On counterpossibles

Abstract

The traditional Lewis–Stalnaker semantics treats all counterfactuals with an impossible antecedent as trivially or vacuously true. Many have regarded this as a serious defect of the semantics. For intuitively, it seems, counterfactuals with impossible antecedents—counterpossibles—can be non-trivially true and non-trivially false. Whereas the counterpossible "If Hobbes had squared the circle, then the mathematical community at the time would have been surprised" seems true, "If Hobbes had squared the circle, then sick children in the mountains of Afghanistan at the time would have been thrilled" seems false. Many have proposed to extend the Lewis–Stalnaker semantics with impossible worlds to make room for a non-trivial or non-vacuous treatment of counterpossibles. Roughly, on the extended Lewis–Stalnaker semantics, we evaluate a counterfactual of the form "If A had been true, then C would have been true" by going to closest world—whether possible or impossible—in which A is true and check whether C is also true in that world. If the answer is "yes", the counterfactual is true; otherwise it is false. Since there are impossible worlds in which the mathematically impossible happens, there are impossible worlds in which Hobbes manages to square the circle. And intuitively, in the closest such impossible worlds, sick children in the mountains of Afghanistan are not thrilled—they remain sick and unmoved by the mathematical developments in Europe. If so, the counterpossible "If Hobbes had squared the circle, then sick children in the mountains of Afghanistan at the time would have been thrilled" comes out false, as desired. In this paper, I will critically investigate the extended Lewis–Stalnaker semantics for counterpossibles. I will argue that the standard version of the extended semantics, in which impossible worlds correspond to maximal, logically inconsistent entities, fails to give the correct semantic verdicts for many counterpossibles. In light of the negative arguments, I will then outline a new version of the extended Lewis–Stalnaker semantics that can avoid these problems.

Appendix

In this appendix, I give the proofs of (Satisfiability) and (Inc).

Proof of (Satisfiability)

Let $\Upgamma$ be any set of sentences that satisfies (i) and (ii) in (Satisfiability). We want to show that there is a propositional evaluation function ν that makes each sentence A in $\Upgamma$ true. To this end, we stipulate an interpretation $\mathcal{I}$ such that for all atomic A:

• $\mathcal{I}(A) = T$ iff $A \in \Upgamma$.

• $\mathcal{I}(A) = F$ iff $A \notin \Upgamma$.

This is a possible stipulation because $\mathcal{I}$ cannot assign both T and F to any atomic A. We then need to show that every sentence in $\Upgamma$ is true under this interpretation. I do this by induction on the length of a sentence, where the length of a sentence is given by the number of symbols it contains:

• Base case: Assume for atomic A that $A \in \Upgamma$. We want to show that ν(A) = T. We get the result immediately. By definition of $\mathcal{I}, A \in \Upgamma$ iff $\mathcal{I}(A) = T$. By ($\nu\mathcal{I}$), $\mathcal{I}(A) = T$ iff ν (A) = T. So $A \in \Upgamma$ iff ν(A) = T. So ν(A) = T.

• Inductive step: Assume for the induction hypothesis that every sentence in $\Upgamma$ that is shorter than $\neg A$ and (A ∧ B) is true under the evaluation ν based on $\mathcal{I}$. We want to show that if $\neg A \in \Upgamma$, then $\nu(\neg A) = T$, and if $(A \wedge B) \in \Upgamma$, then ν (A ∧ B) = T. There are two cases to consider:

• Case 1: Assume $\neg A \in \Upgamma$. By (i) in (Satisfiability), $\neg A \in \Upgamma$ iff $A \notin \Upgamma$. By induction hypothesis, $A \notin \Upgamma$ iff ν(A) = F. By ($\nu\neg$) in (Evaluation), ν(A) = F iff $\nu(\neg A) = T$. So $\nu(\neg A) = T$.

• Case 2: Assume $(A \wedge B) \in \Upgamma$. By (ii) in (Satisfiability), $(A \wedge B) \in \Upgamma$ iff $A \in \Upgamma$ and $B \in \Upgamma$. By induction hypothesis, $A \in \Upgamma$ and $B \in \Upgamma$ iff ν(A) = T and ν(B) = T. By (ν∧) in (Evaluation), ν(A) = T and ν(B) = T iff ν(A ∧ B) = T. So ν(A ∧ B) = T. □

Proof of (Inc)

Let $\Updelta$ be any maximal, logically inconsistent set of sentences. By (Satisfiability), $\Updelta$ will fail to satisfy either (i) or (ii) and hence contain at least one of the following inconsistent pairs or triples of sentences:

• Case 1: $\Updelta$ may be inconsistent because it fails to satisfy (i) of (Satisfiability), in which case $\Updelta$ contains an inconsistency of the form $\{A, \neg A\}$. That is, $\Updelta$ contains an instance of a LNC-inconsistency.

• Case 2: $\Updelta$ may be inconsistent because it fails to satisfy (ii) of (Satisfiability), in which case $\Updelta$ contains either an inconsistency of the form $\{\neg A, (A \wedge B)\}, \{\neg B, (A \wedge B)\}$, or $\{\neg A, \neg B, (A \wedge B)\}$, or an inconsistency of the form $\{A, B, \neg(A \wedge B)\}$. That is, $\Updelta$ contains either an instance of a CF-inconsistency or an instance of a NCF-inconsistency. □