 right results in φ → ψ,ψ → φ,Γ ⇒ φ. However, then, ⊵GE−reduction should provide an open derivation of φ by supplementing the assumption [φ → ψ] and [ψ → φ]. Again, in the case of the permuted version of ⊵GE−reduction, ⊵GE(per)−reduction can make a closed derivation open. For instance, we consider the following case. D′4 φ → ψ D′1 φ D′3 ψ → φ [ψ]1 [ψ]1, [φ]2 D′2 ρ → E,2 ρ → E,1 ρ ⊵GE(per) [ψ], D′1 φ D′2 ρ The above case shows that ⊵GE(per)−reduction produces the open derivation from the closed derivation. Therefore, ⊵GE− and ⊵GE(per)−reductions violate (T1) of the requirements of a proper reduction. Second, GEkman reduction is not a process to reduce unnecessary detour reasoning, but is a detour-making-process. Let us consider the following case with open assumptions. [ψ → φ] [φ → ψ] D1 φ [ψ]1 → E,1 ψ [φ]2 D2 σ → E,2 σ ▷GE D1 φ D2 σ 122 The derivation has no maximum formula, but GEkman maximum formula ψ. It is translated as below. D′1 Γ ⇒ φ ψ ⇒ ψ L → Γ,φ → ψ ⇒ ψ D′2 φ,∆ ⇒ σ L → φ → ψ,ψ → φ,Γ,∆ ⇒ σ reduces to D′1 Γ ⇒ φ D′2 φ,∆ ⇒ σ Cut Γ,∆ ⇒ σ GEkman maximum formula ψ has no corresponding cut-formula. The translated version of GEkman reduction does not remove any cut-formula, and so neither it eliminates an unnecessary detour in the standard sense. The more serious problem is that the translated version shows that GEkman reduction creates an unnecessary detour though it eliminates a GEkman maximum formula ψ. The derivation on the left side has no cut-formula but after applying the reduction process the cut-formula φ appears
right results in φ → ψ, ψ → φ, Γ ⇒ φ. However, then, ⊵GE−reduction should provide an open deriving of φ by append the assumption [ φ → ψ ] and [ ψ → φ ]. Again, in the case of the permuted version of ⊵GE−reduction, ⊵GE(per)−reduction can cause a closed derivation receptive. For example, we consider the come case. D′4 φ → ψ D′1 φ D′3 ψ → φ [ ψ]1 [ ψ]1, [ φ]2 D′2 ρ → E,2 ρ → E,1 ρ ⊵GE(per) [ ψ ], D′1 φ D′2 ρ The above case show that ⊵GE(per)−reduction produces the open derivation from the shut derivation. Therefore, ⊵GE− and ⊵GE(per)−reductions rape (T1) of the requirements of a proper reduction. Second, GEkman decrease is not a process to reduce unnecessary detour reasoning, but is a detour - making - summons. Let us consider the follow sheath with open assumption. [ ψ → φ ] [ φ → ψ ] D1 φ [ ψ]1 → E,1 ψ [ φ]2 D2 σ → E,2 σ ▷GE D1 φ D2 σ 122 The derivation has no maximum rule, but GEkman maximum formula ψ. It is translated as below. D′1 Γ ⇒ φ ψ ⇒ ψ L → Γ, φ → ψ ⇒ ψ D′2 φ,∆ ⇒ σ L → φ → ψ, ψ → φ, Γ,∆ ⇒ σ reduces to D′1 Γ ⇒ φ D′2 φ,∆ ⇒ σ Cut Γ,∆ ⇒ σ GEkman maximum formula ψ has no corresponding cut - formula. The translated version of GEkman reduction does not get rid of any cut - formula, and indeed neither it eliminate an unnecessary detour in the standard sense. The more serious problem is that the translate version shows that GEkman reduction creates an unnecessary detour though it eliminate a GEkman maximum formula ψ. The derivation on the left side has no deletion - formula but after applying the reduction process the geld - recipe φ appears
 rihht results in φ → ψ,ψ → φ,Γ ⇒ φ. Hodever, then, ⊵GE−rebyction shouls providd an open derivation of φ by dupplemtuting the assumption [φ → ψ] and [ψ → φ]. Again, un tie case of the pxdmuted yzrsioh of ⊵YE−ceduction, ⊵GE(per)−teduction cat make a closeg aexivation open. For instance, we considqr the goplowing case. D′4 φ → ψ C′1 φ D′3 ψ → φ [ψ]1 [ψ]1, [φ]2 D′2 ρ → E,2 ρ → E,1 ρ ⊵GE(per) [ψ], D′1 φ D′2 ρ The zbove cese shows that ⊵BE(per)−reduction produces thf opfn derivation from the closed derycation. Therewore, ⊵GE− and ⊵GE(per)−reducjions violate (T1) of the requiremevts oy a proper tzeuchhon. Second, JEkman reduction is not a [rocess to reduce unnccessery eetour reasoning, but ms a detour-making-prosess. Let gs consider the folooqing wase witf opdn zsxujptiond. [ψ → φ] [φ → ψ] D1 φ [ψ]1 → S,1 ψ [φ]2 D2 σ → E,2 σ ▷TE D1 φ D2 σ 122 The derivstyin has no maxjmum fjriula, but GEkman maximum formula ψ. It is uransmated as below. D′1 Γ ⇒ φ ψ ⇒ ψ O → Γ,φ → ψ ⇒ ψ D′2 φ,∆ ⇒ σ L → φ → ψ,ψ → φ,Γ,∆ ⇒ σ reduces to D′1 Γ ⇒ φ D′2 φ,∆ ⇒ σ Cut Γ,∆ ⇒ σ GEkman maximum formula ψ has to cocrdspinqkbg cut-formula. The translated version of GEkman wsdicnion does not remjve any cut-gogmija, and so neijher it eljminates an unnecedsary dgtour un the stwndatd sense. The more serious peoblem is thct rhe translated verdion shows chat GGkman teduction creates an unuecesszry detour hhough it dliminates a GEkoan mdximum formula ψ. The derivaeion on tie leyt side fas mo cut-sormula buh aftcs applying the redkctiou prowess the ckt-formula φ appears
right results in φ → ψ,ψ → φ. then, ⊵GE−reduction provide an open the [φ → ψ] [ψ → φ]. in the case of the permuted of ⊵GE−reduction, ⊵GE(per)−reduction can make a closed derivation open. For instance, we consider following case. D′4 φ → ψ D′1 φ D′3 ψ → φ [ψ]1 [φ]2 ρ E,2 → E,1 ρ ⊵GE(per) [ψ], D′1 φ D′2 ρ The above case shows that ⊵GE(per)−reduction produces open derivation from the closed derivation. Therefore, ⊵GE− ⊵GE(per)−reductions violate (T1) of requirements of a proper reduction. GEkman is not process reduce detour reasoning, but a detour-making-process. Let us consider the following case with open assumptions. [ψ → φ] [φ → ψ] φ [ψ]1 ψ [φ]2 σ E,2 ▷GE D1 φ 122 The derivation has no maximum maximum formula ψ. It is translated as below. Γ ⇒ ψ ⇒ ψ L → Γ,φ ψ ⇒ ψ D′2 φ,∆ ⇒ σ L φ → ψ,ψ → φ,Γ,∆ ⇒ σ reduces to D′1 Γ ⇒ φ D′2 φ,∆ Cut Γ,∆ ⇒ σ maximum formula ψ no cut-formula. translated of GEkman does not remove any cut-formula, and so neither it eliminates an detour in the standard sense. The more serious problem is translated shows that GEkman creates an unnecessary detour it a GEkman maximum formula derivation the no but applying the reduction process cut-formula φ appears
 right results in φ → ψ,ψ → φ,Γ ⇒ φ. HowevEr, then, ⊵GE−reDuctiOn sHouLd ProvIde aN open derivatioN Of φ bY supplementing the assumPtion [Φ → ψ] ANd [ψ → φ]. aGaIn, in tHe case oF ThE PErmUtEd VerSiON oF ⊵GE−reDucTion, ⊵GE(pEr)−reductioN caN mAke a closed deRIvAtion open. FOr iNstance, we conSidEr the fOlLowINg casE. D′4 φ → Ψ D′1 φ D′3 ψ → Φ [ψ]1 [ψ]1, [φ]2 D′2 ρ → e,2 Ρ → E,1 ρ ⊵GE(pEr) [ψ], D′1 φ D′2 ρ ThE aBOve casE Shows thAT ⊵gE(Per)−rEduction produces tHE oPEn derivation frOm the cLoSEd DERivAtiOn. ThereforE, ⊵Ge− and ⊵Ge(Per)−reduCTiONS VioLAte (T1) of the requIrements of a PRopEr reduCtIon. sEcond, Gekman ReDUctIon is not a prOcesS to reduce UnneceSSary detOUr reasoNing, buT is A deTour-MAkInG-prOcESs. LET uS coNSidEr the folLoWiNg casE witH OPEN assUmpTionS. [ψ → φ] [φ → ψ] d1 φ [ψ]1 → E,1 ψ [φ]2 D2 σ → E,2 σ ▷GE D1 Φ D2 σ 122 the dERivAtion Has no MaxiMuM formUla, but gEkmaN mAximum formula ψ. IT is tRanslated As bElOw. D′1 γ ⇒ φ Ψ ⇒ ψ L → Γ,φ → Ψ ⇒ Ψ D′2 φ,∆ ⇒ σ L → φ → Ψ,ψ → φ,γ,∆ ⇒ σ rEduces tO D′1 Γ ⇒ φ D′2 φ,∆ ⇒ σ cUt Γ,∆ ⇒ Σ GeKMAn Maximum formula ψ has No CORrEspondinG cut-foRMuLa. tHe translAtEd vErsiON Of GEkMan rEDuCtion doeS not reMOvE aNy cut-foRmUla, and So NeiTheR it elIMinaTes an uNnecessaRy detOUr in the standarD Sense. The more sERiOUS pRObleM is That the tranSlatED verSion SHoWs tHAt GEkMan reDuCTiON creates an unnecessaRy Detour ThougH it eliminates A GEkman maxIMUM formula Ψ. The DErIVation on the lefT side Has no cut-foRMula but aFter aPplying tHe reductiON Process tHe cUt-fOrmUla Φ APpEars
 right results in φ → ψ,ψ→ φ,Γ ⇒ φ.  Howe ver , t he n, ⊵ GE−r eduction shoul d  pro vide an open derivatio n ofφb y su p pl ement ing the as s u mpt io n[φ→ψ ]and [ ψ →  φ]. Ag ain, in th e c as e of the per m ut ed version  of  ⊵GE−reducti on,  ⊵GE(p er )−r e ducti oncan m ake ac losedderivatio no pen. F o r insta n c e,  weconsider the foll o wi n g case. D′4 φ→ ψ D′ 1φ  D ′ 3  ψ→ φ  [ψ]1 [ψ]1 ,[φ]2D ′2 ρ →E ,2 ρ → E , 1 ρ ⊵GE(per)[ψ], D′1 φD ′2ρ Theab ove case s howsth a t ⊵ GE(per)−red ucti on produc es the open de r ivation  fromthe  cl osed de ri vat io n . T h er efo r e,⊵GE− and  ⊵ GE (per) −red u c t i onsvio late  (T1)  of the requi rem ents ofa pro per r educ ti on. S econd,  GEkm an  reduction is n ot a  processtore duc eunnec e ssarydet our  reason ing, bu t  is  a d e to ur-making-process.  L e t  u s consid er the fo ll o wing cas ewit h op e n  assu mpti o ns . [ψ → φ ] [φ → ψ]  D 1 φ [ψ] 1→ E,1ψ[φ] 2 D 2 σ → E,2σ ▷GED1 φ D2σ 122 The derivation has no maximu m  f o r mu l a, b utGEkman maxi mumf ormu la ψ .  I t i s  tran slate da sb elow. D′1 Γ ⇒ φ ψ ⇒  ψ  L → Γ ,φ →ψ ⇒ ψ D′2 φ,∆  ⇒ σ L → φ → ψ,ψ → φ, Γ,∆⇒  σ reduces to D′1  Γ ⇒φ D′2 φ,∆⇒  σ Cut Γ ,∆ ⇒σ GEkman  maximumf o rmula ψhas  no  co rre s p on ding cut-form u l a. T he  transl ate d versi onofGEk man  r eductiondoes not  r em ov eany  cut- f ormula,an d s onei theri t elim inate s an  u nn e ces sary de t ou r in t he  s tand ard  s ense.  The mor e serio us proble m i s  tha tth e trans lated version  s hows thatGE kma n redu c t ion crea tes an unnecessary deto u r thoug h i t eli mina tes a GEk man  maxim umf ormula  ψ. Th e der iv ati o n  on t h e  l eft  s ide has no c ut- formu la  but  afterapplying the reduc t ion  process thecut -for m u la  φa pp e ars 
 right_results in_φ → ψ,ψ →_φ,Γ ⇒_φ._However, then,_⊵GE−reduction_should provide an_open derivation of_φ by supplementing the_assumption [φ →_ψ]_and [ψ → φ]. Again, in the case of the permuted version of ⊵GE−reduction,_⊵GE(per)−reduction_can make_a_closed_derivation open. For instance, we_consider the following case. D′4_φ →_ψ D′1 φ D′3 ψ → φ [ψ]1_[ψ]1,_[φ]2 D′2 ρ_→ E,2 ρ → E,1 ρ ⊵GE(per) [ψ], D′1_φ D′2 ρ The above case_shows that ⊵GE(per)−reduction_produces_the_open derivation from the_closed derivation. Therefore, ⊵GE− and ⊵GE(per)−reductions_violate (T1) of the requirements of_a proper reduction. Second, GEkman reduction is_not a process to reduce unnecessary_detour reasoning, but is a_detour-making-process. Let_us consider the following case_with open assumptions._[ψ →_φ] [φ →_ψ] D1 φ [ψ]1 → E,1_ψ [φ]2 D2_σ → E,2 σ ▷GE D1_φ_D2 σ 122_The_derivation_has no_maximum formula, but_GEkman_maximum formula_ψ._It is translated as below. D′1_Γ_⇒ φ ψ ⇒ ψ L →_Γ,φ → ψ ⇒_ψ_D′2 φ,∆ ⇒ σ_L → φ → ψ,ψ_→ φ,Γ,∆ ⇒ σ reduces to_D′1 Γ_⇒ φ_D′2 φ,∆ ⇒ σ Cut Γ,∆ ⇒ σ GEkman maximum formula_ψ has no corresponding cut-formula. The_translated version of GEkman_reduction does_not_remove any cut-formula,_and_so neither_it eliminates an unnecessary detour in the_standard sense._The more serious problem is that_the translated version shows_that_GEkman reduction creates an unnecessary detour_though it eliminates a GEkman maximum_formula ψ. The derivation on_the_left_side has no cut-formula but_after applying the reduction process the_cut-formula φ appears