Is Evidence of Evidence Evidence? Screening-Off vs. No-Defeaters William Roche Department of Philosophy, Texas Christian University, Fort Worth, TX, USA, e-mail: w.roche@tcu.edu ABSTRACT: I argue elsewhere (Roche 2014) that evidence of evidence is evidence under screening-off. Tal and Comesaña (2017) argue that my appeal to screening-off is subject to two objections. They then propose an evidence of evidence thesis involving the notion of a defeater. There is much to learn from their very careful discussion. I argue, though, that their objections fail and that their evidence of evidence thesis is open to counterexample. KEYWORDS: Comesaña, Evidence of Evidence, No-Defeaters Condition, Screening-Off Condition, Tal, Transitivity in Evidence 1 Introduction I argue elsewhere (Roche 2014) that evidence of evidence is evidence under screening-off. I argue, in fact, for four distinct theses to that effect. The last of them can be put as follows: EEESOC If (i) Pr(H1 | E) > Pr(H1) where H1 is the proposition that H2's probability given S's total evidence is high, (ii) Pr(H2 | H1) > Pr(H2), and (iii) Pr(H2 | H1 & E) ≥ Pr(H2 | H1) Pr(H2 | ¬H1 & E) ≥ Pr(H2 | ¬H1), then (iv) Pr(H2 | E) > Pr(H2). (Here the tacit quantification is over any propositions E, H1, and H2 and any subject S. Similar points are true of TIPSOC, EEENDC, EEE*NDC, TIPNDC, EEE**NDC, TIP*NDC, and TIP**NDC below.) Let "evidence-HP" be evidence in the sense of high probability and "evidence-IP" be evidence in the sense of increase in probability.1 Then EEESOC says that if (i) E is evidence-IP for H1 where H1 is the proposition that S's total evidence is evidence-HP 1 These are but two senses of evidence in the literature. See Douven (2011), Roche (2012b, 2015), and Roche and Shogenji (2014b). 2 for H2, (ii) H1 is evidence-IP for H2, and (iii) H1 screens-off E from H2 in that E has no negative impact on H2's probability given the truth or falsity of H1, then (iv) E is evidenceIP for H2.2,3 The argument behind EEESOC is straightforward. EEESOC is a special case of the more general thesis that evidence-IP is transitive under screening-off: TIPSOC If (i) Pr(H1 | E) > Pr(H1), (ii) Pr(H2 | H1) > Pr(H2), and (iii) Pr(H2 | H1 & E) ≥ Pr(H2 | H1) Pr(H2 | ¬H1 & E) ≥ Pr(H2 | ¬H1), then (iv) Pr(H2 | E) > Pr(H2). The difference with TIPSOC, as compared to EEESOC, is that the first condition in TIPSOC's antecedent can hold even if H1 is not the proposition that H2's probability given S's total evidence is high. TIPSOC holds without exception (see Roche 2012a) and thus so too does EEESOC. Imagine a case where initially you have a low credence in: H2 The departmental meeting is scheduled for 3:30 today. You then learn the following about a trusted colleague: E John has a high credence in H2. Now consider the following: H1 John's total evidence is evidence-HP for H2. You do not know all the details of John's total evidence. But, still, it seems clear that E is evidence-IP for H1 and that H1, in turn, is evidence-IP for H2. It also seems clear that E is 2 Moretti (2016b) argues for a related but distinct EEE thesis involving screening-off. 3 EEESOC's screening-off condition is a "negative-impact" screening-off condition. There is also a "no-impact" screening-off condition to the effect that E has no impact on H2's probability given the truth or falsity of H1, and a "positive-impact" screening-off condition to the effect that E has no positive impact on H2's probability given the truth or falsity of H1. See Roche and Shogenji (2014a) for relevant discussion. 3 evidence-IP for H2. There are counterexamples, though, to the thesis that evidence-IP is transitive. Is the case at hand a case in point? If what I argue elsewhere (Roche 2014) is right, then the answer is negative. This is because the case at hand is a case where H1 screens-off E from H2 (in that E has no negative impact on H2's probability given the truth or falsity of H1) and thus the third condition in EEESOC's antecedent holds. Tal and Comesaña (2017), hereafter "T&C", argue that my appeal to screening-off is subject to two objections. They write: First, it is not clear what the evidential relevance of the screening-off conditions is, and so even if they worked flawlessly we would not have a deep understanding of the resulting principle. Second, they do not work flawlessly, for the screening-off conditions are sufficient but not necessary for evidential transitivity. (T&C 2017, p. 107) Here by "screening-off conditions" they have in mind, first, the condition that Pr(H2 | H1 & E) ≥ Pr(H2 | H1) and, second, the condition that Pr(H2 | ¬H1 & E) ≥ Pr(H2 | ¬H1). They call the former "the positive screening-off condition" and the latter "the negative screening-off condition". T&C then propose an alternative EEE thesis. They continue the passage above as follows: In the next section we argue that the right fix for the right version of the EEE principle appeals not to screening-off conditions, but to the notion of a defeater. Defeaters wear their epistemic relevance on their sleeves. Moreover, we will argue that a suitable addition of a no-defeaters condition provides a necessary as well as a sufficient condition for evidence of evidence to be evidence. (T&C 2017, p. 107) Their proposed alternative EEE thesis can be put as follows: EEENDC If (i) Pr(H1 | E) > Pr(H1) to degree α where H1 is the proposition that there is a proposition E* such that E* is true and is evidence-IP for H2 to degree β, (ii) Pr(H2 | H1) > Pr(H2) to degree γ, and (iii) Pr(H2 | H1 & E) > Pr(H2) to degree δ, then (iv) there is a degree ε such that Pr(H2 | E) > Pr(H2) to degree ε. 4 This thesis differs from EEESOC in at least four key respects.4 First, there is no mention in EEENDC of a subject. Second, the evidence at issue in H1 is evidence-IP. Third, there is reference in EEENDC to degrees of evidence-IP.5 Fourth, the third condition in EEENDC's antecedent is a "no-defeaters" condition. There is much to learn from T&C's very careful discussion. I aim to show, though, that their objections to my appeal to screening-off fail and that EEENDC is open to counterexample.6 4 The second condition in EEENDC's antecedent is suppressed in T&C's formulation of their EEE thesis (which they call "Existential EEE1 de dicto no defeat"). See T&C (2017, p. 110, en. 7). 5 See Moretti (2016a) for discussion of a thesis like EEENDC but with no reference to degrees of evidence-IP. 6 I take issue with certain other parts of T&C's discussion. First, consider: What is this case [i.e., Fitelson's case] a counterexample to? It is uncontroversially a counterexample to a de re reading of EEE1 .... But Roche suggests that the case is not a counterexample to EEE1 de dicto .... According to Roche, Fitelson's case is not a counterexample to EEE1 de dicto because we already know the following existential proposition: there is some proposition that John has which is evidence for H'. (T&C 2017, p. 100, emphasis original) T&C then argue that my argument fails. It is not the case, though, that I argue that Fitelson's case is not a counterexample to EEE1 de dicto. I have in mind a thesis in which there is no reference to degrees of evidence-IP. I call the thesis in question "EEEU" (see Roche 2014, p. 120). Given this, and given that there is reference in EEE1 de dicto to degrees of evidenceIP, it follows that I have in mind a thesis distinct from EEE1 de dicto. Second, consider: Roche seems to think that the point holds only under a Williamsonian conception of evidence possession, and thus hedges by saying that Fitelson's case only may be a counterexample to (in effect) EEE1 de dicto. But Roche's point holds (if at all) on any sane conception of evidence-possession, not only a Williamsonian one. (T&C 2017, p. 110, en. 8, emphasis original). What I say (in effect) is that the point in question holds if a Williamsonian conception is assumed. This, of course, does not mean that the point holds only if a Williamsonian conception is assumed. See Roche (2014, p. 121, fn. 6). The reason why I hedge and use the expression "might" was because of a worry to the effect that if S believes E* as a result of an 5 2 Transitivity in evidence-IP: SOC versus NDC There are counterexamples to the thesis that evidence-IP is transitive. It will help in terms of understanding the rationale behind my appeal to screening-off to understand why there are counterexamples to the thesis that evidence-IP is transitive. Consider the following theorem of the probability calculus: Pr(H2 | E)− Pr(H2 ) = Pr(H1 | E)− Pr(H1)[ ] Pr(H2 |H1)− Pr(H2 )[ ] + Pr(¬H1 | E)− Pr(¬H1)[ ] Pr(H2 |¬H1)− Pr(H2 )[ ] + Pr(H1 | E) Pr(H2 |H1 & E)− Pr(H2 |H1)[ ] + Pr(¬H1 | E) Pr(H2 |¬H1 & E)− Pr(H2 |¬H1)[ ] This theorem is due to Shogenji (forthcoming) and thus shall hereafter be referred to as "Shogenji's theorem". Let the first addend on the right be "A1", the second addend on the right be "A2", the third addend on the right be "A3", and the fourth addend on the right be "A4". Suppose that E is evidence-IP for H1 and that H1 is evidence-IP for H2. Then each of A1 and A2 is positive and thus A1+A2 is positive. It does not follow, though, that A1+A2+A3+A4 is positive, because it could be that A3+A4 is negative and greater than or equal to in absolute value A1+A2. All counterexamples to the thesis that evidence-IP is transitive are cases where A1+A2 is positive but A3+A4 is negative and greater than or equal to in absolute value A1+A2. Suppose, to illustrate, that a card is randomly drawn from a standard and well-shuffled deck of cards. Let E be the proposition that the card drawn is not a Heart, H1 be the proposition that the card drawn is a Diamond, and H2 be the proposition that the card drawn is a Red. Then, since E is evidence-IP for H1 and H1 is evidence-IP for H2, A1+A2 is positive: 1 3 − 1 4 ⎡ ⎣⎢ ⎤ ⎦⎥ 1− 1 2 ⎡ ⎣⎢ ⎤ ⎦⎥ + 2 3 − 3 4 ⎡ ⎣⎢ ⎤ ⎦⎥ 1 3 − 1 2 ⎡ ⎣⎢ ⎤ ⎦⎥ = 1 18 A3+A4, however, is negative and greater than or equal to in absolute value A1+A2: 1 3 ⎡ ⎣⎢ ⎤ ⎦⎥ 1−1[ ]+ 2 3 ⎡ ⎣⎢ ⎤ ⎦⎥ 0 − 1 3 ⎡ ⎣⎢ ⎤ ⎦⎥ = − 2 9 inference, then E* is not included in S's total evidence even if there is a proposition E such that E is included in S's total evidence and E* is entailed by E. 6 It follows that A1+A2+A3+A4 is negative and that so too is Pr(H2 | E) – Pr(H2). This case is thus a counterexample to the thesis that evidence-IP is transitive.7 It is clear from Shogenji's theorem than any condition that is sufficient for transitivity in evidence-IP is a condition that guarantees that, when E is evidence-IP for H1 and H1 is evidence-IP for H2, it is not the case that A3+A4 is negative and greater than or equal to in absolute value A1+A2. There is simply no way around it. Now consider the condition: SOC Pr(H2 | H1 & E) ≥ Pr(H2 | H1) and Pr(H2 | ¬H1 & E) ≥ Pr(H2 | ¬H1) This is the screening-off condition in EEESOC and TIPSOC. SOC guarantees that each of A3 and A4 is non-negative and thus guarantees that, when E is evidence-IP for H1 and H1 is evidence-IP for H2, it is not the case that A3+A4 is negative and greater than or equal to in absolute value A1+A2. It follows that TIPSOC holds without exception and that so too does EEESOC. It is important to note that neither conjunct of SOC is sufficient for transitivity in evidence-IP. Return to the card case above. E is evidence-IP for H1, H1 is evidence-IP for H2, and the first conjunct of SOC holds (because each probability equals 1). But it is not the case that E is evidence-IP for H2. Hence the first conjunct of SOC is not sufficient for transitivity in evidence-IP. Now let E be the proposition that the card drawn is a Club, the Ace of Spades, or the Ace of Hearts, H1 be the proposition that the card drawn is not a Spade, and H2 be the proposition that the card drawn is a Red. Then E is evidence-IP for H1, H1 is evidence-IP for H2, and the second conjunct of SOC holds (because each probability equals 0). But, as Pr(H2 | E) = 1/15 < 1/2 = Pr(H2), it is not the case that E is evidence-IP for H2. Hence the second conjunct of SOC is not sufficient for transitivity in evidence-IP. T&C suggest that the second conjunct of SOC be dropped and that the first conjunct be replaced by a no-defeaters condition: NDC Pr(H2 | H1 & E) > Pr(H2) This is the no-defeaters condition in EEENDC. NDC is weaker than the first conjunct of SOC on the assumption that H1 is evidence-IP for H2. Any case where H1 is evidence-IP for H2 and the first conjunct of SOC holds is a case where Pr(H2 | H1 & E) ≥ Pr(H2 | H1) > Pr(H2) and thus is a case where NDC holds. But some cases where H1 is evidence-IP for H2 and 7 It is also a counterexample to Hempel's "Special Consequence Condition" (Hempel 1965) understood in terms of evidence-IP: If (i) E is evidence-IP for H1 and (ii) H1 entails H2, then E is evidence-IP for H2. 7 NDC holds are cases where Pr(H2 | H1) > Pr(H2 | H1 & E) > Pr(H2) and thus are cases where the first conjunct of SOC does not hold. Is NDC sufficient for transitivity in evidence-IP? The answer, clearly, is no. NDC is weaker than the first conjunct of SOC on the assumption that H1 is evidence-IP for H2 and so holds in all cases where the first conjunct of SOC holds and H1 is evidence-IP for H2. But, as noted above, there are cases where E is evidence-IP for H1, H1 is evidence-IP for H2, the first conjunct of SOC holds, and yet it is not the case that E is evidence-IP for H2. All such cases are counterexamples to the thesis that NDC is sufficient for transitivity in evidence-IP. T&C acknowledge that my appeal to SOC works. They claim, though, that it works by "brute force". They want to know why it works. This in effect is their first objection to my appeal to SOC. It should be clear, at this point, why my appeal to SOC works: SOC guarantees that, when E is evidence-IP for H1 and H1 is evidence-IP for H2, it is not the case that A3+A4 is negative and greater than or equal to in absolute value A1+A2. It should also be clear why an appeal to NDC would not work: NDC fails to guarantee that, when E is evidence-IP for H1 and H1 is evidence-IP for H2, it is not the case that A3+A4 is negative and greater than or equal to in absolute value A1+A2. There is simply no getting around the point that any condition that is sufficient for transitivity in evidence-IP is a condition that guarantees that, when E is evidence-IP for H1 and H1 is evidence-IP for H2, it is not the case that A3+A4 is negative and greater than or equal to in absolute value A1+A2. There is more. Consider what can be called the "Trivial Condition": TC Pr(H2 | E) > Pr(H2) This condition, as with SOC, is sufficient for transitivity in evidence-IP. But TC is not at all interesting. TC is impotent to help explain in a given case why E confirms H2 and it is impotent to help in verifying in a given case that E confirms H2. SOC is different. It can help explain in a given case why E confirms H2 and it can help in verifying in a given case, for example, the case of John and the departmental meeting, that E confirms H2.8,9 8 See Roche and Shogenji (2014b) for an argument to the effect that SOC holds in the context of Moore's "proof" of the existence of a material world. 9 Consider the following conditions (the first of which is the so-called "Dragging Condition" from Kotzen 2012): 8 What about T&C's charge that SOC is not necessary for transitivity in evidence-IP? How exactly do T&C mean for this charge to be understood? One answer is that they mean for it to be understood as the charge that SOC is not necessary for the relation of "being evidence-IP for" to be transitive, that is, for it to be such that any case where E is evidence-IP for H1 and H1 is evidence-IP for H2 is a case where E is evidence-IP for H2. If this answer is right, then T&C's charge is clearly correct. This can be seen by considering what can be called the "Weaker Condition": WC It is not the case that A3+A4 is negative and greater than or equal to in absolute value A1+A2. This condition is weaker than SOC. Yet it is sufficient for transitivity in evidence-IP. It turns out, though, that T&C mean for their charge to be understood differently. Their worry is that there are cases where E is evidence-IP for H1, H1 is evidence-IP for H2, SOC does not hold, and yet E is evidence-IP for H2.10 They want a condition C such that (i) C is sufficient for transitivity in evidence-IP and (ii) if E is evidence-IP for H1, H1 is evidence-IP for H2, and C does not hold, then E is not evidence-IP for H2. It is true that SOC is not a condition of that sort. But note that the same is true of NDC. This can be seen by appeal to a variant of the card cases above. Let E be the proposition that the card drawn is a Two or a Three, H1 be the proposition that the card drawn is a Three or a Four, and H2 be the proposition that the card drawn is a Two or a Four. Pr(H1 | E) = 1/2 > 2/13 = Pr(H1), so E is evidence-IP for H1. Pr(H2 | H1) = 1/2 > 2/13 = Pr(H2), so H1 is Pr(H2) < Pr(H1 | E) Pr(H2 & ¬H1 | E) ≥ Pr(H2 & ¬H1) Each of these conditions is sufficient for transitivity in evidence-IP in the special case where H1 entails H2. But, unlike SOC, neither of them is sufficient for transitivity in evidence-IP in the general case where it is not required that H1 entails H2. They thus have no application in cases like the case of John and the departmental meeting. See Roche and Shogenji (2014b) for discussion. 10 T&C (2017, p. 111, en. 19) claim in effect that Roche and Shogeniji (2014b) note that the second conjunct of SOC (what T&C call "the negative screening-off condition") is sufficient but not necessary for transitivity in evidence-IP. This is misleading. Roche and Shogenji (2014b, pp. 807-808) have in mind the special case where H1 entails H2. It is not the case, and Roche and Shogenji (2014b) do not suggest otherwise, that the second conjunct of SOC is sufficient for transitivity in evidence-IP. 9 evidence-IP for H2. Pr(H2 | H1 & E) = 0 < 2/13 = Pr(H2), so NDC does not hold. But Pr(H2 | E) = 1/2 > 2/13 = Pr(H2), so E is evidence-IP for H2. There are conditions of the sort that T&C want. One is TC. But this condition, as noted above, is not at all interesting. A better condition is WC. This condition, unlike TC, is distinct from the condition that E is evidence-IP for H2. There are cases where WC holds but, since in part E is not evidence-IP for H1 or H1 is not evidence-IP for H2, it is not the case that E is evidence-IP for H2. I see no need to choose between SOC and WC or between theses involving SOC such as TIPSOC and EEESOC and similar theses involving WC. They all have their place. I turn now to EEENDC and some variants of it. 3 A counterexample to T&C's EEE thesis Consider the following variant of EEENDC: EEE*NDC If (i) Pr(H1 | E) > Pr(H1) where H1 is the proposition that H2's probability given S's total evidence is high, (ii) Pr(H2 | H1) > Pr(H2), and (iii) Pr(H2 | H1 & E) > Pr(H2), then (iv) Pr(H2 | E) > Pr(H2). EEE*NDC is a special case of: TIPNDC If (i) Pr(H1 | E) > Pr(H1), (ii) Pr(H2 | H1) > Pr(H2), and (iii) Pr(H2 | H1 & E) > Pr(H2), then (iv) Pr(H2 | E) > Pr(H2). TIPNDC, though, is false, for it is not the case that NDC is sufficient for transitivity in evidence-IP. Does it follow that EEE*NDC too is false? The answer, it turns out, is negative. It could be that TIPNDC is false but nonetheless holds without exception in special cases where H1 is a proposition about evidence. Consider, for example, the thesis: EEE**NDC If (i) Pr(H1 | E) > Pr(H1) where H1 is the proposition that S has some evidence-IP for H2, 10 (ii) Pr(H2 | H1) > Pr(H2), and (iii) Pr(H2 | H1 & E) > Pr(H2), then (iv) Pr(H2 | E) > Pr(H2). EEE**NDC holds without exception if it is assumed both that all subjects have some propositions (with non-extreme unconditional probabilities) as evidence and that if a subject has a proposition as evidence, then she also has any implication (with a non-extreme unconditional probability) of that proposition as evidence. For, if those things are assumed, then there are no cases where the first condition in EEE**NDC's antecedent holds. This is easy to see. Suppose that S has E* as evidence. Then, since E* entails E* ∨ H2, it follows that S has that disjunction as evidence. Given this, and given that E* ∨ H2 is entailed by H2, it follows that E* ∨ H2 is evidence-IP for H2 (assuming that H2 has a non-extreme unconditional probability). Hence S has some evidence-IP for H2. So, though EEE**NDC is a special case of TIPNDC, and though the latter is false, it is not the case that the former is false.11,12 Hence the mere fact that TIPNDC is false leaves it open that it is not the case that EEE*NDC, which is a special case of TIPNDC, is false. Consider now a different version of the case of John and the departmental meeting. Each of John's credences is either high or not high, either in a proposition for which his total evidence is evidence-HP or not in a proposition for which his total evidence is evidence-HP, and either in a true proposition or not in a true proposition. Suppose that you know that: (a) 3/5 of John's credences are high & in a proposition for which his total evidence is evidence-HP & in a true proposition (b) 1/15 of John's credences are high & in a proposition for which his total evidence is evidence-HP & not in a true proposition (c) 0 of John's credences are high & not in a proposition for which his total evidence is evidence-HP & in a true proposition (d) 2/27 of John's credences are high & not in a proposition for which his total evidence is evidence-HP & not in a true proposition 11 These points about EEE**NDC carry over to an EEE thesis put forward in Roche (2014). The thesis in question is called "(EEE'U)" and is like EEE**NDC except that it has SOC where EEE**NDC has NDC. 12 See Comesaña and Tal (2015) for a similar point about an EEE thesis discussed in Fitelson (2012). 11 (e) 0 of John's credences are not high & in a proposition for which his total evidence is evidence-HP & in a true proposition (f) 0 of John's credences are not high & in a proposition for which his total evidence is evidence-HP & not in a true proposition (g) 1/4 of John's credences are not high & not in a proposition for which his total evidence is evidence-HP & in a true proposition (h) 1/108 of John's credences are not high & not in a proposition for which his total evidence is evidence-HP & not in a true proposition (Note that 3/5 + 1/15 + 0 + 2/27 + 0 + 0 + 1/4 + 1/108 = 1.) Let E be the proposition that John has a high credence in H2, H1 be the proposition that John's total evidence is evidenceHP for H2, and H2 be the proposition that the departmental meeting is scheduled for 3:30 today. Suppose that H2 is one of the propositions in which John has a credence, and that, from your perspective, John's credence in H2 is a random member of the set of propositions in which he has a credence. Suppose, further, and consistent with all this, that your probability for H2 is 0.85. Then you should have the following probability distribution: E H1 H2 Pr T T T 35 T T F 115 T F T 0 T F F 227 F T T 0 F T F 0 F F T 14 F F F 1108 It follows that: (i) Pr(H1 | E) = 9/10 > 2/3 = Pr(H1) (j) Pr(H2 | H1) = 9/10 > 85/100 = Pr(H2) (k) Pr(H2 | H1 & E) = 9/10 > 85/100 = Pr(H2) 12 (l) Pr(H2 | E) = 81/100 < 85/100 = Pr(H2) So, E is evidence-IP for H1 where H1 is the proposition that John's total evidence is evidence-HP for H2, H1 is evidence-IP for H2, and NDC holds, and yet it is not the case that E is evidence-IP for H2. Hence EEE*NDC, as with TIPNDC, is false. It is unclear at this point whether the same is true of EEENDC. This is because it is unclear at this point how exactly EEENDC is to be understood. T&C never specify how degree of evidence-IP is to be measured. They use expressions like "entailing evidence" and "evidence that raises the probability to at least 4/5". But expressions like those leave it open how exactly evidence-IP is to be measured. Consider the following measures of evidence-IP: ed (H ,E) = Pr(H | E)− Pr(H ) eKO (H ,E) = Pr(E |H )− Pr(E |¬H ) Pr(E |H )+ Pr(E |¬H ) The first of these measures is the "difference" measure. The second is the "KemenyOppenheim" measure.13 Suppose that E entails H and thus is evidence-IP for H (assuming that neither E nor H has an extreme unconditional probability). It follows by ed that whether the degree of evidence-IP is high or low depends on whether Pr(H) is high or low. If Pr(H) is high, then the degree of evidence-IP is low. If Pr(H) is low, then the degree of evidence-IP is high. It follows by eKO, in contrast, that the degree of evidence-IP is high, indeed, is maximal at 1, regardless of whether Pr(H) is high or low. T&C presumably hold that EEENDC is true regardless of which of the leading measures of evidence-IP in the literature is assumed (otherwise, presumably, they would have specified how degree of evidence-IP is to be measured). I shall assume ed and thus shall assume that in the case of John and the departmental meeting the degree to which H1 is evidence-IP for H2 is 0.05. But nothing of importance for my purposes hinges on this, since all the main points below would hold mutatis mutandis if eKO or any alternative measure of evidence-IP (in the literature) were assumed instead. Now, continuing with the second version of the case of John and the departmental meeting, let H* be the proposition that there is a proposition E* such that E* is true and is evidence-IP for H2 to degree 0.05. There is no reference to H* in the probability distribution above. It turns out, though, that H* is logically equivalent (in the context) to H1. (See Appendix for details.) It follows immediately that: 13 See Roche and Shogenji (2014a) for a list of the main measures of evidence-IP (or "confirmation") in the literature. 13 (m) Pr(H* | E) = 9/10 > 2/3 = Pr(H*) (n) Pr(H2 | H*) = 9/10 > 85/100 = Pr(H2) (o) Pr(H2 | H* & E) = 9/10 > 85/100 = Pr(H2) (p) Pr(H2 | E) = 81/100 < 85/100 = Pr(H2) By (m) it follows that E is evidence-IP for H* to degree 7/30 where H* is the proposition that there is a proposition E* such that E* is true and is evidence-IP for H2 to degree 0.05. By (n) it follows that H* is evidence-IP for H2 to degree 0.05. By (o) it follows that H* & E is evidence-IP for H2 to degree 0.05. By (p) it follows that E is not evidence-IP for H2. Hence EEENDC, as with EEE*NDC, is open to counterexample. T&C close their discussion by giving a gloss of EEENDC and writing (where notation has been slightly modified): Evidence that there is ... evidence for H2 is itself evidence for H2 when it is not at the same time a defeater for the support that the proposition that there is evidence for H2 provides to H2. Doesn't quite roll off the tongue, but it has not yet been shown false. (T&C 2017, p. 110) They can no longer say that. Now return to the point from Section 2 that there are counterexamples to the thesis that NDC is sufficient for transitivity in evidence-IP. It turns out that things are different in the special case where E entails H1. The following holds without exception: TIP*NDC If (i) Pr(H1 | E) > Pr(H1), (ii) Pr(H2 | H1) > Pr(H2), (iii) Pr(H2 | H1 & E) > Pr(H2), and (iv) E entails H1, then (v) Pr(H2 | E) > Pr(H2). Suppose that (i)-(iv) in TIP*NDC all hold. It follows from (iv) that E is logically equivalent to H1 & E. Hence, given (iii), it follows that Pr(H2 | E) > Pr(H2). Should this be welcome news to T&C? No. Any case where (iv) in TIP*NDC holds is a case where (v) holds if and only if (iii) holds. It does not matter whether (i) and (ii) hold. This can be seen by noting that the following variant of TIP*NDC also holds without exception: 14 TIP**NDC If (i*) Pr(H1 | E) = Pr(H1), (ii*) Pr(H2 | H1) = Pr(H2), (iii) Pr(H2 | H1 & E) > Pr(H2), and (iv) E entails H1, then (v) Pr(H2 | E) > Pr(H2). TIP*NDC, as with TIP*NDC, is a TIP thesis in name only.14 T&C could modify EEENDC along the lines of TIP*NDC. Then it would not be open to counterexample. But, at the same time, it would be an EEE thesis in name only. A better tack would be to abandon the no-defeaters approach in favor of the screeningoff approach. EEESOC is not open to counterexample and is not an EEE thesis in name only. 4 Conclusion SOC is sufficient for transitivity in evidence-IP. Further, unlike TC (the condition that E is evidence-IP for H2), SOC can help explain in a given case why E confirms H2, and can help in verifying in a given case that E confirms H2. Thus SOC is fit for use in EEE theses. NDC, by contrast, is not sufficient for transitivity in evidence-IP and thus is not fit for use in EEE theses. These points are borne out by the fact that EEESOC holds without exception whereas EEENDC does not. Acknowledgments I wish to thank an anonymous reviewer, an audience at the 12th Annual Formal Epistemology Workshop, Juan Comesaña, Tomoji Shogenji, and Eyal Tal for helpful very feedback. 14 There are cases where TIP**NDC's antecedent holds. Take a card case where E is the proposition that the card drawn is a Red, H1 is the proposition that the card drawn is a Jack or not a Jack, and H2 is the proposition that the card drawn is a Diamond. Here Pr(H1 | E) = 1 = Pr(H1), Pr(H2 | H1) = 1/4 = Pr(H2), Pr(H2 | H1 & E) = 1/2 > 1/4 = Pr(H2), and E entails H1. 15 Appendix Take the probability distribution given above in Section 3. First, it follows that H* is logically equivalent (in the context) to a disjunction with the following disjuncts: (q) H1 (r) (E & H2) ∨ H1 (s) (E & H1) ∨ (E & H2) ∨ (H1 & H2) (t) (E & H1) ∨ (E & H2) ∨ (H1 & ¬H2) (u) (E & H1) ∨ (E & H2) (v) (E & H1) ∨ (H1 & H2) (w) (E & H1) ∨ (H1 & ¬H2) (x) (E & H1) This is because there are exactly 256 distinct Boolean propositions constructible from one or more of E, H1, and H2, each of (q)-(x) is evidence-IP for H2 to degree 0.05, and none of the remaining 248 distinct Boolean propositions constructible from one or more of E, H1, and H2 is evidence-IP for H2 to degree 0.05.15 Second, each of (q)-(x) entails and is entailed by H1. The key here is that H1 entails (in the context) E and is entailed by (in the context) E & H2. QED References Comesaña, J., and Tal, E. 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