Discourse, Diversity, and Free Choice

ABSTRACT

‘You may have beer or wine’ suggests that you may have beer and you may have wine. Following Klinedinst, I argue that this ‘free choice’ effect is a special kind of scalar implicature, arising from the application of an unspecific predicate to a plurality (of worlds). I show that the implicature can be derived from general norms of cooperative communication, without postulating new grammatical rules or hidden lexical items. The derivation calls for an extension to the classical neo-Gricean model. I give independent arguments for this extension.

KEYWORDS:

• free choice
• implicatures
• dynamic semantics

Disclosure Statement

No potential conflict of interest was reported by the author.

Notes

1 Example (10) is from Klinedinst [2007]; see also Eckardt [2007] and Fox [2007].

2 The problem is related to Lewis’s [1979] ‘problem about permission’.

3 A complication: one might argue that even though (18a) states that each of the best hammers is made of steel, its negation states that none of the best hammers are made of steel, due to the homogeneity presupposition triggered by plural definite descriptions. The conjunction of (2a) with ¬(18a) and ¬(18b) would then be inconsistent. To explain the implicature, we might therefore have to assume that the Gricean algorithm invokes the ‘weak falsity’ of (18a) and (18b)—their non-truth—rather than the truth of their negation. From a neo-Gricean perspective, this makes sense: the assumption that (18a) is not true suffices to explain why an informed and cooperative speaker doesn’t utter it, even if the negation of (18a) is not true either. But see Spector and Sudo [2017] for reasons to think that a different analysis might be required.

4 If this isn’t obvious, imagine a context in which (30) is an executive decision, at a point where it is not yet settled who should be sent. The decision might even be revoked later, so that nobody ends up being sent.

5 This is a simplification: see, for instance, Brasoveanu [2011] and Nouwen [2016].

6 Modulo the complication discussed in note 3.

7 More precisely, if we represent (41) as this,$Ac{c}_E\left(W\right)\wedge \left[\mathrm{\forall }w\in W\right]\left(\mathrm{B}\mathrm{a}\mathrm{r}1\right(w)\vee \mathrm{B}\mathrm{a}\mathrm{r}2(w)\vee \mathrm{B}\mathrm{a}\mathrm{r}3(w)\vee \mathrm{B}\mathrm{a}\mathrm{r}4(w\left)\right)$then its formal alternatives include all fourteen sentences of the same form but with fewer disjuncts. Conjoining the negation of these fourteen sentences with the original sentence, and existentially closing, yields the implicature that each bar is a possible location.

8 For simplicity, I ignore conjunctive alternatives like ♢$\left(p\wedge q\right)$ that are irrelevant to free choice.

9 Klinedinst [2007] suggests a different local mechanism to derive free choice. In essence, he suggests that ♢$\left(p\vee q\right)$ should be analysed as$\left[\mathrm{\exists }W:Acc\left(W\right)\right]Exh\left(\left[\mathrm{\forall }w\in W\right]\left(p\left(w\right)\vee q\left(w\right)\right)\right),$where $W$ is a plural variable over possible worlds. Relevant alternatives to the embedded clause $\left[\mathrm{\forall }w\in W\right]\left(p\left(w\right)\vee q\left(w\right)\right)$ are $\left[\mathrm{\forall }w\in W\right]p\left(w\right)$ and $\left[\mathrm{\forall }w\in W\right]q\left(w\right)$. So, $Exh\left(\left[\mathrm{\forall }w\in W\right]\left(p\left(w\right)\vee q\left(w\right)\right)\right)$ is $\left[\mathrm{\forall }w\in W\right]\left(p\left(w\right)\vee q\left(w\right)\right)\wedge \mathrm{\neg }\left[\mathrm{\forall }w\in W\right]p\left(w\right)\wedge \mathrm{\neg }\left[\mathrm{\forall }w\in W\right]q\left(w\right)$. Hence, ♢$\left(p\vee q\right)$ entails that some accessible worlds are $p$-worlds and others are $q$-worlds. Klinedinst’s proposal and my proposal make very similar predictions.

10 Here is why. Following the reasoning for two disjuncts, the relevant alternatives to Exh(♢$\left(p\vee q\vee r\right)$) should be Exh(♢$p$)$=$♢$p\wedge$¬♢$q$ ¬♢$r$, Exh(♢$q$)$=$♢$q\wedge$¬♢$p\wedge$¬♢$r$, and Exh(♢$r$)$=$♢$r\wedge$¬♢$p\wedge$¬♢$q$. Conjoining the negation of these alternatives with ♢$\left(p\vee q\vee r\right)$ yields a statement that is true iff at least two of ♢$p$, ♢$q$, and ♢$r$ are true.

11 Thanks to Paolo Santorio and three anonymous referees for helpful comments on earlier versions.