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 x2 (x1 x3 x4) = x1 x3 (x1 x2 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.
Assume H2: x0 x2.
Assume H3: x0 x3.
Assume H4: x0 x4.
Assume H5: x0 x5.
Assume H6: x0 x6.
Apply unknownprop_2ce9a82c8ef9efc0240c60d5f07d019e2f7a44da8d6114bc529d6fb2d8f3a783 with
x0,
x1,
x2,
x3,
x4,
x5,
x6,
λ x7 x8 . x8 = x1 x4 (x1 x5 (x1 x2 (x1 x3 x6))) leaving 8 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.
Apply unknownprop_2ce9a82c8ef9efc0240c60d5f07d019e2f7a44da8d6114bc529d6fb2d8f3a783 with
x0,
x1,
x3,
x4,
x5,
x2,
x6 leaving 7 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 H2.
The subproof is completed by applying H6.