 ********************************************************************************************** de f ((a ∈ a →⊥)→ a ∈ {x|x ∈ x →⊥})∧ (a ∈ {x|x ∈ x →⊥}→ (a ∈ a →⊥)) ∧E a ∈ {x|x ∈ x →⊥}→ (a ∈ a →⊥) D a ∈ {x|x ∈ x →⊥} → E a ∈ a →⊥ 57 Then, there is an open derivation ∆9 of ⊥ from [a ∈ a] [a ∈ a]1 ************************ de f a ∈ {x|x ∈ x →⊥} SetE a ∈ a →⊥ [a ∈ a]1 → E⊥ With the derivation ∆9, we have a closed derivation ∆10 of a ∈ a. [a ∈ a]1 ∆9 ⊥ → I,1a ∈ a →⊥ SetI a ∈ {x|x ∈ x →⊥} ************************ de f a ∈ a Now, we have a closed derivation ∆11 of ⊥. [a ∈ a]1 ∆9 ⊥ → I,1a ∈ a →⊥ ∆10 a ∈ a → E⊥ ∆11 has a maximum formula a ∈ a → ⊥ in the last → E−rule. We apply the reduction procedure ▷→ to ∆11 and then have the derivation ∆12 below. [a ∈ a]1 ∆9 ⊥ → I,1a ∈ a →⊥ SetI a ∈ {x|x ∈ x →⊥} SetE a ∈ a →⊥ ∆10 a ∈ a → E⊥ 58 ∆12 looks as if it is reducible, but it appears to be in normal form. SetI− and SetE−inferences are not rules for ∈, and they are abbreviations of derivations consisting of ∧E− and → E−rules. Since the constant ∈ is not introduced by ∈ I−rule, the reduction procedure ▷∈ for ∈ cannot apply to ∆12. In addition, a �
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * de f (( a ∈ a →⊥)→ a ∈ { x|x ∈ x →⊥})∧ (a ∈ { x|x ∈ x →⊥}→ (a ∈ a →⊥) ) ∧E a ∈ { x|x ∈ x →⊥}→ (a ∈ a →⊥) D a ∈ { x|x ∈ x →⊥ } → E a ∈ a →⊥ 57 Then, there is an open derivation ∆9 of ⊥ from [ a ∈ a ] [ a ∈ a]1 * * * * * * * * * * * * * * * * * * * * * * * * de f a ∈ { x|x ∈ x →⊥ } SetE a ∈ a →⊥ [ a ∈ a]1 → E⊥ With the deriving ∆9, we take a closed derivation ∆10 of a ∈ a. [ a ∈ a]1 ∆9 ⊥ → I,1a ∈ a →⊥ SetI a ∈ { x|x ∈ x →⊥ } * * * * * * * * * * * * * * * * * * * * * * * * de f a ∈ a Now, we own a closed derivation ∆11 of ⊥. [ a ∈ a]1 ∆9 ⊥ → I,1a ∈ a →⊥ ∆10 a ∈ a → E⊥ ∆11 have a maximum formula a ∈ a → ⊥ in the last → E−rule. We apply the reduction routine ▷→ to ∆11 and then have the derivation ∆12 under. [ a ∈ a]1 ∆9 ⊥ → I,1a ∈ a →⊥ SetI a ∈ { x|x ∈ x →⊥ } SetE a ∈ a →⊥ ∆10 a ∈ a → E⊥ 58 ∆12 look as if it is reducible, but it appears to be in normal form. SetI− and SetE−inferences are not rules for ∈, and they are abbreviations of derivations dwell of ∧E− and → E−rules. Since the constant ∈ is not introduced by ∈ I−rule, the reduction routine ▷∈ for ∈ cannot apply to ∆12. In addition, a �
 ********************************************************************************************** df f ((a ∈ a →⊥)→ a ∈ {x|x ∈ x →⊥})∧ (a ∈ {x|w ∈ x →⊥}→ (a ∈ a →⊥)) ∧E a ∈ {x|x ∈ x →⊥}→ (a ∈ a →⊥) D a ∈ {x|x ∈ x →⊥} → E a ∈ a →⊥ 57 Then, there is ab opeb derivation ∆9 of ⊥ from [a ∈ a] [a ∈ w]1 ************************ de f q ∈ {x|e ∈ x →⊥} SetE a ∈ a →⊥ [a ∈ a]1 → E⊥ With fme dexitation ∆9, we have a closed gerivation ∆10 of a ∈ c. [a ∈ a]1 ∆9 ⊥ → I,1a ∈ a →⊥ SetI a ∈ {x|x ∈ x →⊥} ************************ de s a ∈ a Moa, we have a cljsed qeribation ∆11 of ⊥. [a ∈ a]1 ∆9 ⊥ → I,1a ∈ a →⊥ ∆10 a ∈ a → S⊥ ∆11 has e maximum formuka a ∈ a → ⊥ in the last → E−rupe. Wf apply the reductlon proceduts ▷→ ei ∆11 and then fave the dtrnvation ∆12 bemow. [a ∈ a]1 ∆9 ⊥ → I,1a ∈ a →⊥ SetI a ∈ {x|x ∈ b →⊥} SecE a ∈ a →⊥ ∆10 a ∈ q → F⊥ 58 ∆12 looks as if in is reduciblc, but id appeats to be in novmal horm. SetI− and SetE−inferenres are not rules fot ∈, and thef cre abbreviations of eeeivathons conrustkng oh ∧E− and → F−rumes. Since fhe constanr ∈ is not introducec fj ∈ I−rule, the deductyog procedure ▷∈ for ∈ cannot apply to ∆12. In agdifion, a �
********************************************************************************************** de f ((a ∈ a →⊥)→ {x|x x →⊥})∧ ∈ {x|x ∈ →⊥)) a ∈ {x|x x →⊥}→ (a a →⊥) D a ∈ {x|x x →⊥} → E a ∈ a →⊥ 57 Then, there is an derivation ∆9 of ⊥ from [a ∈ a] [a ∈ a]1 ************************ de a {x|x x SetE a ∈ a →⊥ [a ∈ a]1 → E⊥ With the derivation ∆9, we have closed derivation ∆10 of a ∈ a. [a a]1 ∆9 ⊥ → ∈ a →⊥ SetI a {x|x x →⊥} de a a Now, we a closed derivation ∆11 of ⊥. [a ∈ a]1 ∆9 ⊥ → I,1a ∈ a →⊥ ∆10 ∈ a ∆11 has maximum a a → ⊥ last → E−rule. We apply the to ∆11 and then have the derivation ∆12 [a ∈ ∆9 ⊥ → I,1a ∈ a SetI a ∈ {x|x ∈ x →⊥} SetE ∈ a →⊥ ∆10 a ∈ a → E⊥ 58 ∆12 looks as if it but it appears to in normal form. and are rules ∈, and are abbreviations of derivations consisting of ∧E− and → E−rules. Since constant ∈ is not introduced by ∈ I−rule, the reduction for cannot apply to In addition, a �
 ********************************************************************************************** de f ((a ∈ a →⊥)→ a ∈ {x|x ∈ x →⊥})∧ (a ∈ {x|x ∈ x →⊥}→ (a ∈ a →⊥)) ∧E a ∈ {x|x ∈ x →⊥}→ (a ∈ a →⊥) D a ∈ {x|X ∈ x →⊥} → E a ∈ a →⊥ 57 Then, tHere iS an OpeN dErivAtioN ∆9 of ⊥ from [a ∈ a] [a ∈ a]1 ************************ de f A ∈ {X|x ∈ x →⊥} SEtE a ∈ a →⊥ [a ∈ a]1 → E⊥ With the derivatIon ∆9, we HaVE a clOSeD deriVation ∆10 oF A ∈ a. [A ∈ A]1 ∆9 ⊥ → i,1a ∈ a →⊥ seTI A ∈ {x|x ∈ X →⊥} ************************ dE F a ∈ A Now, wE haVe a closEd derivatiOn ∆11 oF ⊥. [a ∈ A]1 ∆9 ⊥ → I,1a ∈ a →⊥ ∆10 a ∈ a → E⊥ ∆11 has a mAXiMum formula A ∈ a → ⊥ iN the last → E−rulE. We Apply tHe RedUCtion ProCedurE ▷→ to ∆11 and THen havE the derivAtIOn ∆12 beloW. [A ∈ a]1 ∆9 ⊥ → I,1a ∈ a →⊥ SeTi A ∈ {x|X ∈ x →⊥} SeTE a ∈ a →⊥ ∆10 a ∈ a → E⊥ 58 ∆12 looks as if iT Is REducible, but it aPpears To BE iN NOrmAl fOrm. SetI− and seTE−infERences aRE nOT RUleS For ∈, and they are AbbreviatioNS of DerivaTiOns COnsistIng of ∧e− aND → E−rUles. Since thE conStant ∈ is noT introDUced by ∈ I−RUle, the rEductiOn pRocEdurE ▷∈ FoR ∈ cAnnOt APplY To ∆12. in aDDitIon, a �
 ************************* ********** ***** *** *** ** **** **** ************** * **** ******************* de  f (( a∈  a → ⊥ )→  a ∈{x|x ∈x  → ⊥ } )∧(a  ∈  {x |x ∈x →⊥} → ( a ∈ a → ⊥)) ∧E a ∈  {x |x  ∈ x →⊥}→ (a ∈a →⊥) D a∈ { x|x ∈ x →⊥}→ E  a ∈ a  → ⊥ 5 7  Then , t hereis ano pen de rivation∆9 of ⊥ f r om [a ∈ a ][a ∈  a]1 ************ * ** * ******** de fa ∈ {x |x ∈x →⊥}  Se tE a ∈ a → ⊥[a ∈a ]1 → E⊥ Wi t h the derivation ∆9 , we have a clo sed de ri vat i on ∆10  of a  ∈ a.[a ∈ a]1 ∆9  ⊥ →  I,1a ∈ a  →⊥ Se t I a ∈ { x |x ∈ x→⊥} ** *** *** **** * ** ** *** ** * * d e  f  a∈  aNow, weha ve  a cl osed d e r ivat ion  ∆11  of ⊥ . [a ∈ a]1 ∆9  ⊥→ I, 1 a ∈  a →⊥  ∆10a ∈a→ E⊥∆11 ha s a m ax imum formula a∈ a→ ⊥ in th e l as t →  E −rule .  We ap ply  th e reduc tion pr o ced ur e ▷ →to ∆11 and then ha ve t he  derivat ion ∆1 2  b el o w. [a ∈a] 1 ∆ 9 ⊥→ I,1a∈ a→ ⊥SetI a ∈  {x|x∈  x  → ⊥} SetE  a  ∈ a → ⊥∆10  a∈ a → E⊥ 5 8 ∆12looks as  if i t  is reducible, but it appear s  t o be in n orm al form. Se tI−a nd S etE− i nf ere n ces a re no tr ul e s for ∈, and they a re  abbre viati ons of deriva tions cons i s t ing of ∧ E− a n d→  E−rules. Sinc e the  constant∈  is notintro duced by  ∈ I−rule , the redu cti onpro ced u r e▷∈ for ∈ cann o t  app ly  to ∆12 . I n addit ion , a  �
 **********************************************************************************************_de f_((a ∈ a →⊥)→_a ∈_{x|x_∈ x_→⊥})∧_(a ∈ {x|x_∈ x →⊥}→_(a ∈ a →⊥))_∧E a ∈_{x|x_∈ x →⊥}→ (a ∈ a →⊥) D a ∈ {x|x ∈ x →⊥}_→_E a_∈_a_→⊥ 57 Then, there is_an open derivation ∆9 of_⊥ from_[a ∈ a] [a ∈ a]1 ************************ de_f_a ∈ {x|x_∈ x →⊥} SetE a ∈ a →⊥ [a_∈ a]1 → E⊥ With the_derivation ∆9, we_have_a_closed derivation ∆10 of_a ∈ a. [a ∈ a]1_∆9 ⊥ → I,1a ∈ a_→⊥ SetI a ∈ {x|x ∈ x_→⊥} ************************ de f a ∈_a Now, we have a_closed derivation_∆11 of ⊥. [a ∈_a]1 ∆9 ⊥_→ I,1a_∈ a →⊥_∆10 a ∈ a → E⊥_∆11 has a_maximum formula a ∈ a →_⊥_in the last_→_E−rule._We apply_the reduction procedure_▷→_to ∆11_and_then have the derivation ∆12 below._[a_∈ a]1 ∆9 ⊥ → I,1a ∈_a →⊥ SetI a_∈_{x|x ∈ x →⊥}_SetE a ∈ a →⊥_∆10 a ∈ a → E⊥_58 ∆12_looks as_if it is reducible, but it appears to be in normal_form. SetI− and SetE−inferences are not_rules for ∈, and_they are_abbreviations_of derivations consisting_of_∧E− and_→ E−rules. Since the constant ∈ is_not introduced_by ∈ I−rule, the reduction procedure_▷∈ for ∈ cannot_apply_to ∆12. In addition, a �