## The independence of free choice alternatives

It is well-known that disjunctive possibility and necessity statements appear to imply the possibility of the disjuncts:

(FC) $$\Diamond(p \lor q) \Rightarrow \Diamond p \land \Diamond q$$.
(RP) $$\Box(p \lor q) \Rightarrow \Diamond p \land \Diamond q$$.

The first kind of inference is known as a "free choice" inference, the second is "Ross's Paradox".

For example, (1a) seems to imply (1b) and (1c):

(1a) Alice might [or: must] have gone to the party or to the concert.
(1b) Alice might have gone to the party.
(1c) Alice might have gone to the concert.

In chapter 3 of his dissertation, Booth (2022), Richard Booth points out that (FC) and (RP) underdescribe the true effect.

Suppose we are sure that Alice went to the concert, and unsure whether she also went to the party (afterwards). In that case, we would not say (1a). The disjunctive possibility statement (1a) seems appropriate only if both disjuncts are "independently" possible, meaning that it is possible that either is true without the other.

Schematically, the true effect appears to be something like this:

(FC') $$\Diamond(p \lor q) \Rightarrow \Diamond(p \land \neg q) \land \Diamond(\neg p \land q)$$.
(RP') $$\Box(p \lor q) \Rightarrow \Diamond(p \land \neg q) \land \Diamond(\neg p \land q)$$.

(RP'), but not (RP), predicts the badness of the following inference (from Booth):

(2a) Alice must have mailed the letter.
(2b) Alice might have used the phone [unrelatedly].
(2c) So, Alice must have either mailed the letter or used the phone.

Similarly, (RP'), but not (RP), predicts the apparent validity of the following:

(3a) Alice ought to either write an essay or give a presentation.
(3b) So, if she doesn't write an essay, she ought to give a presentation.

Booth claims that existing accounts of (FC) and (RP) cannot explain (FC') and (RP'), and proposes a new account based on inquisitive semantics.

Booth doesn't really explain why existing accounts of (FC) and (RP) are unable to explain (FC') and (RP'). It surely depends on what these accounts say.

According to the account I have proposed in Schwarz (2021), the (FC) inference is a kind of implicature. The reasoning from (1a) to (1b) and (1c), for example, is supposed to go roughly like this:

The speaker described some epistemically possible worlds as "party or concert" worlds. If these worlds were all "party" worlds it would have been simpler and more informative to describe them as such. So they are not all "party" worlds. For the same reason, they are not all "concert" worlds. So some of them are "party" worlds and some are "concert" worlds. So some epistemically possible worlds are "party" worlds, and some are "concert" worlds.

This immediately predicts not just the (FC) inference, but also the (FC') inference.

To be fair, Booth doesn't mention my proposal. (As far as I can tell, my paper has gone entirely unnoticed.) And I think he is right that some other accounts of (FC) don't immediately predict (FC').

Let's look at the popular "double-exhaustification" account, due to Fox (2007). On this account, the LF of $$\Diamond(p \lor q)$$ is $$\text{exh}(\text{exh}(\Diamond(p \lor q)))$$, where the semantic effect of '$$\text{exh}(\phi)$$' is to conjoin $$\phi$$ with the negation of its ("innocently excludable") alternatives. Plausibly, the alternatives to $$\text{exh}(\Diamond(p \lor q))$$ include $$\text{exh}(\Diamond p)$$ and $$\text{exh}(\Diamond q)$$. Assuming that $$\Diamond q$$ and $$\Diamond p$$ are (the only relevant) alternatives of one another, $$\text{exh}(\Diamond p)$$ is equivalent to $$\Diamond p \land \neg\Diamond q$$, and $$\text{exh}(\Diamond q)$$ to $$\Diamond q \land \neg\Diamond p$$. Since $$\text{exh}(\text{exh}(\Diamond(p \lor q)))$$ negates $$\text{exh}(\Diamond p)$$ and $$\text{exh}(\Diamond q)$$, it entails $$\neg(\Diamond p \land \neg\Diamond q) \land \neg(\Diamond q \land \neg\Diamond p)$$, which is equivalent (by propositional logic) to $$\Diamond p \leftrightarrow \Diamond q$$. It also entails $$\Diamond(p \lor q)$$ and thus $$\Diamond p \lor \Diamond q$$. And $$\Diamond p \leftrightarrow \Diamond q$$ together with $$\Diamond p \lor \Diamond q$$ entails $$\Diamond p \land \Diamond q$$.

Informally, the reasoning goes something like this (compare Kratzer and Shimoyama (2002)):

The speaker said $$\Diamond(p \lor q)$$. Why didn't she say $$\Diamond p$$? There are two possibilities: (i) she thinks $$\Diamond p$$ is false; (ii) she thinks $$\Diamond p$$ would falsely convey that only $$p$$ is possible, that $$\neg\Diamond q$$. So either $$\neg\Diamond p$$ or $$\Diamond q$$. The same reasoning with $$q$$ yields $$\neg\Diamond q$$ or $$\Diamond p$$. It follows that $$\Diamond p \land \Diamond q$$.

This really doesn't seem to predict (FC'). We would get (FC') if we assumed that the alternatives to $$\text{exh}(\Diamond(p \lor q))$$ include $$\text{exh}(\Diamond p \land \neg\Diamond q)$$ and $$\text{exh}(\Diamond q \land \neg\Diamond p)$$. But this isn't licensed by standard accounts of alternatives (e.g., Fox and Katzir (2011)).

We can get some way towards (FC') by assuming – plausibly – that $$\Box p$$ is an alternative to $$\Diamond (p \lor q)$$. (Informally, if all accessible worlds are p worlds, why would you say that some are p or q worlds?) The double-exhaustification account then still doesn't predict (FC'), but at least it predicts (FC''):

(FC'') $$\Diamond(p \lor q) \Rightarrow \Diamond p \land \Diamond q \land \Diamond \neg p \land \Diamond \neg q$$.

A parallel move for (RP') might give us

(RP'') $$\Box(p \lor q) \Rightarrow \Diamond p \land \Diamond q \land \Diamond \neg p \land \Diamond \neg q$$.

And that's enough to explain much of the data. For example, (FC'') is enough to explain why (1a) is inappropriate if we know that Alice went to the concert. And (RP'') is enough to predict the badness of the inference in (2).

But I suspect that other data calls for the full strength of (FC') and (RP').

We can get (FC') in the double-exhaustification account if we assume that the alternatives to $$\Diamond(p \lor q)$$ include $$\Diamond(p \land q)$$. We then still get $$\Diamond p$$ and $$\Diamond q$$, by essentially the same reasoning as above. Conjoined with $$\neg \Diamond(p \land q)$$, these entail $$\Diamond (p \land \neg q)$$ and $$\Diamond (\neg p \land q)$$.

Is $$\Diamond(p \land q)$$ an alternative to $$\Diamond(p \lor q)$$? In many contexts, $$\Diamond(p \lor q)$$ does indeed convey that $$\Diamond(p \land q)$$ is false. In such contexts, the assumption is plausible. But I don't think it is always true.

In sum, I'm not sure if there is a problem here for the double-exhaustification account. One would need to check if the full strength of (FC'), and not just (FC''), is needed in cases where $$\Diamond(p \lor q)$$ does not convey $$\neg \Diamond(p \land q)$$.

Booth, Richard Jefferson. 2022. “Underspecificity in Modal Contexts.” PhD thesis, Columbia University.
Fox, Danny. 2007. “Free Choice and the Theory of Scalar Implicatures.”
Fox, Danny, and Roni Katzir. 2011. “On the Characterization of Alternatives.” Natural Language Semantics 19: 87–107.
Kratzer, Angelika, and Junko Shimoyama. 2002. “Indeterminate Pronouns: The View from Japanese.” In Proceedings of the 3rd Tokyo Conference on Psycholinguistics, 1–25. Tokyo: Hituzi Syobo.
Schwarz, Wolfgang. 2021. “Discourse, Diversity, and Free Choice.” Australasian Journal of Philosophy 99 (1): 48–67. doi.org/10.1080/00048402.2020.1736108.