Let x0 of type ι → ο be given.

Let x1 of type ι → ι → ι be given.

Assume H0: ∀ x2 x3 . x0 x2 ⟶ x0 x3 ⟶ x0 (x1 x2 x3).

Assume H1: ∀ x2 x3 x4 . x0 x2 ⟶ x0 x3 ⟶ x0 x4 ⟶ x1 x2 (x1 x3 x4) = x1 x3 (x1 x2 x4).

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.

Let x9 of type ι be given.

Let x10 of type ι be given.

Let x11 of type ι be given.

Let x12 of type ι be given.

Assume H2: x0 x2.

Assume H3: x0 x3.

Assume H4: x0 x4.

Assume H5: x0 x5.

Assume H6: x0 x6.

Assume H7: x0 x7.

Assume H8: x0 x8.

Assume H9: x0 x9.

Assume H10: x0 x10.

Assume H11: x0 x11.

Assume H12: x0 x12.

Apply unknownprop_7e908b40c46b73fe6e5f778b7a7c4240e8f2018f37d884c13073c8fe98783893 with x0, x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, λ x13 x14 . x14 = x1 x7 (x1 x8 (x1 x9 (x1 x10 (x1 x11 (x1 x2 (x1 x3 (x1 x4 (x1 x5 (x1 x6 x12))))))))) leaving 14 subgoals.

The subproof is completed by applying H0.

The subproof is completed by applying H1.

The subproof is completed by applying H2.

The subproof is completed by applying H3.

The subproof is completed by applying H4.

The subproof is completed by applying H5.

The subproof is completed by applying H6.

The subproof is completed by applying H7.

The subproof is completed by applying H8.

The subproof is completed by applying H9.

The subproof is completed by applying H10.

The subproof is completed by applying H11.

The subproof is completed by applying H12.

Apply unknownprop_71482150118f9bbc66b43fb6522636e329499b09fb3294b389fd1fcc0b9cfe14 with x0, x1, x3, x4, x5, x6, x7, x8, x9, x10, x11, x2, x12 leaving 13 subgoals.