Let x0 of type ι be given.
Let x1 of type ι be given.
Let x2 of type ι → ι be given.
Apply H0 with
bij x1 x0 (inv x0 x2).
Assume H1:
and (∀ x3 . x3 ∈ x0 ⟶ x2 x3 ∈ x1) (∀ x3 . x3 ∈ x0 ⟶ ∀ x4 . x4 ∈ x0 ⟶ x2 x3 = x2 x4 ⟶ x3 = x4).
Apply H1 with
(∀ x3 . x3 ∈ x1 ⟶ ∃ x4 . and (x4 ∈ x0) (x2 x4 = x3)) ⟶ bij x1 x0 (inv x0 x2).
Assume H2: ∀ x3 . x3 ∈ x0 ⟶ x2 x3 ∈ x1.
Assume H3: ∀ x3 . x3 ∈ x0 ⟶ ∀ x4 . x4 ∈ x0 ⟶ x2 x3 = x2 x4 ⟶ x3 = x4.
Assume H4:
∀ x3 . x3 ∈ x1 ⟶ ∃ x4 . and (x4 ∈ x0) (x2 x4 = x3).
Apply and3I with
∀ x3 . x3 ∈ x1 ⟶ (λ x4 . prim0 (λ x5 . and (x5 ∈ x0) (x2 x5 = x4))) x3 ∈ x0,
∀ x3 . x3 ∈ x1 ⟶ ∀ x4 . x4 ∈ x1 ⟶ (λ x5 . prim0 (λ x6 . and (x6 ∈ x0) (x2 x6 = x5))) x3 = (λ x5 . prim0 (λ x6 . and (x6 ∈ x0) (x2 x6 = x5))) x4 ⟶ x3 = x4,
∀ x3 . x3 ∈ x0 ⟶ ∃ x4 . and (x4 ∈ x1) ((λ x5 . prim0 (λ x6 . and (x6 ∈ x0) (x2 x6 = x5))) x4 = x3) leaving 3 subgoals.
Let x3 of type ι be given.
Assume H6: x3 ∈ x1.
Apply L5 with
x3,
(λ x4 . prim0 (λ x5 . and (x5 ∈ x0) (x2 x5 = x4))) x3 ∈ x0 leaving 2 subgoals.
The subproof is completed by applying H6.
Assume H7:
prim0 (λ x4 . and (x4 ∈ x0) (x2 x4 = x3)) ∈ x0.
Assume H8:
x2 (prim0 (λ x4 . and (x4 ∈ x0) (x2 x4 = x3))) = x3.
The subproof is completed by applying H7.
Let x3 of type ι be given.
Assume H6: x3 ∈ x1.
Let x4 of type ι be given.
Assume H7: x4 ∈ x1.
Assume H8:
(λ x5 . prim0 (λ x6 . and (x6 ∈ x0) (x2 x6 = x5))) x3 = (λ x5 . prim0 (λ x6 . and (x6 ∈ x0) (x2 x6 = x5))) x4.
Apply L5 with
x3,
x3 = x4 leaving 2 subgoals.
The subproof is completed by applying H6.
Assume H9:
(λ x5 . prim0 (λ x6 . and (x6 ∈ x0) (x2 x6 = x5))) x3 ∈ x0.
Assume H10:
x2 ((λ x5 . prim0 (λ x6 . and (x6 ∈ x0) ...)) ...) = ....