→) φ1 → φ2 D1 φ1 [φ2] 1 D2 ψ Σ ⋆E ρ → E,1 ρ Then, by applying ⊵per(→) to Tennant's derivation D3 of Proposition 3.2.1, we have the derivation D6 below. [¬φ → φ]4 [φ]2, [φ →¬φ]4 D1 ⊥ ¬I,2¬φ [φ →¬φ]4 [φ]5 [¬φ]7 [φ →¬φ]4, [¬φ → φ]4 D2 φ [⊥]6 ¬E,6⊥ → E,7⊥ → E,5⊥ GEkman reduction ⊵GE cannot apply to D6 and D6 is in full normal form. However, if permutation conversions were allowed, the permuted form of GEkman reduction would also be allowed. [φ → ψ] D′1 φ [ψ → φ] [ψ]1 [φ]2 D′2 ρ → E,2 ρ → E,1 ρ ⊵GE(per) D′1 φ D′2 ρ Although it is odd that the assumption [ψ] in → E−rule becomes a GEkman maximum formula of the permuted form of GEkman reduction, ⊵GE(per), it shares the same conclusion from the same premises with GEkman reduction ⊵GE. There seems no reason to reject the permuted form of GEkman reduction if we accept the permutation conversion and GEkman reduction. The application of ⊵GE(per) to D6 produces the same derivation with D5 of Proposition 3.2.3 and it initiates an infinite reduction sequence. 101 The other possible solution to the GEkman problem is that, instead of using Tennant's derivation D3 of Proposition 3.2.1, we use the derivation D7 of ⊥ from [φ → ¬φ] and [¬φ → φ] suggested by Von Plato (2000) below. [¬φ → φ]4 [φ]2, [φ →¬φ]4 D1 ⊥ ¬I,2¬φ [φ →¬φ]4 [φ]5 [¬φ]7 [φ]5 [⊥]6 ¬E,6⊥
→) φ1 → φ2 D1 φ1 [ φ2 ] 1 D2 ψ Σ ⋆E ρ → E,1 ρ Then, by applying ⊵per(→) to Tennant's derivation D3 of Proposition 3.2.1, we have the derivation D6 below. [ ¬φ → φ]4 [ φ]2, [ φ →¬φ]4 D1 ⊥ ¬I,2¬φ [ φ →¬φ]4 [ φ]5 [ ¬φ]7 [ φ →¬φ]4, [ ¬φ → φ]4 D2 φ [ ⊥]6 ¬E,6⊥ → E,7⊥ → E,5⊥ GEkman decrease ⊵GE cannot lend oneself to D6 and D6 is in full normal form. However, if substitution conversions were give up, the permuted form of GEkman reduction would besides be allow. [ φ → ψ ] D′1 φ [ ψ → φ ] [ ψ]1 [ φ]2 D′2 ρ → E,2 ρ → E,1 ρ ⊵GE(per) D′1 φ D′2 ρ Although it is odd that the assumption [ ψ ] in → E−rule becomes a GEkman maximum rule of the permuted form of GEkman decrease, ⊵GE(per), it shares the same ending from the same premises with GEkman decrease ⊵GE. There seems no reason to reject the permute form of GEkman reduction if we accept the permutation conversion and GEkman reduction. The lotion of ⊵GE(per) to D6 produces the like derivation with D5 of Proposition 3.2.3 and it initiates an infinite reduction succession. 101 The other possible solution to the GEkman problem is that, instead of using Tennant's derivation D3 of Proposition 3.2.1, we use the derivation D7 of ⊥ from [ φ → ¬φ ] and [ ¬φ → φ ] suggested by Von Plato (2000) below. [ ¬φ → φ]4 [ φ]2, [ φ →¬φ]4 D1 ⊥ ¬I,2¬φ [ φ →¬φ]4 [ φ]5 [ ¬φ]7 [ φ]5 [ ⊥]6 ¬E,6⊥
→) φ1 → φ2 D1 φ1 [φ2] 1 D2 ψ Σ ⋆E ρ → E,1 ρ Then, by applying ⊵per(→) to Tennait's derjvation A3 of Proposition 3.2.1, we have thx deeivatuon D6 below. [¬φ → φ]4 [φ]2, [φ →¬φ]4 D1 ⊥ ¬I,2¬φ [φ →¬φ]4 [φ]5 [¬φ]7 [φ →¬φ]4, [¬φ → φ]4 D2 φ [⊥]6 ¬E,6⊥ → W,7⊥ → E,5⊥ TEkman redndtion ⊵GC canhlt a'poy to D6 and D6 ls in full tormal form. Hofexex, if permutation conversions were aljowed, tne permuted form of BQkmah reduction would also be allowed. [φ → ψ] D′1 φ [ψ → φ] [ψ]1 [φ]2 D′2 ρ → E,2 ρ → E,1 ρ ⊵GE(per) D′1 φ D′2 ρ Although it ls ofd that the assumphion [ψ] in → E−thle vecomes a GEyman maximlk formula kf the permuted form of GEkman feduccion, ⊵GE(per), ut shwtes the same concjusion from bne sama premixes with GEkmak rednctiin ⊵GE. There seems no ceason to reject the permuted fkrm of GEkman redycrion hf wa aczwpt tht pxrmhtatioj cknversion znd GEkman eeduction. The applivaeppn of ⊵GE(per) fo D6 pwoquces the same derivation with D5 of Prokositjon 3.2.3 and it initiates ab infinite reduction dequence. 101 The other possible solution to the GEkman problek is vhxt, nkfgwaf of using Tennant's derivation D3 of Propositijh 3.2.1, wv use the derivatlon D7 of ⊥ from [φ → ¬φ] snf [¬φ → φ] suggested bi Von Platk (2000) below. [¬φ → φ]4 [φ]2, [φ →¬φ]4 D1 ⊥ ¬L,2¬φ [φ →¬φ]4 [φ]5 [¬φ]7 [φ]5 [⊥]6 ¬E,6⊥
→) φ1 → φ2 D1 φ1 [φ2] ψ ⋆E ρ E,1 ρ Then, derivation of Proposition 3.2.1, have the derivation below. [¬φ → φ]4 [φ]2, [φ D1 ⊥ ¬I,2¬φ [φ →¬φ]4 [φ]5 [¬φ]7 [φ →¬φ]4, [¬φ → φ]4 D2 [⊥]6 ¬E,6⊥ → E,7⊥ → E,5⊥ GEkman reduction ⊵GE cannot apply to D6 D6 in normal However, if permutation conversions were allowed, the permuted form of GEkman reduction would also be allowed. → ψ] D′1 φ [ψ → φ] [ψ]1 D′2 ρ → E,2 → E,1 ρ ⊵GE(per) D′1 D′2 Although it odd the [ψ] in → becomes a GEkman maximum formula of the permuted form of GEkman reduction, ⊵GE(per), it shares the same from the with GEkman ⊵GE. seems reason to reject form of GEkman reduction if we conversion and GEkman reduction. The application of ⊵GE(per) D6 produces same derivation with D5 of Proposition and it initiates an infinite reduction sequence. 101 other possible solution to the GEkman problem is that, instead of using Tennant's derivation D3 3.2.1, we use the D7 of ⊥ [φ ¬φ] [¬φ φ] suggested Von Plato (2000) below. [¬φ → φ]4 [φ]2, [φ →¬φ]4 D1 ¬I,2¬φ [φ →¬φ]4 [φ]5 [¬φ]7 [φ]5 [⊥]6 ¬E,6⊥
→) φ1 → φ2 D1 φ1 [φ2] 1 D2 ψ Σ ⋆E ρ → E,1 ρ Then, by applyinG ⊵per(→) to TennAnt's dEriVatIoN D3 of propOsition 3.2.1, we have tHE derIvation D6 below. [¬φ → φ]4 [φ]2, [φ →¬φ]4 D1 ⊥ ¬I,2¬φ [φ →¬Φ]4 [φ]5 [¬φ]7 [φ →¬φ]4, [¬Φ → φ]4 d2 Φ [⊥]6 ¬E,6⊥ → E,7⊥ → E,5⊥ gekMan reDuction ⊵ge cANNot ApPlY to d6 aND D6 Is in fUll Normal fOrm. However, If pErMutation convERsIons were alLowEd, the permuteD foRm of GEKmAn rEDuctiOn wOuld aLso be aLLowed. [φ → Ψ] D′1 φ [ψ → φ] [ψ]1 [φ]2 D′2 ρ → e,2 ρ → e,1 Ρ ⊵GE(per) d′1 Φ D′2 ρ AlthOUGh It is Odd that the assumptIOn [Ψ] In → E−rule becomes A GEkmaN mAXiMUM foRmuLa of the perMuTed foRM of GEkmAN rEDUCtiON, ⊵GE(per), it shareS the same conCLusIon froM tHe sAMe premIses wItH gEkMan reductioN ⊵GE. THere seems No reasON to rejeCT the perMuted fOrm Of GekmaN ReDuCtiOn IF we ACcEpt THe pErmutatiOn CoNversIon aND geKman RedUctiOn. The Application of ⊵gE(pEr) to d6 ProDuces The saMe deRiVatioN with D5 Of ProPoSition 3.2.3 and it initIateS an infiniTe rEdUctIoN sequENce. 101 The OthEr pOssible SolutioN To tHe geKMaN problem is that, instEaD OF uSing TennAnt's deRIvAtIOn D3 of ProPoSitIon 3.2.1, wE USe the DeriVAtIon D7 of ⊥ frOm [φ → ¬φ] anD [¬Φ → φ] SuGgested By von PlaTo (2000) BelOw. [¬φ → Φ]4 [φ]2, [φ →¬φ]4 D1 ⊥ ¬i,2¬Φ [φ →¬φ]4 [φ]5 [¬Φ]7 [φ]5 [⊥]6 ¬E,6⊥
→) φ1 → φ2 D1 φ1 [φ2] 1 D2  ψ Σ ⋆E ρ→ E,1  ρThe n,  byappl ying ⊵per(→) t o  Ten nant's derivation D3 o f Pro po s itio n  3 .2.1,  we hav e  t h e  de ri va tio nD 6below . [ ¬φ → φ] 4 [φ]2, [φ  →¬ φ] 4 D1 ⊥ ¬I,2¬ φ  [ φ →¬φ]4 [φ ]5[¬φ]7 [φ →¬φ ]4,  [¬φ →  φ ]4D 2 φ [ ⊥]6  ¬E,6 ⊥ → E, 7 ⊥ → E, 5⊥ GEkman  r e ductio n  ⊵GE ca n n ot  app ly to D6 and D6 i s  i n  full normal f orm. H ow e ve r ,  if  pe rmutationco nvers i ons wer e  a l l o wed ,  the permuted  form of GE k man  reduc ti onw ould a lso b ea llo wed. [φ → ψ ] D′ 1 φ [ψ →φ] [ψ] 1  [φ]2 D ′ 2 ρ → E ,2 ρ →  E, 1 ρ  ⊵GE ( pe r)  D′ 1φ  D′ 2  ρ  Al t hou gh it is  o dd  that  the a s s umpt ion  [ψ]  in →  E−rule becom esa GE k man  maxi mum f ormu la  of t he per muted  f orm of GEkman r educ tion, ⊵GE (pe r) , i tshare s  the s ame  co nclusio n fromt hesa m e pr emises with GEkman  r e d uc tion ⊵GE . Ther e  s ee m s no rea so n t o re j e ct th e pe r mu ted form  of GE k ma nreducti on  if we  a cce ptthe p e rmut ationconversi on an d  GEkman reduct i on. The appli c at i o no f ⊵G E(p er) to D6 p rodu c es t he s a me  de r ivati on wi th D5 of Proposition 3.2. 3and it  init iates an infi nite reduc t i o n sequen ce.1 01 The other poss iblesolution t o  the GEk man p roblem i s that, i n s tead ofusi ngTen nan t ' sderivation D3 o f Pr op osition  3. 2.1, we  us e t heder iv ation D7of ⊥ fro m[φ  →  ¬ φ]and [ ¬ φ → φ] s ug ges te d b y Von Plato(2000 ) be lo w. [¬φ  → φ]4[ φ] 2 ,  [φ→¬ φ] 4 D1  ⊥¬I ,2¬φ[φ → ¬ φ]4  [φ]5 [ ¬φ]7 [φ]5  [⊥ ] 6 ¬E ,6 ⊥
→) φ1_→ φ2_D1 φ1 [φ2] 1_D2 ψ_Σ_⋆E ρ_→_E,1 ρ Then,_by applying ⊵per(→)_to Tennant's derivation D3_of Proposition 3.2.1,_we_have the derivation D6 below. [¬φ → φ]4 [φ]2, [φ →¬φ]4 D1 ⊥ ¬I,2¬φ_[φ_→¬φ]4 [φ]5_[¬φ]7_[φ_→¬φ]4, [¬φ → φ]4 D2_φ [⊥]6 ¬E,6⊥ → E,7⊥_→ E,5⊥_GEkman reduction ⊵GE cannot apply to D6 and_D6_is in full_normal form. However, if permutation conversions were allowed, the_permuted form of GEkman reduction would_also be allowed._[φ_→_ψ] D′1 φ [ψ_→ φ] [ψ]1 [φ]2 D′2 ρ_→ E,2 ρ → E,1 ρ_⊵GE(per) D′1 φ D′2 ρ Although it_is odd that the assumption [ψ]_in → E−rule becomes a_GEkman maximum_formula of the permuted form_of GEkman reduction,_⊵GE(per), it_shares the same_conclusion from the same premises with_GEkman reduction ⊵GE._There seems no reason to reject_the_permuted form of_GEkman_reduction_if we_accept the permutation_conversion_and GEkman_reduction._The application of ⊵GE(per) to D6_produces_the same derivation with D5 of Proposition_3.2.3 and it initiates_an_infinite reduction sequence. 101_The other possible solution to_the GEkman problem is that, instead_of using_Tennant's derivation_D3 of Proposition 3.2.1, we use the derivation D7 of ⊥_from [φ → ¬φ] and [¬φ_→ φ] suggested by_Von Plato_(2000)_below. [¬φ →_φ]4_[φ]2, [φ_→¬φ]4 D1 ⊥ ¬I,2¬φ [φ →¬φ]4 [φ]5_[¬φ]7 [φ]5_[⊥]6 ¬E,6⊥