Tal and Comesaña on evidence of evidence Luca Moretti l.moretti@abdn.ac.uk University of Aberdeen & Munich Center for Mathematical Philosophy R. Feldman defends a general principle about evidence the slogan form of which says that 'evidence of evidence is evidence' (cf. 2014: 284–99, 2011 and 2007: 194-214). B. Fitelson (2012: 85–88) considers three renditions of this principle and contends they are all falsified by counterexamples. Against both Feldman and Fitelson, J. Comesaña and E. Tal (2015: 557-59) show that the third rendition––the one actually endorsed by Feldman––isn't affected by Fitelson's counterexamples, but only because it is trivially true and thus uninteresting. Tal and Comesaña (2015) defend a fourth version of Feldman's principle, which––they claim––'has not yet been shown false' (p. 16). Against Tal and Comesaña, I will show that this new version of Feldman's principle is in fact false. The third version of Feldman's principle considered by Fitelson (2014) is this: (EEE3) If S1 possesses evidence, E1, that supports the proposition that S2 possesses evidence, E2, that supports P, then S1 possesses evidence, E3, that supports P. EEE3 has been defended by Feldman (2011). Furthermore, Feldman (2014: 292) endorses a restatement of this principle that is only unimportantly different. Here is Fitelson's alleged counterexample to EEE3: S1's background information says that a card c will be picked out randomly from a standard deck. S1 is then told that S2 knows which card c is exactly, and that: (E1) c is a black card. In these circumstances, E1 gives S1 some support for the proposition that S2 possesses the following information: (E2) c is the ace of spades. Furthermore, E2 entails and supports the proposition: (P) c is an ace. 2 In this setting, upon learning E1, S1 acquires evidence that supports the proposition that S2 possesses evidence E2 that supports P. So EEE3's antecedent is satisfied. However––Fitelson contends––S1 doesn't have any evidence E3 that supports P. For we can stipulate that in this scenario all evidence S1 possesses about c is constituted by E1, the proposition that S2 knows which card c is exactly, and any consequence of these two propositions. But none of these propositions is––according to Fitelson––evidence for P. Since EEE3's antecedent is satisfied but not its consequent, EEE3 is false. Comesaña and Tal (2015) retort that this is no counterexample to EEE3. For in this scenario––pace Fitelson––S1 has some evidence E3 supporting P. For example, S1 believes the trivial consequence of E1, c is not the Jack of hearts, which supports P. Comesaña and Tal emphasize that this upshot doesn't actually help Feldman because: For any pair of propositions E and Q (about which the subject in question is not already certain), something entailed by E supports Q: for instance, the disjunction either E or Q. Therefore, Feldman's EEE3 is only trivially true, and so the fact that it is not refuted by Fitelson's case is irrelevant. (2015: 559, edited) The moral is that Feldman can reject Fitelson's contention that EEE3 has a counterexample, but this is a Pyrrhic victory because EEE3's truth is immaterial to the general epistemological thesis that Feldman would like to substantiate. I endorse this conclusion. To rescue the evidence-of-evidence-is-evidence principle from the triviality problem and other difficulties, Tal and Comesaña (2015: 14) propose replacing EEE3 with this principle: (EEE4) For all E and Q, if (i) E is evidence that there is some evidence for Q and (ii) E is not a defeater for the support that the proposition that there is evidence for Q provides for Q, then E is evidence for Q. In EEE4, 'evidence' means any true proposition regardless of its being possessed by a subject. Since Feldman (2014: §15.2) thinks of evidence as a proposition possessed by a subject, EEE4 may be unsuitable to render the principle he has in mind. EEE4 is afflicted by a more serious problem: it is 3 not trivially true but just false. For there are many pairs of ordinary propositions E and Q (about which we are uncertain) that satisfy EEE4's antecedent but not EEE4's consequent. Take E and Q from two disparate domains––for instance, E = 'Aristotle used to snore' and Q = 'There is a mouse in my house'. Even so, E and Q satisfy (i) because E is evidence that there is some evidence for Q––namely, any (uncertain) proposition E* that entails both E and Q (e.g. the conjunction E & Q). This is so because E* entails E. Thus E is evidence for E*. (As E* entails E, E confirms E* in the sense that Pr(E*|E) > Pr(E*), if Pr(E*) > 0 and Pr(E) < 1.) Furthermore, E* entails Q. Thus E* is evidence for Q. But E and Q also satisfy (ii), for it is intuitively true that E is not a defeater for the support that the proposition that there is evidence for Q provides for Q. A way to flesh out this intuition is the following: the existential proposition that there is evidence for Q can be construed as a disjunction each disjunct of which states that [En, and En supports Q] for any relevant En. E would be a defeater for the support that this disjunction provides for Q only if E were a defeater for the support that all or most of these disjuncts individually supply for Q. But we have no reason to believe this is the case. Rather, we have reasons to believe the opposite. Take for example En = 'There are chew marks on the cupboard'. Clearly, E isn't a defeater for the support that [there are chew marks on the cupboard, and the proposition that there are chew marks on the cupboard supports Q] provides for Q. The same result obtains for any other En that stands for typical evidence for Q. The same happens in many cases in which En stands for atypical evidence for Q. Suppose for instance En = E*. E isn't a defeater for the support that [E*, and E* supports Q] provides for Q. For the conjunction E & [E*, and E* supports Q] supports Q. This is so because, since E* entails E, E & [E*, and E* supports Q] is logically equivalent to [E*, and E* supports Q], which supports Q. In conclusion, since E and Q satisfy both (i) and (ii), EEE4's antecedent is satisfied. Nevertheless, since E is not evidence for Q, EEE4's consequent is unsatisfied. Therefore, EEE4 is false. 4 References Comesaña, J. and E. Tal. 2015. 'Evidence of evidence is evidence (trivially)'. Analysis 75: 557-59. Feldman, R. 2007. 'Reasonable religious disagreements'. In Philosophers Without God: Meditations on Atheism and the Secular Life, ed. L. Antony, 194–214. Oxford: OUP. Feldman, R. 2011. 'Evidence of evidence is evidence'. In Keynote Lecture at Feldmania: A Conference in Honor of Richard Feldman, UT-San Antonio, Texas, 19 February 2011. Feldman, R. 2014. 'Evidence of evidence is evidence'. In The Ethics of Belief, ed. J. Mattheson and R. Vitz, 284–99. Oxford: OUP. Fitelson, B. 2012. 'Evidence of evidence is not (necessarily) evidence'. Analysis 72: 85–88. Tal, E. and J. Comesaña. 2015. 'Is Evidence of Evidence Evidence?' Nous. doi: 10.1111/nous.12101.