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A note on Gibbard’s proof

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Abstract

A proof by Allan Gibbard (Ifs: Conditionals, beliefs, decision, chance, time. Reidel, Dordrecht, 1981) seems to demonstrate that if indicative conditionals have truth conditions, they cannot be stronger than material implication. Angelika Kratzer's theory that conditionals do not denote two-place operators purports to escape this result [see Kratzer (Chic Linguist Soc 22(2):1–15, 1986, 2012)]. In this note, I raise some trouble for Kratzer’s proposed method of escape and then show that her semantics avoids this consequence of Gibbard’s proof by denying modus ponens. I also show that the same holds for Anthony Gillies’ semantics (Philos Rev 118(3):325–349, 2009) and argue that this consequence of these theories is not obviously prohibitive—hence, both remain viable theories of indicative conditionals.

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Notes

  1. Material conditional theorists include (Grice 1989; Lewis 1976; Jackson 1979). Non-propositionalists include (Adams 1975; Gibbard 1981; Edgington 1995; Bennett 2003).

  2. For instance, though it has a false antecedent, the following conditional seems false:

    1. (i)

      If the moon is made of cheese, there are several thousand moon-rats eating away at it.

    The standard defense of the material conditional theory holds that (i) is unassertable rather than false. I’m doubtful such defenses will actually succeed, but even if they do, there’s an often overlooked problem for the material conditional theory that doesn’t seem defensible in the same way [pointed out by Grice (1989, p. 85)]. The problem is that according to such a theory, the negation of if p, q entails the truth of p and the falsity of q, but that’s not correct:

    1. (ii)

      I don’t know whether a Republican won the election, but of course, since Anderson didn't win, I know that it's not the case that if a Republican won, Anderson did.

    If the material conditional theory were true, (ii) should be infelicitous (just as (iii) is), but it's not.

    1. (iii)

      #I don't know whether a Republican won the election, but of course, since Reagan won, I know a Republican won.

    But if if p, q is stronger than material implication, its falsity does not entail that p is true and q false, and hence no such implausible result is predicted.

  3. The material conditional \(p\supset q\) is true iff p is false or q is true.

  4. Variably strict conditionals have the resources to deny this equivalence [see for instance (Stalnaker 1975)], though it’s not clear they should.

  5. What follows is a crib of the version of Gibbard's proof in Gillies (2009).

  6. The inspiration for this view comes from a proposal in Lewis (1975) in which he discusses the interaction between if-clauses and adverbial quantifiers.

  7. Some conditionals seem to involve a covert habitual (or generic) quantifier over events or situations, such as:

    1. (i)

      If John is in town, he visits Rudy’s bar.

    The indicative conditionals that are the targets of Gibbard’s proof are ones like (8), and not (i) so I’ll set these aside for now.

  8. In the presentation of the original Gibbard’s proof, I assumed that ‘\(\supset\)’ and ‘∨’ were propositional operators. In what follows below, I assume they are instead sentential operators.

  9. Stalnaker (1975); Warmbrod (1983); Weatherson (2001, 2009); Stephenson (2007); Gillies (2004, 2009, 2010) also endorse the thesis that indicative conditionals are epistemic. Strictly speaking, Kratzer rejects identifying the domain of epistemic modals with those worlds compatible with the knowledge of some group but instead identifies it with the worlds compatible with some body of information (see Kratzer (2009)). This information may be encoded in the contents of a closed filing cabinet, for instance. This detail won’t matter for our purposes.

  10. We’re ignoring a host of complications here that don’t concern us. The only relevant feature of epistemic modals needed here is that they involve universal quantification over a set of worlds.

  11. This is a consequence of the fact that it validates (B1) and assigns indicative conditionals truth conditions stronger than material implication. Here’s a simple counterexample to modus ponens with embedded indicatives:

    • \(E(c,w) = \{w, w_1, w_2, w_3\}, [\![p]\!]^{c} = \{w, w_2\}, [\![q]\!]^{c} = \{w_2,w_3\}, [\![r]\!]^{c} = \{w_2\}\)

    • Then \([\![if\,p,\,then\,if\,q,\,then\,r]\!]^{c,w} = 1\) and \([\![p]\!]^{c,w} = 1\).

    • But \([\![if\,q,\,then\,r]\!]^{c,w} = 0;\) since \(w_3\in E(c,w)\cap[\![q]\!]^{c}\) and \(w_3\not\in[\![r]\!]^{c}\).

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Acknowledgements

I’m grateful to Aaron Norby, Zoltán Gendler Szabó, and an anonymous reviewer for Philosophical Studies for helpful comments and suggestions that led to significant improvements on an earlier draft.

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Correspondence to Justin Khoo.

Appendix

Appendix

For readability, I reprint (B1)–(B3) here:

  1. (B)

    For all contexts c:

    1. 1.

      \([\![if\,p,\,then\,if\,q\,then\,r]\!]^{c} = [\![if\,p\,and\,q,\,then\,r]\!]^{c}\)

    2. 2.

      \([\![if\,p,\,q]\!]^{c}\subseteq [\![p \supset q]\!]^{c}\)

    3. 3.

      Either \(([\![p]\!]^{c} \not\subseteq [\![q]\!]^{c})\) or (\(\wp(W) \subseteq [\![if\,p,q]\!]^{c}\))

Here is a reconstruction of Gibbard’s original presentation of his proof that \([\![p\supset q]\!]^{c}\subseteq [\![if\,p, q]\!]^{c}\), for any context c, using (B1)–(B3). Consider the following conditional:

  1. (G)

    if (\(p\supset q\)), then if p, q

In what follows, we’ll make use of an arbitrary context c. By (B1), \([\![if\,(p\supset\,q),\,then\,if\,p,\,q]\!]^{c} = [\![if (p\supset q)\,and\,p,\,then\,r]\!]^{c}\). Since \([\![(p\supset q)\,and\,p ]\!]^{c}\subseteq [\![q]\!]^{c}\) , it follows from (B3) that \(\wp(W)\subseteq\,[\![if\,(p\supset q)\,and\,p,\,then\,r]\!]^{c}\) and hence that \(\wp(W)\subseteq\,[\![if (p\supset q),\,then\,if\,p,\,q]\!]^{c}\). That is, \([\![({\text{G}})]\!]^{c,w} = 1\) for any w. Now, by substituting appropriately into (B2), we get that \([\![if\,(p\supset q), then\,if\,p,\,q]\!]^{c}\subseteq [\![(p\supset q)\supset if\,p,\,q]\!]^{c}\). But since \([\![({\text{G}})]\!]^{c}\) is necessarily true, and since \([\![({\text{G}})]\!]^{c}\) entails \((p\supset q)\supset if\,p,\,q\), then the latter is also necessarily true. But that means that there are no worlds in which \((p\supset q)\) is true and if p, q false—hence, \([\![p\supset q]\!]^{c} \subseteq [\![ if\,p,\,q]\!]^{c}\).

In this version of the proof, the key step is similar to Gillies’ version discussed earlier—it is the step where we substitute into (B2). Kratzer’s semantics validates the reasoning up to the step where \(\wp(W)\subseteq [\![if\,(p\supset q),\,then\,if\,p,\, q]\!]^{c}\). But the semantics allows for models in which there is a world w such that \([\![if (p\supset q),\,then\,if\,p, q]\!]^{c,w} = 1\) and \([\![(p\supset q)\supset if\,p, q]\!]^{c,w} = 0\). Here is one: \(E(c,w) = \{w,w_1,w_2\}, [\![p]\!]^{c} = \{w_1\}, [\![q]\!]^{c} = \{w\}\), and \([\![p\supset q]\!]^{c} = \{w,w_2\}\). Then:

(20)

a.

\([\![if (p\supset q), then\,if\,p,\,q]\!]^{c,w} = 1\)

-Since \(\forall w'\in (E(c,w)\cap [\![p\supset q]\!]^{c} \cap [\![p]\!]^{c}): [\![q]\!]^{c,w'} = 1\), vacuously

-Since {ww 1w 2} ∩ {ww 2} ∩ {w 1} = ∅

 

b.

\([\![(p\supset q)\supset if \, p, q]\!]^{c,w} = 0\)

-Since \([\![p\supset q]\!]^{c,w} = 1\) and \([\![if\,p, q]\!]^{c,w} = 0\)

-The latter holds because \(\exists w'\in E(c,w)\cap [\![p]\!]^{c}: [\![q]\!]^{c,w'} = 0\); the world is w 1

Hence, by predicting that the required substitution instance of (B2) is in fact false, Kratzer’s semantics avoids the conclusion of Gibbard’s original presentation of the proof as well.

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Khoo, J. A note on Gibbard’s proof. Philos Stud 166 (Suppl 1), 153–164 (2013). https://doi.org/10.1007/s11098-012-0062-x

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