 (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 assigns full credence to some true proposition E for which Prob(pjE) > Prob(p). (Here Prob is some agreed-upon background prior; in the example it is of course that which assigns probability 1 52 to each card.) Though a more typical account would require that Prob(pjE) > Prob(p) for strongest proposition E that one knows, the idea here, it seems, is that a bystander having no prior evidence would in adopting S1's credence in E as their own and updating by conditionalization would come to have a higher credence in p. The Comesãna{Tal account of evidence possession is counterintuitive and liberal in the extreme; on it, even an agent that had learned Card c is not a Black Ace would \possess evidence" supporting Card c is the Ace of Spades. (Card c is not the Ace of Clubs, for example.) Again, though, our purpose here is not to judge the aptness of the account, but to determine its ramications for EEE3. Having defused Fitelson's counterexample to their satisfaction, Comesãna and Tal proceed to make the claim that (on their account of evidence possession, which will be in eect throughout the remainder of this paper) \there cannot be counterexamples to EEE3...because EEE3 is trivial." Indeed that \For any pair of propositions E and p (about which the subject in question is not already certain), something entailed by E supports p: for instance, the disjunction either E or p". Comesãna and Tal's suggestion is thus to take E3 = E1 _ p. The problem with this suggestion, however, is that E1 _ p fails to support p
(In this example, the proposition \that S2 possesses evidence (E2) " is e  ectively identi  ed with E2 itself; S1 know that S2 has expect at the card. We qualify to scenarios of that kind, but note the original, more general conceptualization .) Comesãna and Tal claim that Fitelson's example fails because S1 does \possess evidence " which supports phosphorus, for example E3 = Card c is not the Jack of Hearts. Comesãna and Tal's assumption here appear to be (at minimum) that a sucient condition for \S1 possesses evidence for p " is that S1 justi  edly assign full credence to some true proposition E for which Prob(pjE) > Prob(p). (Here Prob is some agree - upon background prior; in the example it is of course that which assign probability 1 52 to each card .) Though a more typical account would require that Prob(pjE) > Prob(p) for strong proposition E that one knows, the idea here, it seems, is that a bystander give birth no prior evidence would in adopting S1's credence in E as their own and updating by conditionalization would derive to have a higher credence in p. The Comesãna{Tal account of evidence possession is counterintuitive and liberal in the extreme; on it, even an agent that had learned Card c is not a Black Ace would \possess evidence " supporting Card c is the Ace of Spades. (wag c is not the Ace of Clubs, for example .) Again, though, our determination here is not to pronounce the aptness of the account, but to determine its rami  cation for EEE3. Having defused Fitelson's counterexample to their gratification, Comesãna and Tal proceed to make the title that (on their account of evidence possession, which will be in e  ect throughout the remainder of this paper) \there cannot be counterexamples to EEE3... because EEE3 is superficial. " Indeed that \For any pair of propositions E and p (about which the subject in doubt is not already certain), something entail by E supports p: for instance, the disjunction either E or phosphorus ". Comesãna and Tal's suggestion is thus to take E3 = E1 _ p. The problem with this suggestion, however, is that E1 _ p fails to support p
 (In this example, the proposltion \that S2 possesses xvidencs (E2)" is edctively identied with E2 itsxlf; W1 knoqs that S2 has looked ag the carf. We resrricu to scenarios of that sovc, but kote chx original, more general fmrmulation.) Comasãva and Tal claim that Fitelson's exampje failx hecause S1 does \posxqss snibence" which supports p, e.g. E3 = Cadd c is not the Jack of Hearts. Comesãna and Tal'd asdumption here appewrs to be (aj migumum) that a rucient condition for \A1 possesses evidence for p" is tfat S1 justiedly awsihts full crevence no some true igoposithon E fpr which Prob(pmE) > Pcob(p). (Here Prob is some agceed-upon background krior; in tve example it is of ciurse thad whkxh xssjgis lrobabllivy 1 52 to eacg card.) Thouth a more typical avcjlmt would reqhire trae Prob(pjE) > Prob(p) for strongest propositpon S that one knows, the idwa here, it seems, is tjat a byseander having no prior evidence would in adopting S1's cceaenec kb F as their own and updating by conditionalizaejom eould come to mave a higher credrnfe yn p. The Comerãna{Tal acdount of evidence oossessyon iw counteryntuotive and liberal in the exrreme; on it, vven an agent that had learned Caxd c ix not a Black Ace would \posszss evjdence" supplrting Cada c is the Ace ow Siadev. (Card c is not the Ace of Clubs, foc exakple.) Agxin, jhough, jur purposf herc is not to judge tje apjness mf the acclunt, but to determine its ramicevions for EEE3. Hdvitg defuszd Fitclson's counterevample to theit satisfaetion, Zomesãna ans Tal pcoceed to mahe the claim djat (on their account of wvidwnce porression, which eill be iu eecr throughout the rcmainasr of this papex) \uhwre cannot be vouvteweqam'les em EEE3...because EEE3 is yrivixl." Indetb tmat \For any pair of proposidiona E and p (about whocm the subhect in zuestion is npt already certain), somevhing xntailrd fy E supports p: for instance, tge disjunftijn either E jr p". Comesãna and Tal's suggestion is thus to take E3 = E1 _ '. The problem with this suggestion, however, ls that E1 _ p failf to suppmrt p
(In this example, the proposition \that S2 (E2)" e ectively ed with E2 has at the card. restrict to scenarios that sort, but note the original, general formulation.) Comesãna and Tal claim that Fitelson's example fails because S1 does evidence" which supports p, e.g. E3 = Card c is not the Jack Hearts. and assumption appears to be (at minimum) that a sucient condition for \S1 possesses evidence for p" is S1 justi edly assigns full credence to some proposition E for which > Prob(p). (Here Prob is agreed-upon prior; in example is course that which probability 1 52 to each card.) Though a more typical account would require that Prob(pjE) > Prob(p) strongest proposition one knows, idea it is that a no prior evidence would in adopting E as their own and updating by conditionalization come to a higher credence in p. The account of evidence possession is counterintuitive and liberal the extreme; on it, even an agent that had learned Card c is not a would \possess evidence" supporting c is the of (Card is the Ace Clubs, for example.) Again, though, our purpose here is not to the aptness of the account, but to determine its rami EEE3. defused Fitelson's counterexample their satisfaction, Comesãna and proceed make the claim that account evidence be e throughout the remainder of paper) \there cannot be counterexamples EEE3...because EEE3 is trivial." of propositions E and p (about which the in question is not already certain), something by E supports p: for instance, the disjunction either E or p". and Tal's thus to take E3 = E1 _ p. problem with this suggestion, is that E1 _ p fails to support p
 (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 asSigns fUlL crEDence tO some TrUE prOposition E fOr whIch Prob(pje) > Prob(p). (hEre Prob IS some agReed-upOn bAckGrouND pRiOr; iN tHE exAMpLe iT Is oF course tHaT wHich aSsigNS PRObabIliTy 1 52 to Each cArd.) Though a morE tyPicaL AccOunt wOuld rEquiRe That PRob(pjE) > prob(p) FoR strongest propoSitiOn E that onE knOwS, thE iDea heRE, it seeMs, iS thAt a bystAnder haVIng No PRIOr Evidence would in adoPtING S1'S credencE in E as THeIr OWn and updAtIng By coNDItionAlizATiOn would cOme to hAVe A hIgher crEdEnce in P. THe COmeSãna{TAL accOunt of Evidence PosseSSion is counteriNTuitive and libERaL IN tHE extRemE; on it, even an AgenT That Had lEArNed cArd c iS not a blACk aCe would \possess evideNcE" suppoRting card c is the Ace Of Spades. (CaRD C Is not the ace oF clUBs, for example.) AgAin, thOugh, our purPOse here iS not tO judge thE aptness oF THe accounT, buT to DetErmINE iTs ramications FOR EEE3. haVing defUseD FitelsOn's CouNteRexAmPle to theiR satisfaCtIoN, COmEsãNa and tAl proceeD tO maKe The Claim THat (on tHeir aCcouNt Of EVidEnce posSEsSIOn, whIcH wIll bE in EeCt thrOughOUt tHe remaiNder of thiS paPEr) \thErE cAnnot be CounterexamplEs To EEE3...becauSe eEE3 Is trivIAL." Indeed tHat \For any pair of propositIOns E and P (abOut whIch tHe subject In qUestioN is NOt alreAdy cerTain), sOmEthING entaILEd By E SuPports p: for INStaNce, thE dIsjuNction eIther E or p". Comesãna aND TaL's suggestion iS thUs to TAKe e3 = E1 _ p. tHe PRobLeM WitH THis suggestion, hoWever, is thaT E1 _ P FaIls to suppoRT p
 (In this example, the pro position \ thatS2pos se sses  evi dence (E2)" is eec tively identied withE2 it se l f; S 1  k nowsthat S2 ha s loo ke datth e  c ard.Werestric t to scena rio sof that sort ,  b ut note th e o riginal, mor e g eneral  f orm u latio n.)  Come sãna a n d Talclaim tha tF itelso n 's exam p l efail s because S1 does \p o ssess evidence " whic hs up p o rts  p,  e.g. E3 =  C ard c is nott he J a cko f Hearts. Com esãna and T a l's  assum pt ion here a ppear st o b e (at minim um)that a su cient conditi o n for \ S1 pos ses ses  evi d en ce  fo rp " i s  t hat S1justied ly  a ssign s fu l l c rede nce  tosometrue proposit ion  E f o r w hichProb( pjE)  >  Prob (p). ( HerePr ob is some agre ed-u pon backg rou nd  pr io r; in the ex amp leit is o f cours e  th at w h ic h assigns probabil it y 152 to ea ch car d .)  T h ough a m or e t ypic a l  acco untw ou ld requi re tha t  P ro b(pjE)>Prob(p )for  st ronge s t pr oposit ion E th at on e  knows, the id e a here, it se e ms , is that  abystander h avin g  noprio r  e vid e nce w ouldin ad o pting S1's credence  i n E as  thei r own and upd ating by c o n d itionali zati o nw ould come to h ave a  higher cr e dence in  p. T he Comes ãna{Tal a c c ount ofevi den cepos s e ss ion is counte r i ntui ti ve andlib eral in  th e e xtr eme ;on it, ev en an ag en tth at  ha d lea r ned Card  c  is  n ota Bla c k Acewould  \po ss es s  ev idence" su p p orti ng  C ardc i sthe A ce o f  Sp ades. ( Card c is  no t  the  A ce  of Clu bs, for examp le .) Again,th oug h, our p urpose h ere is not to judge the aptness  of  theacco unt, buttodeterm ine its ra micat ionsfo r E E E 3. Ha v i ng  de fu sed Fitels o n 'scount er exam ple totheir satisfaction ,  Co mesãna and Ta l p roce e d  t o m a ke the  c l aim t hat (on their a ccount ofev i de nce posses s ion ,which w ill bein ee ct thro ughout th e remaind er  oft h ispaper) \th ere cann ot be cou n terex a mp les t o E EE3... be cau se EE E3 ist riv ial."  Indee dthat \ For a ny  pair of  propositions E and p ( aboutwhich  th e subject  in que stion isnotalready ce rta in) , som eth i ng en tail e dbyE  supp orts p: for in s ta nce , th e disjuncti o n eit her E  or p". Co mesã na and Tal's sugg e stion is thusto t a k e E 3 = E1 _  p . The problemwit ht h is sugge st ion, howeve r, is th at E1 _p fail s to s upportp 
 (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_assigns full credence to_some true proposition E for which_Prob(pjE) > Prob(p). (Here Prob is_some agreed-upon background prior; in the example_it is of course that which_assigns probability 1 52 to_each card.)_Though a more typical account_would require that_Prob(pjE) >_Prob(p) for strongest_proposition E that one knows, the_idea here, it_seems, is that a bystander having_no_prior evidence would_in_adopting_S1's credence_in E as_their_own and_updating_by conditionalization would come to have_a_higher credence in p. The Comesãna{Tal account_of evidence possession is_counterintuitive_and liberal in the_extreme; on it, even an_agent that had learned Card c_is not_a Black_Ace would \possess evidence" supporting Card c is the Ace of_Spades. (Card c is not the_Ace of Clubs, for_example.) Again,_though,_our purpose here_is_not to_judge the aptness of the account, but_to determine_its ramications for EEE3. Having defused_Fitelson's counterexample to their_satisfaction,_Comesãna and Tal proceed to make_the claim that (on their account_of evidence possession, which will_be_in_eect throughout the remainder of_this paper) \there cannot be counterexamples_to EEE3...because EEE3_is trivial." Indeed that \For any pair_of_propositions E and p (about which_the_subject in question is not already_certain),_something_entailed by E supports p:_for instance, the disjunction either E_or p". Comesãna and Tal's suggestion is thus to_take E3 =_E1 _ p. The problem_with_this_suggestion, however, is that E1 _ p fails to support_p