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.
Let x7 of type ι be given.
Let x8 of type ι be given.
Let x9 of type ι be given.
Let x10 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.
Assume H7: x0 x7.
Assume H8: x0 x8.
Assume H9: x0 x9.
Assume H10: x0 x10.
Apply H1 with
x2,
x3,
x1 x4 (x1 x5 (x1 x6 (x1 x7 (x1 x8 (x1 x9 x10))))),
λ x11 x12 . x12 = x1 x3 (x1 x4 (x1 x5 (x1 x6 (x1 x7 (x1 x8 (x1 x9 (x1 x2 x10))))))) leaving 4 subgoals.
The subproof is completed by applying H2.
The subproof is completed by applying H3.
Apply H0 with
x4,
x1 x5 (x1 x6 (x1 x7 (x1 x8 (x1 x9 x10)))) leaving 2 subgoals.
The subproof is completed by applying H4.
Apply H0 with
x5,
x1 x6 (x1 x7 (x1 x8 (x1 x9 x10))) leaving 2 subgoals.
The subproof is completed by applying H5.
Apply H0 with
x6,
x1 x7 (x1 x8 (x1 x9 x10)) leaving 2 subgoals.
The subproof is completed by applying H6.
Apply H0 with
x7,
x1 x8 (x1 x9 x10) leaving 2 subgoals.
The subproof is completed by applying H7.
Apply H0 with
x8,
x1 x9 x10 leaving 2 subgoals.
The subproof is completed by applying H8.
Apply H0 with
x9,
x10 leaving 2 subgoals.
The subproof is completed by applying H9.
The subproof is completed by applying H10.
set y11 to be x1 x3 (x1 x2 (x1 x4 (x1 x5 (x1 x6 (x1 x7 (x1 x8 (x1 x9 x10)))))))
set y12 to be x2 x4 (x2 x5 (x2 x6 (x2 x7 (x2 x8 (x2 x9 (x2 x10 (x2 x3 y11)))))))
Claim L11: ∀ x13 : ι → ο . x13 y12 ⟶ x13 y11
Let x13 of type ι → ο be given.
Assume H11: x13 (x3 x5 (x3 x6 (x3 x7 (x3 x8 (x3 x9 (x3 x10 (x3 y11 (x3 x4 y12)))))))).
set y14 to be λ x14 . x13
Apply unknownprop_e032662aa31d0d7c6cd21c7ec8f978be1dee32c5d8585622c751063506e260f9 with
x2,
x3,
x4,
x6,
x7,
x8,
x9,
x10,
y11,
y12,
λ x15 x16 . y14 (x3 x5 x15) (x3 x5 x16) leaving 11 subgoals.
The subproof is completed by applying H0.
The subproof is completed by applying H1.
The subproof is completed by applying H2.
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.
The subproof is completed by applying H11.
Let x13 of type ι → ι → ο be given.
Apply L11 with
λ x14 . x13 x14 y12 ⟶ x13 y12 x14.
Assume H12: x13 y12 y12.
The subproof is completed by applying H12.