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Proofgold Proof

pf
Let x0 of type ιο be given.
Let x1 of type ιιι be given.
Let x2 of type ιιι be given.
Assume H0: ∀ x3 x4 . x0 x3x0 x4x0 (x1 x3 x4).
Assume H1: ∀ x3 x4 . x0 x3x0 x4x0 (x2 x3 x4).
Assume H2: ∀ x3 x4 x5 . x0 x3x0 x4x0 x5x1 x3 (x1 x4 x5) = x1 x4 (x1 x3 x5).
Assume H3: ∀ x3 x4 x5 . x0 x3x0 x4x0 x5x1 (x1 x3 x4) x5 = x1 x3 (x1 x4 x5).
Assume H4: ∀ x3 x4 x5 . x0 x3x0 x4x0 x5x2 x3 (x1 x4 x5) = x1 (x2 x3 x4) (x2 x3 x5).
Assume H5: ∀ x3 x4 x5 . x0 x3x0 x4x0 x5x2 (x1 x3 x4) x5 = x1 (x2 x3 x5) (x2 x4 x5).
Let x3 of type ιι be given.
Assume H6: ∀ x4 . x0 x4x3 x4 = x2 x4 x4.
Let x4 of type ι be given.
Let x5 of type ι be given.
Let x6 of type ι be given.
Let x7 of type ι be given.
Assume H7: x0 x4.
Assume H8: x0 x5.
Assume H9: x0 x6.
Assume H10: x0 x7.
Apply H6 with x1 x4 (x1 x5 (x1 x6 x7)), λ x8 x9 . x9 = x1 (x3 x4) (x1 (x2 x4 x5) (x1 (x2 x4 x6) (x1 (x2 x4 x7) (x1 (x2 x5 x4) (x1 (x3 x5) (x1 (x2 x5 x6) (x1 (x2 x5 x7) (x1 (x2 x6 x4) (x1 (x2 x6 x5) (x1 (x3 x6) (x1 (x2 x6 x7) (x1 (x2 x7 x4) (x1 (x2 x7 x5) (x1 (x2 x7 x6) (x3 x7))))))))))))))) leaving 2 subgoals.
Apply H0 with x4, x1 x5 (x1 x6 x7) leaving 2 subgoals.
The subproof is completed by applying H7.
Apply H0 with x5, x1 x6 x7 leaving 2 subgoals.
The subproof is completed by applying H8.
Apply H0 with x6, x7 leaving 2 subgoals.
The subproof is completed by applying H9.
The subproof is completed by applying H10.
Apply H6 with x4, λ x8 x9 . x2 (x1 x4 (x1 x5 (x1 x6 x7))) (x1 x4 (x1 x5 (x1 x6 x7))) = x1 x9 (x1 (x2 x4 x5) (x1 (x2 x4 x6) (x1 (x2 x4 x7) (x1 (x2 x5 x4) (x1 (x3 x5) (x1 (x2 x5 x6) (x1 (x2 x5 x7) (x1 (x2 x6 x4) (x1 (x2 ... ...) ...))))))))) leaving 2 subgoals.
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