Let x0 of type ι → ι → ο be given.
Let x1 of type ι → ι → ο be given.
Assume H0: ∀ x2 x3 . x0 x2 x3 ⟶ x1 x2 x3.
Let x2 of type ι be given.
Let x3 of type ι be given.
Let x4 of type ι → ι → ο be given.
Assume H2: ∀ x5 x6 . x1 x5 x6 ⟶ x4 x5 x6.
Assume H3: ∀ x5 . x4 x5 x5.
Assume H4:
∀ x5 x6 x7 x8 . x4 x5 x6 ⟶ x4 x7 x8 ⟶ x4 (57d6a.. x5 x7) (57d6a.. x6 x8).
Assume H5:
∀ x5 x6 . ∀ x7 x8 : ι → ι . x4 x5 x6 ⟶ (∀ x9 . x4 (x7 x9) (x8 x9)) ⟶ x4 (bcddf.. x5 x7) (bcddf.. x6 x8).
Assume H6:
∀ x5 x6 . ∀ x7 x8 : ι → ι . x4 x5 x6 ⟶ (∀ x9 . x4 (x7 x9) (x8 x9)) ⟶ x4 (d7cf0.. x5 x7) (d7cf0.. x6 x8).
Apply H1 with
x4 leaving 5 subgoals.
Let x5 of type ι be given.
Let x6 of type ι be given.
Assume H7: x0 x5 x6.
Apply H2 with
x5,
x6.
Apply H0 with
x5,
x6.
The subproof is completed by applying H7.
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.