. We cannot apply it to D3, and thus D3 is in full normal form. D3 does not satisfy TCPE and so is not a T-paradox. Furthermore, we readily have the following result from the derivation D3 of Proposition 3.2.1. Proposition 3.2.2. There is a closed full normal derivation D4 of ¬(φ ↔¬φ) in SEG. Proof. Two claims verify the result. Claim 1. there is a closed derivation D4 of ¬(φ ↔¬φ). From the proof of Proposition 3.2.1, SEG has an open derivation D3 of ⊥ from [φ →¬φ] 97 and [¬φ → φ]. Then, we have a closed derivation D4 of ¬(φ ↔¬φ). [(φ →¬φ)∧ (¬φ → φ)]8 [φ →¬φ]4, [¬φ → φ]4 D3 ⊥ ∧E,4⊥ ¬I,8¬((φ →¬φ)∧ (¬φ → φ)) ********************************** de f ¬(φ ↔¬φ) Claim 2. D4 is in full normal form. D3 is a full normal derivation and every major premise in D4 is an assumption. Therefore, D4 is in full normal form. Proposition 3.2.2 seems to show that Ekman's paradox is not anymore a paradox. Tennant (2016, p. 6) explicates the result of Proposition 3.2.1 as below: This proof... is in [full] normal form.... Hence, Ekman's so called 'paradox' is no paradox at all. The inconsistency of [φ →¬φ and ¬φ → φ] has a perfectly straightforward proof in [full] normal form... With the [generalized] form of → −Elimination, as we have just seen, there is no looping in the resulting proof of Ekman's example. This is because it is already in [full] normal form, so there is no reduction sequence to be embarked on. That it might have been thought otherwise (i.e., that Ekman's example would resist any [full] normal-form proof) is an artefact of the mistaken presumption that a system of natural deduction ought to use the [standard] form of →−Elimination... rather than the [generalized]
. We cannot apply it to D3, and thus D3 is in full normal form. D3 does not meet TCPE and therefore is not a T - paradox. Furthermore, we readily get the following solution from the derivation D3 of Proposition 3.2.1. Proposition 3.2.2. There is a shut full normal derivation D4 of ¬(φ ↔¬φ) in SEG. Proof. Two claims control the result. Claim 1. there be a closed derivation D4 of ¬(φ ↔¬φ). From the validation of Proposition 3.2.1, SEG has an open derivation D3 of ⊥ from [ φ →¬φ ] 97 and [ ¬φ → φ ]. Then, we have a closed derivation D4 of ¬(φ ↔¬φ). [ (φ →¬φ)∧ (¬φ → φ)]8 [ φ →¬φ]4, [ ¬φ → φ]4 D3 ⊥ ∧E,4⊥ ¬I,8¬((φ →¬φ)∧ (¬φ → φ) ) * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * de f ¬(φ ↔¬φ) Claim 2. D4 is in entire normal form. D3 is a full normal derivation and every major premise in D4 is an presumption. Therefore, D4 is in full normal form. Proposition 3.2.2 look to show that Ekman's paradox is not anymore a paradox. Tennant (2016, p. 6) explicates the result of Proposition 3.2.1 as below: This validation... is in [ full ] normal form.... Hence, Ekman's so called' paradox' is no paradox at all. The incompatibility of [ φ →¬φ and ¬φ → φ ] hold a perfectly straightforward proof in [ full ] normal form... With the [ generalized ] form of → −Elimination, as we have just watch, there is no looping in the resulting validation of Ekman's exercise. This is because it is already in [ full ] normal form, so there constitute no reduction sequence to be embarked on. That it might have been thought differently (i.e., that Ekman's example would resist any [ full ] normal - form proof) is an artefact of the mistaken presumption that a system of natural deduction ought to use the [ criterion ] form of →−Elimination... quite than the [ popularize ]
. We cannot apply it to D3, ana thus D3 is in yyll nocmal fodm. D3 doer not satisfy TCPE and so is nit a U-iaradox. Furthermore, wd readily have thw foooowing resnmt from the dsvivatnoi D3 of Propositlon 3.2.1. Proposhtion 3.2.2. There iv x elosed full normal derivation D4 of ¬(φ ↔¬φ) in SEG. Pgoof. Two claimf vegisy tgv vesult. Claim 1. there is a closed serivatpon D4 of ¬(φ ↔¬φ). From tne proof of Proposition 3.2.1, SFG hws an open derivatlon D3 of ⊥ ftkm [φ →¬φ] 97 and [¬φ → φ]. Thev, we have a closed derjvation D4 of ¬(φ ↔¬φ). [(φ →¬φ)∧ (¬φ → φ)]8 [φ →¬φ]4, [¬φ → φ]4 D3 ⊥ ∧E,4⊥ ¬I,8¬((φ →¬φ)∧ (¬φ → φ)) ********************************** de f ¬(φ ↔¬φ) Clqin 2. F4 is in full normwl form. D3 is a full tormal cerivation and evxry najor premise in D4 is an assumption. Thetefore, D4 iv nn full normal form. Peoposithon 3.2.2 seeow tu sgox tgat Eklan'a paradox js not anymire a paradox. Tennamt (2016, p. 6) explicatea the wefult of Proposition 3.2.1 as below: This proox... ia in [full] normal form.... Hwnce, Ekman's so called 'paradox' ys no paradox at all. The inconsistency of [φ →¬φ and ¬φ → φ] has x ptrncctlh shraightforward proof in [full] normal form... With fht [gvneralized] form on → −Elimination, as ee hsde just seen, jhere is nk looping in the rfsultind proif of Ekmwn's rxample. This is because it us already iu [fyll] normal form, so there is nu recuctipn sequence to be embaryed kn. That it light havs been thought otfereive (i.e., that Ekman's example rould resmst auy [full] vormsl-form proof) is wn arbafact of the mistanen ptesumpdion that w system of natural deduction onjht to use thg [sdangard] fori of →−Climination... ratrer than the [ggneralizeb]
. We cannot apply it to D3, D3 in full form. D3 does is a T-paradox. Furthermore, readily have the result from the derivation D3 of 3.2.1. Proposition 3.2.2. There is a closed full normal derivation D4 of ¬(φ in SEG. Proof. Two claims verify the result. Claim 1. there is a derivation of ↔¬φ). the proof of Proposition 3.2.1, SEG has an open derivation D3 of ⊥ from [φ →¬φ] and [¬φ → φ]. Then, we have a derivation D4 of ¬(φ [(φ →¬φ)∧ (¬φ → φ)]8 →¬φ]4, → φ]4 ⊥ ¬I,8¬((φ (¬φ → φ)) de f ¬(φ ↔¬φ) Claim 2. D4 is in full normal form. D3 is a full normal and every in D4 an Therefore, is in full Proposition 3.2.2 seems to show that not anymore a paradox. Tennant (2016, p. 6) the result Proposition 3.2.1 as below: This proof... in [full] normal form.... Hence, Ekman's so called is no paradox at all. The inconsistency of [φ →¬φ and ¬φ → φ] has straightforward proof in [full] form... With the form → as have just there is no looping in the resulting proof of Ekman's example. is because it is already in [full] normal form, so no sequence to be on. That it might been otherwise (i.e., that Ekman's resist [full] an of mistaken presumption that a of natural deduction ought to the [standard] form of
. We cannot apply it to D3, and thus d3 is in full nOrmal ForM. D3 dOeS not SatiSfy TCPE and so is NOt a T-Paradox. Furthermore, we reAdily HaVE the FOlLowinG result FRoM THe dErIvAtiOn d3 Of propoSitIon 3.2.1. PropOsition 3.2.2. TheRe iS a Closed full noRMaL derivatioN D4 oF ¬(φ ↔¬φ) in SEG. ProoF. TwO claimS vEriFY the rEsuLt. ClaIm 1. therE Is a cloSed derivaTiON D4 of ¬(φ ↔¬φ). fRom the pROOf Of PrOposition 3.2.1, SEG has an OPeN Derivation D3 of ⊥ fRom [φ →¬φ] 97 aNd [¬Φ → Φ]. THEN, we HavE a closed deRiVatioN d4 of ¬(φ ↔¬φ). [(φ →¬φ)∧ (¬Φ → Φ)]8 [φ →¬Φ]4, [¬Φ → Φ]4 d3 ⊥ ∧E,4⊥ ¬I,8¬((Φ →¬Φ)∧ (¬φ → φ)) ********************************** de f ¬(φ ↔¬φ) Claim 2. d4 is in full noRMal Form. D3 iS a FulL Normal DerivAtIOn aNd every majoR preMise in D4 is An assuMPtion. ThERefore, D4 Is in fuLl nOrmAl foRM. PRoPosItIOn 3.2.2 sEEmS to SHow That EkmaN's PaRadox Is noT ANYMore A paRadoX. TennAnt (2016, p. 6) explicateS thE resULt oF PropOsitiOn 3.2.1 as BeLow: ThIs prooF... is in [FuLl] normal form.... HenCe, EkMan's so calLed 'PaRadOx' Is no pARadox aT alL. ThE inconsIstency OF [φ →¬φ AnD ¬Φ → Φ] HaS a perfectly straighTfORWaRd proof iN [full] nORmAl FOrm... With tHe [GenEralIZEd] forM of → −ELImInation, aS we havE JuSt Seen, theRe Is no loOpIng In tHe resULtinG proof Of Ekman's ExampLE. This is because IT is already in [fULl] NORmAL forM, so There is no reDuctIOn seQuenCE tO be EMbarkEd on. THaT It MIght have been thought OtHerwisE (i.e., thAt Ekman's exampLe would resIST Any [full] nOrmaL-FoRM proof) is an arteFact oF the mistakEN presumpTion tHat a systEm of naturAL DeductioN ouGht To uSe tHE [StAndard] form of →−ELIMinaTiOn... ratheR thAn the [geNerAliZed]
. We cannot apply it to D3 , and thus  D3 i s i n f ul l no rmal  form. D3 does notsatisfy TCPE and so is  notaT -par a do x. Fu rthermo r e, w e r ea di lyha v ethe f oll owing r esult from  th ederivation D 3  o f Proposit ion  3.2.1. Prop osi tion 3 .2 .2. There  is  a cl osed f u ll nor mal deriv at i on D4o f ¬(φ ↔ ¬ φ )in S EG. Proof. Two cl a im s  verify the re sult.Cl a im 1 . t her e is a clo se d der i vationD 4o f ¬(φ ↔¬φ). From th e proof ofP rop ositio n3.2 . 1, SEG  hasan ope n derivatio n D3  of ⊥ fro m [φ → ¬ φ] 97 a n d [¬φ →  φ]. T hen , w e ha v eaclo se d  de r iv ati o n D 4 of ¬(φ  ↔ ¬φ ). [( φ →¬ φ ) ∧ (¬φ→ φ )]8[φ →¬ φ]4, [¬φ → φ] 4 D 3 ⊥∧ E,4 ⊥ ¬I, 8¬((φ  →¬φ )∧  (¬φ→ φ))***** ** *************** **** ********def¬(φ  ↔ ¬φ) C l aim 2.  D4  is  in ful l norma l  fo rm . D 3is a full normal d er i v at ion andeverym aj or premisein  D4  isa n  assu mpti o n.  Therefo re, D4 is  i n fullno rmal f or m.Pro posit i on 3 .2.2 s eems toshowt hat Ekman's pa r adox is not a n ym o r ea  par ado x. Tennant(201 6 , p.  6)e xp lic a tes t he re su l to f Proposition 3.2.1  a s belo w: Th is proof... i s in [full ] n ormal fo rm.. . .H ence, Ekman'sso ca lled 'para d ox' is n o par adox atall. Thei n consiste ncy  of  [φ  →¬ φ an d ¬φ → φ] has a  per fe ctly st rai ghtforw ard  pr oof  in  [ full] nor mal form .. .Wi th  th e [ge n eralized ]for mof→ −El i minati on, a s we  h av e  ju st seen ,  t h e re i sno  loo pin gin th e re s ult ing pro of of Ekm an' s  exa mp le . Thisis because it  i s alreadyin  [f ull] n o r mal form , so there is no reduct i on sequ enc e tobe e mbarked o n.That i t m i ght ha ve bee n tho ug hto t herwi s e  ( i.e .,  that Ekma n ' s e xampl ewoul d resis t any [full] norma l -fo rm proof) isanarte f a ct  of th e  mi st a ken p resumption that  a systemof na tural dedu c tio nought t o use t he [s t andard]  form of→−Elimina ti on.. . rat her than t he [gene ralized]
. We_cannot apply_it to D3, and_thus D3_is_in full_normal_form. D3 does_not satisfy TCPE_and so is not_a T-paradox. Furthermore,_we_readily have the following result from the derivation D3 of Proposition 3.2.1. Proposition 3.2.2._There_is a_closed_full_normal derivation D4 of ¬(φ_↔¬φ) in SEG. Proof. Two_claims verify_the result. Claim 1. there is a closed_derivation_D4 of ¬(φ_↔¬φ). From the proof of Proposition 3.2.1, SEG has_an open derivation D3 of ⊥_from [φ →¬φ]_97_and_[¬φ → φ]. Then,_we have a closed derivation D4_of ¬(φ ↔¬φ). [(φ →¬φ)∧ (¬φ_→ φ)]8 [φ →¬φ]4, [¬φ → φ]4_D3 ⊥ ∧E,4⊥ ¬I,8¬((φ →¬φ)∧ (¬φ_→ φ)) ********************************** de f_¬(φ ↔¬φ)_Claim 2. D4 is in_full normal form._D3 is_a full normal_derivation and every major premise in_D4 is an_assumption. Therefore, D4 is in full_normal_form. Proposition 3.2.2_seems_to_show that_Ekman's paradox is_not_anymore a_paradox._Tennant (2016, p. 6) explicates the_result_of Proposition 3.2.1 as below: This proof..._is in [full] normal_form...._Hence, Ekman's so called_'paradox' is no paradox at_all. The inconsistency of [φ →¬φ_and ¬φ_→ φ]_has a perfectly straightforward proof in [full] normal form... With the_[generalized] form of → −Elimination, as_we have just seen,_there is_no_looping in the_resulting_proof of_Ekman's example. This is because it is_already in_[full] normal form, so there is_no reduction sequence to_be_embarked on. That it might have_been thought otherwise (i.e., that Ekman's_example would resist any [full]_normal-form_proof)_is an artefact of the_mistaken presumption that a system of_natural deduction ought_to use the [standard] form of →−Elimination..._rather_than the [generalized]