Propositional discourse logic

Abstract

A novel normal form for propositional theories underlies the logic pdl, which captures some essential features of natural discourse, independent from any particular subject matter and related only to its referential structure. In particular, pdlallows to distinguish vicious circularity from the innocent one, and to reason in the presence of inconsistency using a minimal number of extraneous assumptions, beyond the classical ones. Several, formally equivalent decision problems are identified as potential applications: non-paradoxical character of discourses, admissibility of arguments in argumentation networks, propositional satisfiability, and the existence of kernels of directed graphs. Directed graphs provide the basis for the semantics of pdl and the paper concludes by an overview of relevant graph-theoretical results and their applications in diagnosing paradoxical character of natural discourses.

Appendix: Proofs

Proposition 3.4

Given a graph $\mathsf{G}$, we have:

• (a) If $L \in Lk(\mathsf{G})$ then $\alpha _{\overline{L}} \models _{{{\L }}}\mathcal{D}(\mathsf{G})$ and;

• (b) If $\alpha \models _{{{\L }}}\mathcal{D}(\mathsf{G})$ for $\alpha : G{\rightarrow }\{\mathbf{1 , \mathbf 0 , \bot }\}$ then $\overline{\varnothing }\subseteq \alpha ^\mathbf 1 \in Lk(\mathsf{G})$

Proof

• (a) Assume $L \in Lk(\mathsf{G})$ and consider arbitrary $x \leftrightarrow \bigwedge _{y \in E(x)} \lnot y \in \mathcal{D}(\mathsf{G})$. Assume towards contradiction that $\alpha _{\overline{L}} \not \models _{{{\L }}}x \leftrightarrow \bigwedge _{y \in E(x)} \lnot y$. We let $\overline{\alpha }_{\overline{L}}$ denote the evaluation of complex formulae obtained from $\alpha _{\overline{L}}$ according to tables (3.3). Then we have $\alpha _{\overline{L}}(x) \not = \overline{\alpha }_{\overline{L}}(\bigwedge _{y \in E(x)} \lnot y)$. If $\alpha _{\overline{L}}(x) = \mathbf 1$ this inequality means that there is one $y \in E(x)$ such that $\alpha _{\overline{L}}(y) \in \{\mathbf{1 ,\bot }\}$, impossible by the fact that $\overline{L}$ is a local kernel (which requires, for all $y \in E(x)$, $y \in E^\smallsmile \!(\overline{L})$, i.e. $\alpha _{\overline{L}}(y) = \mathbf 0$). If $\alpha _{\overline{L}}(x) = \mathbf 0$ we must then have, for every $y \in E(x)$, $\alpha _{\overline{L}}(y) \in \{\mathbf{0 ,\bot }\}$ but this is also ruled out by the fact that $\overline{L}$ is a local kernel (which requires existence of some $y \in E(x)$ such that $\alpha _{\overline{L}}(y) = \mathbf 1$). The last possibility is that $\alpha _{\overline{L}}(x) = \bot$ in which case there are two possibilities. (1) We have some $y \in E(x)$ such that $\alpha _{\overline{L}}(y) = \mathbf 1$. This contradicts $x \not \in E^\smallsmile \!(L)$ (required since we have $\alpha _{\overline{L}}(x) = \bot$). (2) For all $y \in E(x)$ we have $\alpha _{\overline{L}}(y) = \mathbf 0$. This means $x \in sinks(\mathsf{G}\setminus (\overline{L} \cup E^\smallsmile \!(\overline{L}))$, impossible by Definition 2.2 of $\overline{L}$.

• (b) Assume $\alpha \models _{{{\L }}}\mathcal{D}(\mathsf{G})$. We show that $\alpha ^\mathbf 1$ is a local kernel. We show first that $\alpha ^\mathbf 1$ is independent. Assume towards contradiction that it is not. Then there are $x,y \in \alpha ^\mathbf 1$ with $y \in E(x)$. So we have $\alpha (x) = \alpha (y) = \mathbf 1$ and from inspecting the tables (3.3) we see that $\overline{\alpha }(\bigwedge _{y \in E(x)} \lnot y) = \mathbf 0$. In particular, we have $\alpha (x) \not = \overline{\alpha }(\bigwedge _{y \in E(x)} \lnot y)$, contrary to hypothesis. Assume towards contradiction that $\alpha ^\mathbf 1$ is not locally absorbing. Then there is some $x \in \alpha ^\mathbf 1$ with $y \in E(x)$ such that $E(y) \cap \alpha ^\mathbf 1 = \varnothing$. Since $\alpha (z) \not = \mathbf 1$ for all $z \in E(x)$, by $\alpha ^\mathbf 1$ being independent, this means that $\alpha (y) = \bot$ and that $\overline{\alpha }(\bigwedge _{y \in E(x)} \lnot y) = \bot \not = \alpha (x) = \mathbf 1$, contrary to hypothesis. To show that $\overline{\varnothing }\subseteq \alpha ^\mathbf{1 }$ is a simple proof by induction over definition 2.2. For the basis, if $x \in sinks(\mathsf{G})$, i.e. $x \in \varnothing _1$, then $x \leftrightarrow \mathbf 1 \in \mathcal{D}(\mathsf{G})$. Then is is clear that $x \in \alpha ^\mathbf 1$. The inductive step is also trivial.

$\square$

Theorem 3.5

$\models \langle \Gamma : \mathsf{G}\rangle$ iff there is some $\alpha : G{\rightarrow }\{\mathbf{1 , \mathbf 0 , \bot }\}$ such that $\alpha \models _{{{\L }}}\mathcal{D}(\mathsf{G})$ and $\alpha \models _{{{\L }}}\Gamma$.

Proof

$\Rightarrow$) Assume that $\models \langle \Gamma : \mathsf{G}\rangle$. Then by Definition 3.1 there is some local kernel $L \in Lk(\mathsf{G})$ such that $L \models \langle \Gamma : \mathsf{G}\rangle$. We have from Proposition 3.4.(a) that $\alpha _{\overline{L}} \models _{{{\L }}}\mathcal{D}(\mathsf{G})$. We show $\alpha _{\overline{L}} \models _{{{\L }}}\Gamma$ by induction on the complexity of $\Gamma$. We take its complexity to be the sum of the complexity of its formulae divided by $|\Gamma |$. The basis is for $\Gamma$ a collection of literals. Then $\Gamma ^+ \subseteq L$ and $\Gamma ^- \subseteq E^\smallsmile \!(L)$, so for all $x \in \Gamma ^+$ we have $\alpha _{\overline{L}}(x) = \mathbf 1$ and for all $y \in \Gamma ^-$ we have $\alpha _{\overline{L}}(y) = \mathbf 0$ (remember that $L \subseteq \overline{L}$). It follows from inspecting tables (3.3) that $\alpha _{\overline{L}} \models \Gamma$. The inductive steps are easy. For instance, if $\models \langle \Gamma : \mathsf{G}\rangle$ and there is $\lnot \lnot A \in \Gamma$ that has maximal complexity among formulae of $\Gamma$, then we form $\Gamma ^{\prime }$ which is like $\Gamma$ except that $A$ replaces $\lnot \lnot A$. $\Gamma ^{\prime }$ has lower complexity than $\Gamma$ and, obviously from Definition 3.1, $\models \langle \Gamma ^{\prime } : \mathsf{G}\rangle$. So by IH we get $\models _{{{\L }}}\Gamma ^{\prime }$. Consulting tables (3.3) we see that this gives us $\models _{{{\L }}}\Gamma$ so we are done. The cases for $\lnot (A \wedge B)$ and $A \wedge B$ are equally easy.

$\Leftarrow$) Assume $\alpha \models _{{{\L }}}\mathcal{D}(\mathsf{G})$ and $\alpha \models _{{{\L }}}\Gamma$. We have $\alpha ^\mathbf 1 \in Lk(\mathsf{G})$ from 3.4.(b) and obtain $\alpha ^\mathbf 1 \models \langle \Gamma : \mathsf{G}\rangle$ by induction on the complexity of $\Gamma$, measured as in the proof of $\Rightarrow$). From this $\models \langle \Gamma : \mathsf{G}\rangle$ follows by Definition 3.1. The basis is for $\Gamma$ a collection of literals. Consulting tables (3.3), we see that for all $x \in \Gamma ^+$, we have $\alpha (x) = \mathbf 1$ so $x \in \alpha ^\mathbf 1$. For all $y \in \Gamma ^-$, on the other hand, we have $\alpha (y) = \mathbf 0$. Since $\alpha \models _{{{\L }}}\mathcal{D}(\mathsf{G})$, we have $\alpha \models _{{{\L }}}y \leftrightarrow \bigwedge _{z \in E(y)} \lnot z$, meaning $\alpha (y) = \overline{\alpha }(\bigwedge _{z \in E(y)} \lnot z)$ (where $\overline{\alpha }$ is the evaluation of $\alpha$ according to tables (3.3)). From tables (3.3) we see that there must then be some $z \in E(y)$ such that $\alpha (z) = \mathbf 1$, meaning $y \in E^\smallsmile \!(\alpha ^\mathbf 1 )$. It follows from Definition 3.1 that $\alpha ^\mathbf 1 \models \langle \Gamma : \mathsf{G}\rangle$. The inductive steps are straightforward. For instance, if there is some $A \wedge B \in \Gamma$ that has maximal complexity among formulae of $\Gamma$, we form $\Gamma ^{\prime }$ which is like $\Gamma$ except that we replace $A \wedge B$ by $A$ and $B$. Then $\models _{{{\L }}}\Gamma ^{\prime }$ and $\Gamma ^{\prime }$ has smaller complexity than $\Gamma$ so by IH $\alpha ^\mathbf 1 \models \langle \Gamma ^{\prime } : \mathsf{G}\rangle$. It follows immediately from Definition 3.1 that $\alpha ^\mathbf 1 \models \langle \Gamma : \mathsf{G}\rangle$ and we are done. The cases of $\lnot \lnot A$ and $\lnot (A \wedge B)$ are equally easy. $\square$

1.1 Soundness and completeness of pdl

Soundness and completeness follow easily from the following simple lemma giving us the compositionality we need with respect to admissibility in graphs.

Lemma 7.1

For any graph $\mathsf{G}$ and $a \in G$ we have:

• (1) $\models \langle \Gamma , a : \mathsf{G}\rangle$ iff $\models \langle \Gamma , \{{\lnot b \mid b \in E(a)}\} : \mathsf{G}\setminus out(a)\rangle$

• (2) $\models \langle \Gamma , \lnot a : \mathsf{G}\rangle$ iff for some $b \in E(a)$, $\models \langle \Gamma , b : \mathsf{G}\rangle$

Proof

(1) $\Rightarrow$) Assume $\Gamma , a$ is admissible in $\mathsf{G}$ and let $L \subseteq G$ be a local kernel witnessing to $\Gamma$ and containing $a$. Clearly, $L$ is a local kernel also in $\mathsf{G}\setminus out(a)$. Now, since $a \in L$ it follows that $E(a) \subseteq E^\smallsmile \!(L)$, so $\Gamma \cup \{{\lnot b \mid b \in E(a)}\}$ is indeed admissible (in both $\mathsf{G}$ and $\mathsf{G}\setminus out(a)$)

$\Leftarrow$) Assume $\Gamma \cup \{{\lnot b \mid b \in E(a)}\}$ is admissible in $\mathsf{G}\setminus out(a)$ and let $L \subseteq G$ be an arbitrary local kernel in $\mathsf{G}\setminus out(a)$ witnessing to this fact. Then for every $b \in E(a)$ we have $E(b) \cap L \not = \varnothing$ so $L \cup \{a\}$ is a local kernel in $\mathsf{G}$ (as well as in $\mathsf{G}\setminus out(a)$)

(2) $\Rightarrow$) Let $L \subseteq G$ be a local kernel witnessing to the admissibility of $\Gamma , \lnot a$ in $\mathsf{G}$. Then, for some $b \in E(a)$, we have $b \in L$. So $\Gamma , b$ is admissible in $\mathsf{G}$.

$\Leftarrow$) Assume that there is some $b \in E(a)$ such that $\Gamma , b$ is admissible. Let $L \subseteq G$ be a witness. Then $L$ also witness to the admissibility of $\Gamma , \lnot a$ in $\mathsf{G}$. $\square$

This lemma establishes soundness and invertibility of the only rules from pdl that are not essentially classical. The rest is easily verified, yielding

Theorem 7.2

pdl is sound and all its rules are invertible.

Proof

The standard sequent rules for the composite formulae in $\Theta \vdash \Phi$ are trivially invertible, as are the rules for non-atomic basic $\langle \Gamma :\mathsf{G}\rangle$ (which form a one-sided sequent system for propositional logic). Lemma 7.1 established soundness and invertibility of the four rules for literals in $\Gamma$. We only have to show that the two axiom schemata are valid:

• (1) $\Theta , \langle \Gamma , \lnot a : \mathsf{G}\rangle \vdash \Phi$ for some $a \in sinks(\mathsf{G})$.

To show $\Theta , \langle \Gamma , \lnot a : \mathsf{G}\rangle \models \Phi$, it suffices to show that $\not \models \langle \lnot a : \mathsf{G}\rangle$, by Definition 3.1. By Definition 3.1, this amounts to the nonexistence of a local kernel $L$ of $\mathsf{G}$ containing a successor of $a$. But since $a$ is a sink in $\mathsf{G}$, no such $L$ exists.

• (2) $\Theta \vdash \langle \Gamma : \mathsf{G}\rangle , \Phi$ for some $\Gamma \subseteq sinks(\mathsf{G})$.

To show $\Theta \models \langle \Gamma : \mathsf{G}\rangle , \Phi$, it suffices to show $\models \langle \Gamma : \mathsf{G}\rangle$. Since $\Gamma$ is a collection of atomic expressions this amounts to showing that there is a local kernel $L$ in $\mathsf{G}$ such that $\Gamma \subseteq L$. But $sinks(\mathsf{G})$ is such a local kernel in $\mathsf{G}$ so the claim follows. $\square$

Completeness of pdl follows now by the standard line of reasoning, demonstrating invalidity of any unprovable sequent. We say that a sequent $\Theta \vdash \Phi$ is reduced when $\Theta$ and $\Phi$ contain only atomic formulae, i.e., every $\langle \Gamma : \mathsf{G}\rangle \in \Theta \cup \Phi$ contains only literals and, moreover, literals over sinks of $\mathsf{G}$, i.e.,

$$\begin{aligned} \Gamma = \{a\mid a\in \Gamma ^+\subseteq sinks(\mathsf{G})\} \cup \{{\lnot b\mid b \in \Gamma ^-\subseteq sinks(\mathsf{G})}\}. \end{aligned}$$

We first argue that, for any sequent, the rules suffice to create a proof-tree with all leafs reduced.

Trivially, the top level rules and rules for composite $\Gamma$ suffice to create a proof-tree where all leafs have the form $\Theta \vdash \Phi$ with $\Theta$ and $\Phi$ being collections of atomic expressions $\langle \Gamma : \mathsf{G}\rangle$, i.e., each $\Gamma$ being a collection of literals. Now, we employ the rules for literals, as long as there is some $a \in \Gamma$ or $\lnot a \in \Gamma$ with $E(a) \not = \varnothing$, i.e. as long as the sequent is not reduced. For any finite graph $\mathsf{G}$, it is clear that by employing these rules we will eventually reach a stage where all sequents have been reduced. If (i) $a\in \Gamma$ is not a sink, an application of the rule $(\vdash {\!a})$, resp., $(a\!\vdash )$, makes it a sink. If (ii) $\lnot a\in \Gamma$ is not a sink, then an application of the rule $(\vdash {\!\lnot })$, resp., $(\lnot \!\vdash )$, replaces it by all its out-neighbours with positive polarity, for which case (i) applies in the next round.

Theorem 3.10

System pdl is sound and complete: $\Theta \vdash \Phi$ iff $\Theta \models \Phi$.

Proof

We show that reduced, non-axiomatic $\Theta \vdash \Phi$, is invalid. We have:

(1) $\forall \langle \Gamma _T : \mathsf{G}_T\rangle \in \Theta : \Gamma _T^- = \varnothing$ and (2) $\forall \langle \Gamma _F : \mathsf{G}_F\rangle \in \Phi : \Gamma _F^- \not = \varnothing .$

(1) follows since $\Theta \vdash \Phi$ is reduced, so for all $\langle \Gamma _T : \mathsf{G}_T\rangle \in \Theta : \Gamma _T^+ \cup \Gamma _T^- \subseteq sinks(\mathsf{G}_T)$. Since the sequent is not axiomatic, we must have $\Gamma _T^- = \varnothing$. Consequently, $\Gamma _T \subseteq sinks(\mathsf{G}_T)$ and, since $sinks(\mathsf{G}_T)\in Lk(\mathsf{G}_T)$, so $\models \langle \Gamma _T : \mathsf{G}_T\rangle$.

(2) follows since, as before, $\Gamma _F^+ \cup \Gamma _F^- \subseteq sinks(\mathsf{G}_F)$ and, since the sequent is not axiomatic, $\Gamma _F \not \subseteq sinks(\mathsf{G}_F)$. Consequently, $\Gamma _F^- \not = \varnothing$ (since atoms from this set are negated in $\Gamma _F$). So there is some $a \in \Gamma _F^- \subseteq sinks(\mathsf{G}_F)$, i.e. $\lnot a \in \Gamma _F$ while $a \in sinks(\mathsf{G}_F)$. It follows that $\not \models \langle \Gamma _F : \mathsf{G}_F\rangle$.

Having obtained $\models \langle \Gamma _T : \mathsf{G}_T\rangle$ for all $\langle \Gamma _T : \mathsf{G}_T\rangle \in \Theta$ and $\not \models \langle \Gamma _F : \mathsf{G}_F\rangle$ for all $\langle \Gamma _F : \mathsf{G}_F\rangle \in \Phi$, we conclude by Definition 3.1 that $\Theta \not \models \Phi$. Invertibility of all the rules ensures that if such a reduced sequent is obtained as a leaf in a proof tree from some initial sequent $S$, then also $S$ is invalid. Invertibility was shown in Theorem 7.2 and here we also established soundness of the system. $\square$