Let x0 of type ι → ο be given.
Let x1 of type (ι → ι) → ο be given.
Assume H0: ∀ x2 : ι → ι . x1 x2 ⟶ ∀ x3 . x0 x3 ⟶ x0 (x2 x3).
Let x2 of type ι → ι be given.
Let x3 of type ι → ι be given.
Let x4 of type ι → ι be given.
Let x5 of type ι → ι be given.
Let x6 of type ι → ι be given.
Let x7 of type ι → ι be given.
Let x8 of type ι → ι be given.
Let x9 of type ι → ι be given.
Let x10 of type ι → ι be given.
Let x11 of type ι → ι be given.
Let x12 of type ι → ι be given.
Let x13 of type ι → ι be given.
Let x14 of type ι → ι be given.
Let x15 of type ι → ι be given.
Let x16 of type ι → ι be given.
Let x17 of type ι → ι be given.
Assume H1: x1 x2.
Assume H2: x1 x3.
Assume H3: x1 x4.
Assume H4: x1 x5.
Assume H5: x1 x6.
Assume H6: x1 x7.
Assume H7: x1 x8.
Assume H8: x1 x9.
Assume H9: x1 x10.
Assume H10: x1 x11.
Assume H11: x1 x12.
Assume H12: x1 x13.
Assume H13: x1 x14.
Assume H14: x1 x15.
Assume H15: x1 x16.
Assume H16: x1 x17.
Let x18 of type ι be given.
Assume H17: x0 x18.
Apply H0 with
x17,
x16 (x15 (x14 (x13 (x12 (x11 (x10 (x9 (x8 (x7 (x6 (x5 (x4 (x3 (x2 x18)))))))))))))) leaving 2 subgoals.
The subproof is completed by applying H16.
Apply unknownprop_2bdb4c5556be2b32d7566dc3af4be72cefa9996183118366c05e6a28b5f3d57a with
x0,
x1,
x2,
x3,
x4,
x5,
x6,
x7,
x8,
x9,
x10,
x11,
x12,
x13,
x14,
x15,
x16,
x18 leaving 17 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 H3.
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.
The subproof is completed by applying H12.
The subproof is completed by applying H13.
The subproof is completed by applying H14.
The subproof is completed by applying H15.
The subproof is completed by applying H17.