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
surj x0 x1 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)) ⟶ surj x0 x1 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 andI with
∀ x3 . x3 ∈ x0 ⟶ x2 x3 ∈ x1,
∀ x3 . x3 ∈ x1 ⟶ ∃ x4 . and (x4 ∈ x0) (x2 x4 = x3) leaving 2 subgoals.
The subproof is completed by applying H2.
The subproof is completed by applying H4.