 G. Though Σ1 is a similar version of Π1, it cannot be the clue to block GEkman reduction process.2 There are two standard elimination rules for ∧I−rule and so are two reduction procedures for ∧. As the derivation Π1 uses the standard ∧E1−rule but does not have any application of ∧E2−rule, only the left conjunct φ and its derivation D1 are picked by the reduction for ∧. On the other hand, ∧I−rule has only one generalized elimination rule and its reduction procedure uses both derivations of two conjuncts. Hence, the application of the reduction process of Σ1 uses both derivations D1 and D2 of φ and provides the same derivation with Σ2. So to speak, Schroeder-Heister and Tranchini's example stated in generalized form does not show that GEkman reduction trivializes the identity of proofs. 2For the permuted derivation of Σ1, we have the same result by applying ▷GE(per). 111 Whether a new reduction passes Triviality test is dependent on a choice of the form of elimination rules, and so on a given system. One may say that we can freely choose one of both derivations when we use ▷∧−reduction in generalized form. However, then, regardless of GEkman reduction process, we can readily show that ▷∧−reduction trivializes the identity of proofs because by such free choice of derivations we have the following three different derivations from Σ1. D1 φ D1 φ ∧I φ ∧φ D1 φ D2 φ ∧I φ ∧φ D2 φ D2 φ ∧I φ ∧φ Thus, if the free choice of derivations in ▷∧−reduction is acceptable, according to Triviality test, ▷∧−reduction is not a proper reduction. However, Schroeder-Heister and Tranchini will not want this unwelcome conclusion. It is not a good answer to the above approach that we take two generalized elimination rules for ∧. φ ∧ψ [φ]1 D1 σ ∧E1,1 σ φ ∧ψ [ψ]1 D2 σ ∧E2,1 σ ∧I−rule with the two generalized elimination rules above may have two distinct reduction procedures for left and right conjuncts. If ∧I−rule has the two generalized elimination rules, we readily
G. Though Σ1 is a similar version of Π1, it cannot be the clue to block GEkman decrease process.2 There be two standard elimination rules for ∧I−rule and so are two decrease procedures for ∧. As the ancestry Π1 uses the standard ∧E1−rule but does not have any lotion of ∧E2−rule, entirely the left conjunct φ and its deriving D1 are picked by the reduction for ∧. On the early hand, ∧I−rule has only one generalized elimination rule and its decrease procedure uses both derivation of two conjuncts. Hence, the application of the decrease process of Σ1 uses both derivations D1 and D2 of φ and provide the same derivation with Σ2. thus to speak, Schroeder - Heister and Tranchini's model stated in generalized form does not show that GEkman reduction trivializes the identity of proof. 2For the permuted derivation of Σ1, we have the same result by applying ▷GE(per). 111 Whether a newfangled reduction passes Triviality test is dependent on a choice of the form of elimination rules, and so on a given arrangement. One may say that we can freely choose one of both derivations when we practice ▷∧−reduction in generalized human body. However, then, regardless of GEkman decrease process, we can readily show that ▷∧−reduction trivializes the identity of proofs because by such free choice of derivations we accept the following three different derivations from Σ1. D1 φ D1 φ ∧I φ ∧φ D1 φ D2 φ ∧I φ ∧φ D2 φ D2 φ ∧I φ ∧φ Thus, if the free choice of derivations in ▷∧−reduction is acceptable, accord to Triviality test, ▷∧−reduction is not a proper reduction. However, Schroeder - Heister and Tranchini will not want this unwelcome stopping point. It is not a dear answer to the above approach that we take two generalized elimination rules for ∧. φ ∧ψ [ φ]1 D1 σ ∧E1,1 σ φ ∧ψ [ ψ]1 D2 σ ∧E2,1 σ ∧I−rule with the two generalized elimination dominion above may have two distinct reduction routine for left and right conjuncts. If ∧I−rule has the two generalized elimination rule, we readily
 G. Hhough Σ1 is a similar verrion of Π1, it canuit be vhe clus to blozk GEkman reduction process.2 Vherw are two standard eliminatkon rules for ∧I−ruoe aid so are two revhction ixocedhves fmc ∧. As the derivstion Π1 usev the standard ∧E1−fupe but does not have any applicatiog of ∧E2−rilf, only the lefj conktnct φ and its derivation D1 are picked gy the geduction for ∧. On the other hand, ∧I−rule has lnly one generalized epimination thle qnd its reduztion proctdbre uses bojh derivations of two conjuncts. Fence, the applixarioj of the rednction process of Σ1 uses bodh derifations D1 and C2 oh φ abd provides the same verivation with Σ2. So jo speak, Swhxoeder-Heister and Trabcyini's exakple wtaged ii gsnerallzev form does not show tyat GEkman reductiom egovializes ths ideneiey of proofs. 2For the permuted derivatiot or Σ1, we have the same resylt by applying ▷GE(per). 111 Whether a new reduction passes Triviality test is dependant oi x ckiice ud hhe form of elimination rules, and so on a givqh xyxtem. One may scy that we can fteflu choose one ow both dedivations when we kse ▷∧−redoction in generwlizrd form. However, then, regardoess of GEkmcn eeduction process, ce can readiuy snow tnat ▷∧−reduction trivializzs the identity ov proofs gdcause by such ffee cvoice of derivations we hade the foolownng thred digferene derivatilns fvmm Σ1. D1 φ D1 φ ∧I φ ∧φ D1 φ D2 φ ∧I φ ∧φ D2 φ D2 φ ∧I φ ∧φ Tjus, if the free choice of derivevions in ▷∧−reduvthon is acce'table, according to Triviality text, ▷∧−reduetion ks not a pgoper rednction. Howevqr, Schroeder-Halster and Trenchini wyll bot qant thkr unwelcome comclusion. Pt is bot a good answer bo thg zbove approach ckau qe take two gemerxlisef xlimigdtion rules xor ∧. φ ∧ψ [φ]1 C1 σ ∧E1,1 σ φ ∧ψ [ψ]1 D2 σ ∧T2,1 σ ∧L−ruue woth the two generalised slimination rules sbjve may yave two distinct rediction procedures vor lxft anv righy cjnjuncts. If ∧I−rule has the two feneralizfd climination rtles, we readily
G. Though Σ1 is a similar version it be the to block GEkman standard rules for ∧I−rule so are two procedures for ∧. As the derivation uses the standard ∧E1−rule but does not have any application of ∧E2−rule, only left conjunct φ and its derivation D1 are picked by the reduction for On other ∧I−rule only one generalized elimination rule and its reduction procedure uses both derivations of two conjuncts. Hence, application of the reduction process of Σ1 uses derivations D1 and D2 φ and provides the same with So to Schroeder-Heister Tranchini's stated in generalized does not show that GEkman reduction trivializes the identity of proofs. 2For the permuted derivation of Σ1, have the by applying 111 a reduction passes Triviality dependent on a choice of the rules, and so on a given system. One say that can freely choose one of both when we use ▷∧−reduction in generalized form. However, regardless of GEkman reduction process, we can readily show that ▷∧−reduction trivializes the identity of by such free choice derivations we have following different from D1 φ φ ∧I φ ∧φ D1 φ D2 φ ∧I φ ∧φ φ D2 φ ∧I φ ∧φ Thus, if the free derivations ▷∧−reduction is acceptable, to Triviality test, ▷∧−reduction not proper reduction. However, Schroeder-Heister will want It not good answer to the approach that we take two elimination rules for ∧. ∧E1,1 σ φ ∧ψ [ψ]1 D2 σ ∧E2,1 ∧I−rule with the two generalized elimination rules may have two distinct reduction procedures for left and right conjuncts. If has the elimination rules, we readily
 G. Though Σ1 is a similar version Of Π1, it cannoT be thE clUe tO bLock gEkmAn reduction proCEss.2 THere are two standard elimInatiOn RUles FOr ∧i−rule And so arE TwO REduCtIoN prOcEDuRes foR ∧. As The deriVation Π1 useS thE sTandard ∧E1−rule BUt Does not havE anY application Of ∧E2−Rule, onLy The LEft coNjuNct φ aNd its dERivatiOn D1 are picKeD By the rEDuction FOR ∧. ON the Other hand, ∧I−rule has ONlY One generalized EliminAtIOn RULe aNd iTs reductioN pRocedURe uses bOTh DERIvaTIons of two conjUncts. Hence, tHE apPlicatIoN of THe reduCtion PrOCesS of Σ1 uses botH derIvations D1 And D2 of Φ And provIDes the sAme derIvaTioN witH σ2. SO tO spEaK, schROeDer-hEisTer and TrAnChIni's eXampLE STAted In gEnerAlizeD form does not sHow That gekmAn redUctioN triViAlizeS the idEntitY oF proofs. 2For the peRmutEd derivatIon Of σ1, we HaVe the SAme resUlt By aPplying ▷gE(per). 111 WhETheR a NEW ReDuction passes TriviAlITY tEst is depEndent ON a ChOIce of the FoRm oF eliMINatioN rulES, aNd so on a gIven sySTeM. ONe may saY tHat we cAn FreEly ChoosE One oF both dErivatioNs wheN We use ▷∧−reduction IN generalized fORm. hOWeVEr, thEn, rEgardless of gEkmAN redUctiON pRocESs, we cAn reaDiLY sHOw that ▷∧−reduction trivIaLizes tHe ideNtity of proofs Because by sUCH Free choiCe of DErIVations we have tHe folLowing threE DifferenT deriVations fRom Σ1. D1 φ D1 φ ∧I Φ ∧Φ d1 φ D2 φ ∧I φ ∧φ D2 Φ D2 φ ∧i φ ∧φ thuS, if THE fRee choice of deRIVatiOnS in ▷∧−reduCtiOn is accEptAblE, acCorDiNg to TriviAlity tesT, ▷∧−rEdUcTiOn iS not a PRoper redUcTioN. HOweVer, ScHRoeder-heistEr anD TRaNChiNi will nOT wANT thiS uNwElcoMe cOnClusiOn. It IS noT a good aNswer to thE abOVe apPrOaCh that wE take two generAlIzed eliminAtIon Rules fOR ∧. Φ ∧ψ [φ]1 D1 σ ∧E1,1 σ φ ∧Ψ [ψ]1 D2 σ ∧E2,1 σ ∧I−rule with the two geNEralizeD elIminaTion Rules abovE maY have tWo dIStinct ReductIon prOcEduRES for lEFT aNd rIgHt conjunctS. iF ∧I−rUle haS tHe twO generaLized elimination ruLEs, wE readily
 G. Though Σ1 is a similar  version o f Π1,  it  ca nn ot b e th e clue to bloc k  GEk man reduction process. 2 The re aret wo  stan dard el i mi n a tio nru les  f o r∧I−ru leand soare two re duc ti on procedure s  f or ∧. As t hederivation Π 1 u ses th esta n dard∧E1 −rule  but d o es not  have any  a p plicat i on of ∧ E 2 −r ule,  only the left co n ju n ct φ and its d erivat io n  D 1 are  pi cked by th ereduc t ion for ∧. O n  th e  other hand,∧I−rule has onl y onege ner a lizedelimi na t ion  rule and i ts r eductionproced u re uses both de rivati ons  of  two co nj unc ts .  He n ce , t h e a pplicati on  o f the  red u c t i on p roc essof Σ1  uses both de riv atio n s D 1 and  D2 o f φan d pro videsthe s am e derivation wi th Σ 2. So tospe ak , S ch roede r -Heist erand  Tranch ini's e x amp le s t at ed in generalizedfo r m  d oes notshow t h at  G E kman red uc tio n tr i v ializ es t h eidentity  of pr o of s.  2For t he  permu te d d eri vatio n  ofΣ1, we  have th e sam e  result by app l ying ▷GE(per) .  1 1 1  W h ethe r a  new reduct ionp asse s Tr i vi ali t y tes t isde p en d ent on a choice ofth e form  of e limination ru les, and s o o n a give n sy s te m . One may saythatwe can fre e ly choos e one  of both  derivati o n s when w e u se▷∧− red u c ti on in general i z ed f or m. Howe ver , then,  re gar dle ssof  GEkman r eduction  p ro ce ss , w e can readilysh owth at▷∧−re d uction  triv iali ze st heidentit y  o f proo fs  b ecau seby  such  fre e  ch oice of  derivati ons we h av ethe fol lowing threedi fferent de ri vat ions f r o m Σ1. D1  φ D1 φ ∧I φ ∧φ D1 φ D2 φ ∧I φ∧φD2 φD2 φ  ∧I φ ∧φThu s, ifthe free c hoiceof de ri vat i o ns in ▷ ∧− red uc tion is ac c e pta ble,ac cord ing toTriviality test, ▷ ∧ −re duction is no t a  pro p e rred u ct i on.  H o wev e r , Schroeder-Hei ster and T ra n ch ini will n o t w an t thisunwelco me co n clusion . It is n ot a good  a nswe r tothe aboveapproach  that wet ake t w ogener ali zed el im ina tionrulesf or∧. φ∧ψ [φ] 1D1 σ ∧ E1,1σφ ∧ψ [ψ] 1 D2 σ ∧E2,1 σ ∧I−rulewith t he tw o g eneralize d e l imi nation ru lesabove mayhav e t wo di sti n ct re duct i on  pr o cedur es f o r left an d  r igh t co njuncts. If ∧ I −ru le ha s t h e twogene ralized eliminati o n rules, we re adil y 
 G._Though Σ1_is a similar version_of Π1,_it_cannot be_the_clue to block_GEkman reduction process.2_There are two standard_elimination rules for_∧I−rule_and so are two reduction procedures for ∧. As the derivation Π1 uses the_standard_∧E1−rule but_does_not_have any application of ∧E2−rule,_only the left conjunct φ_and its_derivation D1 are picked by the reduction for_∧._On the other_hand, ∧I−rule has only one generalized elimination rule and_its reduction procedure uses both derivations_of two conjuncts._Hence,_the_application of the reduction_process of Σ1 uses both derivations_D1 and D2 of φ and_provides the same derivation with Σ2. So_to speak, Schroeder-Heister and Tranchini's example_stated in generalized form does_not show_that GEkman reduction trivializes the_identity of proofs._2For the_permuted derivation of_Σ1, we have the same result_by applying ▷GE(per)._111 Whether a new reduction passes_Triviality_test is dependent_on_a_choice of_the form of_elimination_rules, and_so_on a given system. One may_say_that we can freely choose one of_both derivations when we_use_▷∧−reduction in generalized form._However, then, regardless of GEkman_reduction process, we can readily show_that ▷∧−reduction_trivializes the_identity of proofs because by such free choice of derivations we_have the following three different derivations_from Σ1. D1 φ_D1 φ_∧I_φ ∧φ D1_φ_D2 φ_∧I φ ∧φ D2 φ D2 φ_∧I φ_∧φ Thus, if the free choice_of derivations in ▷∧−reduction_is_acceptable, according to Triviality test, ▷∧−reduction_is not a proper reduction. However,_Schroeder-Heister and Tranchini will not_want_this_unwelcome conclusion. It is not_a good answer to the above_approach that we_take two generalized elimination rules for ∧._φ_∧ψ [φ]1 D1 σ ∧E1,1 σ_φ_∧ψ [ψ]1 D2 σ ∧E2,1 σ_∧I−rule_with_the two generalized elimination rules_above may have two distinct reduction_procedures for left and right conjuncts. If ∧I−rule has_the two generalized_elimination rules, we readily