The Meaning of Epistemic Modality and the Absence of Truth

Abstract

When one asserts the disjunction ‘the keys might be in the drawer, or they might be in the car,’ the speaker seems committed to both of the disjuncts, ‘the keys might be in the drawer’ and ‘they might be in the car’ (Kamp, Proc Aristotelian Soc N S 74:57–74 (1973), [12]). Namely, ‘or’ behaves like a conjunction ‘and’ when it meets epistemic modality ‘might’. It has been noted that it is very difficult to explain this phenomenon in terms of conversational implicature (Zimmermann, Nat Lang Seman 8:255–290 (2000), [19]); a semantic explanation is worth pursuing. This paper proposes the first semantics that explains the conjunctive ‘or’ as a semantic phenomenon and still preserves classical logic when ‘might’ is absent, all done without ad hoc case distinctions. The truth-conditional approach to semantics has not been able to do that. Instead of truth conditions, the proposed semantics provides acceptability conditions. To be more specific, information states are modeled by sets of possible worlds, and each sentence is compositionally evaluated at each information state as: acceptable, deniable, or undecided. Working with acceptability conditions does not mean that we abandon truth conditions altogether. In fact, we can employ a sentence’s acceptability condition to determine whether it has a truth condition. Epistemic modals turn out to lack truth conditions, while sentences like “snow is white” can have truth conditions if you wish. Although the above may appear to be a mere case study in linguistics, the result points to a new, general semantic framework for addressing a central issue in philosophical logic and meta-ethics: Which types of declarative sentences lack truth conditions, especially epistemic modals, indicative conditionals, and moral claims?

Appendices

Arguement for ‘Or’-Introduction in Ordinary Cases

Consider the following conversation.

• X: “Everyone in the party got drunk or overate.”¹⁶

• Y: “Really?!”

• X: “Yeah. Alice, Bob, and Charles got drunk, and they have almost nothing to eat because Dorothy and I ate too much.”

The general claim entails the truth of the instance “Alice got drunk or overate.” But the speaker knows that Alice did not overate, as can be seen from the above conversation. So, if Zimmermann’s genuineness condition is correct, the instance “Alice got drunk or overate” is false and, hence, the general claim is false—but that is counterintuitive.¹⁷ Furthermore, the speaker X uses his second claim to justify his first claim, and the justification is naturally understood as follows:

Alice got drunk, so (by ‘or’-introduction) she got drunk or overate. Similarly, Bob, Charles, Dorothy, and I got drunk or overate. So everyone in the party got drunk or overate.

That is why I insist on the classical ‘or’-introduction rule of inference, the very rule of inference that contradicts Zimmermann’s genuineness condition. In general, classical inferences should be preserved as much as possible—that is why I take (B) as a feature.

Proofs

Proof of Proposition 1 For $(\Longrightarrow )$, suppose that $[\![\Diamond \phi _{1} \vee \Diamond \phi _{2} ]\!]^{I} = \mathsf{Acceptable}$. Then, by the acceptability conditions of disjunctions, I equals $I_{1} \cup I_{2}$ for some sets $I_{1}, I_{2}$ such that $[\![\Diamond \phi _{i} ]\!]^{I_{i}} = \mathsf{Acceptable}$ for $i = 1, 2$. Then, by the acceptability conditions of epistemic modals, $I_{i}$ has a nonempty subset $I_{i}'$ such that $[\![\phi ]\!]^{I_{i}'} = \mathsf{Acceptable}$. It follows that I has a nonempty subset, namely $I_{i}'$, such that $[\![\phi ]\!]^{I_{i}'} = \mathsf{Acceptable}$ (because $I_{i}' \subseteq I_{i} \subseteq I$). So, by the acceptability conditions of epistemic modals, $[\![\Diamond \phi _{i} ]\!]^{I} = \mathsf{Acceptable}$, for $i = 1, 2$.

For $(\Longleftarrow )$, suppose that $[\![\Diamond \phi _{1} ]\!]^{I} = [\![\Diamond \phi _{2} ]\!]^{I} = \mathsf{Acceptable}$. Then, since $I = I \cup I$, it follows from the acceptability conditions of disjunctions that $[\![\Diamond \phi \vee \Diamond \psi ]\!]^{I} = \mathsf{Acceptable}$. $\Box$

Proof of Proposition 2 Prove by induction on the complexity of $\phi$ as follows. Inductive basis: suppose that $\phi$ is an atomic sentence $\alpha$ that has truth condition $|\alpha |$. Then the proposition holds by the acceptability and deniability conditions of $\alpha$. Inductive step for ($\lnot$): suppose that $\phi$ is a negation $\lnot \psi$. If I is empty, then the derivation is almost trivial:

$$\begin{aligned}{}[\![\lnot \psi ]\!]^{I} = \mathsf{Acceptable}\Leftrightarrow & {} [\![\psi ]\!]^{I} = \mathsf{Acceptable} \\\Leftrightarrow & {} I \subseteq |\psi | \\\Leftrightarrow & {} I \subseteq |\lnot \psi |\;\text{(since } \text{ the } \text{ empty } \text{ set } I \text{ is } \text{ included } \text{ in } \text{ every } \text{ set) }. \end{aligned}$$

$$\begin{aligned}{}[\![\lnot \psi ]\!]^{I} = \mathsf{Deniable}\Leftrightarrow & {} [\![\psi ]\!]^{I} = \mathsf{Deniable} \\\Leftrightarrow & {} I \cap |\psi | = \varnothing \text{ and } I \not = \varnothing \;\text{(which } \text{ is } \text{ impossible) } \\\Leftrightarrow & {} I \cap |\lnot \psi | = \varnothing \text{ and } I \not = \varnothing \;\text{(which } \text{ is } \text{ impossible, } \text{ too) }. \end{aligned}$$

If I is nonempty, then:

$$\begin{aligned}{}[\![\lnot \psi ]\!]^{I} = \mathsf{Acceptable}\Leftrightarrow & {} [\![\psi ]\!]^{I} = \mathsf{Deniable} \\\Leftrightarrow & {} I \cap |\psi | = \varnothing \text{ and } I \not = \varnothing \\\Leftrightarrow & {} I \cap |\psi | = \varnothing \\\Leftrightarrow & {} I \subseteq |\lnot \psi |. \end{aligned}$$

$$\begin{aligned}{}[\![\lnot \psi ]\!]^{I} = \mathsf{Deniable}\Leftrightarrow & {} [\![\psi ]\!]^{I} = \mathsf{Acceptable} \\\Leftrightarrow & {} I \subseteq |\psi | \\\Leftrightarrow & {} I \cap |\lnot \psi | = \varnothing \\\Leftrightarrow & {} I \cap |\lnot \psi | = \varnothing \text{ and } I \not = \varnothing . \end{aligned}$$

Inductive step for ($\wedge$): suppose that $\phi$ is a conjunction $\phi _{1} \wedge \phi _{2}$. Then:

$$\begin{aligned}{}[\![\phi _{1} \wedge \phi _{2} ]\!]^{I} = \mathsf{Acceptable}\Leftrightarrow & {} [\![\phi _{1} ]\!]^{I} = \mathsf{Acceptable} \text{ and } [\![\phi _{2} ]\!]^{I} = \mathsf{Acceptable} \\\Leftrightarrow & {} I \subseteq |\phi _{1}| \text{ and } I \subseteq |\phi _{2}| \\\Leftrightarrow & {} I \subseteq |\phi _{1}| \cap |\phi _{2}| \\\Leftrightarrow & {} I \subseteq |\phi _{1} \wedge \phi _{2}|. \end{aligned}$$

$$\begin{aligned}{}[\![\phi _{1} \wedge \phi _{2} ]\!]^{I} = \mathsf{Deniable}\Leftrightarrow & {} \text{ for } \text{ each } I', \text{ if } I' \text{ is } I \text{ or } \text{ a } \text{ nonempty } \text{ subset } \text{ of } I \\&\text{ then } [\![\phi _{1} \wedge \phi _{2} ]\!]^{I'} \not = \mathsf{Acceptable} \\\Leftrightarrow & {} \text{ for } \text{ each } I'\text{, } \text{ if } I' \text{ is } I \text{ itself } \text{ or } \text{ a } \text{ nonempty } \text{ subset } \text{ of } I \\&\text{ then } [\![\phi _{1} ]\!]^{I'} \not = \mathsf{Acceptable} \text{ or } [\![\phi _{1} ]\!]^{I'} \not = \mathsf{Acceptable} \\\Leftrightarrow & {} \text{ for } \text{ each } I'\text{, } \text{ if } I' \text{ is } I \text{ itself } \text{ or } \text{ a } \text{ nonempty } \text{ subset } \text{ of } I \\&\text{ then } I' \not \subseteq |\phi _{1}| \text{ or } I' \not \subseteq |\phi _{2}| \\\Leftrightarrow & {} \text{ for } \text{ each } I'\text{, } \text{ if } I' \text{ is } I \text{ itself } \text{ or } \text{ a } \text{ nonempty } \text{ subset } \text{ of } I \\&\text{ then } \text{ it } \text{ is } \text{ not } \text{ the } \text{ case } \text{ that } I' \subseteq |\phi _{1}| \text{ and } I' \subseteq |\phi _{2}| \\\Leftrightarrow & {} \text{ for } \text{ each } I'\text{, } \text{ if } I' \text{ is } I \text{ itself } \text{ or } \text{ a } \text{ nonempty } \text{ subset } \text{ of } I \\&\text{ then } \text{ it } \text{ is } \text{ not } \text{ the } \text{ case } \text{ that } I' \subseteq |\phi _{1}| \cap |\phi _{2}| \\\Leftrightarrow & {} \text{ for } \text{ each } I'\text{, } \text{ if } I' \text{ is } I \text{ itself } \text{ or } \text{ a } \text{ nonempty } \text{ subset } \text{ of } I \\&\text{ then } \text{ it } \text{ is } \text{ not } \text{ the } \text{ case } \text{ that } I' \subseteq |\phi _{1} \wedge \phi _{2}| \\\Leftrightarrow & {} I \not = \varnothing \text{ and } I \cap |\phi _{1} \wedge \phi _{2}| = \varnothing \end{aligned}$$

Inductive step for ($\vee$): suppose that $\phi$ is a disjunction $\phi _{1} \vee \phi _{2}$. Then:

To establish the $(\Rightarrow )$ side of (a), it suffices to note that, if $I_{i} \subseteq |\phi _{i}|$ for $i = 1, 2$, then $I_{1} \cup I_{2} \subseteq |\phi _{1}| \cup |\phi _{2}|$. To establish the $(\Leftarrow )$ side of (a), it suffices to let $I_{i} = I \cap |\phi _{i}|$ for $i = 1, 2$.

To establish the $(\Rightarrow )$ side of (b), suppose that the left hand side is true. If $I = \varnothing$, then the left hand side has a counterexample: $I' = I_{1} = I_{2} = \varnothing$. So $I \not = \varnothing$. If I is not disjoint from $|\phi _{1}|$, then the left hand side has a counterexample: $I' = I_{1} = I \cap |\phi _{1}|, I_{2} = \varnothing$. So I is disjoint from $|\phi _{1}|$. By symmetry, I is disjoint from $|\phi _{2}|$. To establish the $(\Leftarrow )$ side of (a), suppose (for reductio) that the right hand side is true and the left hand side is false. Since $I \not = \varnothing$, it follows from the falsity of the left hand side that there exist subsets $I', I_{1}, I_{2}$ of I such that $I' \not = \varnothing$, $I = I_{1} \cup I_{2}$, and $I_{i} \subseteq |\phi _{i}|$ for $i = 1, 2$. Since $I'$ is nonempty and $I' = I_{1} \cup I_{2}$, $I_{j}$ is nonempty for some $j \in \{1, 2\}$. Since $I_{j}$ is a nonempty subset both of I and of $|\phi _{j}|$, I is not disjoint from $|\phi _{j}|$, which contradicts the right hand side. $\Box$