Let x0 of type ι be given.
Let x1 of type ι be given.
Assume H0: x1 ⊆ x0.
Let x2 of type ι be given.
Assume H1:
x2 ∈ {x3 ∈ setexp x0 x0|and (bij x0 x0 (λ x4 . ap x3 x4)) (∀ x4 . x4 ∈ x1 ⟶ ap x3 x4 = x4)}.
Let x3 of type ι be given.
Assume H2:
x3 ∈ {x4 ∈ setexp x0 x0|and (bij x0 x0 (λ x5 . ap x4 x5)) (∀ x5 . x5 ∈ x1 ⟶ ap x4 x5 = x5)}.
Apply SepE with
setexp x0 x0,
λ x4 . and (bij x0 x0 (λ x5 . ap x4 x5)) (∀ x5 . x5 ∈ x1 ⟶ ap x4 x5 = x5),
x2,
(λ x4 x5 . lam x0 (λ x6 . ap x5 (ap x4 x6))) x2 x3 ∈ {x4 ∈ setexp x0 x0|and (bij x0 x0 (λ x5 . ap x4 x5)) (∀ x5 . x5 ∈ x1 ⟶ ap x4 x5 = x5)} leaving 2 subgoals.
The subproof is completed by applying H1.
Assume H4:
and (bij x0 x0 (ap x2)) (∀ x4 . x4 ∈ x1 ⟶ ap x2 x4 = x4).
Apply H4 with
(λ x4 x5 . lam x0 (λ x6 . ap x5 (ap x4 x6))) x2 x3 ∈ {x4 ∈ setexp x0 x0|and (bij x0 x0 (λ x5 . ap x4 x5)) (∀ x5 . x5 ∈ x1 ⟶ ap x4 x5 = x5)}.
Assume H5:
bij x0 x0 (λ x4 . ap x2 x4).
Assume H6:
∀ x4 . x4 ∈ x1 ⟶ ap x2 x4 = x4.
Apply SepE with
setexp x0 x0,
λ x4 . and (bij x0 x0 (λ x5 . ap x4 x5)) (∀ x5 . x5 ∈ x1 ⟶ ap x4 x5 = x5),
x3,
(λ x4 x5 . lam x0 (λ x6 . ap x5 (ap x4 x6))) x2 x3 ∈ {x4 ∈ setexp x0 x0|and (bij x0 x0 (λ x5 . ap x4 x5)) (∀ x5 . x5 ∈ x1 ⟶ ap x4 x5 = x5)} leaving 2 subgoals.
The subproof is completed by applying H2.
Assume H9:
and (bij x0 x0 (ap x3)) (∀ x4 . x4 ∈ x1 ⟶ ap x3 x4 = x4).
Apply H9 with
(λ x4 x5 . lam x0 (λ x6 . ap x5 (ap x4 x6))) x2 x3 ∈ {x4 ∈ setexp x0 ...|...}.