Let x0 of type (ι → ο) → ο be given.
Assume H0:
∀ x1 : ι → ο . (λ x2 : ι → ο . ∀ x3 : (ι → ο) → ο . (∀ x4 : ι → ο . x3 x4 ⟶ x3 ((λ x5 : ι → ο . λ x6 . and (x5 x6) (x6 = prim0 (λ x7 . x5 x7) ⟶ ∀ x7 : ο . x7)) x4)) ⟶ (∀ x4 : (ι → ο) → ο . (∀ x5 : ι → ο . x4 x5 ⟶ x3 x5) ⟶ x3 (Descr_Vo1 x4)) ⟶ x3 x2) x1 ⟶ x0 x1 ⟶ x0 ((λ x2 : ι → ο . λ x3 . and (x2 x3) (x3 = prim0 (λ x4 . x2 x4) ⟶ ∀ x4 : ο . x4)) x1).
Assume H1:
∀ x1 : (ι → ο) → ο . (∀ x2 : ι → ο . x1 x2 ⟶ ∀ x3 : (ι → ο) → ο . (∀ x4 : ι → ο . x3 x4 ⟶ x3 ((λ x5 : ι → ο . λ x6 . and (x5 x6) (x6 = prim0 (λ x7 . x5 x7) ⟶ ∀ x7 : ο . x7)) x4)) ⟶ (∀ x4 : (ι → ο) → ο . (∀ x5 : ι → ο . x4 x5 ⟶ x3 x5) ⟶ x3 (Descr_Vo1 x4)) ⟶ x3 x2) ⟶ (∀ x2 : ι → ο . x1 x2 ⟶ x0 x2) ⟶ x0 (Descr_Vo1 x1).
Let x1 of type ι → ο be given.
Assume H2:
(λ x2 : ι → ο . ∀ x3 : (ι → ο) → ο . (∀ x4 : ι → ο . x3 x4 ⟶ x3 ((λ x5 : ι → ο . λ x6 . and (x5 x6) (x6 = prim0 (λ x7 . x5 x7) ⟶ ∀ x7 : ο . x7)) x4)) ⟶ (∀ x4 : (ι → ο) → ο . (∀ x5 : ι → ο . x4 x5 ⟶ x3 x5) ⟶ x3 (Descr_Vo1 x4)) ⟶ x3 x2) x1.
Claim L3:
and ((λ x2 : ι → ο . ∀ x3 : (ι → ο) → ο . (∀ x4 : ι → ο . x3 x4 ⟶ x3 ((λ x5 : ι → ο . λ x6 . and (x5 x6) (x6 = prim0 (λ x7 . x5 x7) ⟶ ∀ x7 : ο . x7)) x4)) ⟶ (∀ x4 : (ι → ο) → ο . (∀ x5 : ι → ο . x4 x5 ⟶ x3 x5) ⟶ x3 (Descr_Vo1 x4)) ⟶ x3 x2) x1) (x0 x1)Apply H2 with
λ x2 : ι → ο . and ((λ x3 : ι → ο . ∀ x4 : (ι → ο) → ο . (∀ x5 : ι → ο . x4 x5 ⟶ x4 ((λ x6 : ι → ο . λ x7 . and (x6 x7) (x7 = prim0 (λ x8 . x6 x8) ⟶ ∀ x8 : ο . x8)) x5)) ⟶ (∀ x5 : (ι → ο) → ο . (∀ x6 : ι → ο . x5 x6 ⟶ x4 x6) ⟶ x4 (Descr_Vo1 x5)) ⟶ x4 x3) x2) (x0 x2) leaving 2 subgoals.
Let x2 of type ι → ο be given.
Assume H3:
and ((λ x3 : ι → ο . ∀ x4 : (ι → ο) → ο . (∀ x5 : ι → ο . ... ⟶ x4 ((λ x6 : ι → ο . λ x7 . and (x6 x7) (... ⟶ ∀ x8 : ο . x8)) ...)) ⟶ (∀ x5 : (ι → ο) → ο . (∀ x6 : ι → ο . x5 x6 ⟶ x4 x6) ⟶ x4 (Descr_Vo1 x5)) ⟶ x4 x3) ...) ....
Apply L3 with
x0 x1.
Assume H4:
∀ x2 : (ι → ο) → ο . (∀ x3 : ι → ο . x2 x3 ⟶ x2 (λ x4 . and (x3 x4) (x4 = prim0 (λ x5 . x3 x5) ⟶ ∀ x5 : ο . x5))) ⟶ (∀ x3 : (ι → ο) → ο . (∀ x4 : ι → ο . x3 x4 ⟶ x2 x4) ⟶ x2 (Descr_Vo1 x3)) ⟶ x2 x1.
Assume H5: x0 x1.
The subproof is completed by applying H5.