 Zb ∈ Zb [⊥]3 ¬E,3⊥ ∈Z E,2⊥ Then, there is an open derivation Σ2 of Zb ∈ Zb from [Zb ∈ b]. [Zb ∈ b]4 [Zb ∈ Zb]1 Σ1 ⊥ ¬I,1¬Zb ∈ Zb ∈Z IZb ∈ {x ∈ b|¬x ∈ x}************************** de f Zb ∈ Zb Now, we have an open derivation Σ3 of ⊥ from [Zb ∈ b]. [Zb ∈ Zb]5 Σ1 ⊥ ¬I,5¬Zb ∈ Zb [Zb ∈ b]4 Σ2 Zb ∈ Zb [⊥]6 ¬E,6⊥ Claim 2. Σ3 initiates a non-terminating reduction sequence and so is not fully normalizable. 125 By applying ▷¬ to Σ3, we have the derivation Σ4 below. [Zb ∈ b]4 [Zb ∈ Zb]1 Σ1 ⊥ ¬I,1¬Zb ∈ Zb ∈Z IZb ∈ {x ∈ b|¬x ∈ x} [¬Zb ∈ Zb]2 [Zb ∈ b]2 [¬Zb ∈ Zb]2 ∈Z IZb ∈ {x ∈ b|¬x ∈ x}************************** de f Zb ∈ Zb [⊥]3 ¬E,3⊥ ∈Z E,2⊥ Since Zb ∈ {x ∈ b|¬x ∈ x} in ∈Z E−rule is not an assumption, the application of ▷∈Z− reduction produces the same derivation with Σ3. Therefore, Σ3 generates a non-terminating reduction sequence and so is not fully normalizable. We call the result of Proposition 3.4.1 Crabbé's case. Furthermore, from the derivation Σ3, we readily obtain a closed derivation Σ5 of ¬∃y(Zy ∈ y) as follows. [∃y(Zy ∈ y)]7 [Zb ∈ b]4 Σ3 ⊥ ∃
Zb ∈ Zb [ ⊥]3 ¬E,3⊥ ∈Z E,2⊥ Then, there is an open derivation Σ2 of Zb ∈ Zb from [ Zb ∈ b ]. [ Zb ∈ b]4 [ Zb ∈ Zb]1 Σ1 ⊥ ¬I,1¬Zb ∈ Zb ∈Z IZb ∈ { x ∈ b|¬x ∈ x } * * * * * * * * * * * * * * * * * * * * * * * * * * de f Zb ∈ Zb Now, we get an loose derivation Σ3 of ⊥ from [ Zb ∈ b ]. [ Zb ∈ Zb]5 Σ1 ⊥ ¬I,5¬Zb ∈ Zb [ Zb ∈ b]4 Σ2 Zb ∈ Zb [ ⊥]6 ¬E,6⊥ Claim 2. Σ3 initiates a non - terminating reduction succession and so is not fully normalizable. 125 By applying ▷¬ to Σ3, we hold the derivation Σ4 below. [ Zb ∈ b]4 [ Zb ∈ Zb]1 Σ1 ⊥ ¬I,1¬Zb ∈ Zb ∈Z IZb ∈ { x ∈ b|¬x ∈ x } [ ¬Zb ∈ Zb]2 [ Zb ∈ b]2 [ ¬Zb ∈ Zb]2 ∈Z IZb ∈ { x ∈ b|¬x ∈ x } * * * * * * * * * * * * * * * * * * * * * * * * * * de f Zb ∈ Zb [ ⊥]3 ¬E,3⊥ ∈Z E,2⊥ Since Zb ∈ { x ∈ b|¬x ∈ x } in ∈Z E−rule is not an premise, the application of ▷∈Z− reduction produce the same derivation with Σ3. consequently, Σ3 generates a non - terminating reduction sequence and so is not fully normalizable. We call the result of Proposition 3.4.1 Crabbé's event. Furthermore, from the derivation Σ3, we readily obtain a shut derivation Σ5 of ¬∃y(Zy ∈ y) as follows. [ ∃y(Zy ∈ y)]7 [ Zb ∈ b]4 Σ3 ⊥ ∃
 Zb ∈ Zb [⊥]3 ¬E,3⊥ ∈Z E,2⊥ Then, there ir an open derivcrion Σ2 mf Zb ∈ Zb from [Zb ∈ b]. [Zb ∈ b]4 [Zb ∈ Zb]1 Σ1 ⊥ ¬I,1¬Zb ∈ Zu ∈Z UZb ∈ {z ∈ b|¬x ∈ x}************************** de f Zb ∈ Zb Nuw, we havv an open dermvation Σ3 of ⊥ from [Zb ∈ b]. [Dy ∈ Zb]5 Σ1 ⊥ ¬I,5¬Zy ∈ Zb [Zb ∈ b]4 Σ2 Zb ∈ Zb [⊥]6 ¬E,6⊥ Clahm 2. Σ3 initiates a nln-terminating reduction sequence anq so is nlt fully normajizanje. 125 Gj cpplying ▷¬ to Σ3, we have the derivztion Σ4 uelow. [Zb ∈ b]4 [Zb ∈ Zb]1 Σ1 ⊥ ¬I,1¬Zb ∈ Zb ∈Z IZb ∈ {x ∈ b|¬x ∈ x} [¬Xb ∈ Zb]2 [Zb ∈ b]2 [¬Zb ∈ Zh]2 ∈Z IZb ∈ {x ∈ b|¬x ∈ z}************************** de f Zb ∈ Zc [⊥]3 ¬E,3⊥ ∈Z E,2⊥ Spuce Zb ∈ {x ∈ g|¬x ∈ x} in ∈Z E−rule is not an assuoptiou, the applixarioj of ▷∈Z− reducvion pgoduces the same derieation eith Σ3. Thereforc, Σ3 geierares a non-terminating ceduction sequence agd so is toc fully normalizable. Qe call the resjot uf Kro'osjtion 3.4.1 Crebbé's case. Fhrthermore, drom the derivation Σ3, rv readily obtzin a sljsed derivation Σ5 of ¬∃y(Zy ∈ y) as follows. [∃y(Ey ∈ y)]7 [Zb ∈ b]4 Σ3 ⊥ ∃
Zb ∈ Zb [⊥]3 ¬E,3⊥ ∈Z E,2⊥ is open derivation of Zb ∈ [Zb b]4 [Zb ∈ Σ1 ⊥ ¬I,1¬Zb Zb ∈Z IZb ∈ {x ∈ ∈ x}************************** de f Zb ∈ Zb Now, we have an open derivation of ⊥ from [Zb ∈ b]. [Zb ∈ Zb]5 Σ1 ⊥ ¬I,5¬Zb ∈ [Zb b]4 Zb Zb [⊥]6 ¬E,6⊥ Claim 2. Σ3 initiates a non-terminating reduction sequence and so is not fully 125 By applying ▷¬ to Σ3, we have derivation Σ4 below. [Zb b]4 [Zb ∈ Zb]1 Σ1 ¬I,1¬Zb Zb ∈Z ∈ ∈ ∈ x} [¬Zb Zb]2 [Zb ∈ b]2 [¬Zb ∈ Zb]2 ∈Z IZb ∈ {x ∈ b|¬x ∈ x}************************** de f ∈ Zb ∈Z E,2⊥ Zb {x b|¬x ∈ x} E−rule is not an assumption, the reduction produces the same derivation with Σ3. Therefore, generates a reduction sequence and so is not normalizable. We call the result of Proposition 3.4.1 case. Furthermore, from the derivation Σ3, we readily obtain a closed derivation Σ5 of ¬∃y(Zy as follows. [∃y(Zy ∈ [Zb ∈ b]4 ⊥
 Zb ∈ Zb [⊥]3 ¬E,3⊥ ∈Z E,2⊥ Then, there is an open dErivation Σ2 Of Zb ∈ ZB frOm [ZB ∈ b]. [zb ∈ b]4 [ZB ∈ Zb]1 Σ1 ⊥ ¬i,1¬Zb ∈ Zb ∈Z IZb ∈ {x ∈ b|¬x ∈ x}************************** dE F Zb ∈ ZB Now, we have an open derivaTion Σ3 Of ⊥ FRom [ZB ∈ B]. [ZB ∈ Zb]5 Σ1 ⊥ ¬I,5¬zb ∈ Zb [Zb ∈ b]4 σ2 zb ∈ zB [⊥]6 ¬e,6⊥ ClAiM 2. Σ3 IniTiATeS a non-TerMinatinG reduction SeqUeNce and so is noT FuLly normaliZabLe. 125 By applying ▷¬ To Σ3, We have ThE deRIvatiOn Σ4 Below. [zb ∈ b]4 [Zb ∈ ZB]1 σ1 ⊥ ¬I,1¬Zb ∈ Zb ∈z IZb ∈ {x ∈ b|¬x ∈ x} [¬ZB ∈ ZB]2 [zb ∈ b]2 [¬Zb ∈ ZB]2 ∈z IZb ∈ {x ∈ b|¬x ∈ X}************************** DE f zb ∈ Zb [⊥]3 ¬e,3⊥ ∈Z E,2⊥ Since Zb ∈ {x ∈ b|¬x ∈ x} in ∈Z e−RuLE is not an assumpTion, thE aPPlICAtiOn oF ▷∈Z− reductioN pRoducES the samE DeRIVAtiON with Σ3. TherefoRe, Σ3 generateS A noN-termiNaTinG ReductIon seQuENce And so is not fUlly NormalizaBle. We cALl the reSUlt of PrOpositIon 3.4.1 craBbé's CAsE. FUrtHeRMorE, FrOm tHE deRivation σ3, wE rEadilY obtAIN A ClosEd dErivAtion σ5 of ¬∃y(Zy ∈ y) as follOws. [∃Y(Zy ∈ y)]7 [zB ∈ b]4 Σ3 ⊥ ∃
 Zb ∈ Zb [⊥]3 ¬E,3⊥ ∈Z E,2 ⊥ Then, th ere i s a n o pe n de riva tion Σ2 of Zb∈  Zbfrom [Zb ∈ b]. [Zb ∈ b ]4 [Z b∈  Zb] 1  Σ 1 ⊥ ¬ I,1¬Zb∈  Z b ∈ZIZ b∈ { x∈  b |¬x ∈  x} ******* ********** *** ** **** de f Zb ∈Zb Now, we  ha ve an open d eri vation  Σ 3 o f  ⊥ fr om[Zb ∈  b]. [ Z b ∈ Zb ]5 Σ1 ⊥ ¬ I, 5 ¬Zb ∈Z b [Zb ∈ b ]4  Σ2Zb ∈ Zb [⊥]6 ¬E,6 ⊥  C l aim 2. Σ3 init iatesan on - t erm ina ting reduc ti on se q uence a n ds o isn ot fully norm alizable. 1 2 5 B y appl yi ng▷ ¬ to Σ 3, we  h a vethe derivat ionΣ4 below.  [Zb ∈ b]4 [Zb ∈ Zb]1Σ1 ⊥ ¬ I,1 ¬Zb  ∈ Z b  ∈ ZIZb  ∈ {x∈  b |¬x ∈ x } [¬Zb ∈  Z b] 2 [Zb  ∈ b ] 2 [ ¬Zb∈ Z b]2∈Z IZ b ∈ {x ∈ b|¬x  ∈x}** * *** ***** ***** **** ** ****de f Z b ∈ Z b[⊥]3 ¬E,3⊥ ∈Z E ,2⊥Since Zb∈ { x∈ b |¬ x ∈ x }  in ∈Z  E− rul e is no t an as s ump ti o n ,  t he application of▷∈ Z −  r eduction  produ c es  t h e same d er iva tion w ith Σ 3. T h er efore, Σ 3 gene r at es  a non- te rminat in g r edu ction sequ ence a nd so is  notf ully normaliza b le. We call t h er e su l t of  Pr oposition 3 .4.1 Crab bé's ca se. Furth ermor e, fr o m the derivation Σ3 ,we rea dilyobtain a clos ed derivat i o n  Σ5 of ¬ ∃y(Z y  ∈ y) as follows.  [∃y( Zy ∈ y)]7[ Zb ∈ b]4  Σ3 ⊥  ∃
 Zb_∈ Zb_[⊥]3 ¬E,3⊥ ∈Z E,2⊥_Then, there_is_an open_derivation_Σ2 of Zb_∈ Zb from_[Zb ∈ b]. [Zb_∈ b]4 [Zb_∈_Zb]1 Σ1 ⊥ ¬I,1¬Zb ∈ Zb ∈Z IZb ∈ {x ∈ b|¬x ∈ x}**************************_de_f Zb_∈_Zb_Now, we have an open_derivation Σ3 of ⊥ from_[Zb ∈_b]. [Zb ∈ Zb]5 Σ1 ⊥ ¬I,5¬Zb ∈_Zb_[Zb ∈ b]4_Σ2 Zb ∈ Zb [⊥]6 ¬E,6⊥ Claim 2. Σ3_initiates a non-terminating reduction sequence and_so is not_fully_normalizable._125 By applying ▷¬_to Σ3, we have the derivation_Σ4 below. [Zb ∈ b]4 [Zb_∈ Zb]1 Σ1 ⊥ ¬I,1¬Zb ∈ Zb_∈Z IZb ∈ {x ∈ b|¬x_∈ x} [¬Zb ∈ Zb]2_[Zb ∈_b]2 [¬Zb ∈ Zb]2 ∈Z_IZb ∈ {x_∈ b|¬x_∈ x}************************** de_f Zb ∈ Zb [⊥]3 ¬E,3⊥_∈Z E,2⊥ Since_Zb ∈ {x ∈ b|¬x ∈_x}_in ∈Z E−rule_is_not_an assumption,_the application of_▷∈Z−_reduction produces_the_same derivation with Σ3. Therefore, Σ3_generates_a non-terminating reduction sequence and so is_not fully normalizable. We_call_the result of Proposition_3.4.1 Crabbé's case. Furthermore, from_the derivation Σ3, we readily obtain_a closed_derivation Σ5_of ¬∃y(Zy ∈ y) as follows. [∃y(Zy ∈ y)]7 [Zb ∈_b]4 Σ3 ⊥ ∃