Let x0 of type ι → ο be given.
Let x1 of type ι → ι → ι be given.
Let x2 of type ι → ι → ι be given.
Assume H0: ∀ x3 x4 . x0 x3 ⟶ x0 x4 ⟶ x0 (x1 x3 x4).
Assume H1: ∀ x3 x4 x5 . x0 x3 ⟶ x0 x4 ⟶ x0 x5 ⟶ x2 (x1 x3 x4) x5 = x1 (x2 x3 x5) (x2 x4 x5).
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 x3.
Assume H3: x0 x4.
Assume H4: x0 x5.
Assume H5: x0 x6.
Assume H6: x0 x7.
Assume H7: x0 x8.
Apply H1 with
x3,
x1 x4 (x1 x5 (x1 x6 x7)),
x8,
λ x9 x10 . x10 = x1 (x2 x3 x8) (x1 (x2 x4 x8) (x1 (x2 x5 x8) (x1 (x2 x6 x8) (x2 x7 x8)))) leaving 4 subgoals.
The subproof is completed by applying H2.
Apply unknownprop_b48d4480a5526e51a91293fec1b0b9440be4280265441ce358bda14cced12479 with
x0,
x1,
x4,
x5,
x6,
x7 leaving 5 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.
set y9 to be x1 (x2 x3 x8) (x2 (x1 x4 (x1 x5 (x1 x6 x7))) x8)
set y10 to be x2 (x3 x4 y9) (x2 (x3 x5 y9) (x2 (x3 x6 y9) (x2 (x3 x7 y9) (x3 x8 y9))))
Claim L8: ∀ x11 : ι → ο . x11 y10 ⟶ x11 y9
Let x11 of type ι → ο be given.
Assume H8: x11 (x3 (x4 x5 y10) (x3 (x4 x6 y10) (x3 (x4 x7 y10) (x3 (x4 x8 y10) (x4 y9 y10))))).
set y12 to be λ x12 . x11
Apply unknownprop_f8cd2cc9fa3527ece23bf769ef6e4562a4db8193110bd03a6fde0a07ab4c38b8 with
x2,
x3,
x4,
x6,
x7,
x8,
y9,
y10,
λ x13 x14 . y12 (x3 (x4 x5 y10) x13) (x3 (x4 x5 y10) x14) leaving 8 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.
Let x11 of type ι → ι → ο be given.
Apply L8 with
λ x12 . x11 x12 y10 ⟶ x11 y10 x12.
Assume H9: x11 y10 y10.
The subproof is completed by applying H9.