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Proofgold Proof

pf
Let x0 of type ιο be given.
Let x1 of type ιιι be given.
Assume H0: ∀ x2 x3 . x0 x2x0 x3x0 (x1 x2 x3).
Assume H1: ∀ x2 x3 x4 . x0 x2x0 x3x0 x4x1 x2 (x1 x3 x4) = x1 x3 (x1 x2 x4).
Assume H2: ∀ x2 x3 . x0 x2x0 x3x1 x2 x3 = x1 x3 x2.
Let x2 of type ι be given.
Let x3 of type ι be given.
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.
Let x8 of type ι be given.
Assume H3: x0 x2.
Assume H4: x0 x3.
Assume H5: x0 x4.
Assume H6: x0 x5.
Assume H7: x0 x6.
Assume H8: x0 x7.
Assume H9: x0 x8.
Apply H2 with x7, x8, λ x9 x10 . x1 x2 (x1 x3 (x1 x4 (x1 x5 (x1 x6 x10)))) = x1 x3 (x1 x2 (x1 x4 (x1 x5 (x1 x6 (x1 x8 x7))))) leaving 3 subgoals.
The subproof is completed by applying H8.
The subproof is completed by applying H9.
Apply H1 with x2, x3, x1 x4 (x1 x5 (x1 x6 (x1 x8 x7))) leaving 3 subgoals.
The subproof is completed by applying H3.
The subproof is completed by applying H4.
Apply H0 with x4, x1 x5 (x1 x6 (x1 x8 x7)) leaving 2 subgoals.
The subproof is completed by applying H5.
Apply H0 with x5, x1 x6 (x1 x8 x7) leaving 2 subgoals.
The subproof is completed by applying H6.
Apply H0 with x6, x1 x8 x7 leaving 2 subgoals.
The subproof is completed by applying H7.
Apply H0 with x8, x7 leaving 2 subgoals.
The subproof is completed by applying H9.
The subproof is completed by applying H8.