 [¬φ → φ] in SE. First, we have an open derivation D1 of ⊥ from [φ →¬φ] and [φ]. [φ →¬φ]1 [φ]2 → E¬φ [φ]2 ¬E⊥ 93 With the derivation D1, we have an open derivation D2 of φ from [φ →¬φ] and [¬φ → φ]. [¬φ → φ]3 [φ →¬φ]1, [φ]2 D1 ⊥ ¬I,2¬φ ¬E φ Now, we have the following open derivation Π1 of ⊥ from [φ →¬φ] and [¬φ → φ]. [φ →¬φ]1, [φ]2 D1 ⊥ ¬I,2¬φ [φ →¬φ]1, [¬φ → φ]3 D2 φ ¬E⊥ Claim 2. If an Ekman reduction process is used for Π1, then Π1 generates a non-terminating reduction sequence and is not normalizable. Π1 has a maximum formula ¬φ in ¬E−rule. By applying ¬−reduction, we have the following derivation Π2. [φ →¬φ]1 [¬φ → φ]3 [φ →¬φ]1, [φ]2 D1 ⊥ ¬I,2¬φ → E φ → E¬φ [φ →¬φ]1, [¬φ → φ]3 D2 φ ¬E⊥ The derivation Π2 has no maximum formula but it has an Ekman maximum formula. By applying Ekman reduction, we obtain the same derivation with Π1 which we started. Therefore, if the set of reductions of SE includes Ekman reduction, Π1 initiates a non-terminating reduction sequence and so is not normalizable. 94 We call the result of Proposition 3.1.1 Ekman's paradox. Schroeder-Heister and Tranchini (2017, pp. 570-571) appear to think that the derivation Π1 uses id est inference since Π1 has inferences from φ to ¬φ and ¬φ to φ. Also, Π1 is a derivation of ⊥ and generates a reduction loop. Hence, according to TCPE, Π1 is a T-paradox. Thus, Ekman's paradox can be a counterexample to
[ ¬φ → φ ] in SE. First, we have an open derivation D1 of ⊥ from [ φ →¬φ ] and [ φ ]. [ φ →¬φ]1 [ φ]2 → E¬φ [ φ]2 ¬E⊥ 93 With the deriving D1, we take an open derivation D2 of φ from [ φ →¬φ ] and [ ¬φ → φ ]. [ ¬φ → φ]3 [ φ →¬φ]1, [ φ]2 D1 ⊥ ¬I,2¬φ ¬E φ Now, we own the following open ancestry Π1 of ⊥ from [ φ →¬φ ] and [ ¬φ → φ ]. [ φ →¬φ]1, [ φ]2 D1 ⊥ ¬I,2¬φ [ φ →¬φ]1, [ ¬φ → φ]3 D2 φ ¬E⊥ Claim 2. If an Ekman decrease process is used for Π1, then Π1 generates a non - terminating reduction sequence and is not normalizable. Π1 suffer a maximum convention ¬φ in ¬E−rule. By practice ¬−reduction, we have the following derivation Π2. [ φ →¬φ]1 [ ¬φ → φ]3 [ φ →¬φ]1, [ φ]2 D1 ⊥ ¬I,2¬φ → E φ → E¬φ [ φ →¬φ]1, [ ¬φ → φ]3 D2 φ ¬E⊥ The derivation Π2 accept no maximum formula but it receive an Ekman maximum formula. By applying Ekman decrease, we obtain the like derivation with Π1 which we started. Therefore, if the set of reduction of SE includes Ekman reduction, Π1 initiates a non - terminating reduction sequence and therefore is not normalizable. 94 We call the result of Proposition 3.1.1 Ekman's paradox. Schroeder - Heister and Tranchini (2017, pp. 570 - 571) appear to think that the derivation Π1 uses i d est inference since Π1 has inferences from φ to ¬φ and ¬φ to φ. Also, Π1 is a deriving of ⊥ and generates a reduction closed circuit. therefore, accord to TCPE, Π1 is a T - paradox. Thus, Ekman's paradox can be a counterexample to
 [¬φ → φ] in SE. First, we have an upen derivation D1 of ⊥ hrom [φ →¬φ] znd [φ]. [φ →¬φ]1 [φ]2 → E¬φ [φ]2 ¬E⊥ 93 With the derivation D1, ww havt an open derivatiov D2 of φ fgom [φ →¬φ] and [¬φ → φ]. [¬φ → φ]3 [φ →¬φ]1, [φ]2 D1 ⊥ ¬I,2¬φ ¬X φ Now, wc havs the hollowing open cerivation Π1 of ⊥ from [φ →¬φ] atd [¬φ → φ]. [φ →¬φ]1, [φ]2 D1 ⊥ ¬I,2¬φ [φ →¬φ]1, [¬φ → φ]3 D2 φ ¬E⊥ Claim 2. If an Ekian redichion process if ustd sor Π1, nhtn Π1 generates a non-terminating resuction sequence and is not normalizable. Π1 has w madimum formula ¬φ in ¬F−rule. By apkmyigt ¬−reduction, de have tht yollowing dgrivation Π2. [φ →¬φ]1 [¬φ → φ]3 [φ →¬φ]1, [φ]2 D1 ⊥ ¬I,2¬φ → E φ → E¬φ [φ →¬φ]1, [¬φ → φ]3 D2 φ ¬E⊥ The detncatlmn Π2 has no naximlm formula bub it hav an Ekkan maximum fovmula. By applying Ekman reducvion, we obtain the swme derivdtnon with Π1 which we stqrred. Tverexore, uf ghe sxt kf redkctmons of SE jncludes Eknan reduction, Π1 initoaevx a non-termihating rqduction sequence and so is not normalieable. 94 We call the result of Proposition 3.1.1 Ekman's karadox. Scrroeder-Heister and Tranchini (2017, pp. 570-571) appear to think that ghe dcrivxriln Π1 uses id est inference since Π1 has inferencqa grpm φ to ¬φ and ¬φ tj φ. Also, Π1 is a drtivation of ⊥ avd genzdafes a reduction lolp. Hencg, accoeding to UCPE, Π1 is a T-paradox. Thus, Ekman's paradox can ye q counterexample tl
[¬φ → φ] in SE. First, we open D1 of from [φ →¬φ] → [φ]2 ¬E⊥ 93 the derivation D1, have an open derivation D2 of from [φ →¬φ] and [¬φ → φ]. [¬φ → φ]3 [φ →¬φ]1, [φ]2 ⊥ ¬I,2¬φ ¬E φ Now, we have the following open derivation Π1 of from →¬φ] [¬φ φ]. [φ →¬φ]1, [φ]2 D1 ⊥ ¬I,2¬φ [φ →¬φ]1, [¬φ → φ]3 D2 φ ¬E⊥ Claim If an Ekman reduction process is used for then Π1 generates a reduction sequence and is not Π1 a maximum ¬φ ¬E−rule. applying ¬−reduction, we the following derivation Π2. [φ →¬φ]1 [¬φ → φ]3 [φ →¬φ]1, [φ]2 D1 ⊥ ¬I,2¬φ → E → E¬φ [¬φ → D2 ¬E⊥ derivation Π2 has formula but it has an Ekman applying Ekman reduction, we obtain the same derivation Π1 which started. Therefore, if the set of of SE includes Ekman reduction, Π1 initiates a reduction sequence and so is not normalizable. 94 We call the result of Proposition 3.1.1 Schroeder-Heister and Tranchini (2017, 570-571) appear to that derivation uses est inference Π1 has inferences from φ to ¬φ and ¬φ to φ. Π1 is a derivation of ⊥ and generates a reduction according TCPE, Π1 is T-paradox. Thus, Ekman's paradox be counterexample to
 [¬φ → φ] in SE. First, we have an open deRivation D1 oF ⊥ from [Φ →¬φ] aNd [φ]. [Φ →¬φ]1 [Φ]2 → E¬φ [φ]2 ¬e⊥ 93 WitH the derivation d1, We haVe an open derivation D2 of φ From [φ →¬Φ] aND [¬φ → φ]. [¬φ → Φ]3 [Φ →¬φ]1, [Φ]2 D1 ⊥ ¬I,2¬φ ¬E Φ Now, we hAVe THE foLlOwIng OpEN dErivaTioN Π1 of ⊥ froM [φ →¬φ] and [¬φ → φ]. [φ →¬φ]1, [Φ]2 D1 ⊥ ¬I,2¬Φ [φ →¬Φ]1, [¬φ → φ]3 D2 φ ¬E⊥ Claim 2. IF An ekman reducTioN process is usEd fOr Π1, theN Π1 GenERates A noN-termInatinG ReductIon sequenCe ANd is noT NormaliZABlE. Π1 haS a maximum formula ¬φ IN ¬E−RUle. By applying ¬−rEductiOn, WE hAVE thE foLlowing derIvAtion π2. [Φ →¬φ]1 [¬φ → φ]3 [φ →¬φ]1, [φ]2 d1 ⊥ ¬i,2¬φ → e Φ → e¬Φ [φ →¬φ]1, [¬Φ → Φ]3 D2 φ ¬E⊥ The derivaTion Π2 has no mAXimUm formUlA buT It has aN EkmaN mAXimUm formula. By ApplYing Ekman ReductIOn, we obtAIn the saMe deriVatIon With π1 WhIcH we StARteD. thEreFOre, If the set Of ReDuctiOns oF se INcluDes ekmaN reduCtion, Π1 initiatEs a Non-tERmiNatinG reduCtioN sEquenCe and sO is noT nOrmalizable. 94 We caLl thE result of proPoSitIoN 3.1.1 EkmaN'S paradOx. SChrOeder-HeIster anD traNcHINI (2017, pP. 570-571) appear to think that ThE DErIvation Π1 Uses id ESt InFErence siNcE Π1 hAs inFERenceS froM Φ tO ¬φ and ¬φ to Φ. Also, Π1 IS a DeRivatioN oF ⊥ and geNeRatEs a ReducTIon lOop. HenCe, accordIng to tcPE, Π1 is a T-paradoX. thus, Ekman's parADoX CAn BE a coUntErexample to
 [¬φ → φ] in SE. First, we  have an o pen d eri vat io n D1  of⊥ from [φ →¬φ] and[φ]. [φ →¬φ]1 [φ]2 → E ¬φ [φ ]2 ¬E⊥9 3Withthe der i va t i onD1 ,weha v ean op enderivat ion D2 ofφ f ro m [φ →¬φ] an d  [ ¬φ → φ]. [ ¬φ→ φ]3 [φ →¬φ ]1,  [φ]2D1  ⊥¬ I,2¬φ  ¬E  φ No w, weh ave th e followi ng open d e rivatio n Π1  of⊥ from [φ →¬φ] an d  [ ¬ φ → φ]. [φ →¬φ ]1, [φ ]2 D1 ⊥  ¬I ,2¬ φ [φ →¬φ]1 ,[¬φ → φ]3 D2φ  ¬ E ⊥ Cla i m 2. If an Ek man reducti o n p rocess  i s u s ed for  Π1,th e n Π 1 generates  a n on-termin atingr eductio n  sequen ce and  is  no t no r ma li zab le .  Π1 ha s a max imum for mu la  ¬φ i n ¬E − r u l e. B y a pply ing ¬ −reduction, w e h avet hefollo wingderi va tionΠ2. [φ  →¬φ] 1[¬φ → φ]3 [φ →¬ φ]1,  [φ]2 D1⊥ ¬ I, 2¬φ  →  E φ→  E¬φ [ φ → ¬φ] 1, [¬φ→ φ]3 D 2  φ¬E ⊥ T he  derivation Π2 has  n o ma ximum fo rmulab ut  i t  has anEk man  max i m um fo rmul a .By apply ing Ek m an  r eductio n,  we ob ta inthe  same deri vation  with Π1  whic h  we started. T h erefore, if t h es e to f re duc tions of SE  inc l udes  Ekm a nred u ction , Π1in i ti a tes a non-terminati ng  reduc tionsequence andso is notn o r malizabl e. 9 4  W e  call the resu lt of  Propositi o n 3.1.1Ekman 's parad ox. Schro e d er-Heist erand  Tr anc h i ni  (2017, pp. 5 7 0 -571 )appeartothink t hat  th e d eri va tion Π1 u ses id e st  i nf er enc e sin c e Π1 has  i nfe re nce s fro m  φ to¬φ an d ¬φ  t oφ . A lso, Π1 is a  der iv at ionof⊥and g ener a tes  a redu ction loo p.H ence ,ac cording  to TCPE, Π1is  a T-parad ox . T hus, E k m an's par adox can be a counterex a mple to 
 [¬φ_→ φ]_in SE. First, we_have an_open_derivation D1_of_⊥ from [φ_→¬φ] and [φ]._[φ →¬φ]1 [φ]2 →_E¬φ [φ]2 ¬E⊥_93_With the derivation D1, we have an open derivation D2 of φ from [φ_→¬φ]_and [¬φ_→_φ]._[¬φ → φ]3 [φ →¬φ]1,_[φ]2 D1 ⊥ ¬I,2¬φ ¬E_φ Now,_we have the following open derivation Π1 of_⊥_from [φ →¬φ]_and [¬φ → φ]. [φ →¬φ]1, [φ]2 D1 ⊥_¬I,2¬φ [φ →¬φ]1, [¬φ → φ]3_D2 φ ¬E⊥_Claim_2._If an Ekman reduction_process is used for Π1, then_Π1 generates a non-terminating reduction sequence_and is not normalizable. Π1 has a_maximum formula ¬φ in ¬E−rule. By_applying ¬−reduction, we have the_following derivation_Π2. [φ →¬φ]1 [¬φ →_φ]3 [φ →¬φ]1,_[φ]2 D1_⊥ ¬I,2¬φ →_E φ → E¬φ [φ →¬φ]1,_[¬φ → φ]3_D2 φ ¬E⊥ The derivation Π2_has_no maximum formula_but_it_has an_Ekman maximum formula._By_applying Ekman_reduction,_we obtain the same derivation with_Π1_which we started. Therefore, if the set_of reductions of SE_includes_Ekman reduction, Π1 initiates_a non-terminating reduction sequence and_so is not normalizable. 94 We_call the_result of_Proposition 3.1.1 Ekman's paradox. Schroeder-Heister and Tranchini (2017, pp. 570-571) appear_to think that the derivation Π1_uses id est inference_since Π1_has_inferences from φ_to_¬φ and_¬φ to φ. Also, Π1 is a_derivation of_⊥ and generates a reduction loop._Hence, according to TCPE,_Π1_is a T-paradox. Thus, Ekman's paradox_can be a counterexample to