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 H1 with
x4,
x5,
x6,
λ x7 x8 . x1 x2 (x1 x3 x8) = x1 x6 (x1 x3 (x1 x2 (x1 x5 x4))) leaving 4 subgoals.
The subproof is completed by applying H5.
The subproof is completed by applying H6.
The subproof is completed by applying H7.
Apply H2 with
x4,
x6,
λ x7 x8 . x1 x2 (x1 x3 (x1 x5 x8)) = x1 x6 (x1 x3 (x1 x2 (x1 x5 x4))) leaving 3 subgoals.
The subproof is completed by applying H5.
The subproof is completed by applying H7.
Apply unknownprop_01fcd6886950395663334639e7fa0ba571c79753a62fd50292d889e3bc03b9f3 with
x0,
x1,
x2,
x3,
x5,
x6,
x4 leaving 7 subgoals.
The subproof is completed by applying H0.
The subproof is completed by applying H1.
The subproof is completed by applying H3.
The subproof is completed by applying H4.
The subproof is completed by applying H6.
The subproof is completed by applying H7.
The subproof is completed by applying H5.