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).
Assume H2: ∀ x2 x3 . x0 x2 ⟶ x0 x3 ⟶ x1 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.
Assume H3: x0 x2.
Assume H4: x0 x3.
Assume H5: x0 x4.
Assume H6: x0 x5.
Assume H7: x0 x6.
Apply H2 with
x5,
x6,
λ x7 x8 . x1 x2 (x1 x3 (x1 x4 x8)) = x1 x3 (x1 x2 (x1 x4 (x1 x6 x5))) leaving 3 subgoals.
The subproof is completed by applying H6.
The subproof is completed by applying H7.
Apply H1 with
x2,
x3,
x1 x4 (x1 x6 x5) leaving 3 subgoals.
The subproof is completed by applying H3.
The subproof is completed by applying H4.
Apply H0 with
x4,
x1 x6 x5 leaving 2 subgoals.
The subproof is completed by applying H5.
Apply H0 with
x6,
x5 leaving 2 subgoals.
The subproof is completed by applying H7.
The subproof is completed by applying H6.