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).
Assume H2: ∀ x3 x4 x5 . x0 x3 ⟶ x0 x4 ⟶ x0 x5 ⟶ x2 x3 (x1 x4 x5) = x1 (x2 x3 x4) (x2 x3 x5).
Assume H3: ∀ x3 x4 . x0 x3 ⟶ x0 x4 ⟶ x0 (x2 x3 x4).
Assume H4: ∀ x3 x4 . x0 x3 ⟶ x0 x4 ⟶ x2 x3 x4 = x2 x4 x3.
Assume H5: ∀ x3 x4 x5 . x0 x3 ⟶ x0 x4 ⟶ x0 x5 ⟶ x1 (x1 x3 x4) (x1 x4 x5) = x1 x3 (x1 (x2 2 x4) x5).
Let x3 of type ι be given.
Let x4 of type ι be given.
Assume H6: x0 x3.
Assume H7: x0 x4.
Apply unknownprop_72d123fd4d496da4fbf16cba456f7da0b762c033cc9f0bd9adf4e316000f1ad6 with
x0,
x1,
x2,
x3,
x4,
x3,
x4,
λ x5 x6 . x6 = x1 (x2 x3 x3) (x1 (x2 2 (x2 x3 x4)) (x2 x4 x4)) 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 H6.
The subproof is completed by applying H7.
The subproof is completed by applying H6.
The subproof is completed by applying H7.
Apply H4 with
x4,
x3,
λ x5 x6 . x1 (x1 (x2 x3 x3) (x2 x3 x4)) (x1 x6 (x2 x4 x4)) = x1 (x2 x3 x3) (x1 (x2 2 (x2 x3 x4)) (x2 x4 x4)) leaving 3 subgoals.
The subproof is completed by applying H7.
The subproof is completed by applying H6.
Apply H5 with
x2 x3 x3,
x2 x3 x4,
x2 x4 x4 leaving 3 subgoals.
Apply H3 with
x3,
x3 leaving 2 subgoals.
The subproof is completed by applying H6.
The subproof is completed by applying H6.
Apply H3 with
x3,
x4 leaving 2 subgoals.
The subproof is completed by applying H6.
The subproof is completed by applying H7.
Apply H3 with
x4,
x4 leaving 2 subgoals.
The subproof is completed by applying H7.
The subproof is completed by applying H7.