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Proofgold Proof

pf
Let x0 of type ι be given.
Let x1 of type ι be given.
Let x2 of type ιι be given.
Assume H0: bij x0 x1 x2.
Apply H0 with ccad8.. x0 x1.
Assume H1: and (∀ x3 . x3x0x2 x3x1) (∀ x3 . x3x0∀ x4 . x4x0x2 x3 = x2 x4x3 = x4).
Apply H1 with (∀ x3 . x3x1∃ x4 . and (x4x0) (x2 x4 = x3))ccad8.. x0 x1.
Assume H2: ∀ x3 . x3x0x2 x3x1.
Assume H3: ∀ x3 . x3x0∀ x4 . x4x0x2 x3 = x2 x4x3 = x4.
Assume H4: ∀ x3 . x3x1∃ x4 . and (x4x0) (x2 x4 = x3).
Let x3 of type ο be given.
Assume H5: ∀ x4 . and (and (∀ x5 . x5x0∃ x6 . and (x6x1) (KPair_alt7 x5 x6x4)) (∀ x5 . x5x1∃ x6 . and (x6x0) (KPair_alt7 x6 x5x4))) (∀ x5 x6 x7 x8 . KPair_alt7 x5 x6x4KPair_alt7 x7 x8x4iff (x5 = x7) (x6 = x8))x3.
Apply H5 with {KPair_alt7 x4 (x2 x4)|x4 ∈ x0}.
Apply and3I with ∀ x4 . x4x0∃ x5 . and (x5x1) (KPair_alt7 x4 x5{KPair_alt7 x6 (x2 x6)|x6 ∈ x0}), ∀ x4 . x4x1∃ x5 . and (x5x0) (KPair_alt7 x5 x4{KPair_alt7 x6 (x2 x6)|x6 ∈ x0}), ∀ x4 x5 x6 x7 . KPair_alt7 x4 x5{KPair_alt7 x8 (x2 x8)|x8 ∈ x0}KPair_alt7 x6 x7{KPair_alt7 x8 (x2 x8)|x8 ∈ x0}iff (x4 = x6) (x5 = x7) leaving 3 subgoals.
Let x4 of type ι be given.
Assume H6: x4x0.
Let x5 of type ο be given.
Assume H7: ∀ x6 . and (x6x1) (KPair_alt7 x4 x6{KPair_alt7 x7 (x2 x7)|x7 ∈ x0})x5.
Apply H7 with x2 x4.
Apply andI with x2 x4x1, KPair_alt7 x4 (x2 x4){KPair_alt7 x6 (x2 x6)|x6 ∈ x0} leaving 2 subgoals.
Apply H2 with x4.
The subproof is completed by applying H6.
Apply ReplI with x0, λ x6 . KPair_alt7 x6 (x2 x6), x4.
The subproof is completed by applying H6.
Let x4 of type ι be given.
Assume H6: x4x1.
Apply H4 with x4, ∃ x5 . and (x5x0) (KPair_alt7 x5 x4{KPair_alt7 x6 (x2 ...)|x6 ∈ x0}) leaving 2 subgoals.
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