Persuasive Argumentation and Epistemic Attitudes

Abstract

This paper studies the relation between persuasive argumentation and the speaker’s epistemic attitude. Dung-style abstract argumentation and dynamic epistemic logic provide the necessary tools to characterize the notion of persuasion. Within abstract argumentation, persuasive argumentation has been previously studied from a game-theoretic perspective. These approaches are blind to the fact that, in real-life situations, the epistemic attitude of the speaker determines which set of arguments will be disclosed by her in the context of a persuasive dialogue. This work is a first step to fill this gap. For this purpose we extend one of the logics of Schwarzentruber et al. with dynamic operators, designed to capture communicative phenomena. A complete axiomatization for the new logic via reduction axioms is provided. Within the new framework, a distinction between actual persuasion and persuasion from the speaker’s perspective is made. Finally, we explore the relationship between the two notions.

The research activity of Antonio Yuste-Ginel is supported by MECD-FPU 2016/04113. Carlo Proietti gratefully acknowledges funding received from the European Commission (Marie Skłodowska-Curie Individual Fellowship 2016, 748421) and Sveriges Riksbanken (P16-0596:1) for his research.

Appendix: Notes and Proofs

Notes

Note 1

As mentioned, we have assumed that 1 is always the speaker while 2 is the hearer. Nevertheless, this logical setting can be easily generalized in order to model argumentative dialogues (where the role speaker/hearer changes in turns). Definition 8 should be extended to obtain a different operation $(\cdot )^{a}_i$ for each $i\in \mathsf {Ag}$ where the precondition is, consequently, $a \in \mathsf {owns}_i(w)$. Furthermore, the language must also be extended with the clause $[+_i a] \varphi$ for each $a \in A$, $i\in \mathsf {Ag}$. The completeness result presented below can be easily extended for such generalization.

Note 2

($(\cdot )^{+b}$ as a global update). An alternative way of understanding the action $(\cdot )^{+b}$ (see Definition 7) such that the update becomes world-independent, i.e. $(\cdot )^{+b}$ goes from models to models, works as follows. First note that we still need a suitable notion of actual world so that we can express the precondition according to which the speaker has to be aware of b in the actual world for $(\cdot )^{+b}$ to have any effect. Hence, our former notion of pointed model is now simply called a model, i.e., a model is a tuple $M=(W,w,\mathcal {R},\mathcal {D})$. A pointed model is now a pair $((W,w,\mathcal {R},\mathcal {D}),v)$ where $v \in W$. Here w represents the actual world. The definition of $(\cdot )^{+b}$ stays the same (see Definition 8), but note that now the function $(\cdot )^{+b}$ goes from models to models. Finally, the notion of truth is redefined in pointed models as usual for the rest of the operators and it is the following one for $[+b]$:

$$(W,w,\mathcal {R},\mathcal {D}),v\vDash [+b] \varphi \qquad \text {iff} \qquad (W,w,\mathcal {R},\mathcal {D})^{+b},v\vDash \varphi $$

Proofs

All the proofs follow standard methods. We just include some of them here for illustration.

Proposition 1

Proof

Let $\mathcal {G}=(A, \rightsquigarrow , A_i, A_j)$ be a 2-AF model, let (M, w) be a pointed model for $\mathcal {G}$, i.e., $A_i=\mathcal {D}_i (w)$ and $A_j=\mathcal {D}_j(w)$. Suppose, for the sake of contradiction, that $A_i \nsubseteq A_j$ but $\mathcal {R}_j$ is reflexive. We have that there is an argument $a \in A$ s.t. $a\in A_i$ but $a\notin A_j$ or equivalently $a \in \mathcal {D}_i(w)$ but $a \notin \mathcal {D}_j(w)$. But since $\mathcal {R}_j$ is reflexive, we know that $w \mathcal {R}_j w$. The last two assertions contradict Condition 2 of Definition 7.

Completeness of the Static Logics. Definition of deduction from assumptions ($\varGamma \vdash _{*} \varphi$), consistent set $(\varGamma \nvdash _{*} \perp )$ (where $*\in \{\mathsf {A}, \mathsf {AK}, \mathsf {AB}\}$) and maximal consistency are standard [3]. Let us denote by $\mathfrak {MC^{*}}$ the class of all maximal consistent sets in $*\in \{\mathsf {A}, \mathsf {AK}, \mathsf {AB}\}$. When the context is clear or irrelevant, we just write $\mathfrak {MC}$.

For the proof of the next two claims, the reader is referred to [3] (p. 199).

Proposition 5

(Properties of MC-sets). Let $\varGamma \in \mathfrak {MC}$:

• For every $\varphi \in \mathcal {L}(A)$: $\varphi \in \varGamma$ or $\lnot \varphi \in \varGamma$

• If $\varGamma \vdash \varphi$, then $\varphi \in \varGamma$

Lemma 1

(Lindenbaum). Let $\varGamma \subseteq \mathcal {L}(A)$, if $\varGamma \nvdash \perp$, then there is a $\varGamma ' \in \mathfrak {MC}$ s.t. $\varGamma \subseteq \varGamma '$.

Definition 11 (Canonical Model)

Given a language $\mathcal {L}(A)$, define the canonical model $M^{c}=(W^{c},\mathcal {R}^{c},\mathcal {D}^{c})$ as:

$$ \begin{aligned}&\;W^{c}:=\{\varGamma \subseteq \mathcal {L}(A) \mid \varGamma \in \mathfrak {MC}\}\\ \varGamma \mathcal {R}^{c}_i&\varDelta \quad \text {iff} \quad \{\varphi \in \mathcal {L}(A) \mid \square _i \varphi \in \varGamma \}\subseteq \varDelta \\&\;a \in \mathcal {D}^{c}_i(\varGamma ) \quad \text {iff}\quad \mathsf {owns}_i(a)\in \varGamma \end{aligned}$$

Lemma 2 (Canonicity)

Given $\mathcal {L}(A)$, its canonical model $M^{c}$ is a model.

Proof

All we need to show is that conditions 1 and 2 of Definition 7 are satisfied by $M^{c}$. For Condition 1, let us suppose that $\varGamma \mathcal {R}_i^{c} \varDelta$ (*). Suppose that $a \in \mathcal {D}_i^{c}(\varGamma )$, by definition this is equivalent to $\mathsf {owns}_i(a) \in \varGamma$. From this we obtain $\varGamma \vdash \mathsf {owns}_i(a)$ and note that, by monotonicity of $\vdash$ and Ax (PI) we have that $\varGamma \vdash \mathsf {owns}_i(a) \rightarrow \square _i \mathsf {owns}_i(a)$. Applying modus ponens we get $\varGamma \vdash \square _i \mathsf {owns}_i(a)$. By Lemma 5 we have that $\square _i \mathsf {owns}_i(a) \in \varGamma$. This, together with (*) and the definition of $\mathcal {R}_i^{c}$ implies $\mathsf {owns}_i(a) \in \varDelta$ which is equivalent by definition of $\mathcal {D}^{c}$ to $a \in \mathcal {D}_i^{c}(\varDelta )$. We have proven the left to right inclusion, for the right to left, the proof is analogous to Condition 2, where (GNI) is applied with $i = j$.

As for Condition 2, suppose $\varGamma \mathcal {R}_i^{c} \varDelta$ (*). Suppose $a \in \mathcal {D}_j^{c}(\varDelta )$. The latter is equivalent by definition to $\mathsf {owns}_j(a)\in \varDelta$, which implies $\varDelta \vdash \mathsf {owns}_j(a)$. Now, suppose, reasoning by contradiction, that $a \notin \mathcal {D}_i^{c}(\varGamma )$. This is equivalent by definition to $\mathsf {owns}_i(a)\notin \varGamma$. By Proposition 5 we have that $\lnot \mathsf {owns}_i(a)\in \varGamma$ which implies $\varGamma \vdash \lnot \mathsf {owns}_i(a)$. Using axiom (GNI) and MP we have that $\varGamma \vdash \square _i \lnot \mathsf {owns}_j (a)$. This, together with (*) and the definition of $\mathcal {R}^{c}$, implies $\lnot \mathsf {owns}_j(a) \in \varDelta$ and hence $\varDelta \vdash \lnot \mathsf {owns}_j(a)$ that contradicts the consistency of $\varDelta$. Therefore, $a \in \mathcal {D}_i^{c}(\varGamma )$.

Lemma 3

(Existence Lemma). If $\lozenge _i \varphi \in \varGamma$, then there is a $\varDelta \in W^{c}$ s.t. $\varGamma \mathcal {R}_i^{c} \varDelta$ and $\varphi \in \varDelta$.

Since $\square _i$ is a normal modal operator, the proof is completely standard. The reader is referred to [3] (pp. 200–201).

Lemma 4 (Truth Lemma)

For each $\varGamma \in W^{c}$ and and each $\varphi \in \mathcal {L}(A)$:

$$\varphi \in \varGamma \quad \text {iff}\quad M^{c},\varGamma \vDash \varphi $$

Proof

The proof is by induction on the construction of $\varphi$ (see [3] Chap. 4).

Finally Theorem 1 follows from the Truth Lemma by the typical argument.

Completeness for the Dynamic Extensions. For the completeness of $\mathsf {A}^{!+}$, $\mathsf {AK}^{!+}$ and $\mathsf {AB}^{!+}$, we apply the general method described in [15]. Let us show some of the details.

Proposition 2

Proof

Proving that SE preserves validity can be done by induction of the construction on $\varphi$. The proof is simple but long, so we leave it for the reader. Note that in standard awareness logic [11], this is not generally the case, since awareness sets do not need to be closed under logical equivalence. In $\mathcal {L}$, however, the range of awareness operators is restricted to arguments, and therefore soundness of SE is guaranteed.

As for the validity of the reduction axioms, let us just show two cases. Let (M, w) be a pointed model:

• $\vDash [+a]\varphi \leftrightarrow (\mathsf {owns}_1 (a) \rightarrow [a!] \varphi ) \wedge (\lnot \mathsf {owns}_1(a) \rightarrow \varphi )$

  “$\rightarrow$”. Suppose $M,w \vDash [+a] \varphi$. This is true iff $M^{+a},w \vDash \varphi$. Let us reason by cases. If $a \notin \mathcal {D}_1(w)$, then the first conjunct is trivially true. For the second conjunct, note that if $a \notin \mathcal {D}_1(w)$, then $M^{+a},w=M,w$. We can then substitute $M^{+a},w$ by M, w and obtain $M,w\vDash \varphi$ which implies $M,w \vDash \lnot \mathsf {owns}_1(a) \rightarrow \varphi$. Now, if $a \in \mathcal {D}_1(a)$ then the second conjunct is trivially true. For the first conjunct we have that if $a \in \mathcal {D}_1(a)$, then $M^{+a},w=M^{a!},w$. By substituting equals in the hypothesis we obtain $M^{a!},w\vDash \varphi$ which is equivalent by the semantic definition of [a!] to $M,w\vDash [a!] \varphi$.

  “$\leftarrow$”. This direction is analogous, each of the cases ($a \in \mathcal {D}_1(w)$ and $a \notin \mathcal {D}_1(w)$) makes one of the conjuncts trivially true and allows us to obtain the true consequent of the other. With that information is easy to deduce $M,w\vDash [+a]\varphi$.

• $\vDash [a!]\square _i \varphi \leftrightarrow \square _i [a!]\varphi$

  Suppose $M,w \vDash [a!]\square _i \varphi$. This is true iff $M^{a!},w \vDash \square _i \varphi$ (Definition 9) iff $M^{a!},w' \vDash \varphi$ for every $w'$ s.t. $w \mathcal {R}_i w'$ (Definition 7) iff $M,w'\vDash [a!]\varphi$ for every $w'$ s.t. $w \mathcal {R}_i w'$ (substituting equivalents of Definition 9 in the last assertion) iff $M,w \vDash \square _i [a!] \varphi$ (Definition 7).

Definition 12

(Complexity measures).

$+$-depth:

Define ${\mathop {n}\limits ^{+}}: \mathcal {L}^{!+}(A)\rightarrow \mathbb {N}$ that returns the number of nested $[+a]$ in $\varphi$ for any $a \in A$. More detailed: ${\mathop {n}\limits ^{+}}(\mathsf {owns}_i(a)):=0$,    ${\mathop {n}\limits ^{+}}(\star \varphi ):={\mathop {n}\limits ^{+}}(\varphi )$ where $\star \in \{\lnot , \square _i, [a!]\}$,    ${\mathop {n}\limits ^{+}}(\varphi \wedge \psi ):=max({\mathop {n}\limits ^{+}}(\varphi ),{\mathop {n}\limits ^{+}}(\psi ))$ and ${\mathop {n}\limits ^{+}}([+a]\varphi ):=1+{\mathop {n}\limits ^{+}}(\varphi )$.

Depth:

Define $d:\mathcal {L}^{!}(A) \rightarrow \mathbb {N}$ as $d(\mathsf {owns}_i(a))=0$, $d(\star \varphi )=1+d(\varphi )$ where $\star \in \{\lnot , \square _i, [a!]\}$ and $d(\varphi \wedge \psi )=max(d(\varphi ),d(\psi ))$.

O-depth:

Define $Od:\mathcal {L}^{!}(A) \rightarrow \mathbb {N}$ that returns the number of nested [a!] in $\varphi$ for any $a \in A$. More detailed: $Od(\mathsf {owns}_i(a))=0$,    $Od(\lnot \varphi )=Od(\square _i \varphi ):= Od(\varphi )$,    $Od(\varphi \wedge \psi ):=max(Od(\varphi ),Od(\psi ))$, and    $Od([a!]\varphi )=1+Od(\varphi )$.

Ord:

Define $Ord:\mathcal {L}^{!}(A)\rightarrow \mathbb {N}$ that returns the depth of the outermost occurrence of [a!]. More detailed: $Ord(\mathsf {owns}_i(a)):=0$,    $Ord(\lnot \varphi )=Ord(\square _i \varphi ):=Ord(\varphi )$,    $Ord(\varphi \wedge \psi ):= max(Ord(\varphi ), Ord (\psi ))$,    $Ord([a!]\varphi )=1 + d(\varphi )$.

Lemma 5

(From $\mathcal {L}^{!+}(A)$ to $\mathcal {L}^{!}(A)$). For every $\varphi \in \mathcal {L}^{!+}(A)$, there is a $\psi \in \mathcal {L}^{!}(A)$ s.t. $\vdash _{\mathsf {A}^{!+}} \varphi \leftrightarrow \psi$.

Proof

By induction on ${\mathop {n}\limits ^{+}}(\varphi )$. If ${\mathop {n}\limits ^{+}}(\varphi )=0$, we have that $\varphi \in \mathcal {L}^{!}(A)$, and by (Taut) we have that $\vdash \varphi \leftrightarrow \varphi$, so we are done.

Assume as induction hypothesis that for every $\varphi \in \mathcal {L}^{!+}(A)$ s.t. ${\mathop {n}\limits ^{+}}(\varphi )\le k$, there is a $\psi \in \mathcal {L}^{!}(A)$ s.t.: $\vdash _{\mathsf {A}^{+!}}\varphi \leftrightarrow \psi$. Suppose ${\mathop {n}\limits ^{+}}(\varphi )=k+1$. Take every $\delta _i \in sub(\varphi )$ s.t. ${\mathop {n}\limits ^{+}}(\delta _i)\le k$. Note that by induction hypothesis we have that there is a $\delta _i' \in \mathcal {L}^{!}(A)$ s.t.: $\vdash _{\mathsf {A}^{+!}}\delta _i \leftrightarrow \delta _i'$. By SE we have that $\vdash \varphi \leftrightarrow \varphi [\delta _i/ \delta _i']$. Note that ${\mathop {n}\limits ^{+}}(\varphi [\delta _i/ \delta _i'])=1$. It is easy to see that (Def+) and SE assures the existence of a formula $\psi \in \mathcal {L}^{!}(A)$ for every $\varphi$ s.t ${\mathop {n}\limits ^{+}}(\varphi )=1$ satisfying $\vdash _{\mathsf {A}^{+!}}\varphi \leftrightarrow \psi$. In particular, we have that there is a $\psi \in \mathcal {L}^{!}(A)$ s.t.: $\vdash _{\mathsf {A}^{+!}} \varphi [\delta _i/ \delta _i'] \leftrightarrow \psi$. By transitivity of $\leftrightarrow$ we have that $\vdash _{\mathsf {A}^{+!}} \varphi \leftrightarrow \psi$.

Remark 8

For every $\varphi \leftrightarrow \psi \in \{(Atoms^{=})-(Box)\}$ it holds that $Ord(\varphi )> Ord(\psi )$.

Lemma 6

For every $\varphi \in \mathcal {L}^{!}(A)$ s.t. $Od(\varphi )=1$ there is a $\psi \in \mathcal {L}(A)$ s.t. $\vdash _{\mathsf {A}^{+!}}\varphi \leftrightarrow \psi$.

Proof

Suppose $Od(\varphi )=1$, the rest of the proof is by induction on $Ord(\varphi )$.

For the basic case, suppose $Ord(\varphi )=0$, then $\varphi \in \mathcal {L}(A)$ and $\vdash \varphi \leftrightarrow \varphi$, so we are done.

Suppose, as induction hypothesis, that for every $\varphi \in \mathcal {L}^{!}(A)$ s.t. $Ord(\varphi )\le k$ there is a $\psi \in \mathcal {L}(A)$ s.t. $\vdash _{\mathsf {A}^{+!}}\varphi \leftrightarrow \psi$. Now, suppose $Ord(\varphi )=k+1$. We have, by definition of Ord, that there is $[a!]\delta \in sub(\varphi )$ s.t. $Ord([a!]\delta )=k+1$. Note that, since $Od(\varphi )=1$, then $Od([a!]\delta )=1$ (there are no nested announcements in $[a!]\delta$) and therefore there is an axiom in Table 2 of the form $\vdash _{\mathsf {A}^{+!}}[a!]\delta \leftrightarrow \delta '$. By Remark 8, $Ord([a!]\delta )>Ord(\delta ')$. By the induction hypothesis we have that there is a $\sigma \in \mathcal {L}(A)$ s.t. $\vdash _{\mathsf {A}^{+!}} \delta ' \leftrightarrow \sigma$. By transitivity of $\leftrightarrow$ we have that $\vdash _{\mathsf {A}^{+!}}[a!]\delta \leftrightarrow \sigma$ and, by SE, $\vdash _{\mathsf {A}^{+!}} \varphi \leftrightarrow \varphi [[a!]\delta /\sigma ]$. We can repeat the same argument for every $\delta _i \in sub(\varphi )$ s.t. $Ord(\delta _i)=k+1$. It is clear than the resulting formula $\psi$ is in $\mathcal {L}(A)$ and that $\vdash _{\mathsf {A}^{+!}}\varphi \leftrightarrow \psi$.

Lemma 7

(From $\mathcal {L}^{!}(A)$ to $\mathcal {L}(A)$). For every $\varphi \in \mathcal {L}^{!}(A)$ there is a $\psi \in \mathcal {L}(A)$ s.t. $\vdash _{\mathsf {A}^{+!}} \varphi \leftrightarrow \psi$.

Proof

By induction on $Od(\varphi )$. The atomic case is straightforward since, if $Od(\varphi )=0$, then $\varphi \in \mathcal {L}(A)$ and we are done. As for the inductive step, suppose as induction hypothesis that for every $\varphi \in \mathcal {L}^{!}(A)$ s.t $Od(\varphi )\le k$ there is a $\psi \in \mathcal {L}(A)$ s.t. $\vdash _{\mathsf {A}^{+!}} \varphi \leftrightarrow \psi$. Suppose $Od(\varphi )=k+1$. Then, there is a $\delta \in sub(\varphi )$ s.t. $Od(\delta )\le k$. By the induction hypothesis we have that there is a $\delta ' \in \mathcal {L}(A)$ s.t. $\vdash _{\mathsf {A}^{+!}}\delta \leftrightarrow \delta '$ and by SE it holds that $\vdash _{\mathsf {A}^{+!}} \varphi \leftrightarrow \varphi [\delta /\delta ']$. We can repeat the same argument for every $\delta _i \in sub(\varphi )$ s.t. $Od(\delta _i)\le k$. Note that, since $Od(\delta _i')=0$, we have that $Od(\varphi [\delta _i/\delta _i'])=1$ and by SE $\vdash _{\mathsf {A}^{+!}} \varphi \leftrightarrow \varphi [\delta _i/\delta _i']$. By Lemma 6 we have that there is a $\psi \in \mathcal {L}(A)$ s.t. $\vdash _{\mathsf {A}^{+!}} \varphi [\delta _i/\delta _i'] \leftrightarrow \psi$ and by transitivity of $\leftrightarrow$ it holds that $\vdash _{\mathsf {A}^{+!}}\varphi \leftrightarrow \psi$.

Theorem 2

Proof

We prove completeness for $\mathsf {A}^{!+}$ w.r.t. $\mathcal {M}$, the other two cases are completely analogous. Let $\varphi \in \mathcal {L}^{!+}(A)$, suppose $\vDash \varphi$. By Lemmas 5 and 7 and transitivity of $\leftrightarrow$ we have that there is a $\psi \in \mathcal {L}(A)$ s.t. $\vdash _{\mathsf {A}^{+!}}\varphi \leftrightarrow \psi$. From soundness of $\mathsf {A}^{+!}$ and the initial hypothesis it follows that $\vDash \psi$ and, by completeness of $\mathsf {A}$ we have that $\vdash _{\mathsf {A}}\psi$. Since $\mathsf {A}^{+!}$ is an extension of $\mathsf {A}$, we have that $\vdash _{\mathsf {A}^{+!}} \psi$. By SE we obtain $\vdash _{\mathsf {A}^{+!}} \varphi$.