Let x0 of type ι → ι → ο be given.

Let x1 of type ι → ι → ο be given.

Let x2 of type ι → ι → ο be given.

Let x3 of type ι be given.

Let x4 of type ι be given.

Let x5 of type ι → ι be given.

Let x6 of type ι → ι be given.

Assume H0: f6435.. x3.

Assume H1: 6fe8d.. x0 x1 x2 (d7cf0.. x4 x6) x3.

Assume H2: ∀ x7 . 6fe8d.. x0 x1 (a603a.. x2 x7 x4) (x5 x7) (x6 x7).

Let x7 of type (ι → ι → ο) → ι → ι → ο be given.

Assume H3: ∀ x8 : ι → ι → ο . ∀ x9 x10 . x0 x9 x10 ⟶ x7 x8 x9 x10.

Assume H4: ∀ x8 : ι → ι → ο . ∀ x9 x10 . x8 x9 x10 ⟶ x7 x8 x9 x10.

Assume H5: ∀ x8 : ι → ι → ο . x7 x8 32d20.. 6915e...

Assume H6: ∀ x8 : ι → ι → ο . ∀ x9 x10 . ∀ x11 : ι → ι . f6435.. x9 ⟶ x7 x8 x10 32d20.. ⟶ (∀ x12 . x7 (a603a.. x8 x12 x10) (x11 x12) x9) ⟶ x7 x8 (d7cf0.. x10 x11) x9.

Assume H7: ∀ x8 : ι → ι → ο . ∀ x9 x10 x11 . ∀ x12 : ι → ι . x7 x8 x9 (d7cf0.. x11 x12) ⟶ x7 x8 x10 x11 ⟶ x7 x8 (57d6a.. x9 x10) (x12 x10).

Assume H8: ∀ x8 : ι → ι → ο . ∀ x9 x10 . ∀ x11 x12 : ι → ι . f6435.. x9 ⟶ x7 x8 (d7cf0.. x10 x12) x9 ⟶ (∀ x13 . x7 (a603a.. x8 x13 x10) (x11 x13) (x12 x13)) ⟶ x7 x8 (bcddf.. 236c6.. x11) (d7cf0.. x10 x12).

Assume H9: ∀ x8 : ι → ι → ο . ∀ x9 x10 . ∀ x11 x12 : ι → ι . f6435.. x9 ⟶ x7 x8 (d7cf0.. x10 x12) x9 ⟶ (∀ x13 . x7 (a603a.. x8 x13 x10) (x11 x13) (x12 x13)) ⟶ x7 x8 (bcddf.. x10 x11) (d7cf0.. x10 x12).

Assume H10: ∀ x8 : ι → ι → ο . ∀ x9 x10 x11 x12 x13 . f6435.. x9 ⟶ x7 x8 x10 x11 ⟶ x7 x8 x12 x9 ⟶ d3ec1.. x1 x11 x13 ⟶ d3ec1.. x1 x12 x13 ⟶ x7 x8 x10 x12.

Apply H9 with x2, x3, x4, x5, x6 leaving 3 subgoals.

The subproof is completed by applying H0.

Apply H1 with x7 leaving 8 subgoals.

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.

Let x8 of type ι be given.

Apply H2 with x8, x7 leaving 8 subgoals.