Unsettling Preferential Semantics

Abstract

This paper is concerned with removing the identity schema from the axiomatic basis of deontic conditionals. This is in order to allow a stipulated ideal to be contrary or opposite in nature to the fact it is predicated upon. It is desirable, or so it is argued, to retain the order-theoretic orientation of preferential semantics towards the analysis of deontic conditionals, more specifically of maximality semantics in the tradition from Bengt Hansson. So understood, the problem involves abstracting away the settledness assumption that is built in to maximality semantics. This is the assumption that what is optimal given ϕ is that which all the best ϕ-states have in common, notably ϕ itself. We propose a solution based on a strict and finite preference relation over which deontic conditionals are evaluated by letting ϕ-states evolve freely, as fate or fortune would have it, into different possibly ensuing optima that may but need not be ϕ-states themselves. The result is a deontic conditional that does not have identity. This new conditional is shown to be a proper generalization of the Hansson conditional. Hansson’s conditional can be recovered in the new idiom as a special case. Indeed, the new semantics is general enough to cover several apparently very different conceptions of deontic conditionality. For instance, the input/output logic known as basic output is a sublogic of the new system. This is somewhat surprising and suggests that there may yet be unity to be had in the field of deontic logic.

Appendix A: Proof of Selected Theorems

1.1 A.1 The Completeness of KL5

The weak completeness result for KL5 is based on constructing a finite canonical countermodel for every finite set Σ such that $\bigwedge {\Sigma }$ is a non-theorem. Finiteness is ensured by filtration through Σ. This proof strategy is standard (and discussed at length in [4]). The only adjustments that need to be made for KL5 is to design the canonical model to fit with the interpretation of the global modality [U] (Definition 17), and to verify that there are consequently always enough states available to falsify $\bigwedge {\Sigma }$ (Lemma 1). The crucial step is where the interaction schema I, which is the only schema of KL5 that doesn’t already belong to Gödel-Löb logic, is used to generalize from ≺-accessible states to all states, so as to generate the required contradiction.

Definition 14 (ϕ-formula)

For a given sentence ϕ a modal sentence ψ is a ϕ-formula if and only if ψ is a subsentence of ϕ or ψ is the negation of a subsentence in ϕ.

Definition 15 (KL5-consistence)

A set of ϕ-formulae Σ is KL5-consistent iff $\nvdash _{KL5} \neg (\bigwedge {\Sigma })$

Definition 16 (Maximal KL5-consistent set)

A set of ϕ-formulae Σ is maximally KL5-consistent wrt. ϕ iff for every subsentence ψ of ϕ either ψ or ¬ψ is in Σ and Σ is KL5-consistent.

Definition 17 (Canonical model induced by ϕ)

For any formula ϕ, $\mathbb {M}^{\phi }=\langle \mathbb {S}^{\phi }, R^{\phi }, V^{\phi }\rangle$ called the canonical model induced by ϕ is defined as follows:

• 1. $\mathbb {S}^{\phi } =$ the set of maximal KL5-consistent ϕ formulae satisfying the constraint that for all $s, s^{\prime }\in \mathbb {S}$, $\{\psi | [ \texttt {U}] \psi \in s\}\subseteq s^{\prime }$.

• 2. $R^{\phi } s s^{\prime }$ iff v

  • (a) for all [≺]ϕ ∈ s, both [≺]ϕ and ϕ are in $s^{\prime }$, and

  • (b) there is a $[\prec ]\phi \in s^{\prime }$ such that [≺]ϕ∉s.

• 3. V^ϕ(p) = {s ∈ s^ϕ|p ∈ s}

Lemma 1 (Existence lemma)

Suppose$s\in \mathbb {S}^{\phi }$and〈≺〉ψ ∈ s.Then there is a$s^{\prime }\in \mathbb {S}^{\phi }$with$R^{\phi } s s^{\prime }$and$\phi \in s^{\prime }$.

Proof

We need to find a set $s^{\prime }$ which is maximally KL5-consistent wrt. ϕ and that satisfies the constraint in Definition 17 (1) as well as 2 (a) and (b). Put

$$s^{\prime} := \{\delta |[ \texttt{U}] \delta \in s\} \cup \{[\prec] \phi, \phi| [\prec]\phi \in s\}\cup \{\psi, [\prec] \neg \psi\}$$

Note that [≺]¬ϕ∉s since 〈≺〉ψ ∈ s and s is consistent. Thus, if $s^{\prime }$ is consistent and satisfies the constraint in clause 1, then $R^{\phi } s s^{\prime }$. Suppose for reduction that $s^{\prime }$ is inconsistent. Then there is a finite inconsistent subset δ₁ ∧… ∧ δ_k ∧ ϕ₁ ∧ [≺]ϕ₁ ∧… ∧ ϕ_n ∧ [≺]ϕ_n of {δ|[U]δ ∈ s}∪{[≺]ψ,ψ|[≺]ψ ∈ s}. For simplicity we assume that k = 1 = n, since the general case follows by a straightforward induction from this. Thus we assume there is a set δ ∧ ϕ ∧ [≺]ϕ which together with ψ and [≺]¬ψ generates a contradiction. In other words, for ⊩_KL5¬(δ ∧ ϕ ∧ [≺]ϕ ∧ [≺]¬ψ ∧ ψ). It follows by the principle of exportation that $\vdash _{KL5} \neg (\delta \wedge \phi \wedge [\prec ] \phi \wedge [\prec ] \neg \psi \rightarrow \psi )$ whence we obtain $\vdash _{KL5} \delta \wedge \phi \wedge [\prec ] \phi \wedge [\prec ] \neg \psi \rightarrow \neg \psi$ by resolving the negation. Applying exportation once more, we have $\vdash _{KL5} (\delta \wedge \phi \wedge [\prec ] \phi ) \rightarrow ([\prec ] \neg \psi \rightarrow \neg \psi )$. By necessitation and the K-axiom for [≺] it follows that $\vdash _{KL5} ([\prec ] \delta \wedge [\prec ]\phi \wedge [\prec ][\prec ] \phi ) \rightarrow [\prec ] ([\prec ] \neg \psi \rightarrow \neg \psi )$. By the interaction schema I it follows that $\vdash _{KL5} ([ \texttt {U}] \delta \wedge [\prec ] \phi \wedge [\prec ][\prec ] \phi ) \rightarrow ([\prec ] \delta \wedge [\prec ]\phi \wedge [\prec ][\prec ] \phi )$ so $\vdash _{KL5} ([ \texttt {U}] \delta \wedge [\prec ] \phi \wedge [\prec ][\prec ] \phi )\rightarrow [\prec ] ([\prec ] \neg \psi \rightarrow \neg \psi )$. Noting that the top-level consequent is the antecedent of the Löb axiom for [≺] it follows that $\vdash _{KL5} ([ \texttt {U}] \delta \wedge [\prec ] \phi \wedge [\prec ][\prec ] \phi )\rightarrow [\prec ]\neg \psi$. By transitivity and the rule of replacement we may infer $\vdash _{KL5} [ \texttt {U}]\delta \wedge [\prec ] \phi \rightarrow [\prec ] \neg \psi$. By construction of s we have [≺]ϕ, [U]δ ∈ s, and since s is closed under modus ponens, therefore, [≺]¬ψ ∈ s contradicting 〈≺〉ψ ∈ s and the consistency of s. $\ \Box$

1.2 A.2 Proofs for Section 3

Proof

(Theorem 3) It suffices to show that $max_{\prec }^{\mathbb {M}}(\phi )= {\lvert \lvert \phi \wedge [\prec ] \neg \phi \rvert \rvert }^{{\mathbb {M}}}$. So suppose $s\in max_{\prec }^{\mathbb {M}}(\phi )$. Then $\mathbb {M},s \vDash \phi$ and $\neg (\exists s^{\prime } \in \mathbb {S}) (s\prec s^{\prime } \& \mathbb {M}, s^{\prime } \vDash \phi )$ by Definition 8. From the former we have $s\in {\lvert \lvert \phi \rvert \rvert }^{{\mathbb {M}}}$ and from the latter $s\in {\lvert \lvert [\prec ]\neg \phi \rvert \rvert }^{{\mathbb {M}}}$. Hence $s\in {\lvert \lvert \phi \rvert \rvert }^{{\mathbb {M}}}\cap {\lvert \lvert [\prec ]\neg \phi \rvert \rvert }^{{\mathbb {M}}} = {\lvert \lvert \phi \wedge [\prec ] \neg \phi \rvert \rvert }^{{\mathbb {M}}}$. The converse direction is essentially similar. $\ \Box$

Proof

(Theorem 4) For the right-to-left direction suppose ${\tau _{{\prec }}^{{\mathbb {M}}}}(\phi )\subseteq {\lvert \lvert \psi \rvert \rvert }^{{\mathbb {M}}}$ and that s is an arbitrarily chosen state in $\mathbb {S}$. It suffices to show that $\mathbb {M}, s\vDash \phi \rightarrow [\prec ] ([\prec ] \!\perp \rightarrow \psi )$. If $\mathbb {M}, s \nvDash \phi$ then it is vacuously true, so assume the opposite. Suppose for reduction that $\mathbb {M}, s\nvDash [\prec ] ([\prec ]\!\perp \rightarrow \psi )$. Then $\mathbb {M}, s\vDash {\langle \prec \rangle } ([\prec ] \!\perp \wedge \neg \psi )$. Thus, there is an $s^{\prime }\in \mathbb {S}$ such that $s\prec s^{\prime }$ with $\mathbb {M}, s^{\prime }\vDash [\prec ]\!\perp \wedge \neg \psi$. Since $\mathbb {M}, s^{\prime }\vDash [\prec ]\!\perp$ it follows that $s^{\prime }$ is a terminal state, and since $\mathbb {M}, s\vDash \phi$$s^{\prime }$ is also a ϕ-successor. It follows by Definition 6 that $s^{\prime }\in {\tau _{{\prec }}^{{\mathbb {M}}}}(\phi )$. But then $\mathbb {M}, s^{\prime }\vDash \psi$ by the assumption that ${\tau _{{\prec }}^{{\mathbb {M}}}}(\phi )\subseteq {\lvert \lvert \psi \rvert \rvert }^{{\mathbb {M}}}$, contradicting $\mathbb {M}, s^{\prime }\vDash \neg \psi$.

For the converse direction suppose $\mathbb {M}, s\vDash [ \texttt {U}] (\phi \rightarrow [\prec ] ([\prec ]\! \perp \rightarrow \psi ))$. Then $\mathbb {M} \vDash [ \texttt {U}] (\phi \rightarrow [\prec ] ([\prec ]\! \perp \rightarrow \psi ))$, i.e. the formula is satisfied at any point $s^{\prime }\in \mathbb {S}$. We may choose an $s^{\prime }$ such that $\mathbb {M}, s^{\prime }\vDash \phi$, since if there is no such $s^{\prime }$ then ${\tau _{{\prec }}^{{\mathbb {M}}}}(\phi ) = \emptyset$ so the theorem holds trivially. By the T-schema for [U] we have $\mathbb {M}, s^{\prime } \vDash \phi \rightarrow [\prec ]([\prec ]\! \perp \rightarrow \psi )$ so $\mathbb {M}, s^{\prime } \vDash [\prec ] ([\prec ]\! \perp \rightarrow \psi )$ by modus ponens. Let $s^{\prime \prime }$ be any endpoint on a chain from $s^{\prime }$. By transitivity we have $s^{\prime }\prec ss^{\prime \prime }$. Therefore $\mathbb {M}, ss^{\prime \prime }\vDash [\prec ]\!\perp \wedge \psi$. It follows that $ss^{\prime \prime }\in {\tau _{{\prec }}^{{\mathbb {M}}}}(\phi )\cap {\lvert \lvert \psi \rvert \rvert }^{{\mathbb {M}}}$. Since $s^{\prime }$ was arbitrarily chosen, this holds for any endpoint on a chain running through a ϕ-state, so ${\tau _{{\prec }}^{{\mathbb {M}}}}(\phi )\subseteq {\lvert \lvert \psi \rvert \rvert }^{{\mathbb {M}}}$ as desired. $\ \Box$

Proof

(Theorem 5) Suppose that $\mathbb {M}, s\vDash \bigcirc (\psi /\phi )$ and $\mathbb {M}, s\vDash \phi$. It follows from the former that every terminal successor $s^{\prime }$ of $s^{\prime }$ is such that $\mathbb {M}, s^{\prime } \vDash \psi$. Since Definition 6 counts any terminal state that satisfies ϕ an optimal ϕ-successor, there is always at least one such $s^{\prime }$. Since it is terminal we have $\mathbb {M}, s^{\prime }\vDash [\prec ]\! \perp$. Taking the two together yields $\mathbb {M}, s^{\prime }\vDash [\prec ] \!\perp \rightarrow \psi$. Now, let $ss^{\prime \prime }$ be any non-terminal successor of s. If s has no non-terminal successors, then we are done, so we may assume that $ss^{\prime \prime }$ exists. Then $\mathbb {M}, ss^{\prime \prime }\nvDash [\prec ] \perp$ so $\mathbb {M}, ss^{\prime \prime }\vDash [\prec ] \perp \rightarrow \psi$ by propositional principles. Taking stock we have that $[\prec ] \perp \rightarrow \psi$ is true in all terminal and non-terminal successors of s which means that it is true in all successor of s. We may conclude therefore that $\mathbb {M},s\vDash [\prec ]([\prec ] \perp \rightarrow \psi )$, which by Df○^m entails $\mathbb {M},s\vDash \bigcirc (\phi )$. $\ \Box$

Proof

(Theorem 6) For brevity of expression, put 𝜖 := δ ∧○(ψ/γ) ∧⊙(ψ/ϕ) and consider the canonical model $\mathbb {M}^{\epsilon } = \langle \mathbb {S}^{\epsilon }, R^{\epsilon }, V^{\epsilon }\rangle$ induced by 𝜖 (cf. Definition 17). Clearly, both of δ and ○ (ψ/γ) are subformulae of 𝜖. Therefore, since 𝜖 is assumed to be consistent {δ,○(ψ/γ)} can be expanded to a maximal consistent set of 𝜖-formulae, that is, to a state $s_{1}\in \mathbb {S}^{\epsilon }$. Suppose for reduction that $\delta \vdash _{DKL5} \bigcirc (\psi /\gamma )\rightarrow \odot (\psi /\phi )$ but $\delta \nvdash _{DKL5}\odot (\psi /\phi )$. Since ⊙ (ψ/ϕ) is a 𝜖-formula, it follows from the weak completeness of KL5 that {δ,〈U〉((ϕ ∧ [≺]¬ϕ) ∧¬ψ)} can be expanded to a maximal consistent 𝜖-set $s_{2}\in \mathbb {S}^{\epsilon }$. From clause 1 of the canonical model construction this means that there is a state $s_{3}\in \mathbb {S}^{\epsilon }$ s.t.

$$ \mathbb{M}^{\epsilon}, s_{3}\vDash ((\phi \wedge [\prec] \neg\phi) \wedge \neg \psi) $$

(1)

Since conjunction is associative we have $\mathbb {M}^{\epsilon }, s_{3}\vDash \phi \wedge [\prec ] \neg \phi$ from which we may conclude that s₃ is a ϕ-maximal state. Now, by the construction of s₁ we have $\mathbb {M}^{\epsilon }, s_{1}\vDash \bigcirc (\psi /\gamma )$. Moreover, since $\mathbb {M}^{\epsilon }, s_{1} \vDash \delta$ and $\delta \vdash _{DKL5} \bigcirc (\psi /\gamma )\rightarrow \odot (\psi /\phi )$ by the supposition of the theorem, we may infer that $\mathbb {M}^{\epsilon }, s_{1} \vDash \odot (\psi /\phi )$. But since s₃ is ϕ-maximal this means that $\mathbb {M}^{\epsilon } , s_{3}\vDash \psi$ contradicting Eq. 16. $\ \Box$

Proof

(Theorem 8) We prove only the left to right direction, the converse being essentially similar. Suppose $s\in max_{{\prec }}^{{\mathbb {M}}}(\phi )$. Then $\mathbb {M}, s\vDash \phi$ and by Definition 8 $\neg (\exists s^{\prime } \in \mathbb {S}) (s\prec s^{\prime } \& \mathbb {M}, s^{\prime }\vDash \phi )$. From the former we have $s\in \mathbb {S}_{|\phi }$, from the latter we have $s\in {\lvert \lvert [\prec ] \perp \rvert \rvert }^{{\mathbb {M}_{|\psi }}}$. To show that $s\in {\lvert \lvert [\prec ] \!\perp \rvert \rvert }^{{\mathbb {M}_{|\phi }}} \cap (\uparrow \! {\lvert \lvert \phi \rvert \rvert }^{{\mathbb {M}_{|\phi }}} \cup {\lvert \lvert \phi \rvert \rvert }^{{\mathbb {M}_{|\phi }}})$ it now suffices to show that $s\in {\lvert \lvert \phi \rvert \rvert }^{{\mathbb {M}_{|\phi }}}$. By the definition of $\mathbb {M}_{|\phi }$ we have $\mathbb {S}_{|\phi } = {\lvert \lvert \phi \rvert \rvert }^{{\mathbb {M}}}$, so $s\in {\lvert \lvert \phi \rvert \rvert }^{{\mathbb {M}_{|\phi }}}$ by $s\in {\lvert \lvert \phi \rvert \rvert }^{{\mathbb {M}}}$. $\ \Box$

1.3 A.3 Proofs for Section 6

Proof

(Theorem 15) This proof is straightforward. We provide the derivation of conjunctive consequents as an example.

• 1. ○ (ψ₁/ϕ) ∧○(ψ₂/ϕ)

• 2. $[ \texttt {U}] (\phi \rightarrow [\prec ] ([\prec ] \perp \rightarrow \psi _{1}))\wedge [ \texttt {U}] (\phi \rightarrow [\prec ] ([\prec ] \perp \rightarrow \psi _{2}))$, by df. of ○ (/)

• 3. $[ \texttt {U}] (\phi \rightarrow [\prec ] ([\prec ] \perp \rightarrow \psi _{1}) \wedge (\phi \rightarrow [\prec ] ([\prec ] \perp \rightarrow \psi _{2}))$, by CM for [U]

• 4. $[ \texttt {U}] (\phi \rightarrow [\prec ] ([\prec ] \perp \rightarrow \psi _{1}) \wedge [\prec ] ([\prec ] \perp \rightarrow \psi _{2}))$, by RCM

• 5. $[ \texttt {U}] (\phi \rightarrow [\prec ] (([\prec ] \perp \rightarrow \psi _{1}) \wedge ([\prec ] \perp \rightarrow \psi _{2}))$, by CM for [≺]

• 6. $[ \texttt {U}] (\phi \rightarrow [\prec ] ([\prec ] \perp \rightarrow \psi _{1} \wedge \psi _{2}))$, by RCM

• 7. ○ (ψ₁ ∧ ψ₂/ϕ), by df. of ○ (/).

$\ \Box$

Proof

(Theorem 16) In the limiting case that G = ∅ we have G(Γ) = ∅ so ψ = ⊤ whence the biconditional holds trivially. Suppose therefore that G is non-empty.

For the right-to-left, suppose ψ ∈ out₂(G,Γ). Then (Γ,ψ) is derivable from G using the rules of basic output. Since Γ finite it follows that (ϕ,ψ) is derivable from G for some conjunction ϕ of elements of Γ. There is thus a derivation Δ of (ϕ,ψ) from a finite subset (ϕ₁,ψ₁),…, (ϕ_n,ψ_n) ∈ G. Now ○ (ψ₁/ϕ₁),…,○(ψ_n/ϕ_n) ∈ G^[≺], by the construction of G^[≺]. Moreover each one of the rules of basic output have a corresponding axiom in DKL5 so Δ can be simulated in DKL5 to yield G ⊩_DKL5 ○ (ψ/ϕ). Since ϕ is a conjunction of elements of Γ it follows that G,Γ ⊩_DKL5ϕ ∧○(ψ/ϕ). Applying Theorem 5 we have G,Γ ⊩_DKL5 ○ (ψ).

For the converse, suppose that ψ∉out₂(G,Γ). We need to show that $G^{[\prec ]}, {\Gamma }\nvdash _{DKL5} \bigcirc (\psi )$. From the former assumption, there is a complete superset Γ⁺ of Γ such that ψ∉G(Γ⁺). Since Γ is consistent, Γ⁺ can be assumed to be consistent too. It follows from this that we may select two boolean valuations u and v with u(Γ) = 1 and v(G(Γ⁺)) = 1 but v(ψ) = 0. We build a relational model $\mathbb {M} = (W, \preceq , v)$ by putting W = {u,v}, u ≼ v and defining the valuation function V (ϕ) =_df{v ∈ W|v(ϕ) = 1}. As so defined $\mathbb {M}$ is a transitive and converse well-founded model, so it suffices to show that $\mathbb {M}, u\vDash \delta$ for every δ ∈ G^[≺] ∪Γ but $\mathbb {M}, u\nvDash \bigcirc (\psi )$. If δ ∈Γ, then $\mathbb {M}, u\vDash \delta$ by the selection of v. If on the other hand δ ∈ G^[≺], then δ has the form $[ \texttt {U}] (\gamma _{1}\rightarrow [\prec ]([\prec ] \perp \rightarrow \gamma _{2}))$ for some (γ₁,γ₂) ∈ G. Since v is a dead end we have $\mathbb {M}, v\vDash [\prec ]([\prec ] \perp \rightarrow \gamma _{2})$ and therefore $\mathbb {M}, v \vDash \gamma _{1}\rightarrow [\prec ]([\prec ] \perp \rightarrow \gamma _{2})$ by propositional logic. To show that $\mathbb {M}, u \vDash [ \texttt {U}](\gamma _{1}\rightarrow [\prec ]([\prec ] \perp \rightarrow \gamma _{2}))$ it now suffices to show that $\mathbb {M}, u \vDash \gamma _{1}\rightarrow [\prec ]([\prec ] \perp \rightarrow \gamma _{2})$. We have $\mathbb {M}, v\vDash [\prec ] \perp$, since v is a dead end, and $M, v\vDash \gamma _{2}$ by the definition of v. Therefore $\mathbb {M}, v\vDash [\prec ] \perp \rightarrow \gamma _{2}$. The only point that u can see is v so $\mathbb {M}, u \vDash [\prec ]([\prec ] \perp \rightarrow \gamma _{2})$ whence $\mathbb {M}, u \vDash (\gamma _{1}\rightarrow [\prec ]([\prec ] \perp \rightarrow \gamma _{2})$.

It remains only to show that that $\mathbb {M}, v\nvDash \bigcirc (\psi )$ which, by the definition of ○ (ψ), is the same as $\mathbb {M}, v\nvDash [\prec ]([\prec ] \perp \rightarrow \psi )$. The valuation v was selected so that v(ψ) = 0 so $\mathbb {M}, v\nvDash \psi$. Since $\mathbb {M}, v\vDash [\prec ] \perp$ it follows that $\mathbb {M}, v\nvDash [\prec ] \perp \rightarrow \psi$. Therefore $\mathbb {M}, v\nvDash [\prec ]([\prec ] \perp \rightarrow \psi )$ as desired. $\ \Box$