A Boolean Extension of KLM-Style Conditional Reasoning

Abstract

Propositional KLM-style defeasible reasoning involves extending propositional logic with a new logical connective that can express defeasible (or conditional) implications, with semantics given by ordered structures known as ranked interpretations. KLM-style defeasible entailment is referred to as rational whenever the defeasible entailment relation under consideration generates a set of defeasible implications all satisfying a set of rationality postulates known as the KLM postulates. In a recent paper Booth et al. proposed PTL, a logic that is more expressive than the core KLM logic. They proved an impossibility result, showing that defeasible entailment for PTL fails to satisfy a set of rationality postulates similar in spirit to the KLM postulates. Their interpretation of the impossibility result is that defeasible entailment for PTL need not be unique. In this paper we continue the line of research in which the expressivity of the core KLM logic is extended. We present the logic Boolean KLM (BKLM) in which we allow for disjunctions, conjunctions, and negations, but not nesting, of defeasible implications. Our contribution is twofold. Firstly, we show (perhaps surprisingly) that BKLM is more expressive than PTL. Our proof is based on the fact that BKLM can characterise all single ranked interpretations, whereas PTL cannot. Secondly, given that the PTL impossibility result also applies to BKLM, we adapt the different forms of PTL entailment proposed by Booth et al. to apply to BKLM.

A Appendix

1.1 A.1 Proofs of Lemmas in Sect. 3

Lemma 2. Let $\mathscr {R}$ be any ranked interpretation. Then there exists a formula $ch(\mathscr {R}) \in \mathcal {L}^b$ with $\mathscr {R}$ as its unique model.

Proof

Consider the following knowledge bases.

• 1. $\mathcal {K}_\prec = \{ \hat{u}< \hat{v} : u \prec _\mathscr {R}v \} \cup \{ \hat{u} \not < \hat{v} : u \not \prec _\mathscr {R}v \}$

• 2.

By Lemma 1, $\mathscr {R}$ satisfies $\mathcal {K}= \mathcal {K}_\prec \cup \mathcal {K}_\infty$. To show that it is the unique model of $\mathcal {K}$, consider any $\mathscr {R}^* \in \textsc {Mod}(\mathcal {K})$. Since $\mathscr {R}^*$ satisfies $\mathcal {K}_\infty$, $\mathscr {R}^*(u) = \infty$ iff $\mathscr {R}(u) = \infty$ for any $u \in \mathcal {U}$. Now consider any $u, v \in \mathcal {U}$, and suppose that $\mathscr {R}(u) < \infty$. Then $u \prec _\mathscr {R}v$ iff $\mathcal {K}_\prec$ contains $\hat{u} < \hat{v}$. But $\mathscr {R}^*$ satisfies $\mathcal {K}_\prec$, so this is true iff $u \prec _{\mathscr {R}^*} v$ as $\mathscr {R}^*(u) < \infty$. On the other hand, if $\mathscr {R}(u) = \infty$, then $u \not \prec _\mathscr {R}v$ and $u \not \prec _{\mathscr {R}^*} v$. Hence $\prec _\mathscr {R}= \prec _{\mathscr {R}^*}$, which implies that $\mathscr {R}= \mathscr {R}^*$ by Proposition 2. We conclude the proof by letting $\text {ch}(\mathscr {R}) = \bigwedge _{\alpha \in \mathcal {K}} \alpha$.    $\square$

Lemma 3. Let $\mathscr {R}$ be a ranked interpretation, and $u \in \mathcal {U}^\mathscr {R}$ a valuation with $\mathscr {R}(u) < \infty$. Then for all $\alpha \in \mathcal {L}^\bullet$ we have $\mathscr {R}\Vdash tr_u(\alpha )$ if and only if $u \Vdash _\mathscr {R}\alpha$.

Proof

We will prove the result by structural induction on the cases in Definition 4:

• 1. Suppose that $\mathscr {R}\Vdash \text {tr}_u(p)$, i.e. . This is true iff $u \models p$, which is equivalent by definition to $u \Vdash _\mathscr {R}p$. Cases 2 and 3 are similar.

• 4. Suppose that $\mathscr {R}\Vdash \text {tr}_u(\lnot \alpha )$, i.e. $\mathscr {R}\Vdash \lnot \text {tr}_u(\alpha )$. This is true iff $\mathscr {R}\not \Vdash \text {tr}_u(\alpha )$, which by the induction hypothesis is equivalent to $u \not \Vdash _\mathscr {R}\alpha$. But this is equivalent to $u \Vdash _\mathscr {R}\lnot \alpha$ by definition. Case 5 is similar.

• 6. Suppose there exists an $\alpha \in \mathcal {L}^\bullet$ such that $\mathscr {R}\Vdash \text {tr}_u(\bullet \alpha )$ but $u \not \Vdash _\mathscr {R}\bullet \alpha$. Then either $u \not \Vdash _\mathscr {R}\alpha$, which by the induction hypothesis is a contradiction since $\mathscr {R}\Vdash \text {tr}_u(\alpha )$, or there is some $v \in \mathcal {U}$ with $v \prec _\mathscr {R}u$ such that $v \Vdash _\mathscr {R}\alpha$. But by Lemma 1, $v \prec _\mathscr {R}u$ is true only if $\mathscr {R}\Vdash \hat{v} < \hat{u}$. We also have, by the induction hypothesis, that $\mathscr {R}\Vdash \text {tr}_v(\alpha )$ since $v \Vdash _\mathscr {R}\alpha$. Hence $\mathscr {R}\Vdash (\hat{v} < \hat{u}) \wedge \text {tr}_v(\alpha )$, which implies that one of the clauses in $\text {tr}_u(\bullet \alpha )$ is false. This is a contradiction, so we conclude that $\mathscr {R}\Vdash \text {tr}_u(\bullet \alpha )$ implies $u \Vdash _\mathscr {R}\bullet \alpha$.

  Conversely, suppose that $u \Vdash _\mathscr {R}\bullet \alpha$. Then $u \Vdash _\mathscr {R}\alpha$, and hence $\mathscr {R}\Vdash \text {tr}_u(\alpha )$ by the induction hypothesis. We also have that if $v \prec _\mathscr {R}u$ then $v \not \Vdash _\mathscr {R}\alpha$, which is equivalent to $\mathscr {R}\Vdash \lnot \text {tr}_v(\alpha )$ by the induction hypothesis. But by Lemma 1, $v \prec _\mathscr {R}u$ iff $\mathscr {R}\Vdash \hat{v} < \hat{u}$. We conclude that $\mathscr {R}\Vdash (\hat{v} < \hat{u}) \rightarrow \lnot \text {tr}_v(\alpha )$ for all $v \in \mathcal {U}$, and hence $\mathscr {R}\Vdash \text {tr}_u(\bullet \alpha )$.    $\square$

Fig. 1.

Ranked models of .

Lemma 4. For all $\alpha \in \mathcal {L}^\bullet$ and any ranked interpretation $\mathscr {R}$, $\mathscr {R}$ satisfies $\alpha$ iff $\mathscr {R}$ satisfies $tr(\alpha )$.

Proof

Suppose $\mathscr {R}\Vdash \alpha$. Then for all $u \in \mathcal {U}$, either $\mathscr {R}(u) = \infty$ or $u \Vdash _\mathscr {R}\alpha$. The former implies by Lemma 1, and the latter implies $\mathscr {R}\Vdash \text {tr}_u(\alpha )$ by Lemma 3. Thus for all $u \in \mathcal {U}$, which proves $\mathscr {R}\Vdash \text {tr}(\alpha )$ as required. Conversely, suppose $\mathscr {R}\Vdash \text {tr}(\alpha )$. Then for any $u \in \mathcal {U}$, either and hence $\mathscr {R}(u) = \infty$ by Lemma 1, or and hence $\mathscr {R}\Vdash \text {tr}_u(\alpha )$ by hypothesis. But then $\mathscr {R}\Vdash \alpha$ by Lemma 3.    $\square$

1.2 A.2 Proofs of Lemmas in Sect. 4

Lemma 5. There is no BKLM entailment relation $\mid \approx _?$ satisfying Ampliativity, Typical Entailment and the Single Model property.

Proof

Suppose that $\mid \approx _?$ is such an entailment relation, and consider the knowledge base . Both interpretations in Fig. 1, $\mathscr {R}_1$ and $\mathscr {R}_2$, are models of $\mathcal {K}$. $\mathscr {R}_1$ satisfies and not , whereas $\mathscr {R}_2$ satisfies and not . Thus, by the Typical Entailment property, and . On the other hand, by Ampliativity we get . A single ranked interpretation cannot satisfy all three of these assertions, however, and hence no such entailment relation can exist.    $\square$

Lemma 7. Given any knowledge base $\mathcal {K}\subseteq \mathcal {L}^b$, $\textsc {Mod}(\mathcal {K}) \subseteq M_n$, where $n = ind({\mathcal {K}})$.

Proof

An easy induction on step 5 of the algorithm proves that $M_n = \{ \mathscr {R}\in \textsc {RI}: \text {ind}(\mathscr {R}) \ge n \}$. By hypothesis, $\text {ind}(\mathscr {R}) \ge n$ for all $\mathscr {R}\in \textsc {Mod}(\mathcal {K})$, and hence $\textsc {Mod}(\mathcal {K}) \subseteq M_n$.    $\square$

Lemma 8. Given any knowledge base $\mathcal {K}\subseteq \mathcal {L}^b$, $Cn_?(\mathcal {K}) = sat({\mathscr {R}_n})$, where $n = ind({\mathcal {K}})$.

Proof

For all A, $\mathcal {K}_n \mid \approx _R A$ iff $\mathscr {R}\Vdash A$ for all $\mathscr {R}\in \textsc {Mod}(\mathcal {K}_n) = M_n$. But by Lemma 7, $\textsc {Mod}(\mathcal {K}) \subseteq M_n$ and hence $\text {Cn}_R(\mathcal {K}_n) \subseteq \text {Cn}_R(\mathcal {K})$. On the other hand, $\mathscr {R}_n \in \textsc {Mod}(\mathcal {K})$ by hypothesis and hence $\mathscr {R}_n \Vdash A$ for all $A \in \mathcal {K}$. By the definition of step 4 of the algorithm we have $\text {sat}(\mathscr {R}_n) = \text {Cn}_?(\mathcal {K}_n)$, and thus $\mathcal {K}\subseteq \text {Cn}_?(\mathcal {K}_n)$. Applying $\text {Cn}_R$ to each side of this inclusion (using the monotonicity of rank entailment), we get $\text {Cn}_R(\mathcal {K}) \subseteq \text {Cn}_R(\text {Cn}_?(\mathcal {K}_n)) = \text {Cn}_?(\mathcal {K}_n)$, with the last equality following from Lemma 6. Putting it all together, we have $\text {Cn}_R(\mathcal {K}_n) \subseteq \text {Cn}_R(\mathcal {K}) \subseteq \text {Cn}_?(\mathcal {K}_n)$, and hence by Cumulativity we conclude $\text {Cn}_?(\mathcal {K}) = \text {Cn}_?(\mathcal {K}_n) = \text {sat}(\mathscr {R}_n)$.    $\square$