CLS 43 [Session: Syntax & Semantics]
May 3, 2007, University of Chicago *
Indicative Conditionals Have Relative Truth Conditions Tamina Stephenson Massachusetts Institute of Technology web: http://web.mit.edu/tamina/www/; e-mail:
[email protected] Goals of talk: Propose a view of indicative conditionals that treats if-clauses as restrictors of covert epistemic modals gives a relativist semantics for epistemic modals Show that this solves two longstanding puzzles about conditionals: Gibbardian standoffs The or-to-if inference
Indicative Conditionals Have Relative Truth Conditions
The good news: Epistemic modals and conditionals are parallel, supporting the restrictor view. The bad news: Standard accounts of epistemic modals cannot account for (2)! 1.1. The problem with the contextualist view1 (Contextualist accounts include DeRose 1991, von Fintel & Gillies 2006, and others.) Contextualist hypothesis: might φ = it’s compatible with X’s knowledge that φ [for some particular X] But: • X ≠ “me” [speaker-indexical] as seen by Gibbardian standoffs. • X ≠ some contextually relevant group: (4)
Sam thinks that the cook might be the murderer.
1. Puzzle #1: Gibbardian Standoffs
might ≈ compatible with Sam’s knowledge/beliefs
A variation on examples by Gibbard (1981):
might ≠ compatible with the knowledge of the group we think is relevant.
(1)
[A murder investigation. Sam has narrowed down the suspects to the cook or the butler, and Mary has narrowed them down to the butler or the maid, both making only correct inferences.] Sam: Mary:
If the butler isn’t the murderer, the cook is. (No! / Nuh-uh!) If the butler isn’t the murderer, the maid is.
The puzzle: • Sam and Mary seem to be disagreeing / contradicting each other. • At the same time: they both seem to be justified in making their assertions. Essentially the same puzzle holds for epistemic modals (see, e.g., Egan, Hawthorne & Weatherson 2005, Weatherson 2005, MacFarlane 2006): (2)
[Same context as (1).] Sam: Mary:
The cook might be the murderer. (No! / Nuh-uh!) The cook can’t be the murderer.
[Again: Seem to be disagreeing, but both seem to be justified.] Why this is puzzling: Each speaker seems to be expressing their own state of knowledge, but they’re not saying something about their own knowledge state (as would be expressed by an indexical), compare to (3):
• X ≠ “us” [speaker + addressee]: Same reason. In (4), might ≠ compatible with our knowledge.2
2. Puzzle #2: Or-to-If Inference [Note: This plays a crucial role in the so-called direct argument for a non-truth-conditional view of indicative conditionals (see, e.g., Edgington 1986, 1995, Bennett 2003).] Murder mystery scenario: Sam has narrowed down the suspects to the cook or the butler. Thus Sam believes (and could appropriately assert) (5). (5)
[Context: Same as (1).] Sam: I can narrow down the suspects to the cook or the butler. Mary: # (No! / Nuh-uh!) I can narrow down the suspects to the butler or the maid.
*
I would like to thank Kai von Fintel, Irene Heim, Sarah Hulsey, and audiences at Yale and University of Maryland for useful comments and discussion on parts of this work.
The butler is the murderer or the cook is the murderer.
It follows from this that Sam also believes (and could appropriately assert) (6). (6)
If the butler isn’t the murderer, the cook is.
The standard conclusion: (5) entails (6). Generalizing: “not φ or ψ” entails “if φ then ψ.” Meaning that a conditional should be true whenever the antecedent is false or the consequent is true. 1
I’ve given one argument against the contextualist view that I prefer. Other arguments can be found in, e.g., Egan, Hawthorne & Weatherson (2005) and MacFarlane (2006). One might object that this could be a shifting indexical (Schlenker 2003, etc.), but there are two problems with this solution. First, the shifted interpretation of shifting indexicals is generally option, whereas the shifting in (4) is obligatory. Second, examples like (i) are problematic. (i) Sam and Mary disagree. Sam thinks that the murderer might be the cook, and Mary thinks that it can’t be. If the sentence Sam thinks that the murderer might be the cook means Sam thinks that it’s compatible with his own knowledge that the murderer is the cook (and similarly for Mary), then it should not be possible to report this as a disagreement.
2
(3)
2
3
Tamina Stephenson
Indicative Conditionals Have Relative Truth Conditions
[Recall: ¬φ or ψ = φ⊃ψ (the material conditional)] But we know this is wrong! (See, e.g., Bennett 2003, Edgington 2006 for recent summaries of the issues.)
(b)
[[must φ]]
w,j
= 1 iff ∀
∈Epistw,j: [[φ]]
w’,x
=1
Indicative conditionals: Assume that the if-clause restricts a silent epistemic must: (11) (a)
If φ (then) ψ.
(b)
ψ
Some examples of the absurd consequences: (Bennett 2003, Ch. 2): (7) (a)
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[MUST] [REPIST]
If I ate an egg for breakfast this morning, you ate a million eggs. [should be true provided that I didn’t eat an egg]
if φ
(Kratzer 1991, etc.) (b)
If there are no planets anywhere, the solar system has at least eight planets. [should be true since the solar system does have at least eight planets]
3. Relativist / “Judge-Dependent” Account 3.1. Preliminaries Express beliefs using doxastic alternatives (Lewis 1979, Chierchia 1989): (8)
Doxw,x = { : it is compatible with x’s beliefs in w that x (x’s self) is y in w’} [Read: the doxastic alternatives of x in w]
[Needed because people use their beliefs to place themselves within the population of possible individuals, real and imaginary, as well as within possible worlds; e.g., I believe that I’m in Chicago ≠ I believe that the person named Tamina Stephenson is in Chicago, cases of amnesia, etc.] Proposal: Also express knowledge using analogous notion of “Epistemic alternatives”: (9)
Epistw,x = { : it is compatible with x’s knowledge in w that x (x’s self) is y in w’} [Read: the doxastic alternatives of x in w]
(12)
[[if φ then ψ]]w,j = 1 ∀∈Epistw,j: [[φ]]w’,x = 1 ⊃ [[ψ]]w’,x = 1
Belief Reports: attitude predicates operate on the judge parameter: (13)
[[z thinks / believes (that) φ]]w,j = 1 iff ∀∈Doxw,z, [[φ]]w’,y = 1 ↑ ↑ ↑ ↑
Æ In effect: “z thinks that φ” = “z thinks that φ is true as judged by z.” 3.3. Pragmatics (Stephenson 2006, cf. also Egan to appear) Basis: Stalnakerian theory of conversation (see, e.g., Stalnaker 1978, 2002) (14)
Common ground: set of world-judge pairs (instead of worlds); Crucially: every j = the group of conversational participants
(15)
Assertion: proposal to restrict the common ground to world-judge pairs in which the proposition asserted is true.
Key difference: a person’s knowledge cannot rule out the actual individual that they are in the actual world in which they’re located, so Epistw,x must always include itself. [Factivity]
Thus: If A’s assertion that p is accepted, then in effect what is added to the common ground is that p is true as judged by the group of conversational participants.
(Similarly, any doxastic alternatives eliminated “unjustifiably” remain as epistemic alternatives.)
(16)
3.2. Semantics of epistemic modals and conditionals (Stephenson 2005, 2006, extending Lasersohn 2005) Add an individual “judge” parameter to the index, as Lasersohn (2005) proposes for fun/tasty So a proposition = a set of world-individual pairs (type )
⇒ Justified assertion: For speaker A to be justified in asserting a sentence S in w it must be the case that ∀∈Epistw,A: [[S]]w’,x = 1 i.e., A must know that S is true as judged by A (not the whole group)
Intuitive idea: people in conversation together are on a joint venture to place themselves, as a group, in the space of possible groups in different possible worlds, e.g.: • We’re a group of people in a world where it is (currently) raining.
[Note: I’ll be systematically disregarding times, as well as the role of context]
• We’re a group of people whose knowledge is compatible with the butler being the murderer.
Epistemic modals: involve what is known by the judge:
Assertion is an efficient way for an individual to help the group place itself: just propose “let’s establish that this is a group of people who…” and see if people object.
(10) (a)
[[might φ]]w,j = 1 iff ∃∈Epistw,j: [[φ]]w’,x = 1
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Tamina Stephenson
Indicative Conditionals Have Relative Truth Conditions
4. Solution to Puzzle #1 (Gibbardian Standoffs)
5. Solution to Puzzle #2 (Or-to-If Inference)
(17)
[Context: The murder investigation from (1)]
5.1. The issue
Sam: Mary:
(19) (a)
(a) (b)
If the butler isn’t the murderer, the cook is. (No! / Nuh-uh!) If the butler isn’t the murderer, the maid is.
Two issues:
(b)
• Sam and Mary seem be contradicting each other. • Both Sam and Mary are justified in making their assertions. 4.1. Why they are contradicting each other (18)
Propositions expressed in (17):
Observation: If A believes that not φ or ψ [“the disjunction”], then A believes that if φ then ψ [“the conditional”]. Standard Conclusion: the disjunction entails the conditional.
But we know that the standard conclusion is false! (A conditional is not necessarily true when the antecedent is false or the consequent is true.) So this looks like a paradox. My view: predicts that (19a) is true and (19b) is false. This isn’t a paradox because of an assumption implicit in (19): ⇒ Implicit assumption in (19): Entailments are preserved under belief; if believing p is sufficient for believing q, then p entails q.
(a)
(17a): {: ∀: ∈Episw,j & the butler is not the murderer in w’ ⊃ the cook is the murderer in w’}
(20)
(b)
(17b): {: ∀: ∈Episw,j & the butler is not the murderer in w’ ⊃ the maid is the murderer in w’}
(20) doesn’t hold in my system because of the way belief predicates manipulate the judge parameter.
These are disjoint sets provided that (i) there can only be one murderer and (ii)
the common ground does not include any pairs such that the butler is the murderer in all of j’s epistemic alternatives – that is, it is common ground that it has not been established that the butler is the murderer.
6
Need to show: • [Sec. 5.2] The disjunction does NOT entail the conditional. (“not φ or ψ” does not entail “if φ then ψ.”) • [Sec. 5.3] But: Belief of the disjunction DOES entail belief of the conditional. (“A believes not φ or ψ” entails “A believes that if φ then ψ.”) [What’s going on: The disjunction is weaker than the conditional, but they become equivalent under belief.]
[I take (ii) to be a presupposition of the conditional, as a special case of a general prohibition on vacuous satisfaction of universal quantifiers.]
5.2. The disjunction does not entail the conditional
4.2. Why they are justified
Propositions expressed by “not φ or ψ” and “if φ then ψ”:
By assumption: Their reasoning from the evidence was justified enough to constitute knowledge. So:
(21)
• ∀∈Epistw*,Sam: the butler or the cook is the murderer in w* (In all of Sam’s epistemic alternatives, either the butler or the cook is the murderer.)
(22)
• ∀∈Epistw*,Mary: the butler or the maid is the murderer in w* (In all of Mary’s epistemic alternatives, either the butler or the maid is the murderer.) [Assumption: the epistemic alternative relation is transitive.] So according to (16) (norm of justified assertion), both are justified in making their assertions.
not φ or ψ: {: [[φ]] = 0 or [[ψ]] = 1}
=S(21)
if φ then ψ: {: ∀∈Episw,j: [[φ]] = 1 ⊃ [[ψ]] = 1} = {: ∀∈Episw,j: [[φ]] = 0 or [[ψ]] = 1}
=S(22)
Show that S(21) ⊄ S(22) in general with a counterexample. (23)
[worlds w1–w4; one individual A; assume that φ and ψ are non-judge-dependent] w1: φ is true, ψ is true; A’s epistemic alternatives are {, }
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Tamina Stephenson
Indicative Conditionals Have Relative Truth Conditions
w2: φ is true, ψ is false; A’s epistemic alternatives are {, , , } w3: φ is false, ψ is true; A’s epistemic alternatives are {, , , } w4: φ is false, ψ is false; A’s epistemic alternatives are {, }
6.1. Summary I have given new solutions to two longstanding puzzles about conditionals by combining:
• Restrictor analysis of if-clauses.
(24)
“not φ or ψ”: {, , }
=S(24)
(25)
“if φ then ψ”: { , }
=S(25)
S(24) ⊄ S(25) because of 5.3. Belief of the disjunction entails belief of the conditional. Propositions expressed: (26)
A believes that not φ or ψ: {: ∀∈Doxw,A: [[φ]] = 0 or [[ψ]] = 1}
(27)
A believes that if φ then ψ: {: ∀∈Doxw,A : ∀∈Episw”,y: [[φ]] = 1 ⊃ [[ψ]] = 1} = {: ∀∈Doxw,A : ∀∈Episw”,y: [[φ]] = 0 or [[ψ]] = 1} Assume: to believe something is to believe that you know it, so the epistemic alternatives of A’s doxastic alternatives are the same as A’s doxastic alternatives, i.e., {: ∃ [∈Doxw,A and ∈Epistw’,x] } = Doxw,A (b)
6. Conclusion
• Relativist (judge-dependent) view of epistemic modals.
Intuitively: In w1–w4, A believes that φ and ψ have to either be both true or both false. In w1 and w4, A is right (and justified) about this, so A knows this. In w2 and w3, A is wrong about this, so A does not know this.
(a)
8
= {: ∀∈Doxw,A: [[φ]] = 0 or [[ψ]] = 1}
(26) = (27b), so the two are equivalent. Æ So in particular, (26) entails (27). 5.4. What this shows Given that believing p is be sufficient for believing q, it does not follow (in my system) that p entails q. Æ We can capture the or-to-if inference without making the disjunction stronger than the conditional. Also note: we might cut traditional logicians some slack for analyzing conditionals as ⊃, if believing one is the same as believing the other.
6.2. Lessons • Truth vs. assertability. (Gibbard examples may really be telling us something about assertion, not about conditionals.) • Entailments under beliefs: It may not be valid to conclude that p entails q from the fact that believing p is sufficient for believing q. Inference patterns don’t necessarily translate to entailments in the sense of subset relations between sets of indices.
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