he account, but to show that Comesãna and Tal were mistaken; even accepting their account of evidence possession, EEE3 remains false.2 1A previous version of this paper was rejected (in 177 days) by Analysis against the advice of the original referees on the basis of an unambiguously fallacious third report. Four days later I sent  this version to Dialectica, who have sat on it unresponsively for (as of this writing) 796 days. 2Feldman (2014) responds to Fitelson's proposed counterexample by oering still dierent takes on EEE3 as well as a new formulation, running \If S has evidence, E1, supporting the proposition that there is someone who has evidence that supports p, then S has some evidence, E2, that supports p." We are attempting to make a case against neither Feldman nor Fitelson here. 1 MORE TROUBLES FOR COMESÃNA AND TAL 2 (EEE3): If S1 possesses evidence (E1) which supports the claim that S2 possesses evidence (E2) which supports p, then S1 possesses evidence (E3) which supports p. Here is Fitelson's proposed counterexample to EEE3. Let E1 = Card c is black; E2 = Card c is the Ace of Spades; p = Card c is an Ace. It is supposed that S1 knows that Card c (selected uniformly at random from a standard deck of playing cards) has been viewed by S2, and that S1 has evidence E1. Given these assumptions, E1 supports the claim that S2 possesses E2. E2 meanwhile entails p so by EEE3 S1 ought to have evidence supporting p. However, E1 does not support p. So goes the \counterexample". (In this example, the proposition \that S2 possesses evidence (E2)" is eectively identied with E2 itself; S1 knows that S2 has looked at the card. We restrict to scenarios of that sort, but note the original, more general formulation.) Comesãna and Tal claim that Fitelson's example fails because S1 does \possess evidence" which supports p, e.g. E3 = Card c is not the Jack of Hearts. Comesãna and Tal's assumption here appears to be (at minimum) that a sucient condition for \S1 possesses evidence for p" is that S1 justiedly assig