why a priori knowability is closed under semantic presupposition

I encounter all kinds of interesting but, I think, misguided objections to my endorsement of (AK).

(AK) For all Φ, p, q: if Φ expresses p and semantically presupposes q, then p is knowable a priori only if q is too.

In this post, I’ll present my favorite formulation of the argument for (AK). I’ll start by explaining what the notion of semantic presupposition invoked by (AK) amounts to.

1. Semantic Presupposition

The relevant notion of semantic presupposition is this. Let ψ be a formula that expresses Φ’s presupposition q, and let [[Φ]]^c,w be just the truth-value of Φ uttered at context c, evaluated with respect to world w.

• Φ semantically presupposes q iff standard presupposition tests show that q is a presupposition of Φ, and either

1. For all w such that [[ψ]]^c,w ≠ 1, [[Φ]]^c,w is undefined, or
2. For all w such that [[ψ]]^c,w ≠ 1, if w is the world of utterance, [[Φ]]^c is undefined.

Condition (1) involves nothing more than Kai von Fintel’s idea that some semantic presuppositions restrict the set of worlds over which the content of a sentence uttered at a context is defined. The set is restricted to worlds at which the presupposition holds (so, for example, the proposition expressed by ‘the king of France is bald’ is undefined over worlds where there is no king of France or more than one king of France).

Condition (2) involves the idea that some semantic presuppositions encode computability conditions for the meaning of a sentence, such that, for worlds w in which the computability conditions fail to hold, if w is the world in which the sentence is uttered, a meaning for the sentence cannot be computed. Something like this idea is found in Heim and Kratzer, who assign to the definite determiner ‘the’ the following extensional semantic value.

λP_<e,t>λQ_<e,t> . {x: P(x)=1} ⊆ {x: Q(x)=1}
(Where this function is undefined on any function P such that the cardinality of {x: P(x)=1} does not equal 1.)

Using this as the semantic value for ‘the’ will mean that, for any world w where the number of F’s at does not equal 1 — i.e., any world with respect to which a sentence expressing that the existence and uniqueness presuppositions of the definite description is false — if w is the world of utterance, [[the F]]^c is undefined. Under standard computability assumptions (e.g., canonical specifications of rules for computing the meanings of complex constituents), this computability condition will project to any sentence having an NP of the form the F as a constituent. (This actually turns out to be problematic, for reasons that I’ll talk about some other time, but which are unimportant for our purposes here.)

In sum, if Φ expresses p and semantically presupposes q, then either condition (1) or condition (2) holds for Φ, p, and q.

In either case, given certain plausible auxiliary assumptions, (AK) can be seen to follow (or so I’m going to argue).

2. (AK) from Condition (1)

Take an arbitrary Φ uttered in context c and expressing p and presupposing q in virtue of satisfying Condition (1). Let ψ be a formula that expresses the presupposition q. So, for all w such that [[ψ]]^c,w≠1, [[Φ]]^c,w is undefined. So, for all w such that [[Φ]]^c,w=1, [[ψ]]^c,w=1 too. (T1) follows automatically.

(T1) ◻(p → q)

So in the arbitrary case, we know that ◻(p → q). What’s more, if we can know it a priori, i.e., if (T2) is true, then we have an instance of (AK).

(T2) For an arbitrary Φ expressing p and presupposing q in virtue of satisfying Condition (1), it is knowable a priori that ◻(p → q).

We’ll have an instance of (AK) because the relevant instance of (AK) is entailed by the conjunction of (T2) and (T3):

(T3) For any p and q, if it is knowable a priori that ◻(p → q), then p is knowable a priori only if q is too.

(T3) is a modest and extremely plausible closure claim. It says merely that knowability a priori is closed under the relation knowable a priori strict implication.

What about (T2)? (T2) is supportable by appeal to cases (and also by an apparent lack of counterexamples). Consider a sentence Φ containing a definite description the F and which bears the presupposition that there is just one F. It will certainly in general seem to be knowable a priori that ◻[Φ → ∃!x[F(x)]]. For instance, it is plausibly knowable a priori that, necessarily, if the king of France is bald, then there is a unique king of France.

This isn’t, of course, a knockdown argument for (T2), but until I see a plausible counterexample, I’m inclined to think that it’s pretty safe to assume that it’s true.

3. (AK) from Condition (2)

We introduce a rigidifying actually operator ‘@’ (with the usual interpretation) and reason as follows.

Consider an arbitrary Φ uttered at c and expressing p and presupposing q via satisfying Condition (2). So there is a formula ψ expressing q such that, for all w where [[ψ]]^c,w≠1, if w is the world of utterance (i.e., if w is c’s world), then [[Φ]]^c is undefined. Letting c just be an arbitrary context of utterance in the actual world (so w_c is just the actual world), we therefore know that if [[Φ]]^c,w_c = 1, then [[ψ]]^c,w_c = 1. We also know that [[Φ]]^c,w_c = 1 iff @p, and that [[ψ]]^c,w_c = 1 iff @q. So we have that (@p → @q). But then, by a standard logic for the ‘@’ operator, we have @(p → q), and so (T4).

(T4) ◻@(p → q)

Since we know that ◻@(p → q), the argument proceeds roughly as before. If we can know (T4) a priori, i.e., if (T5) is true, then we have an instance of (AK).

(T5) For an arbitrary Φ expressing p and presupposing q in virtue of satisfying Condition (2), it is knowable a priori that ◻@(p → q).

That’s because the relevant instance of (AK) is entailed by the conjunction of (T5) and (T6):

(T6) For any p and q, if it is knowable a priori that ◻@(p → q), then p is knowable a priori only if q is too.

(T6) is extremely plausible, for exactly the same reasons that the earlier closure condition (T3) is plausible.

As was the case with (T2), (T5) is supportable by appeal to cases and by an apparent lack of counterexamples. Let’s look at a case. Consider sentence J…

(J) Julius works at Orange Julius

…uttered at a context c in which ‘Julius’ has been stipulated to name the inventor of the zip. Let j be the proposition expressed by J, let K be a formula expressing the proposition that there is a unique inventor of the zip, and let k be this proposition.

It seems to be knowable a priori (for the philosophically sophisticated stipulator) that: for all w such that [[K]]^c,w≠1, if w is the world in which J is uttered, [[J]]^c is undefined. This is because, for all w such that [[K]]^c,w≠1 (i.e., all worlds where there is not a unique inventor of the zip), if w is the world in which J is uttered, the stipulation has failed to fix a lexical meaning for ‘Julius’, and this is sufficient to make to make [[J]]^c undefined. But then we can reason a priori to the following instance of (T4), using exactly the reasoning by which (T4) was established in the first place.

(T7) ◻@(j → k)

So it seems that (T7)’s truth is knowable a priori. The relevant instance of (AK) follows immediately from this and (T6).

There are, of course, a lot of other ways in which a sentence can semantically presuppose something in virtue of satisfying condition (2). The general task will be to show that, in each of these cases, it is knowable a priori that the sentence and its presupposition do in fact satisfy condition (2), from which it will be easy to show that the relevant instance of (T4) (and, so, the relevant instance of (AK)) can be known a priori. I can’t think of a single reason to think that this general task isn’t feasible; a basic condition of linguistic competence seems to be having some kind of a priori understanding of the conditions under which your utterances will fail to express anything meaningful.

I think these are quite persuasive arguments for (AK). Anyone who wants to reject (AK) needs to say something persuasive against them.