Let x0 of type ι → ο be given.
Let x1 of type ι → ι → ι be given.
Let x2 of type ι → ι → ι be given.
Assume H0: ∀ x3 x4 . x0 x3 ⟶ x0 x4 ⟶ x0 (x1 x3 x4).
Assume H1: ∀ x3 x4 x5 . x0 x3 ⟶ x0 x4 ⟶ x0 x5 ⟶ x2 x3 (x1 x4 x5) = x1 (x2 x3 x4) (x2 x3 x5).
Assume H2: ∀ x3 x4 x5 . x0 x3 ⟶ x0 x4 ⟶ x0 x5 ⟶ x2 (x1 x3 x4) x5 = x1 (x2 x3 x5) (x2 x4 x5).
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.
Assume H3: x0 x3.
Assume H4: x0 x4.
Assume H5: x0 x5.
Assume H6: x0 x6.
Apply H2 with
x3,
x4,
x1 x5 x6,
λ x7 x8 . x8 = x1 (x1 (x2 x3 x5) (x2 x3 x6)) (x1 (x2 x4 x5) (x2 x4 x6)) leaving 4 subgoals.
The subproof is completed by applying H3.
The subproof is completed by applying H4.
Apply H0 with
x5,
x6 leaving 2 subgoals.
The subproof is completed by applying H5.
The subproof is completed by applying H6.
Apply H1 with
x3,
x5,
x6,
λ x7 x8 . x1 x8 (x2 x4 (x1 x5 x6)) = x1 (x1 (x2 x3 x5) (x2 x3 x6)) (x1 (x2 x4 x5) (x2 x4 x6)) leaving 4 subgoals.
The subproof is completed by applying H3.
The subproof is completed by applying H5.
The subproof is completed by applying H6.
Apply H1 with
x4,
x5,
x6,
λ x7 x8 . x1 (x1 (x2 x3 x5) (x2 x3 x6)) x8 = x1 (x1 (x2 x3 x5) (x2 x3 x6)) (x1 (x2 x4 x5) (x2 x4 x6)) leaving 4 subgoals.
The subproof is completed by applying H4.
The subproof is completed by applying H5.
The subproof is completed by applying H6.
Let x7 of type ι → ι → ο be given.
Assume H7: x7 (x1 (x1 (x2 x3 x5) (x2 x3 x6)) (x1 (x2 x4 x5) (x2 x4 x6))) (x1 (x1 (x2 x3 x5) (x2 x3 x6)) (x1 (x2 x4 x5) (x2 x4 x6))).
The subproof is completed by applying H7.