Let x0 of type ι be given.
Let x1 of type ι → ι → ι be given.
Claim L1:
∃ x2 . and (x2 ∈ x0) (and (∀ x3 . x3 ∈ x0 ⟶ and (x1 x2 x3 = x3) (x1 x3 x2 = x3)) (∀ x3 . x3 ∈ x0 ⟶ ∃ x4 . and (x4 ∈ x0) (and (x1 x3 x4 = x2) (x1 x4 x3 = x2))))Apply H0 with
∃ x2 . and (x2 ∈ x0) (and (∀ x3 . x3 ∈ x0 ⟶ and (x1 x2 x3 = x3) (x1 x3 x2 = x3)) (∀ x3 . x3 ∈ x0 ⟶ ∃ x4 . and (x4 ∈ x0) (and (x1 x3 x4 = x2) (x1 x4 x3 = x2)))).
Assume H1:
and (∀ x2 . x2 ∈ x0 ⟶ ∀ x3 . x3 ∈ x0 ⟶ x1 x2 x3 ∈ x0) (∀ x2 . x2 ∈ x0 ⟶ ∀ x3 . x3 ∈ x0 ⟶ ∀ x4 . x4 ∈ x0 ⟶ x1 x2 (x1 x3 x4) = x1 (x1 x2 x3) x4).
Assume H2:
∃ x2 . and (x2 ∈ x0) (and (∀ x3 . x3 ∈ x0 ⟶ and (x1 x2 x3 = x3) (x1 x3 x2 = x3)) (∀ x3 . x3 ∈ x0 ⟶ ∃ x4 . and (x4 ∈ x0) (and (x1 x3 x4 = x2) (x1 x4 x3 = x2)))).
The subproof is completed by applying H2.
Apply Eps_i_ex with
λ x2 . and (x2 ∈ x0) (and (∀ x3 . x3 ∈ x0 ⟶ and (x1 x2 x3 = x3) (x1 x3 x2 = x3)) (∀ x3 . x3 ∈ x0 ⟶ ∃ x4 . and (x4 ∈ x0) (and (x1 x3 x4 = x2) (x1 x4 x3 = x2)))).
The subproof is completed by applying L1.