Logic Programming, Argumentation and Human Reasoning

Abstract

The weak completion semantics, a computational logic approach, has been shown to adequately model various episodes of human reasoning. Since the inception of abstract argumentation in the 1990s, connections between argumentation semantics and logic programming semantics have been studied, but existing work on this connection has not yet covered the weak completion semantics. In this paper we define a novel translation from logic programs to abstract argumentation frameworks and show that under this translation the weak completion semantics corresponds to the grounded semantics of abstract argumentation. Combining this translation with argumentation semantics other than grounded semantics gives rise to novel logic programming semantics. We discuss the potential relevance of these novel semantics to modeling human reasoning and give an outlook on possible future research on this topic.

A Proof of Theorem 34

As mentioned in the Acknowledgements, the proof we present is based on a proof presented by Christian Straßer. It derives Theorem 34 from Theorem 30 with the help of Lemma 43, which establishes a correspondence between the weak completion semantics and the Kripke Kleene semantics.

Definition 42

Given a program $\mathcal{P}$, define $\mathcal{P}_{\mathsf {id}} := \mathcal{P}\cup \{ A \leftarrow A \mid A \in \mathsf {undef}(\mathcal{P}) \}$.

Lemma 43

The WCS model of a program $\mathcal{P}$ is the KK-model of $\mathcal{P}_{\mathsf {id}}$.

Proof

We prove this result by inductively proving that $\varPhi _{\mathcal {P}}^i(\langle \emptyset ,\emptyset \rangle ) = \varPsi _{\mathcal {P}_{\mathsf {id}}}^i(\langle \emptyset ,\emptyset \rangle )$ for any $i \ge 1$ and any program $\mathcal {P}$ (where $\varPhi ^i_{\mathcal {P}}$ respectively $\varPsi ^i_{\mathcal {P}}$ denotes the $i$th iteration of the function based on a program $\mathcal {P}$). In the proof, we will sometimes make use of the following observation:

• $(\star )$ If $A \notin \mathsf {undef}(A)$, $A \leftarrow body \in \mathcal {P}$ iff $A \leftarrow body \in \mathcal {P}_{\mathsf {id}}$.

Let $\langle J_{i}^{\top }, J_i^{\bot } \rangle = \varPsi _{\mathcal {P}_{\mathsf {id}}}^i(\langle \emptyset , \emptyset \rangle )$ and $\langle K_{i}^{\top }, K_i^{\bot } \rangle = \varPhi _{\mathcal {P}}^i(\langle \emptyset , \emptyset \rangle )$. We now show inductively that (i) $J_i^{\top }= K_i^{\top }$ and (ii) $J_i^{\bot } = K_i^{\bot }$.

We make use of a somewhat unusual variant of proof by induction in which both $i = 0$ and $i=1$ are first established as separate base cases, and the induction step is from $i-1$ and i to $i+1$ for any $i > 0$.

$\underline{\mathrm{First\; base\; case:}\ i = 0.}$ In this case, $J_i^{\top }= \emptyset =K_i^{\top }$ and $J_i^{\bot } = \emptyset = K_i^{\bot }$.

$\underline{\mathrm{Second\; base\; case:}\ i = 1.}$ We discuss each point separately. Let $I = \langle \emptyset ,\emptyset \rangle$.

Ad (i). Suppose $A \in J_1^{\top }$. Thus, there is a $A \leftarrow body \in \mathcal {P}_{\mathsf {id}}$ such that $I( body ) = \top$. Since $I = \langle \emptyset ,\emptyset \rangle$, $body$ is $\top$. In view of the definition of $\mathcal {P}_{\mathsf {id}}$, $A \leftarrow \top \,\, \in \mathcal {P} \cap \mathcal {P}_{\mathsf {id}}$ and so $A \in K_1^{\top }$. $K_1^{\top } \subseteq J_1^{\top }$ holds since $\mathcal {P} \subseteq \mathcal {P}_{\mathsf {id}}$.

Ad (ii). Suppose $A \in K_1^{\bot }$. Thus, $A \notin \mathsf {undef}(\mathcal {P})$ and for all $A \leftarrow body \in \mathcal {P}$, $I( body ) = \bot$. By ($\star$), for all $A \leftarrow body \in \mathcal {P}_{\mathsf {id}}$, $I( body ) = \bot$ and hence $A \in J_1^{\bot }$.

Suppose now that $A \in J_1^{\bot }$. So, for all $A \leftarrow body \in \mathcal {P}_{\mathsf {id}}$, $I( body ) = \bot$. Note that $A \notin \mathsf {undef}(\mathcal {P})$ since otherwise $I(A) = \bot$ which is impossible since $I = \langle \emptyset , \emptyset \rangle$. Since $\mathcal {P} \subseteq \mathcal {P}_{\mathsf {id}}$, $A \in K_1^{\bot }$.

$\underline{\mathrm{Induction\; step:}\ i-1, i \Rightarrow i+1 for \;any\; i > 0.}$ Our inductive hypothesis is that $J_{i-1}^{\top } = K_{i-1}^{\top }$, $J_{i-1}^{\bot } = K_{i-1}^{\bot }$, $J_i^{\top } = K_i^{\top }$ and $J_i^{\bot } = K_i^{\bot }$. We again discuss (i) and (ii) separately. Let $I_i = \langle I_{i}^{\top }, I_i^{\bot } \rangle$ where $I_i^{\top } = J_i^{\top } = K_i^{\top }$ and $I_i^{\bot } = J_i^{\bot } = K_i^{\bot }$. Similarly for $I_{i-1}$.

Ad (i). Suppose $A \in K_{i+1}^{\top }$. Thus, there is a $A \leftarrow body \in \mathcal {P}$ for which $I_i( body ) = \top$. Since $\mathcal {P} \subseteq \mathcal {P}_{\mathsf {id}}$, $A \leftarrow body \in \mathcal {P}_{\mathsf {id}}$. Thus, $A \in J_{i+1}^{\top }$.

For the other direction assume $A \in J_{i+1}^{\top }$. Thus, there is a $A \leftarrow body \in \mathcal {P}_{\mathsf {id}}$ for which $I_i( body ) = \top$. Assume for a contradiction that $A \in \mathsf {undef}(\mathcal {P})$. Thus, $body = A$ and $A \in I_i^{\top } = K_i^{\top }$. Thus, there is a $A \leftarrow body ' \in \mathcal {P}$ for which $I_{i-1}( body ') = \top$. This contradicts $A \in \mathsf {undef}(\mathcal {P})$. So $A \notin \mathsf {undef}(\mathcal {P})$ and therefore $A \leftarrow body \in \mathcal {P}$. Thus, $A \in \mathcal {K}_{i+1}^{\top }$.

Ad (ii). Suppose $A \in K_{i+1}^{\bot }$. Thus, $A \notin \mathsf {undef}(\mathcal {P})$ and for all $A \leftarrow body (\mathcal {P})$, $I_i( body ) = \bot$. By ($\star$), for all $A \leftarrow body \in \mathcal {P}_{\mathsf {id}}$, $I_i( body ) = \bot$. Thus, $A \in J_{i+1}^{\bot }$.

Suppose now that $A \in J_{i+1}^{\bot }$. Thus, ($\dagger$) for all $A \leftarrow body \in \mathcal {P}_{\mathsf {id}}$, $I_i( body ) = \bot$. Assume for a contradiction that $A \in \mathsf {undef}(\mathcal {P})$. Then, since $A \leftarrow A \in \mathcal {P}_{\mathsf {id}}$, $I_i(A) = \bot$ and so $A \in I_i^{\bot } = K_i^{\bot }$. Since for all $B \in K_i^{\bot } (= \{ C \mid C \notin \mathsf {undef}(\mathcal {P})$ and for all $C \leftarrow body \in \mathcal {P}, I_{i-1}( body ) = \bot \})$, $B \notin \mathsf {undef}(\mathcal {P})$, this is a contradiction. So, $A \notin \mathsf {undef}(\mathcal {P})$ and by ($\star$) and ($\dagger$), for all $A \leftarrow body \in \mathcal {P}$, $I_i( body ) = \bot$. Thus, $A \in K_{i+1}^{\bot }$.    $\square$