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.
Let x14 of type ι be given.
Let x15 of type ι be given.
Let x16 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.
Assume H14: x0 x14.
Assume H15: x0 x15.
Assume H16: x0 x16.
Apply H1 with
x2,
x3,
x1 x4 (x1 x5 (x1 x6 (x1 x7 (x1 x8 (x1 x9 (x1 x10 (x1 x11 (x1 x12 (x1 x13 (x1 x14 (x1 x15 x16))))))))))),
λ x17 x18 . x18 = x1 x3 (x1 x4 (x1 x5 (x1 x6 (x1 x7 (x1 x8 (x1 x9 (x1 x10 (x1 x11 (x1 x12 (x1 x13 (x1 x14 (x1 x15 (x1 x2 x16))))))))))))) 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 (x1 x13 (x1 x14 (x1 x15 x16)))))))))) 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 (x1 x13 (x1 x14 (x1 x15 x16))))))))) 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 (x1 x13 (x1 x14 (x1 x15 x16)))))))) leaving 2 subgoals.
The subproof is completed by applying H6.
Apply H0 with
x7,
x1 x8 (x1 x9 (x1 x10 (x1 x11 (x1 x12 (x1 x13 (x1 x14 (x1 x15 x16))))))) leaving 2 subgoals.
The subproof is completed by applying H7.
Apply H0 with
x8,
x1 x9 (x1 x10 (x1 x11 (x1 x12 (x1 x13 (x1 x14 (x1 x15 x16)))))) leaving 2 subgoals.
The subproof is completed by applying H8.
Apply H0 with
x9,
x1 x10 (x1 x11 (x1 x12 (x1 x13 (x1 x14 (x1 x15 x16))))) leaving 2 subgoals.
The subproof is completed by applying H9.
Apply H0 with
x10,
x1 x11 (x1 x12 ...) leaving 2 subgoals.