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.
Assume H2: x0 x2.
Assume H3: x0 x3.
Assume H4: x0 x4.
Assume H5: x0 x5.
Assume H6: x0 x6.
Apply unknownprop_6df806693864a23a378ddbca02cda4bb4bc233ff1daa8914d51c06eb72ff2550 with
x0,
x1,
x3,
x4,
x5,
x6,
λ x7 x8 . x1 x2 x8 = x1 x4 (x1 x2 (x1 x5 (x1 x3 x6))) 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 H5.
The subproof is completed by applying H6.
Apply H1 with
x2,
x4,
x1 x5 (x1 x3 x6),
λ x7 x8 . x8 = x1 x4 (x1 x2 (x1 x5 (x1 x3 x6))) leaving 4 subgoals.
The subproof is completed by applying H2.
The subproof is completed by applying H4.
Apply H0 with
x5,
x1 x3 x6 leaving 2 subgoals.
The subproof is completed by applying H5.
Apply H0 with
x3,
x6 leaving 2 subgoals.
The subproof is completed by applying H3.
The subproof is completed by applying H6.
Let x7 of type ι → ι → ο be given.
Assume H7: x7 (x1 x4 (x1 x2 (x1 x5 (x1 x3 x6)))) (x1 x4 (x1 x2 (x1 x5 (x1 x3 x6)))).
The subproof is completed by applying H7.