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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 ι be given.
Let x3 of type ι be given.
Let x4 of type ι be given.
Assume H0: prim1 x4 (0fc90.. (4ae4a.. (4ae4a.. (4ae4a.. (4ae4a.. 4a7ef..)))) (λ x5 . If_i (x5 = 4a7ef..) x0 (If_i (x5 = 4ae4a.. 4a7ef..) x1 (If_i (x5 = 4ae4a.. (4ae4a.. 4a7ef..)) x2 x3)))).
Claim L1: ∃ x5 . and (prim1 x5 (4ae4a.. (4ae4a.. (4ae4a.. (4ae4a.. 4a7ef..))))) (∃ x6 . and (prim1 x6 (If_i (x5 = 4a7ef..) x0 (If_i (x5 = 4ae4a.. 4a7ef..) x1 (If_i (x5 = 4ae4a.. (4ae4a.. 4a7ef..)) ... ...)))) ...)
...
Apply exandE_i with λ x5 . prim1 x5 (4ae4a.. (4ae4a.. (4ae4a.. (4ae4a.. 4a7ef..)))), λ x5 . ∃ x6 . and (prim1 x6 (If_i (x5 = 4a7ef..) x0 (If_i (x5 = 4ae4a.. 4a7ef..) x1 (If_i (x5 = 4ae4a.. (4ae4a.. 4a7ef..)) x2 x3)))) (x4 = aae7a.. x5 x6), ∃ x5 . and (prim1 x5 (4ae4a.. (4ae4a.. (4ae4a.. (4ae4a.. 4a7ef..))))) (∃ x6 . x4 = aae7a.. x5 x6) leaving 2 subgoals.
The subproof is completed by applying L1.
Let x5 of type ι be given.
Assume H2: prim1 x5 (4ae4a.. (4ae4a.. (4ae4a.. (4ae4a.. 4a7ef..)))).
Assume H3: ∃ x6 . and (prim1 x6 (If_i (x5 = 4a7ef..) x0 (If_i (x5 = 4ae4a.. 4a7ef..) x1 (If_i (x5 = 4ae4a.. (4ae4a.. 4a7ef..)) x2 x3)))) (x4 = aae7a.. x5 x6).
Apply exandE_i with λ x6 . prim1 x6 (If_i (x5 = 4a7ef..) x0 (If_i (x5 = 4ae4a.. 4a7ef..) x1 (If_i (x5 = 4ae4a.. (4ae4a.. 4a7ef..)) x2 x3))), λ x6 . x4 = aae7a.. x5 x6, ∃ x6 . and (prim1 x6 (4ae4a.. (4ae4a.. (4ae4a.. (4ae4a.. 4a7ef..))))) (∃ x7 . x4 = aae7a.. x6 x7) leaving 2 subgoals.
The subproof is completed by applying H3.
Let x6 of type ι be given.
Assume H4: prim1 x6 (If_i (x5 = 4a7ef..) x0 (If_i (x5 = 4ae4a.. 4a7ef..) x1 (If_i (x5 = 4ae4a.. (4ae4a.. 4a7ef..)) x2 x3))).
Assume H5: x4 = aae7a.. x5 x6.
Let x7 of type ο be given.
Assume H6: ∀ x8 . and (prim1 x8 (4ae4a.. (4ae4a.. (4ae4a.. (4ae4a.. 4a7ef..))))) (∃ x9 . x4 = aae7a.. x8 x9)x7.
Apply H6 with x5.
Apply andI with prim1 x5 (4ae4a.. (4ae4a.. (4ae4a.. (4ae4a.. 4a7ef..)))), ∃ x8 . x4 = aae7a.. x5 x8 leaving 2 subgoals.
The subproof is completed by applying H2.
Let x8 of type ο be given.
Assume H7: ∀ x9 . x4 = aae7a.. x5 x9x8.
Apply H7 with x6.
The subproof is completed by applying H5.