow (\Rightarrow ^{-1})^{*}c$. But then clearly $c \rightarrow (\Rightarrow ^{-1})^{*}a$ and X does not defend a, a contradiction. Therefore $b\in \mathbf {d}_{\mathcal {A}}(X)$.

Proof of Theorem 3 :

• 1. From right to left. Assume that $\texttt {Stb}^{ns}_{\mathcal {A}(\mathcal {T})}( l^{1})$ . Consider any biconditional $\phi := x_{i} \equiv \bigwedge _{x \in X_{i}} \lnot x \wedge \bigwedge _{x \in Y_{i}} y$ in the theory. There are two cases: (a) $l(x_{i}) = 1$, i.e. $x_{i} \in l^{1}$. An immediate consequence of this, by Theorem 1(4, left-to-right) is that for all attacker x of $x_{i}$: $x \not \in l^{1}$, i.e. $x \in l^{0}$ by the given definition of labelling, and for all supporter y of $x_{i}$: $x \in l^{1}$ (closure of stable sets under support Theorem 1(2)). This suffices to guarantee that $l^{*}(\bigwedge _{x \in X_{i}} \lnot x \wedge \bigwedge _{x \in Y_{i}} y) = 1$ and then $l^{*}(\phi ) = 1$.

  (b) $l(x_{i}) = 0$. Since $l^{0}$ is the complement of $l^{1}$, by Theorem 1(4, left-to-right) either some attacker x of $x_{i}$: $x \in l^{1}$, or some supporter y of $x_{i}$: $y \in l^{0}$. By construction of $\mathcal {A}(\mathcal {T})$ all supporters and attackers figure on the right handside of $\phi$. As a consequence $l^{*}(\bigwedge _{x \in X_{i}} \lnot x \wedge \bigwedge _{x \in Y_{i}} y) = 0$ and then $l^{*}(\phi ) = 1$.

• 2. From left to right. Assume that l is such that $l^{*}(\phi )=1$ for all $\phi \in \mathcal {T}$. In order to show that $\texttt {Stb}^{ns}_{\mathcal {A}(\mathcal {T})}( l^{1})$ we need to prove that $l^{1} = \mathbf {n}^{ns}_{\mathcal {A}}(l^{1})$. We first prove that

  • (a) $l^{1} \subseteq \mathbf {n}^{ns}_{\mathcal {A}}(l^{1})$. Let $x \in l^{1}$. We have three cases. (a.1) x is of the form $\overline{y}$. Then by construction x has no supporters and is only attacked by y. Then $y \in l^{0}$ by condition 1 on labellings. Since, by construction, y is the only (direct or indirect) attacker of x, it follows that $x \in \mathbf {n}^{ns}_{\mathcal {A}}(l^{1})$. (a.2) x appears only on the right hand side of a biconditional. Again, by construction, x has no supporters and is only attacked directly by $\overline{x}$, which however is labelled 0. Ergo $x \in \mathbf {n}^{ns}_{\mathcal {A}}(l^{1})$. Otherwise suppose that (a.3) $x \in l^{1}$ appears on the left hand side of some biconditional. If $x \not \in \mathbf {n}^{ns}_{\mathcal {A}}(l^{1})$ then there is a chain $y_{0} \rightarrow y_{1} \Rightarrow \dots \Rightarrow y_{n} \Rightarrow x$ such that $y_{0}\in l^{1}$, $y_{1}, \dots ,y_{n} \in V(\mathcal {T})$, and at least $y_{2}, \dots ,y_{n}$ appear on the right hand side of some biconditional. This forces $y_{1}, \dots ,y_{n} \in l^{1}$. But then $l^{1} \rightarrow y_{1}$ against condition 1 on labelling. Therefore $x \in \mathbf {n}^{ns}_{\mathcal {A}}(l^{1})$.

  • (b) $\mathbf {n}^{ns}_{\mathcal {A}}(l^{1})\subseteq l^{1}$. For this is sufficient to show that for every $x \not \in l^{1}$ there is an $y \in l^{1}$: $y \rightarrow \Rightarrow ^{*}x$. It is straightforward to prove this for the cases where (b.1) x is of the form $\overline{y}$ or (b.2) x appears only on the right hand side of a biconditional.We consider (b.3) $x = x_{i}\not \in l^{1}$ appears on the left hand side of some biconditional $\phi := x_{i} \equiv \bigwedge _{x \in X_{i}} \lnot x \wedge \bigwedge _{x \in Y_{i}} y$. Since $x_{i} \in l^{0}$ (the complement of $l^{1}$) and $l^{*}(\phi ) = 1$ by assumption, then either one of the conjuncts $z \in X_{i}$ is in $l^{1}$, in which case $x_{i} \not \in \mathbf {n}^{ns}_{\mathcal {A}}(l^{1})$ and we are done, or else one of the conjuncts $y \in Y_{i}$ is in $l^{0}$. If y is as in (b.1) or (b.2) then it is attacked by $l^{1}$ and therefore $x_{i} \not \in \mathbf {n}^{ns}_{\mathcal {A}}(l^{1})$. Otherwise y is either attacked by $l^{1}$ or supported by some $y'$ in $l^{0}$. However the chain of supports cannot go on forever because the support relation is well-founded by assumption. Therefore we should finally find some attacker in $l^{1}$ and $x_{i} \not \in \mathbf {n}^{ns}_{\mathcal {A}}(l^{1})$.