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.
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.
Let x11 of type ι be given.
Assume H10: x0 x11.
Apply H0 with
x10,
x9 (x8 (x7 (x6 (x5 (x4 (x3 (x2 x11))))))) leaving 2 subgoals.
The subproof is completed by applying H9.
Apply unknownprop_6998f1967a37ce79bf145b799234d391f8f1c42a3dd1f19ca25be7477a1f0225 with
x0,
x1,
x2,
x3,
x4,
x5,
x6,
x7,
x8,
x9,
x11 leaving 10 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 H10.