Belief and certainty more

Belief and Certainty Dylan Dodd Northern Institute of Philosophy Department of Philosophy University of Aberdeen dylan.dodd@gmail.com Work in progress! February 16, 2012 1 Introduction People have full beliefs, and they have credences or degrees of belief. Full beliefs – more simply, beliefs – are all-or-nothing. One either believes in God or one doesn’t, in which case one is an atheist or an agnostic. But credences come in degrees. We are more or less confident in various possibilities – we regard them as more or less likely. Say I believe a proposition p. What follows, if anything, about what my credence in p is? In this paper, I’ll defend the following answer to this question: (Certainty) If a subject S believes a proposition p, then S’s credence in p is 1. (Certainty) is an unpopular thesis amongst contemporary epistemologists. For instance, the Lockean Thesis (LT) about belief is accepted by many. According to LT, one believes that p just in case one’s credence in p is higher than some threshold τ (τ < 1).1 LT is obviously incompatible with (Certainty). Many epistemologists reject LT because of what it requires us to say about the Lottery Paradox (Kyburg [19]). Imagine a fair lottery with a sufficient number of tickets so that one’s credence in each ticket Ti that it will lose is greater than τ . According to LT, one believes of each Ti that it LT is defended by Richard Foley [12] and [13], David Christensen [5], and Scott Sturgeon [36], amongst others. 1 1 will lose. Many epistemologists find it plausible that that if one believes p and also believes q, one believes p & q (call this thesis ‘(Closure)’). Say the lottery has n tickets, T1 − Tn . It follows from LT and (Closure) that one believes that T1 & . . . &Tn will all lose, which is absurd since one knows that some ticket will win.2 Defenders of LT respond to the Lottery Paradox by denying (Closure).3 But another response is to reject LT. According to John Pollock [27], Sharon Ryan [29], and Igor Douven [7], amongst many others, one believes that p just in case one’s credence in p is greater than a threshold τ (τ < 1) and one satisfies some further condition C. C is designed to insure, among other things, that one doesn’t believe of a particular lottery ticket that it will lose, and thus enable a solution to the Lottery Paradox that is consistent with (Clousre). Such theories of belief are designed so that their proponents may deny (Certainty) while retaining (Closure). Their defenders assume (Certainty) to be false. Finally, many epistemologists have claimed that it’s a rational requirement that one not have a credence of 1 in any contingent proposition, a thesis often called ‘regularity’.4 When combined with (Certainty), regularity has the absurd consequence that rational subjects have no beliefs about contingencies. Thus regularity represents yet another strand of contemporary epistemology opposed to the thesis I will be defending in this paper. On the other hand, one epistemologist who has defended (Certainty) for some time is Isaac Levi [20], [21].5 Here are the aspects of Levi’s picture of cognition that I like. Being in a doxastic state involves having doxastic commitments. Belief-that-p involves being committed to p’s truth, which in turn involves ruling out the possibility that not-p. But there will be There are different ways of articulating the Paradox. My principle (Closure) could be altered to a principle of about rational belief rather than just belief. Then the absurdity would be that one has a rational belief – or could have a rational belief if one bothered to infer the long conjunction – that T1 & . . . &Tn will all lose. Instead of a principle about believing conjunctions, one might instead claim that it’s not rational to believe each member of a set of propositions that one knows to be inconsistent (call this principle ‘(Consistency)’). One knows, and therefore rationally believes, that one of the lottery tickets will win. Thus one rationally believes not-(T1 will lose & . . . & Tn will lose). On the other hand, it follows from LT that one rationally believes that T1 will lose and rationally believes that T2 will lose and . . . and rationally believes that Tn will lose. Thus, one rationally believes each member of a set of propositions which we may suppose one knows to be inconsistent, violating (Consistency). 3 They also deny the analogue of (Closure), mentioned in the previous footnote, stated in terms of rational belief. And defenders of LT deny (Consistency). 4 For defences of regularity, see Shimony [30], Edwards, et al [10], Stalnaker [32], and Haverkamp, et al [16]. 5 Patrick Maher [25] too. Before them, (Certainty) was defended by Bruno de Finetti [6]. See Arlo-Costa [3]. 2 2 possibilities such that one neither believes nor disbelieves that they are true. Call these the ‘open possibilities.’ For instance, I don’t believe that Spain will repeat as World Cup champions in 2014. But I also don’t believe they won’t. I consider their winning it and their not winning it to both be open possibilities. In addition to one’s beliefs, one has opinions involving the open possibilities. For instance, I regard Spain’s winning the next World Cup and their not winning it to be open possibilities, but I also regard their winning it as more likely than Portugal’s winning the next World Cup (I also regard Portugal’s winning it as an open – though very unlikely – possibility). That is, I have a higher credence in the Spain-will-win possibility than I do in the Portugal-will-win possibility. One’s credence in p is a measure of how likely one regards p.6 In summary, there are a subject S’s beliefs. Then there are possibilities that S’s beliefs leave open. S has credences in at least some of these open possibilities. Like her beliefs, S’s having a credence in a proposition is a doxastic commitment. S’s credences are distributed only over the space of possibilities left open by her beliefs. In this paper I will assume that as an idealized rational believer, S’s credences conform to the probability calculus. Thus, if ω is the set of possibilities left open by S’s beliefs, S’s credence in ω is 1. It follows that S’s credence in each of her beliefs is 1 [= (Certainty)]. I will proceed as follows. In the next section (§2) I’ll give some initial evidence that (Certainty) is true. In §3 I offer an explanation as to why (Certainty) should be true. Finally, in §4 I respond to some objections to (Certainty), before concluding the paper in §5.7 2 Some initial evidence for (Certainty) Say I regard p as ‘probably’ true in the sense that my credence in p is greater than 0.5. I might express my credence in p by thinking or asserting the Just to be clear, having a doxastic commitment regarding how likely it is that p needn’t be construed as a belief about p’s probability. If the notion of probability in question is objective or evidential, I’m skeptical that Spain’s winning even has a probability. Nevertheless, I am of the opinion that Spain is one of the more likely countries to win the World Cup, and has a better chance of winning it than Portugal. When I say that credence is a doxastic commitment about how likely p is I mean this in the sense that one regards p as being more or less likely, not in the sense that one believes that p has a certain likelihood or probability. 7 It’s worth mentioning one argument I won’t be giving for (Certainty), even though I suspect it’s a good argument. Igor Douven and Timothy Williamson [8] offer a powerful argument that on any account of belief that endorses (Closure), (Certainty) will hold. Their argument is strengthened in Smith [31]. Obviously, these arguments provide a way for arguing from (Closure) as a premise to (Certainty) as conclusion. 6 3 sentence ‘Probably p’.8 I might think or assert such sentences as an aid to reasoning. For instance, I might reason as follows: ‘p is probably true. Since p or q follows from p, that means that probably p or q’. This is a good bit of reasoning. Given that my credence in p is greater than 0.5, I’m committed to having a credence in p or q that’s also greater than 0.5. If I regard p as probably true in the sense under discussion, but deny that p or q is probably true as well, my overall doxastic state is inconsistent. My credences violate the probability calculus. My main point here is to introduce a way of using the expression ‘Probably p’: it can be used by a subject to express the fact that her credence in p is greater than 0.5. The subject can use the judgment that ‘Probably p’ to see what other things she’s committed to in virtue of having this credence. Next we’ll examine how the doxastic commitments expressed in such ‘probably’ judgments combine with the commitments involved in other judgments.9 Consider the following three arguments. Assume their use of ‘probably’ is the one just mentioned, in which they express the speaker’s having a credence greater than 0.5 in a proposition, and assume that p and q aren’t logically equivalent.10 Argument 1 Argument 2 Premise 1 (P1): If p then q p or q Probably not-p Premise 2 (P2): Probably p Conclusion (C): Probably q Probably q Argument 3 p Probably q Probably (p & q) Let’s say that someone ‘accepts’ an instance of one of the premises or conclusions in one of these arguments iff (for their first premise (P1)) s/he believes the premise or (for their (P2) or (C)) she has the credence the premise/conclusion speaks of (s/he has a credence greater than 0.5 in the relevant proposition). I think it’s extremely plausible that for each of the Throughout this paper, please read my quotation marks as corner quotes where appropriate. 9 Some readers may favor an expressivist (non-truth conditional) account of the semantics of sentences involving the doxastic use of ‘probably’ I’m focusing on. Such readers will be attracted to expressivist accounts of epistemic modals generally no doubt. Such readers may object to my talk of a judgment that ‘Probably p’, and maybe also object to speaking of an assertion of ‘Probably p’. Here and in what follows, where I talk about judging or asserting sentences containing epistemic modals such as ‘Probably p’, feel free to substitute a different, expressivist-friendly locution for using these sentences (e.g., using the sentence on a particular occasion to express the speaker’s attitude). In spite of my use of words like ‘judging’ and ‘asserting’, in fact my arguments in this paper don’t presuppose that expressivist views about epistemic modals are false (or that they’re true). 10 Argument 1 was discussed by Yalcin [37]. 8 4 three arguments schemas, anyone who accepts its premises is committed to accepting its conclusion.11 Let’s say someone ‘rejects’ the conclusion (C) of one of Arguments 1-3 iff she has a credence in q (for Arguments 1 and 2) or (p & q) (for Argument 3) that is less than or equal to 0.5. I’m claiming that the following is very plausible: (Commitment) Where the ‘probably’ in Arguments 1-3 is used to express that the subject’s credence in the relevant proposition is greater than 0.5: In accepting Arguments 1-3’s premises, one is thereby committed to accepting their conclusions, on pain of inconsistency. In other words, the following set of attitudes is jointly inconsistent: acceptance of (P1) in one of Arguments 1-3, acceptance of (P2) in the same Argument, and rejection of (C) in the same Argument. The plausibility of (Commitment) is revealed when we consider some instances of the argument schemas 1-3. Let’s start with Argument 1. (In the following discussion, imagine that ‘probably’ is used to express that one’s credence is greater than 0.5 in the way explained above.) Imagine Watson says or thinks that if Mr. Smith wasn’t the murderer, then it was Mr. Jones [Argument 1, (P1), or ‘(P1)1 ’]. He also says or thinks that Mr. Smith probably wasn’t the murderer [(P2)1 ]. It seems he’s thereby committed to Mr. Jones probably being the murderer [(C)1 ]. Imagine that Watson actually said the two premises but then denied being thereby committed to the conclusion [‘I deny that it was probably Jones; nevertheless, if it wasn’t Smith it was Jones, and it probably wasn’t Smith’]. We would find what he said exceedingly strange. For surely, we’d respond, if Watson thinks those premises are true, he must conclude that it probably was Jones. He’s being inconsistent. Similarly for Arguments 2 and 3. Say Watson says or thinks that the murderer was either Smith or Jones [(P1)2 ], and then says or thinks that it probably wasn’t Smith [(P2)2 ]. It seems he’s thereby committed to its probably being Jones [(C)2 ]. Were Watson to deny commitment to this conclusion, while affirming the premises [‘It’s not the case that it was probably Jones. Nevertheless, it was either Smith or Jones and probably wasn’t Smith’], we would find what Watson said odd, to say the least. Finally, consider an instance of Argument 3. Imagine someone says or thinks that the Democrats For those who want to analyze these arguments in terms of dynamic logic, where the order of the premises can affect an argument’s validity, this claim is only true if (P1) and (P2) occur in the order in which I present them (i.e., with (P1) first). I’m grateful to Hans Kamp and Johan van Benthem for drawing my attention to this point. I wish to remain neutral on the question of whether these arguments should be analyzed dynamically. 11 5 will have control of the Senate in 2013 [(P1)3 ] while also saying or thinking that the Democrats will probably have control of the House in 2013 [(P2)3 ]. It seems that that person is thereby committed to thinking that probably, the Democrats will have control of both the Senate and the House in 2013 [(C)3 ]. Were someone to deny the conclusion while asserting both premises [‘The Democrats will have control of the Senate, and they’ll probably have control of the House. But I deny that they’ll probably have control of both the House and the Senate’], it would sound to us like this person was contradicting himself. (Commitment) is plausible. As I’ll now show, we can only explain why (Commitment) is true if we assume that in believing/accepting their first premise (P1), the subject has a credence of 1 in that premise. This fact is predicted by (Certainty), the thesis that belief generally involves having a credence of 1. It’s mysterious from the point of view that (Certainty) is false. First I’ll show this for Arguments 2 and 3. I’ll discuss Argument 1 last. 2.1 Argument 2 as Evidence for (Certainty) Why does (Commitment) hold for Argument 2? Our question is: Why is accepting (P1)2 and (P2)2 inconsistent with rejecting (C)2 ? I’ll now show that (Commitment) holds for Argument 1 iff accepting/believing (P1)2 means having a credence of 1 in (P1)2 . Say (P1)2 and (P2)2 are both accepted. To accept (P2)2 is to have a credence greater than 0.5 in not-p (P (¬p) > 0.5). It follows that P (¬p&q) + P (¬p&¬q) > 0.5. First suppose that accepting (P1)2 means having a credence of 1 in (P1)2 : P (p ∨ q) = 1. Then P (¬p&¬q) = 0. Since the acceptance of (P2)2 is equivalent to P (¬p&q) + P (¬p&¬q) > 0.5, it follows that P (¬p&q) > 0.5, and thus that P (q) > 0.5, which means that (C)2 is accepted. What we just saw is that if (P1)2 and (P2)2 are both accepted, and if accepting (P1)2 means having a credence of 1 in (P1)2 , it follows that (C)2 is accepted. (Commitment) holds for Argument 2 if accepting (P1)2 means having a credence of 1 in (P1)2 . Suppose instead that although (P1)2 is accepted, P (p ∨ q) < 1, or equivalently, P (¬p&¬q) > 0. Given this supposition, it’s consistent with the fact that P (¬p&q) + P (¬p&¬q) > 0.5 (i.e., the fact that P (¬p) > 0.5) that P (¬p&q) < 0.5. In short, given that both (P1)2 and (P2)2 are accepted, it’s possible that P (¬p&q) < 0.5. It’s also consistent with accepting (P1)2 and (P2)2 that P (p&q) is as low as we like, even as low as 0. Take accepting (P1)1 . It’s consistent with believing p ∨ q that that one is very confident 6 (and even certain) that p&q is false. I believe a Democrat or a Republican will win the next US Presidential election, but I’m certain it won’t be the case that a Democrat and a Republican both win.12 Now consider accepting (P2)2 (i.e., P (¬p) > 0.5). I take it to be obvious that one can regard not-p as probably true, while having an extremely low confidence in p&q, and even be certain that p&q is false. So we’ve seen that it’s consistent with accepting both (P1)2 and (P2)2 that P (¬p&q) < 0.5, and also that P (p&q) is as low as we like, even as low as 0. It’s also consistent with accepting both (P1)2 and (P2)2 that P (¬p&q) < 0.5 and P (p&q) = 0 at the same time. Therefore, it’s consistent with accepting both (P1)2 and (P2)2 that P (¬p&q) + P (p&q) < 0.5. Since P (q) = P (¬p&q) + P (p&q), it’s consistent with accepting both (P1)2 and (P2)2 that P (q) < 0.5. Therefore, one may consistently accep both (P1)2 and (P2)2 while rejecting (C)2 . Let’s review. First we saw that if accepting/believing (P1)2 means having a credence of 1 in (P1)2 , then accepting (P1)2 and accepting (P2)2 entails (C)2 . Next we saw that if one can accept/believe (P1)2 while having a credence of less than 1 in (P1)2 , then one can consistently accept both (P1)2 and (P2)2 while rejecting (C)2 . Therefore, (Commitment) for Argument 2 is true iff accepting (P1)2 requires having a credence of 1 in (P1)2 . 2.2 Argument 3 as Evidence for (Certainty) Recall that (P1)3 is p, (P2)3 is P (q) > 0.5 [‘Probably q’] and (C)3 is P (p&q) > 0.5 [‘Probably (p & q)’]. P (q) = P (¬p&q) + P (p&q). If P (p) = 1, then P (¬p&q) = 0, and then P (q) = P (p&q). Therefore, if P (q) > 0.5 and P (p) = 1, then P (p&q) > 0.5. Thus, if accepting/believing (P1)3 means having a credence of 1 (P1)3 , then accepting (P1)3 and (P2)3 entails (C)3 . One cannot consistently accept both (P1)3 and (P2)3 while rejecting (C)3 . On the other hand, imagine that P (p) < 1 – equivalently, that P (¬p) = P (¬p&q)+P (¬p&¬q) > 0 – while at the same time both (P1)3 and (P2)3 are accepted. If we’re imagining a genuine possibility, then there’s nothing in this possible situation from disallowing it from also being true that P (¬p&q) > 0. So imagine it’s also true that P (¬p&q) > 0. But if P (¬p&q) > 0, then the fact that (P2)3 is accepted – i.e., the fact that P (q) = P (¬p&q) + P (p&q) > 0.5 – will be consistent with P (p&q) < 0.5. Therefore, if we suppose that it’s possible to believe p (i.e., accept (P1)3 ) while at the same time P (p) < 1, then one can consistently accept (P1)3 and (P2)3 while rejecting (C)3 . Take ‘winning’ an election here to imply that one is actually deemed the winner and takes office. So understood, there can only be one winner. 12 7 We just saw that if accepting/believing (P1)3 requires having a credence of 1 in (P1)3 , then one cannot consistently accept (P1)3 and (P2)3 while rejecting (C)3 . But if one can believe/accept (P1)3 while having a credence below 1 in (P1)3 , then one can consistently accept (P1)3 and (P2)3 while rejecting (C)3 . In short, (Commitment) holds for Argument 3 iff accepting/believing (P1)3 means having a credence of 1 in (P1)3 . 2.3 Argument 1 as Evidence for (Certainty) Recall that (P1)1 is if p then q (p → q), (P2)1 is P (p) > 0.5 (‘Probably p’), and (C)1 is P (q) > 0.5 (‘Probably q’). For Arguments 2 and 3 I argued directly that (Commitment) was true for them iff in accepting/believing each argument’s (P1), one has a credence of 1 in (P1). The problem with arguing in the analogous way for Argument 1 is that in doing so I would have to make an assumption about what the probability of (P1)1 is. But (P1)1 is an indicative conditional, and it’s controversial what the probability of an indicative conditional is. What is the probability of an indicative conditional p → q? We might take indicative conditionals just to be material conditionals (Jackson [17], Rieger [28]). In that case, Argument 1 becomes logically equivalent to Argument 2. Then my argument in §2.1 regarding Argument 2 also shows that (Commitment) is true for Argument 1 iff accepting/believing (P1)1 requires having a credence of 1 in (P1)1 . While the thesis that indicative conditionals are material conditionals isn’t a popular idea amongst contemporary philosophers and linguists,13 it is widely accepted that indicative conditionals entail material conditionals. The plausibility of this thesis follows from the plausibility of the claim that one can refute p → q by showing that p is true and q false. The only philosophers who I can imagine denying that material conditionals entail indicative conditionals are those who think that indicative conditionals don’t have truth conditions, and thus strictly speaking don’t entail anything (I’m assuming here that entailment involves necessary truthpreservation). Another popular and plausible idea is “Stalnaker’s Thesis”: the probability of an indicative conditional p → q is the probability of q conditional on p (Adams [2], Jeffrey [18], Edgington [9], Stalnaker [32]). Lewis [22] is often taken to have shown that Stalnaker’s Thesis can’t be maintained if indicative conditionals have truth-conditions. Those who deny that indicative conditionals have truth conditions typically are motivated by a desire 13 For a nice summary of many of the objections to the material conditional analysis of indicative conditionals, see Abbott [1]. 8 to hang on to Stalnaker’s Thesis. I know of no contemporary philosophers or linguists who deny that indicative conditionals have truth conditions and also deny Stalnaker’s Thesis. So everyone will maintain one of two things about indicative conditionals: (i) they have truth-conditions and p → q entails p ⊃ q; (ii) they don’t have truth conditions and Stalnaker’s Thesis is true. Say I have a credence of 1 in p → q. If (i) is true, then I have a credence of 1 in all of p → q’s entailments.14 Given that indicative conditionals entail material conditionals, one of p → q’s entailments will be p ⊃ q. In that case we can at least test whether one has a credence of 1 in p → q by checking to see if one has a credence of 1 in p ⊃ q. If we accept (ii) instead, then P (p → q) = 1 iff 1 = P (q|p) = P (q&p)/P (p) (if P (p) > 0). First, I’ll argue that on the assumption of Stalnaker’s Thesis, (Commitment) is true for Argument 1 iff P (q|p) = 1. Then we’ll see what happens if indicative conditionals entail material conditionals. Let’s assume Stalnaker’s Thesis. Now suppose that accepting/believing (P1)1 means having a credence of 1 in (P1)1 : P (p → q) = 1. By Stalnaker’s Thesis, this means that P (q|p) = P (q&p)/P (p) = 1. This entails that P (q&p) = P (p). Therefore, if P (p) > 0.5, then P (q&p) > 0.5, which entails that P (q) > 0.5. Therefore, if accepting/believing (P1)1 means having a credence of 1 in (P1)1 , then accepting (P1)1 and (P2)1 entails (C)1 – one cannot consistently accept (P1)1 and (P2)1 and also deny (C)1 . Maintaining our assumption of Stalnaker’s Thesis, let’s instead suppose that (P1)1 is accepted/believed while at the same time one’s credence in (P1)1 is less than 1: P (p → q) = P (q|p) = P (q&p)/P (p) < 1. In that case, P (q&p) < P (p). Now say that (P2)1 is accepted (P (p) > 0.5). Our assumptions so far are consistent with P (q&¬p) having as small a value as we like, even as small as 0. In other words, it’s consistent with believing p → q and regarding p as probably true that one is certain (and as close to certain as we like) that q & not-p is false.15 Since it’s consistent with our assumptions, let’s say P (q&¬p) = 0. P (q) = P (q&p) + P (q&¬p), but now that we’re assuming P (q&¬p) = 0, P (q) = P (q&p). Previously we saw that Here I’m crucially assuming that my credences conform to the probability calculus, from which it follows that I have a credence of 1 in every logical consequence of something I have a credence of 1 in. As usual, this assumption should be thought of as an idealization. 15 Example: imagine someone who believes that every candidates for the next US Presidential election is too conservative. This person thinks that whoever the President will be next year, that person will be too conservative. This person is also certain that Dick Cheney isn’t going to win the next election. Let p be <Dick Cheney will not be the President next year>, and q be <the person who is President next year will be too conservative>. We can imagine that this person believes p → q, regards p as probably true, and is certain that q & not-p is false (since she’s certain p is true and therefore that not-p is false). 14 9 it’s a consequence of our supposition that P (q|p) < 1 that P (q&p) < P (p). So we now have P (q) < P (p). Thus, even though P (p) > 0.5, it’s possible that P (q) ≤ 0.5. Now we see that, given our supposition that P (p → q) = P (q|p) = P (q&p)/P (p) < 1, accepting (P1)1 and (P2)2 is consistent with rejecting (C)1 . What I just showed is that if we assume Stalnaker’s Thesis, then accepting (P1)1 and (P2)2 is inconsistent with rejecting (C)1 iff accepting (P1)1 requires having a credence of 1 in (P1)1 . But I also mentioned that Stalnaker’s Thesis is controversial and is widely regarded as entailing that indicative conditionals lack truth conditions. As I mentioned previously, all those who reject Stalnaker’s Thesis will think that p → q entails p ⊃ q. Let’s now consider that option. P (p ⊃ q) = 1 iff P (p&¬q) = 0. Note that it’s also true that P (q|p) = 1 iff P (p&¬q) = 0.16 In short, P (q|p) = 1 iff P (p ⊃ q) = 1. Earlier we saw that if accepting/believing (P1)1 requires P (q|p) = 1, then accepting (P1)1 and (P2)1 will be inconsistent with denying (C)1 . Now we see that it follows from this conclusion that if accepting/believing (P1)1 requires P (p ⊃ q) = 1, then accepting (P1)1 and (P2)1 will be inconsistent with denying (C)1 . Thus (Commitment) holds for Argument 1 iff accepting/believing (P1)1 requires P (p ⊃ q) = 1. Assuming that p → q entails p ⊃ q, the fact that accepting/believing (P1)1 requires P (p ⊃ q) = 1 is predicted by the thesis that believing/accepting p → q requires P (p → q) = 1 (since if P (P ) = 1, then for every Q that P entails, P (Q) = 1).17 But why would accepting/believing (P1)1 require P (p ⊃ q) = 1 if believing/acceting (P1)1 did not require P (p → q) = 1? It’s hard to see why it would. While the conclusion of the previous paragraph is predicted by the thesis that believing/acceting (P1)1 requires P (p → q) = 1, it’s mysterious from the point of view that believing/accepting (P1)1 does not have this requirement. Call the thesis that (Commmitment) holds for Argument 1 iff one’s credence in (P1)1 is 1 ‘(*)’. Arguing for (*) is more problematic than arguing for its analogues for Arguments 2 and 3, because in order to argue for (*), we have to make some assumptions about the probability of an indicative conditional. Nevertheless, we’ve at least gotten some good evidence for (*). Everyone is willing to grant either that p → q entails p ⊃ q, or else that Stalnaker’s Thesis is true. If Stalnaker’s Thesis holds, then we’ve seen that (*) is true. On the other hand, (Commitment) holds for Argument 1 iff P (q|p) = 1 iff P (q&p) = P (p), and P (q&p) = P (p) iff P (p&¬q) = 0. Again note the idealization. I’m here assuming that one’s credences conform to the probability calculus, which requires logical omniscience. 17 16 10 accepting/believing (P1)1 requires P (p ⊃ q) = 1. If we assume that p → q entails p ⊃ q, then this fact is predicted by (*), but it’s otherwise mysterious. 2.4 Arguments 1 - 3 and (Certainty) Recall the following highly plausible claim about Arguments 1-3: (Commitment) Where the ‘probably’ in Arguments 1-3 is used to express that the subject’s credence in the relevant proposition is greater than 0.5: In being committed to Arguments 1-3’s premises, one is thereby committed to their conclusions, on pain of inconsistency. In other words, when someone believes Argument 1-3’s (P1) and is committed to (P2) in the sense that s/he has a credence greater than 0.5 in the relevant proposition, if s/he fails to be additionally committed to (C) (in the sense of having a credence greater than 0.5 in the relevant proposition), her overall doxastic state is inconsistent. What we saw in §§2.1-2.2 is that (Commitment) is only true for Arguments 2 and 3 iff in accepting/believing their first premise (P1) one has a credence of 1 in (P1). Making the analogous claim for Argument 1 is trickier because of the difficulty of thinking about the probabilities of indicative conditionals. Nevertheless, in §2.3 I showed that if we accept Stalnaker’s Thesis [P (p → q) = P (q|p)], then, as with Arguments 2 and 3, (Commitment) is true for Argument 1 iff accepting/believing Argument 1’s first premise (P1)1 requires having a credence of 1 in (P1)1 . Furthermore, (Commitment) holds for Argument 1 iff accepting/believing (P1)1 means P (p ⊃ q) = 1, a fact predicted by the thesis that believing/accepting (P1)1 requires having a credence of 1 in (P1)1 (assuming that p → q entails p ⊃ q), but which is otherwise mysterious. (Commitment) is very plausible. So we have evidence that believing (P1)1 , (P1)2 , and believing (P1)3 , requires having a credence of 1 in each of these propositions. Why would this be true? The simplest explanation is given by (Certainty): in general, believing p requires having a credence of 1 in p. Deniers of (Certainty) who accept (Commitment) may instead claim the following: • For some reason, when one accepts (P1) and (P2) in Arguments 1-3, believing (P1) requires having a credence of 1 in (P1). However, it’s not generally true that believing a proposition requires having a credence of 1 in it. 11 This response raises a difficult question. Why does a subject’s believing (P1) require having a credence of 1 in (P1) when s/he also accepts (P2), if belief doesn’t generally require having a credence of 1? I don’t know how the denier of (Certainty) will be able to answer this question. Thus, the plausibility of (Commitment) poses a problem for the denier of (Certainty), and provides evidence that (Certainty) is true.18,19 The plausibility of (Commitment) gives us a reason to think (Certainty) is true. But why should (Certainty) be true? Why should believing p require having a credence of 1 in p? I think there’s a straightforward explanation. I said some suggestive things about what that explanation is in the Introduction. Next, in §3, I will say more about why (Certainty) is true. 3 Why (Certainty) is true I just presented some evidence for (Certainty). But why would (Certainty) be true? What is it about belief, or about credence, that would make it the case that believing that p requires having a credence of 1 in p? I think there are two facts, one about belief and one about credence, which together entail (Certainty). The first fact, which I’ll explain and defend in §3.1, is that belief is doxastic necessity: if one believes that p, one’s perspective or doxastic state doesn’t leave open the possibility that not-p. The second fact, which I’ll explain and defend in §3.2, is that one only has a positive credence in propositions that are doxastically possible for one. Those who deny (Certainty) also may reject (Commitment). However, (Commitment) is plausible. Since (Commitment) at least seems plausible, those who deny it in order to deny (Certainty) should explain the source of the illusion. Again, I don’t know how such an explanation would go. 19 Arguments 1-3 are examples of a general phenomenon. I leave it as an exercise for the reader to show that for the following three arguments, the analogue of (Commitment) is true, which like Arguments 1-3 can only be explained by assuming that accepting/believing their first premises (P1) means having a credence of 1 in (P1). Premise 1 (P1): Premise 2 (P2): Conclusion (C): Argument 1 If p then q There’s some probability that p There’s some probability that q Argument 2 p or q There’s some probability that not-p There’s some probability that q 18 Premise 1 (P1): Premise 2 (P2): Conclusion (C): Argument 3 p There’s some probability that q There’s some probability that (p & q) 12 3.1 Belief as doxastic necessity Let’s start by considering Stalnaker [34], [33]’s influential theory of the interaction of assertion and context. A conversational context C is represented as a set of presuppositions which the speaker and audience share, and these presuppositions leave open a set of possible worlds – the set of worlds compatible with the presuppositions. These are the live possibilities for the conversation. When a sentence is asserted in C that expresses the proposition p and the conversational participants accept the assertion, they update the context by removing all the not-p possibilities from from the set of live possibilities.20 Note that after a proposition p is accepted by the conversational participants, all the open possibilities in C will be p-possibilities. In this sense, every accepted proposition is necessary. One benefit of claiming that every proposition that is accepted by the conversational participants is necessary (in the sense explained above) is that then we can explain the infelicity of utterances of the following form (where the right conjunct is used to express the fact that the speaker regards not-p as an open possibility): (Moore) p but possibly not-p. The explanation is this. We update on the left conjunct by ruling out the not-p possibilities. However, the right conjunct is only assertible insofar as the possibility that not-p remains an open one for us. Conversational participants cannot accept both conjuncts in (Moore). (Note the following assumption of the above explanation of (Moore)’s infelicity: the “live” or “open” possibilities posited by Stalnaker’s theory can be referred to by normal English speakers using modal terms like ‘possibly’ and ‘might’. One thing we can mean by ‘p is possible’ or ‘p might be the case’ is true or appropriate21 just in case the Stalnakerian context set contains p-worlds.) Plausibly, an assertion is an outward expression of a belief (Frege [14]). If it is, then it’s plausible that what Stalnaker says about modeling the acceptance of a proposition by participants in a conversation can be extended to a way of modeling the acceptance of a proposition by an individual thinker. In short, it’s plausible that the modeling technique can be applied to beliefs. To come to believe p, we update the set of possibilities the subject regards as open by deleting all the possibilities in which not-p. When S comes to They will update by eliminating other possibilities too, like the ones ruled out by conversational implicatures. I will be ignoring such complications here. 21 I say “or appropriate” to leave open the possibility of an expressivist treatment of these sentences according to which they lack truth-conditions. 20 13 believe that p, S no longer regards any not-p-possibilities as open. In this sense, every proposition believed by S is necessary for S. More precisely, I’m proposing that the following is true: Belief is doxastic necessity (BIDN): If S believes that p, then p is doxastically necessary for S. A positive consequence of (BIDN) is that it enables an explanation of why someone cannot accept a proposition of the form of (Moore), where accepting the right conjunct means regarding not-p as an open possibility. The explanation is this. In order to believe (Moore)’s left conjunct, one must rule out the possibility that not-p, which is incompatible with accepting (Moore)’s right conjunct. So (BIDN) provides a natural explanation of our inability to accept (Moore). Here’s something else (BIDN) can explain. Consider the following arguments, where the ‘possibly P ’ means that P is doxastically possible for the subject: Argument 4 Argument 5 Premise 1 (P1): If p then q p or q Possibly not-p Premise 2 (P2): Possibly p Conclusion (C): Possibly q Possibly q Argument 6 p Possibly q Possibly (p and q) Previously I proposed that (Commitment) was true: for Arguments 1-3, anyone who accepts their premises while rejecting their conclusions is doxastically inconsistent. I think the same thing is true for Arguments 4-6:22 (Commitment3 ) Where the ‘possibly’ in Arguments 4-6 is used to express the fact that the relevant proposition is an open possibility one: In accepting Argument 4-6’s premises, one is thereby committed to accepting conclusions, on pain of inconsistency. In other words, the following set of attitudes is jointly inconsistent: acceptance of (P1) in one Argument 4-6, acceptance of (P2) in the same Argument, and rejection of (C) in the same Argument. As with Arguments 1-3, if we analyze these arguments using dynamic logic, this claim is true only if the arguments occur in the order I presented them in (with (P1) occurring first). Thanks again to Hans Kamp and Johan van Benthem for drawing my attention to this fact. 22 14 The plausibility of (Commitment3 ) is revealed when we consider instances of it. Take Argument 4. I think to myself that if Mustard had access to the key to Jones’ study, then he was Jones’ murderer. Next I think that he might have had access to the key in the sense that I can’t rule out that possibility. It seems I should conclude that Mustard might be the murderer – apparently I can’t rule that out either. Let’s try Argument 5. I think to myself that my wife is either at home or at our daughter’s school picking her up. Then I think that she might not be at home. I’m committed to the thought that she might be picking our daughter up from school. Finally, Argument 6. I believe that Mustard had a violent temper. Then it occurs to me that he might have had a motive to kill Jones. I should conclude that it might be that he both had a violent temper and a motive to kill Jones. (Commitment3 ) is really plausible. I’m going to assume it’s true. What would explain its truth? None of Arguments 4-6 are valid in normal modal logics (e.g., the modal systems K, B, S4, or S5). But if one prefixes a necessity operator to the arguments’ first premise (P1), with wide scope over the whole premise, the arguments all become valid. Thus, if in believing the arguments’ (P1) one rules out the possibility that not-(P1) – if in believing (P1), (P1) becomes doxastically necessary for one – then we can explain why (Commitment3 ) is true. In other words, (BIDN) enables us to explain (Commitment3 ). There may be other ways to explain (Commitment3 ), but I can’t think of any. The ability of (BIDN) to explain (Commitment3 ) provides us with a good reason to think (BIDN) is true. I think (BIDN) is true. In this subsection, I’ve given three reasons to accept it. First, (BIDN) follows if we generalize Stalnaker’s framework for modeling assertion and context to belief. It’s natural to apply this framework to belief, especially if we see assertion as an outward expression of belief. Second, (BIDN) can explain why one is unable to accept propositions of the form of (Moore), where (Moore)’s right conjunct is used to express that not-p is doxastically possible for one. Third, we just saw that (BIDN) can explain (Commitment3 ), and I don’t know how else to explain it. 3.2 Probability, credence, and possibility According to contemporary physics, there is a non-0 physical probability that the egg I just dropped on the floor will suddenly jump back up into my hand and be whole again, as if I never had dropped it. Although such an event has an extremely small probability, its probability is greater than 0. It follows that the splattered egg might jump back into my hand and reconstitute itself. Given that there is a small but non-0 physical probability that the broken 15 egg will spontaneously reconstitute itself, it’s physically possible that the egg will do that.23 The egg example illustrates the following fact: if something has a chance of being true – no matter how small of a chance – then it might be true. Probabilities are distributed over possibilities. The physics example illustrates the general point for physical possibility and probability. But the same thing is true for other kinds of probability and possibility. Although I am extremely confident that Ron Paul won’t be the next President of the United States, if I think there’s at least some probability that he will be, then strictly speaking I think he might be. In general: Credence doxastic possibility link (CDPL): If S has a positive credence in p, then p is doxastically possible for S. The thought behind (CDPL) is that if I give some credence in p, then I haven’t totally ruled out the possibility that p is (or will be) true. I hope that this thought and thus (CDPL) will strike the reader as obvious. Just in case it doesn’t, I will offer a couple of considerations in its favor. First, (CDPL) provides a natural interpretation of how credences are standardly modeled. Contemporary probability theorists and formal epistemologists typically have probabilities distributed over an algebra over a set of possible worlds; in short, probabilities are distributed over possibilities. When credences or subjective probabilities are being modeled, what sort of possibilities should we take the possibilities they’re distributed over to be? Given that credences are a feature of the subject’s doxastic state, the most natural answer seems to be that they’re doxastic possibilities. If we interpret the model in this way, we’re assuming that credences are distributed over doxastic possibilities. In other words, we’re assuming (CDPL). Here’s another consideration in favor of (CDPL). We’ve seen that plausibly, commitment to the premises of Arguments 1-3 commits one to their conclusions, on pain of inconsistency (Commitment). And the same goes for Arguments 4-6 (Commitment3 ). If we look at Arguments 1-3 and 4-6, we notice that they’re very similar. The only difference is that where we find the word ‘probably’ in 1-3, we find ‘possibly’ in 4-6. The fact that 1-3 and 4-6 are so similar makes one think that there should be a uniform explanation of (Commitment) and (Commitment3 ). This leads to a second reason to accept (CDPL): accepting (CDPL) enables us to give a uniform explanation of (Commitment) and (Commitment3 ), as I will now explain. Note that (P1) is the same premise in Arguments 4-6 as it is in Arguments 1-3. In §2 I showed that in order to explain (Commitment), we need to posit 23 The example is taken from Strevens [35]. 16 that in believing Argument 1-3’s first premise (P1) (= Argument 1-6’s (P1)), one has a credence of 1 in (P1). Then in §3.2 we saw that in order to explain (Commitment3 ), we must posit that in believing (P1), (P1) is doxastically necessary for one. It’s a consequence of (CDPL) that if p is doxastically necessary for S, then S has a credence of 1 in p.24 Thus, if we accept (CDPL), the thing required to explain (Commitment3 ) will also explain (Commitment). To repeat, what’s needed to explain (Commitment3 ) is positing that in believing (P1) it’s doxastically necessary for one that (P1). We’ve just seen that if we accept (CDPL) it follows that in believing (P1) one has a credence of 1 in (P1), which, remember, is the assumption we must posit in order to explain (Commitment). Therefore, if we accept (CDPL), we’re able to give a unified explanation of both (Commitment) and (Commitment3 ). (Commitment) and (Commitment3 ) both follow from (CDPL) conjoined with the assumption that in believing Argument 1-6’s (P1), (P1) is doxastically necessary for one. In summary, there are three things to be said in favor of (CDPL). First, at least for me it just seems to be obviously true. Second, it allows for a natural interpretation of how credences are normally modeled by contemporary probability theorists and formal epistemologists. Third, given the similarity of Arguments 1-3 and 4-6, we should expect there to be a unified explanation of (Commitment) and (Commitment3 ). We’ve seen that we can give this explanation if we accept (CDPL). If (CDPL) is true, then the fact needed to explain (Commitment3 ) is also able to explain (Commitment). But I don’t know how we could give a unified explanation of (Commitment) and (Commitment3 ) without accepting (CDPL). 3.3 Putting (BIDN) and (CDPL) together In §3.1 I argued for (BIDN): If S believes that p, then p is doxastically necessary for S. In §3.2 I argued for (CDPL): If S has a positive credence in p, then p is doxastically possible for S. I just explained at the end of §3.2 how it’s a consequence of (CDPL) that if p is doxastically necessary for S, then S’s credence in p is 1. Therefore, it’s a consequence of the conjunction of (BIDN) and (CDPL) that if S believes p, then S’s credence in p is 1. In other words, (BIDN) and (CDPL) jointly entail (Certainty). In §2 I gave some evidence for (Certainty). My goal in this section was to give an explanation as to why (Certainty) should be true, and in the For according to (CDPL), credences are distributed only over doxastic possibilities. So if p is doxastically necessary, S has all of her credence in p and no credence in not-p. Assuming that S’s credences all add up to 1 (as is required by the probability calculus), her credence in p will be 1. 24 17 process further motivate it. Why is (Certainty) true? Because (BIDN) and (CDPL) are. 4 Objections to (Certainty) So far I’ve tried to provide reasons for thinking that (Certainty) is true. But just because there are things to be said in favor of (Certainty) doesn’t mean we should accept it. Maybe there are objections to (Certainty) that are more powerful than the reasons I’ve offered for accepting it. In this section I will respond to three objections to (Certainty). The first two both claim, in different ways, that (Certainty) is psychologically implausible. The first objection contends that I don’t have the betting behavior with respect to my beliefs that (Certainty) predicts. The second objection claims that my credences in my beliefs can’t always be 1 given certain things I know about objective chances. When I’ve just dropped the egg and it’s broken on the floor, I believe it will remain on the floor until I or someone else cleans it up. But as previously noted, according to quantum mechanics there’s a tiny objective chance that the egg will reconstitute itself and jump back up into my hands. Given that I know something about such quantum indeterminacy, my credence that the egg will remain on the floor must be less than 1, strictly speaking, the second objection claims. The third and final objection I’ll discuss objects to (Certainty) on epistemological grounds. It alleges that if (Certainty) were true, then it would never be rational to give up a belief in response to evidence. But sometimes this is the rational response to acquiring new evidence. Therefore, (Certainty) must be false. I can’t definitively refute any of these objections. What I will do is explain why I think they’re inconclusive. 4.1 Objections 1 and 2: (Certainty) is psychologically implausible Traditionally, credences have been cashed out in terms of betting behavior, or in terms of what bets one would regard as fair. I have a credence of n in p only if I would regard a bet on p as fair which paid me $1-n if p, and which I paid $n if not-p. Accordingly, I have a credence of 1 in p only if I would regard a bet as fair that paid me nothing if p, and which I paid any amount you like if not-p! But surely, the first objection points out, I have many beliefs where I would not regard any bet you like on what I believe as fair. I agree, we do have many beliefs like that. The problem with this 18 objection lies in cashing out our credences in terms of what bets we would regard as fair. There are good reasons for thinking that sometimes betting behavior (or what bets I would regard as fair) isn’t even a good measure of credences (Eriksson, et al [11]). One problem is that being presented with a bet may change one’s credences. With regard to (Certainty), maybe I do have a credence of 1 in p (in virtue of believing p), but if you talk to me about a bet on p at extreme odds, that will cause me to rethink how confident I should be in p, and as a result my credence in p will drop to less than 1. For instance, being presented with such a bet may lead me to think, for the first time, of the fact that there are unlikely not-p possibilities that I really can’t rule out. If this would happen frequently in cases where I have a credence of 1, then betting behavior (or what bets I would regard as fair) will be systematically unreliable in measuring my credence in p when it’s 1. So my response to the first objection is that it’s at least possible that being presented with a bet on p at extreme odds might change my credence in p from 1 to less than 1. Therefore, it’s not obvious that the fact that if I were presented with such a bet on p I wouldn’t regard it as fair and wouldn’t take it shows that my credence now, in the absence of being presented with such a bet, is less than 1. My respond to the second objection to (Certainty) is similar. Recall that the second objection says that my knowledge of certain objective chances means that I don’t have a credence of 1 in all my beliefs. For instance, I believe that the egg that I just dropped on the floor will remain there broken until someone cleans it up. Yet my knowledge entails that there’s a tiny chance that it won’t. According to quantum mechanics there’s a non-0 objective probability that the egg will suddenly reconstitute itself and jump back up into my hands. Assuming that my credences should reflect my knowledge of objective probabilities (Miller [26], Lewis [23]), given this knowledge, my credence that the egg will remain broken on the floor either is or should be slightly less than 1. Nevertheless, I do believe that it will remain broken on the floor. Here is my response to this objection.25 Facts such as that there’s a small chance the egg will suddenly reconstitute itself are almost never psychologically salient to me. We go about our day-to-day lives happily ignoring such remote possibilities. Once it’s brought to my attention that there’s a small chance that the broken egg will reconstitute itself, I may adjust my credences accordingly. I may respond by coming to have a slightly-less-than-1 credence that the broken egg will remain broken on the floor until someone cleans it up. According to (Certainty), once 25 I will ignore the response that we have far fewer beliefs than we normally think. 19 I readjust my credence to take into account this remote possibility, I’ll no longer believe that the egg will remain broken on the floor. But that doesn’t mean that my credence that it will remain broken was less than 1 before this possibility was made salient to me. I know things (e.g., facts about quantum mechanics) which entail that many of the beliefs I have have a chance of being false. But that doesn’t mean that I fail to have a credence of 1 in the propositions I believe. I can maintain my credence of 1 in those propositions if I successfully ignore the fact that certain remote possibilities have a small objective chance of obtaining.26 Let p be one of the propositions such that I believe it, even though I know things that entail that there’s a small non-0 objective chance that not-p. I have claimed that as long as I am able to ignore – keep out of my mind – the fact that there’s a non-0 objective chance that not-p, I can successfully maintain a credence of 1 and thus a belief in p. Since I do almost always easily ignore such facts, my knowledge of things like quantum mechanics doesn’t prevent me from having a fair number of beliefs. But should I ignore these facts? Wouldn’t it be more rational – wouldn’t my credences better reflect my evidence – insofar as I had a credence of slightly-less-than-1 in p? To repeat, my knowledge entails that there’s an objective chance that not-p. So even if I nevertheless do have a credence of 1 in p, isn’t it the case that my credence in p should be less than 1? I say: not necessarily. I can’t always think of everything. Ignoring certain remote possibilities at least sometimes enables me to think more efficiently and to focus on the issues that are most important for me to think about. As Gilbert Harman [15] pointed out, thinkers ought not to clutter up their minds with all the facts entailed by what they know and believe. Perhaps my ignoring certain remote possibilities and reasoning as if there’s no chance they’ll obtain is justified because it enables me to de-clutter my mind, focus on what I need to focus on, and reason more efficiently and better. I’ve conjectured that, with respect to some of the propositions p we believe, we can’t appeal to what bets we would accept or regard as fair to determine what our credence in p is, and the fact that I know things that entail that p has a non-0 objective probability of being false doesn’t mean my credence in p either is or should be less than 1. But then, one might ask, how can we ever find out if my credence in p is 1? I think in this paper I’ve given a way. In §2 I showed that when I think to myself without qualification that p or q and Probably not-p (where this expresses the fact that my credence in not-p is greater than 0.5), we can account for the fact that this commits me My response to the first and second objections has much in common with some things some contextualists about ‘knows’ say (e.g., Lewis [24]). 26 20 to Probably q (where this expresses the fact that my credence in q is greater than 0.5) only if we assume that in thinking to myself without qualification that p or q, I have a credence of 1 in p or q. This gives us an answer as to how we can find out that I have a credence of 1 in p or q. I show that I have a credence of 1 in p or q by thinking to myself without qualification that p or q. If, like me, one takes this to be true of p or q in the example I provided because of the fact that in general, believing (and thus being willing to think to oneself without qualification) that p means having a credence of 1 in p, then we can say that we display the fact that we have a credence of 1 in a proposition in our willingness to think it to ourselves without qualification. 4.2 Objection 3: (Certainty) entails that beliefs are indefeasible The third objection to (Certainty) claims that (Certainty) entails that beliefs are indefeasible in the sense that giving up a belief will never be a rational response to evidence. Let’s assume that credences must conform to the probability calculus (synchronic coherence), and that when S acquires evidence E, she should update her credence in p to her prior conditional credence of p on E. It’s a law of probability that if P(p) = 1, then for any E, P(p|E) = 1. (Certainty) says that when S believes p, her credence in p is 1. Given our assumptions, if S believes p, her credence in p conditional on anything will be 1. Therefore, no matter what new evidence she acquires, she’ll be rationally required to maintain a credence of 1 in p when she updates on that evidence. In short, it appears to be a consequence of (Certainty) that beliefs are indefeasible – once they’re acquired, one is rationally obligated to keep them forevermore. Any epistemology that says that beliefs are indefeasible should be rejected. But it’s possible to have beliefs be defeasible even if (Certainty) is true. Instead of assuming that Bayesian conditionalization provides the only way to rationally update our doxastic states in response to evidence, we can develop a belief revision theory, the mechanism of which is not conditionalization. Isaac Levi [21] has developed exactly this view. According to both me and Levi, there are S’s beliefs, and there is a set of doxastic possibilities that are consistent with S’s beliefs. S distributes her credence only over the open possibilities. Since S’s beliefs will be doxastic necessities for her, she will have a credence of 1 in all the propositions she believes. Within this picture we can maintain synchronic coherence as a rational requirement. But with Levi we may posit a belief revision theory on top of all this, which gives a different story of when S must revise her full beliefs in response to 21 evidence.27 5 Conclusion In this paper I’ve argued for (Certainty): believing p entails having a credence of 1 in p. Opposition to (Certainty) is widespread and fierce. While many will remain unconvinced, I hope to have demonstrated that there are good reasons for thinking that (Certainty) is true, and to have made the deniers of (Certainty) aware of some puzzling arguments that they need to explain away.28 The theory of lexicographic probabilities (Blume, et al [4]) also allows for a way of modeling the defeasibility of beliefs in the spirit of Levi’s picture where believing entails having a credence of 1 (as was pointed out to me by Alexandru Baltag). 28 I greatly benefited from feedback on presentations of this material from audiences at the Universities of Aberdeen, Copenhagen, Edinburgh, St. Andrews, Stirling, Texas at San Antonio, and at the EpiConFor Conference at the University of Nancy. 27 22 References [1] Barbara Abbott. Some remarks on indicative conditionals. In K. Watanabe and R. B. Young, editors, Proceedings from Semantics and Linguistic Theory (SALT), Ithaca, NY, 2005. CLC Publications. [2] Ernest Adams. The logic of conditionals. 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