Let x0 of type ι → ι be given.
Let x1 of type ι → (ι → ι) → ι → ι be given.
Let x2 of type ι be given.
Let x3 of type ι → ι → ι be given.
Let x4 of type ι → ι → ι be given.
Assume H0: ∀ x5 . x5 ∈ x2 ⟶ x3 x5 = x4 x5.
Apply xm with
prim3 x2 ∈ x2,
(λ x5 . λ x6 : ι → ι → ι . If_ii (prim3 x5 ∈ x5) (x1 (prim3 x5) (x6 (prim3 x5))) x0) x2 x3 = (λ x5 . λ x6 : ι → ι → ι . If_ii (prim3 x5 ∈ x5) (x1 (prim3 x5) (x6 (prim3 x5))) x0) x2 x4 leaving 2 subgoals.
Assume H1:
prim3 x2 ∈ x2.
Apply H0 with
prim3 x2,
λ x5 x6 : ι → ι . If_ii (prim3 x2 ∈ x2) (x1 (prim3 x2) x6) x0 = If_ii (prim3 x2 ∈ x2) (x1 (prim3 x2) (x4 (prim3 x2))) x0 leaving 2 subgoals.
The subproof is completed by applying H1.
Let x5 of type (ι → ι) → (ι → ι) → ο be given.
The subproof is completed by applying H2.
Apply If_ii_0 with
prim3 x2 ∈ x2,
x1 (prim3 x2) (x3 (prim3 x2)),
x0.
The subproof is completed by applying H1.
Apply If_ii_0 with
prim3 x2 ∈ x2,
x1 (prim3 x2) (x4 (prim3 x2)),
x0.
The subproof is completed by applying H1.
Apply L3 with
λ x5 x6 : ι → ι . If_ii (prim3 x2 ∈ x2) (x1 (prim3 x2) (x3 (prim3 x2))) x0 = x6.
The subproof is completed by applying L2.