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.
Let x11 of type ι be given.
Let x12 of type ι be given.
Let x13 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.
Assume H11: x0 x11.
Assume H12: x0 x12.
Assume H13: x0 x13.
Apply H1 with
x2,
x3,
x1 x4 (x1 x5 (x1 x6 (x1 x7 (x1 x8 (x1 x9 (x1 x10 (x1 x11 (x1 x12 x13)))))))),
λ x14 x15 . x15 = x1 x3 (x1 x4 (x1 x5 (x1 x6 (x1 x7 (x1 x8 (x1 x9 (x1 x10 (x1 x11 (x1 x12 (x1 x2 x13)))))))))) 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 (x1 x10 (x1 x11 (x1 x12 x13))))))) leaving 2 subgoals.
The subproof is completed by applying H4.
Apply H0 with
x5,
x1 x6 (x1 x7 (x1 x8 (x1 x9 (x1 x10 (x1 x11 (x1 x12 x13)))))) leaving 2 subgoals.
The subproof is completed by applying H5.
Apply H0 with
x6,
x1 x7 (x1 x8 (x1 x9 (x1 x10 (x1 x11 (x1 x12 x13))))) leaving 2 subgoals.
The subproof is completed by applying H6.
Apply H0 with
x7,
x1 x8 (x1 x9 (x1 x10 (x1 x11 (x1 x12 x13)))) leaving 2 subgoals.
The subproof is completed by applying H7.
Apply H0 with
x8,
x1 x9 (x1 x10 (x1 x11 (x1 x12 x13))) leaving 2 subgoals.
The subproof is completed by applying H8.
Apply H0 with
x9,
x1 x10 (x1 x11 (x1 x12 x13)) leaving 2 subgoals.
The subproof is completed by applying H9.
Apply H0 with
x10,
x1 x11 (x1 x12 x13) leaving 2 subgoals.
The subproof is completed by applying H10.
Apply H0 with
x11,
x1 x12 x13 leaving 2 subgoals.
The subproof is completed by applying H11.
Apply H0 with
x12,
x13 leaving 2 subgoals.
The subproof is completed by applying H12.
The subproof is completed by applying H13.
set y14 to be ...
set y15 to be ...
Claim L14: ∀ x16 : ι → ο . x16 y15 ⟶ x16 y14
Let x16 of type ι → ο be given.
Assume H14: x16 (x3 x5 (x3 x6 (x3 x7 (x3 x8 (x3 x9 (x3 x10 (x3 x11 (x3 x12 (x3 x13 (x3 y14 (x3 x4 y15))))))))))).
set y17 to be ...
Apply unknownprop_7e908b40c46b73fe6e5f778b7a7c4240e8f2018f37d884c13073c8fe98783893 with
x2,
x3,
x4,
x6,
x7,
x8,
x9,
x10,
x11,
x12,
x13,
...,
...,
... leaving 14 subgoals.
Let x16 of type ι → ι → ο be given.
Apply L14 with
λ x17 . x16 x17 y15 ⟶ x16 y15 x17.
Assume H15: x16 y15 y15.
The subproof is completed by applying H15.