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Proofgold Proof

pf
Let x0 of type ι be given.
Let x1 of type ιι be given.
Let x2 of type ιι be given.
Assume H0: ∀ x3 . x3x0x1 x3 = x2 x3.
Let x3 of type ι be given.
Assume H1: x3lam x0 (λ x4 . x1 x4).
Claim L2: ∃ x4 . and (x4x0) (∃ x5 . and (x5x1 x4) (x3 = lam 2 (λ x6 . If_i (x6 = 0) x4 x5)))
Apply lamE2 with x0, x1, x3.
The subproof is completed by applying H1.
Apply L2 with x3lam x0 x2.
Let x4 of type ι be given.
Assume H3: (λ x5 . and (x5x0) (∃ x6 . and (x6x1 x5) (x3 = lam 2 (λ x7 . If_i (x7 = 0) x5 x6)))) x4.
Apply H3 with x3lam x0 x2.
Assume H4: x4x0.
Assume H5: ∃ x5 . and (x5x1 x4) (x3 = lam 2 (λ x6 . If_i (x6 = 0) x4 x5)).
Apply H5 with x3lam x0 x2.
Let x5 of type ι be given.
Assume H6: (λ x6 . and (x6x1 x4) (x3 = lam 2 (λ x7 . If_i (x7 = 0) x4 x6))) x5.
Apply H6 with x3lam x0 x2.
Assume H7: x5x1 x4.
Assume H8: x3 = lam 2 (λ x6 . If_i (x6 = 0) x4 x5).
Apply H8 with λ x6 x7 . x7lam x0 (λ x8 . x2 x8).
Apply tuple_2_Sigma with x0, λ x6 . x2 x6, x4, x5 leaving 2 subgoals.
The subproof is completed by applying H4.
Apply H0 with x4, λ x6 x7 . x5x6 leaving 2 subgoals.
The subproof is completed by applying H4.
The subproof is completed by applying H7.