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).
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.
Assume H1: x0 x2.
Assume H2: x0 x3.
Assume H3: x0 x4.
Assume H4: x0 x5.
Assume H5: x0 x6.
Assume H6: x0 x7.
Assume H7: x0 x8.
Assume H8: x0 x9.
Assume H9: x0 x10.
Assume H10: x0 x11.
Assume H11: x0 x12.
Assume H12: x0 x13.
Assume H13: x0 x14.
Apply H0 with
x2,
x1 x3 (x1 x4 (x1 x5 (x1 x6 (x1 x7 (x1 x8 (x1 x9 (x1 x10 (x1 x11 (x1 x12 (x1 x13 x14)))))))))) leaving 2 subgoals.
The subproof is completed by applying H1.
Apply unknownprop_b6011236e1a22312171fb30ea3d87ed0b785ea60d02f5f2a289674442a86dda0 with
x0,
x1,
x3,
x4,
x5,
x6,
x7,
x8,
x9,
x10,
x11,
x12,
x13,
x14 leaving 13 subgoals.
The subproof is completed by applying H0.
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.