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 (x1 x2 x3) x4 = x1 x2 (x1 x3 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.
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.
Apply H1 with
x2,
x1 x3 (x1 x4 (x1 x5 (x1 x6 x7))),
x8,
λ x9 x10 . x10 = x1 x2 (x1 x3 (x1 x4 (x1 x5 (x1 x6 (x1 x7 x8))))) leaving 4 subgoals.
The subproof is completed by applying H2.
Apply unknownprop_d7ce6357a8261c6a4be44f579bedcb1c2d65cec14964ea078af8f02cc5aab85a with
x0,
x1,
x3,
x4,
x5,
x6,
x7 leaving 6 subgoals.
The subproof is completed by applying H0.
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.
set y9 to be x1 x2 (x1 (x1 x3 (x1 x4 (x1 x5 (x1 x6 x7)))) x8)
set y10 to be x2 x3 (x2 x4 (x2 x5 (x2 x6 (x2 x7 (x2 x8 y9)))))
Claim L9: ∀ x11 : ι → ο . x11 y10 ⟶ x11 y9
Let x11 of type ι → ο be given.
Assume H9: x11 (x3 x4 (x3 x5 (x3 x6 (x3 x7 (x3 x8 (x3 y9 y10)))))).
set y12 to be λ x12 . x11
Apply unknownprop_b833aa06c1faed5d76aa7b67b54605577d0e95ae20c23cac91bc46748e38d0cb with
x2,
x3,
x5,
x6,
x7,
x8,
y9,
y10,
λ x13 x14 . y12 (x3 x4 x13) (x3 x4 x14) leaving 9 subgoals.
The subproof is completed by applying H0.
The subproof is completed by applying H1.
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.
Let x11 of type ι → ι → ο be given.
Apply L9 with
λ x12 . x11 x12 y10 ⟶ x11 y10 x12.
Assume H10: x11 y10 y10.
The subproof is completed by applying H10.