Apply H0 with
∀ x0 : ι → ο . ∀ x1 : (ι → ο) → (ι → ο) → ο . (∀ x2 : ι → ο . d97e3.. x2 x0 ⟶ ∃ x3 : ι → ο . x1 x2 x3) ⟶ ∃ x2 : (ι → ο) → ι → ο . ∀ x3 : ι → ο . d97e3.. x3 x0 ⟶ x1 x3 (x2 x3).
Let x0 of type ((ι → ο) → ο) → ι → ο be given.
Assume H1: ∀ x1 : (ι → ο) → ο . ∀ x2 : ι → ο . x1 x2 ⟶ x1 (x0 x1).
Let x1 of type ι → ο be given.
Let x2 of type (ι → ο) → (ι → ο) → ο be given.
Assume H2:
∀ x3 : ι → ο . d97e3.. x3 x1 ⟶ ∃ x4 : ι → ο . x2 x3 x4.
Let x3 of type ο be given.
Assume H3:
∀ x4 : (ι → ο) → ι → ο . (∀ x5 : ι → ο . d97e3.. x5 x1 ⟶ x2 x5 (x4 x5)) ⟶ x3.
Apply H3 with
λ x4 : ι → ο . x0 (x2 x4).
Let x4 of type ι → ο be given.
Apply H2 with
x4,
x2 x4 (x0 (x2 x4)) leaving 2 subgoals.
The subproof is completed by applying H4.
Let x5 of type ι → ο be given.
Assume H5: x2 x4 x5.
Apply H1 with
x2 x4,
x5.
The subproof is completed by applying H5.