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Proofgold Proof

pf
Let x0 of type ι be given.
Let x1 of type ιι be given.
Assume H0: ∀ x2 . x2x0∀ x3 . x3x0∀ x4 . x4x1 x2x4x1 x3x2 = x3.
Let x2 of type ο be given.
Assume H1: ∀ x3 : ι → ι . bij (lam x0 (λ x4 . x1 x4)) (famunion x0 (λ x4 . x1 x4)) x3x2.
Apply H1 with λ x3 . ap x3 1.
Apply bijI with lam x0 (λ x3 . x1 x3), famunion x0 (λ x3 . x1 x3), λ x3 . ap x3 1 leaving 3 subgoals.
Let x3 of type ι be given.
Assume H2: x3lam x0 (λ x4 . x1 x4).
Apply famunionI with x0, x1, ap x3 0, ap x3 1 leaving 2 subgoals.
Apply ap0_Sigma with x0, x1, x3.
The subproof is completed by applying H2.
Apply ap1_Sigma with x0, x1, x3.
The subproof is completed by applying H2.
Let x3 of type ι be given.
Assume H2: x3lam x0 (λ x4 . x1 x4).
Let x4 of type ι be given.
Assume H3: x4lam x0 (λ x5 . x1 x5).
Assume H4: ap x3 1 = ap x4 1.
Apply tuple_Sigma_eta with x0, x1, x3, λ x5 x6 . x5 = x4 leaving 2 subgoals.
The subproof is completed by applying H2.
Apply tuple_Sigma_eta with x0, x1, x4, λ x5 x6 . lam 2 (λ x7 . If_i (x7 = 0) (ap x3 0) (ap x3 1)) = x5 leaving 2 subgoals.
The subproof is completed by applying H3.
Claim L5: ap x3 0 = ap x4 0
Apply H0 with ap x3 0, ap x4 0, ap x3 1 leaving 4 subgoals.
Apply ap0_Sigma with x0, x1, x3.
The subproof is completed by applying H2.
Apply ap0_Sigma with x0, x1, x4.
The subproof is completed by applying H3.
Apply ap1_Sigma with x0, x1, x3.
The subproof is completed by applying H2.
Apply H4 with λ x5 x6 . x6x1 (ap x4 0).
Apply ap1_Sigma with x0, x1, x4.
The subproof is completed by applying H3.
Apply L5 with λ x5 x6 . lam 2 (λ x7 . If_i (x7 = 0) x6 (ap x3 1)) = lam 2 (λ x7 . If_i (x7 = 0) (ap x4 0) (ap x4 1)).
Apply H4 with λ x5 x6 . lam 2 (λ x7 . If_i (x7 = 0) (ap x4 0) x6) = lam 2 (λ x7 . If_i (x7 = 0) (ap x4 0) (ap x4 1)).
Let x5 of type ιιο be given.
Assume H6: x5 (lam 2 (λ x6 . If_i (x6 = 0) (ap x4 0) (ap x4 1))) (lam 2 (λ x6 . If_i (x6 = 0) (ap x4 0) (ap x4 1))).
The subproof is completed by applying H6.
Let x3 of type ι be given.
Assume H2: x3famunion x0 (λ x4 . x1 x4).
Apply famunionE_impred with x0, x1, x3, ∃ x4 . and (x4lam x0 (λ x5 . x1 x5)) ((λ x5 . ap x5 1) x4 = ...) leaving 2 subgoals.
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