 (2015) have recently discussed one version of this principle, called EEE3. Fitelson claims that EEE3 is false and oers a counterexample. Comesãna and Tal oer a refutation of this counterexample and claim that EEE3 is true, albeit trivially. It is important to ag the fact that Comesãna and Tal's analysis depends crucially on a novel and contentious account of evidence possession. (If one denies this account of evidence possession, their refutation of Fitelson evaporates.) Our purpose here isn't to judge the aptness of the 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".
(2015) have recently discussed one version of this principle, called EEE3. Fitelson claim that EEE3 is assumed and o  ers a counterexample. Comesãna and Tal o  er a refutation of this counterexample and claim that EEE3 is dependable, albeit trivially. It is important to ag the fact that Comesãna and Tal's analysis count crucially on a fresh and contentious account of evidence possession. (If one denies this account of evidence possession, their falsification of Fitelson evaporates .) Our purpose here isn't to estimate the aptness of the score, but to show that Comesãna and Tal were misguided; even accepting their account of evidence monomania, EEE3 remains false.2 1A previous version of this newspaper was rejected (in 177 days) by Analysis against the advice of the original referees on the basis of an uniquely fallacious third composition. Four days later I sent  this translation to Dialectica, who have sat on it unresponsively for (as of this writing) 796 days. 2Feldman (2014) responds to Fitelson's proposed counterexample by o  ering however di  erent 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 support p, then S has some evidence, E2, that defend p. " We are try to make a case against neither Feldman nor Fitelson here. 1 MORE TROUBLES FOR COMESÃNA AND TAL 2 (EEE3 ): If S1 possess evidence (E1) which supports the claim that S2 possesses evidence (E2) which supports phosphorus, 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 (choose uniformly at random from a standard deck of play cards) has been viewed by S2, and that S1 has evidence E1. Given these assumptions, E1 corroborate the claim that S2 possesses E2. E2 meanwhile entails phosphorus so by EEE3 S1 ought to own evidence supporting p. However, E1 does not support p. indeed goes the \counterexample ".
 (2015) hwve recently discussed oke version of thnw prinriple, czlled EED3. Fitelson claims that EEE3 id dalse and oers a counterexaople. Comedãna and Ral iwr a refutefion of this dluntzrxxample and clalm that EEE3 is true, albeid grnvially. It is important to ag the fast that Clmesãna and Tal'f ansjysia depends crucially on a novel and contenuious account of efidence possession. (If one fenifs this account of evidence pissefwion, their rdfutation of Fitelson gvaporates.) Our purpose here isn't to jbdge the apjuwss mf the acconnt, bun to show thab Comesãta and Yal were mistalen; evwn accepting their acrount of evidence pofsession, AEZ3 remains false.2 1A preciius vgrsiot of rhir pzpxr sas reuecved (in 177 daya) by Analysus against the advive if the originzl refqrqes on the basis of an unambiguously fanladious third report. Four days later I sent  tjis versijn to Dialectica, who have sat on it unresponsivelf for (xs if gyid writing) 796 days. 2Feldman (2014) responds to Fitelson'f ptoiosed counterexamile by oering stilk firtent takes on GEE3 as cslm as a new formulahion, rugning \If S has evicence, E1, supporting the propisition that rhere is someone wko has evideuce thst sulports p, then S has somz evidsnce, E2, that supports o." We are attemptkng tm make a case against neitrer Feldmen nox Fitelsun hgre. 1 MOWE TROUBLED FOR COMESÃNA AND TAL 2 (EFE3): If S1 possesses evidence (E1) which supports the rkaim that S2 kosvesves evidznce (E2) which suppores p, then S1 poxsesses eviddnce (E3) whibh supporvs p. Here is Fitelson's prmkosed counterxxample tj EEW3. Ler E1 = Cafa c is black; E2 = Card c ps the Ace of Spades; p = Gard z is an Ace. It iw sopposed that S1 lnods ehwt Cwsd c (selecteg unkfofkly ag random frio a xtandard deck of plafing cards) has been virwcd by S2, abd that F1 has evidencr E1. Given these asdumptmons, E1 suppprtf the claim that S2 possesses E2. E2 meanwhlle entails p sj by EEE3 S1 ought to have evidence supporting p. However, X1 does not support p. So goes the \counterexcmkle".
(2015) have recently discussed one version of called Fitelson claims EEE3 is false Comesãna Tal o er refutation of this and claim that EEE3 is true, trivially. It is important to ag the fact that Comesãna and Tal's analysis crucially on a novel and contentious account of evidence possession. (If one denies account evidence their of Fitelson evaporates.) Our purpose here isn't to judge the aptness of the account, but to that Comesãna and Tal were mistaken; even accepting account of evidence possession, remains false.2 1A previous version this was rejected 177 by against the advice the original referees on the basis of an unambiguously fallacious third report. Four days later I sent this version who have on unresponsively (as of this days. 2Feldman (2014) responds to Fitelson's o ering still di erent takes on EEE3 well as new formulation, running \If S has E1, supporting the proposition that there is someone has evidence that supports p, then S has some evidence, E2, that supports p." We to make a case neither Feldman nor here. MORE FOR AND TAL (EEE3): If S1 possesses evidence (E1) which supports the claim that possesses evidence (E2) which supports p, then S1 possesses evidence supports Here is Fitelson's counterexample to EEE3. Let = c is black; E2 c the p Card is an Ace. It supposed that S1 knows that c (selected uniformly at of playing cards) has been viewed by S2, that S1 has evidence E1. Given these E1 supports the claim that S2 possesses E2. E2 meanwhile entails p by EEE3 to have evidence supporting p. However, E1 does support p. So goes \counterexample".
 (2015) have recently discussed one vErsion of thIs priNciPle, CaLled eEE3. FItelson claims tHAt EEe3 is false and oers a counteRexamPlE. comeSÃnA and TAl oer a rEFuTATioN oF tHis CoUNtErexaMplE and claIm that EEE3 iS trUe, Albeit triviaLLy. it is importAnt To ag the fact tHat comesãNa And tAl's anAlySis dePends cRUciallY on a novel AnD ContenTIous accOUNt Of evIdence possession. (IF OnE Denies this accoUnt of eViDEnCE PosSesSion, their rEfUtatiON of FiteLSoN EVApoRAtes.) Our purposE here isn't to JUdgE the apTnEss OF the 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".
 (2015) have recently disc ussed oneversi onofth is p rinc iple, called E E E3.Fitelson claims that E EE3 i sf alse an d oe rs a co u nt e r exa mp le . C om e sã na an d T al oer  a refutat ion  o f this count e re xample and  cl aim that EEE 3 i s true ,alb e it tr ivi ally.  It is import ant to ag  t h e fact that Co m e sã na a nd Tal's analysis de p ends crucially  on ano v el a ndcon tentious a cc ounto f evide n ce p o sse s sion. (If one  denies thi s  ac countof  ev i denceposse ss i on,  their refu tati on of Fit elsone vaporat e s.) Our  purpo seher e is n 't  t o j ud g e t h eapt n ess  of theac co unt,butt o s howtha t Co mesãn a and Tal wer e m ista k en;  even  acce ptin gtheir  accou nt of  e vidence possess ion,  EEE3 rem ain sfal se .2 1A previo usver sion of  this p a per  w a s re jected (in 177 day s) b yAnalysis  again s tth e  adviceof  th e or i g inalrefe r ee s on the  basis of  a n unamb ig uously  f all aci ous t h irdreport . Four d ays l a ter I sent  t h is version to Di a l ec t ica,  wh o have saton i t  unr espo n si vel y  for(as o ft hi s  writing) 796 days.  2 Feldma n (20 14) respondsto Fitelso n ' s  propose d co u nt e rexample by oering  still die rent tak es on  EEE3 as  well asa new form ula tio n,run n i ng  \If S has ev i d ence ,E1, sup por ting th e p rop osi tio nthat ther e is som eo ne  w ho  ha s evi d ence tha tsup po rts  p, t h en S h as so me e vi de n ce,  E2, th a ts u ppor ts  p ." W e a re  atte mpti n g t o makea case ag ain s t ne it he r Feldm an nor Fitels on  here. 1 M OR E T ROUBLE S FOR COME SÃNA AND TAL 2 (EEE3):I f S1 po sse ssesevid ence (E1)  wh ich su ppo r ts the  claim  that  S 2 p o s sesse s ev ide nc e (E2) whi c h  su pport sp, t hen S1possesses evidence (E3 ) which suppo rts  p.H e re  is Fi t els on ' s p r o posed counterex ample to E EE 3 .Let E1 = C a rdcis blac k; E2 =  Card c is th e Ace ofSpades; p  =  Car d c i s an Ace.It is su pposed th a t S1k no ws th atCard c  ( sel ected  unifo r mly  at r andomfr om a s tanda rd  deck of  playing cards) has bee n view ed by  S2 , and tha t S 1  ha s evidenc e E1 . Given th ese  as sumpt ion s , E1supp o rt s t h e cla im t h at S2 pos s es ses E 2.  E2 meanwhi l e ent ailsp s o  by EE E3 S 1 ought to have e v idence support ingp .  Ho wev e r, E 1does not suppo rtp. S o goes t he  \counterex ample".
 (2015)_have recently_discussed one version of_this principle,_called_EEE3. Fitelson_claims_that EEE3 is_false and oers_a counterexample. Comesãna and_Tal oer a_refutation_of this counterexample and claim that EEE3 is true, albeit trivially. It is important_to_ag the_fact_that_Comesãna and Tal's analysis depends_crucially on a novel and_contentious account_of evidence possession. (If one denies this account_of_evidence possession, their_refutation of Fitelson evaporates.) Our purpose here isn't to_judge the aptness of the 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".