 {x|¬x ∈ x}), and show that Σ3 is in full normal form. Claim 1. there is a closed derivation Σ3 of ¬∃y(y = {x|¬x ∈ x}). There is an open derivation Σ1 of ⊥ from [a = {x|¬x ∈ x}] and [a ∈ a]. [a = {x|¬x ∈ x}]1 [a ∈ a]5 [a = {x|¬x ∈ x}]1 [¬a ∈ a]2 [a = {x|¬x ∈ x}]1 RD∃!a [¬a ∈ a]2 [a ∈ a]3 [⊥]4 ¬E,4⊥ {}E1,3⊥ {}E2,2⊥ With the open derivation Σ1, we have an open derivation Σ2 of ⊥ from [a = {x|¬x ∈ x}]. [a = {x|¬x ∈ x}]1 [a = {x|¬x ∈ x}]1, [a ∈ a]5 Σ1 ⊥ ¬I,5¬a ∈ a [a = {x|¬x ∈ x}]1 RD∃!a [a = {x|¬x ∈ x}]1, [a ∈ a]6 Σ1 ⊥ {}E1,6⊥ Now, we have a closed derivation Σ3 of ¬∃y(y = {x|¬x ∈ x}). [∃y(y = {x|¬x ∈ x})]7 [a = {x|¬x ∈ x}]1 Σ2 ⊥ ∃E,1⊥ ¬I,7¬∃y(y = {x|¬x ∈ x}) 139 Claim 2. Σ3 is in full normal form. Since all major premises in Σ1, Σ9, and Σ3 are assumptions, Σ3 is in full normal form. The derivation Σ3 of ¬∃y(y = {x|¬x ∈ x}) in Proposition 4.1.1 is in full normal form. However, if we accept an Ekman-type reduction introduced in Section 3.1 of Chapter 3, SF has an open non-full normal derivation of �
{ x|¬x ∈ x }), and show that Σ3 is in full normal form. Claim 1. there be a shut derivation Σ3 of ¬∃y(y = { x|¬x ∈ x }). There is an loose ancestry Σ1 of ⊥ from [ a = { x|¬x ∈ x } ] and [ a ∈ a ]. [ a = { x|¬x ∈ x}]1 [ a ∈ a]5 [ a = { x|¬x ∈ x}]1 [ ¬a ∈ a]2 [ a = { x|¬x ∈ x}]1 RD∃!a [ ¬a ∈ a]2 [ a ∈ a]3 [ ⊥]4 ¬E,4⊥ { } E1,3⊥ { } E2,2⊥ With the open ancestry Σ1, we give birth an open derivation Σ2 of ⊥ from [ a = { x|¬x ∈ x } ]. [ a = { x|¬x ∈ x}]1 [ a = { x|¬x ∈ x}]1, [ a ∈ a]5 Σ1 ⊥ ¬I,5¬a ∈ a [ a = { x|¬x ∈ x}]1 RD∃!a [ a = { x|¬x ∈ x}]1, [ a ∈ a]6 Σ1 ⊥ { } E1,6⊥ Now, we have a shut derivation Σ3 of ¬∃y(y = { x|¬x ∈ x }). [ ∃y(y = { x|¬x ∈ x})]7 [ a = { x|¬x ∈ x}]1 Σ2 ⊥ ∃E,1⊥ ¬I,7¬∃y(y = { x|¬x ∈ x }) 139 Claim 2. Σ3 is in full normal form. Since all major premises in Σ1, Σ9, and Σ3 are assumption, Σ3 is in full normal form. The derivation Σ3 of ¬∃y(y = { x|¬x ∈ x }) in Proposition 4.1.1 is in full normal form. However, if we bear an Ekman - type reduction introduced in incision 3.1 of Chapter 3, SF has an open non - full normal deriving of �
 {x|¬x ∈ x}), and show that Σ3 is in full normal fotm. Claim 1. thers is a cuosed derivation Σ3 of ¬∃y(y = {x|¬x ∈ x}). Thert is an open derivagion Σ1 of ⊥ from [a = {x|¬x ∈ z}] and [a ∈ a]. [a = {x|¬x ∈ w}]1 [a ∈ z]5 [a = {r|¬x ∈ x}]1 [¬a ∈ a]2 [a = {x|¬x ∈ x}]1 RD∃!a [¬a ∈ a]2 [a ∈ a]3 [⊥]4 ¬E,4⊥ {}E1,3⊥ {}E2,2⊥ Wktk the open derivation Σ1, we have an opqn derifahion Σ2 of ⊥ from [a = {q|¬x ∈ x}]. [z = {x|¬x ∈ x}]1 [a = {x|¬x ∈ x}]1, [a ∈ a]5 Σ1 ⊥ ¬I,5¬a ∈ a [a = {s|¬x ∈ x}]1 RV∃!a [a = {x|¬x ∈ x}]1, [a ∈ s]6 Σ1 ⊥ {}E1,6⊥ Now, we have a closed derlvation Σ3 of ¬∃y(y = {x|¬x ∈ x}). [∃y(y = {x|¬x ∈ x})]7 [a = {x|¬x ∈ x}]1 Σ2 ⊥ ∃E,1⊥ ¬I,7¬∃h(y = {x|¬x ∈ x}) 139 Claim 2. Σ3 is jn full normal form. Since all mxjor 'remises in Σ1, Σ9, anf Σ3 are assum'tions, Σ3 is in full normal xorm. Thr derivation Σ3 pf ¬∃b(y = {z|¬x ∈ x}) in Proposition 4.1.1 is in full normal form. Howavzr, if we accept an Eknab-type redgctiub ivtrkdnces in Sfctmon 3.1 of Chalter 3, SF haw an open non-full npriql derivation of �
{x|¬x ∈ x}), and show that Σ3 full form. Claim there is a = ∈ x}). There an open derivation of ⊥ from [a = {x|¬x x}] and [a ∈ a]. [a = {x|¬x ∈ x}]1 [a ∈ a]5 = {x|¬x ∈ x}]1 [¬a ∈ a]2 [a = {x|¬x ∈ x}]1 RD∃!a ∈ [a a]3 ¬E,4⊥ {}E1,3⊥ {}E2,2⊥ With the open derivation Σ1, we have an open derivation Σ2 of ⊥ [a = {x|¬x ∈ x}]. [a = {x|¬x x}]1 [a = {x|¬x x}]1, [a ∈ a]5 Σ1 ¬I,5¬a a [a {x|¬x x}]1 [a = {x|¬x x}]1, [a ∈ a]6 Σ1 ⊥ {}E1,6⊥ Now, we have a closed derivation Σ3 of ¬∃y(y = ∈ x}). {x|¬x ∈ [a {x|¬x x}]1 Σ2 ⊥ = {x|¬x ∈ x}) 139 Claim in full normal form. Since all major premises Σ1, Σ9, Σ3 are assumptions, Σ3 is in normal form. The derivation Σ3 of ¬∃y(y = ∈ x}) in Proposition 4.1.1 is in full normal form. However, if we accept an introduced in Section 3.1 Chapter 3, SF an non-full derivation �
 {x|¬x ∈ x}), and show that Σ3 is in full noRmal form. ClAim 1. thEre Is a ClOsed DeriVation Σ3 of ¬∃y(y = {x|¬x ∈ x}). tHere Is an open derivation Σ1 of ⊥ fRom [a = {x|¬X ∈ x}] ANd [a ∈ a]. [A = {X|¬x ∈ X}]1 [a ∈ a]5 [a = {x|¬X ∈ x}]1 [¬a ∈ a]2 [a = {x|¬x ∈ X}]1 rD∃!A [¬A ∈ A]2 [a ∈ a]3 [⊥]4 ¬e,4⊥ {}E1,3⊥ {}e2,2⊥ WIth ThE OpEn derIvaTion Σ1, we Have an open DerIvAtion Σ2 of ⊥ from [A = {X|¬x ∈ X}]. [a = {x|¬x ∈ x}]1 [a = {x|¬x ∈ x}]1, [a ∈ A]5 Σ1 ⊥ ¬I,5¬A ∈ a [a = {x|¬x ∈ x}]1 RD∃!a [a = {x|¬x ∈ X}]1, [a ∈ a]6 σ1 ⊥ {}E1,6⊥ Now, wE hAve A CloseD deRivatIon Σ3 of ¬∃Y(Y = {x|¬x ∈ x}). [∃y(y = {X|¬x ∈ x})]7 [a = {x|¬x ∈ x}]1 Σ2 ⊥ ∃E,1⊥ ¬i,7¬∃y(Y = {X|¬x ∈ x}) 139 ClaIM 2. Σ3 is in fULL nOrmaL form. Since all majoR PrEMises in Σ1, Σ9, and Σ3 aRe assuMpTIoNS, σ3 is In fUll normal fOrM. The dERivatioN σ3 oF ¬∃Y(Y = {X|¬x ∈ x}) IN Proposition 4.1.1 iS in full normAL foRm. HoweVeR, if WE accepT an EkMaN-TypE reduction iNtroDuced in SeCtion 3.1 oF chapter 3, sf has an oPen non-FulL noRmal DErIvAtiOn OF �
 {x|¬x ∈ x}), and show tha t Σ3 is in  full  no rma lform . Cl aim 1. there i s  a c losed derivation Σ3 of  ¬∃y( y=  {x| ¬ x∈ x}) . There is a n o pe nder iv a ti on Σ1  of  ⊥ from  [a = {x|¬ x ∈  x }] and [a ∈a ].  [a = {x|¬ x ∈  x}]1 [a ∈ a ]5[a = { x| ¬x∈  x}]1  [¬ a ∈ a ]2 [a=  {x|¬x  ∈ x}]1 R D∃ ! a [¬a∈  a]2 [a ∈  a ]3 [ ⊥]4 ¬E,4⊥ {}E1,3⊥ {} E 2,2⊥ With theopen d er i va t i onΣ1,  we have a nopend erivati o nΣ 2 of⊥  from [a = {x |¬x ∈ x}].[ a =  {x|¬x  ∈  x} ] 1 [a =  {x|¬ x∈  x} ]1, [a ∈ a] 5 Σ1  ⊥ ¬I,5¬a  ∈ a [ a  = {x|¬ x  ∈ x}]1  RD∃!a  [a  ={x|¬ x  ∈  x }]1 ,[ a ∈ a] 6 Σ 1  ⊥{}E1,6⊥No w,  we h avea c l osed  de riva tionΣ3 of ¬∃y(y =  {x |¬x∈  x} ). [∃ y(y =  {x| ¬x  ∈ x} )]7 [a  = {x |¬ x ∈ x}]1 Σ2 ⊥ ∃ E,1⊥  ¬I,7¬∃y( y =  { x|¬ x∈ x}) 139 Cl aim  2.  Σ3 isin full nor ma l f or m. Since all major  p r e mi ses in Σ 1, Σ9, an dΣ 3 are as su mpt ions , Σ3 is  inf ul l normal  form. Th ederivat io n Σ3 o f¬∃y (y= {x| ¬ x ∈x}) in  Proposi tion4 .1.1 is in ful l  normal form. Ho w e ve r , if  we  accept anEkma n -typ e re d uc tio n  intr oduce di nS ection 3.1 of Chapt er  3, SF  hasan open non-f ull normal d e rivation  of� 
 {x|¬x_∈ x}),_and show that Σ3_is in_full_normal form._Claim_1. there is_a closed derivation_Σ3 of ¬∃y(y =_{x|¬x ∈ x})._There_is an open derivation Σ1 of ⊥ from [a = {x|¬x ∈ x}] and_[a_∈ a]._[a_=_{x|¬x ∈ x}]1 [a ∈_a]5 [a = {x|¬x ∈_x}]1 [¬a_∈ a]2 [a = {x|¬x ∈ x}]1 RD∃!a_[¬a_∈ a]2 [a_∈ a]3 [⊥]4 ¬E,4⊥ {}E1,3⊥ {}E2,2⊥ With the open_derivation Σ1, we have an open_derivation Σ2 of_⊥_from_[a = {x|¬x ∈_x}]. [a = {x|¬x ∈ x}]1_[a = {x|¬x ∈ x}]1, [a_∈ a]5 Σ1 ⊥ ¬I,5¬a ∈ a_[a = {x|¬x ∈ x}]1 RD∃!a_[a = {x|¬x ∈ x}]1,_[a ∈_a]6 Σ1 ⊥ {}E1,6⊥ Now,_we have a_closed derivation_Σ3 of ¬∃y(y_= {x|¬x ∈ x}). [∃y(y =_{x|¬x ∈ x})]7_[a = {x|¬x ∈ x}]1 Σ2_⊥_∃E,1⊥ ¬I,7¬∃y(y =_{x|¬x_∈_x}) 139_Claim 2. Σ3_is_in full_normal_form. Since all major premises in_Σ1,_Σ9, and Σ3 are assumptions, Σ3 is_in full normal form._The_derivation Σ3 of ¬∃y(y_= {x|¬x ∈ x}) in_Proposition 4.1.1 is in full normal_form. However,_if we_accept an Ekman-type reduction introduced in Section 3.1 of Chapter 3,_SF has an open non-full normal_derivation of �