Let x0 of type ι be given.
Let x1 of type ι be given.
Let x2 of type ι → ι → ι be given.
Let x3 of type ι → ι → ι be given.
Assume H0: ∀ x4 . x4 ∈ x0 ⟶ ∀ x5 . x5 ∈ x0 ⟶ x2 x4 x5 ∈ x0.
Assume H1: ∀ x4 . x4 ∈ x0 ⟶ ∀ x5 . x5 ∈ x0 ⟶ ∀ x6 . x6 ∈ x0 ⟶ x2 x4 (x2 x5 x6) = x2 (x2 x4 x5) x6.
Assume H2: ∀ x4 . x4 ∈ x0 ⟶ ∀ x5 . x5 ∈ x0 ⟶ x2 x4 x5 = x2 x5 x4.
Assume H3: x1 ∈ x0.
Assume H4: ∀ x4 . x4 ∈ x0 ⟶ x2 x1 x4 = x4.
Assume H5:
∀ x4 . x4 ∈ x0 ⟶ ∃ x5 . and (x5 ∈ x0) (x2 x4 x5 = x1).
Assume H6: ∀ x4 . x4 ∈ x0 ⟶ ∀ x5 . x5 ∈ x0 ⟶ x3 x4 x5 ∈ x0.
Assume H7: ∀ x4 . x4 ∈ x0 ⟶ ∀ x5 . x5 ∈ x0 ⟶ ∀ x6 . x6 ∈ x0 ⟶ x3 x4 (x3 x5 x6) = x3 (x3 x4 x5) x6.
Assume H8: ∀ x4 . x4 ∈ x0 ⟶ ∀ x5 . x5 ∈ x0 ⟶ ∀ x6 . x6 ∈ x0 ⟶ x3 x4 (x2 x5 x6) = x2 (x3 x4 x5) (x3 x4 x6).
Assume H9: ∀ x4 . x4 ∈ x0 ⟶ ∀ x5 . x5 ∈ x0 ⟶ ∀ x6 . x6 ∈ x0 ⟶ x3 (x2 x4 x5) x6 = x2 (x3 x4 x6) (x3 x5 x6).
Apply and4I with
and (and (and (and (and (and (∀ x4 . x4 ∈ x0 ⟶ ∀ x5 . x5 ∈ x0 ⟶ x2 x4 x5 ∈ x0) (∀ x4 . x4 ∈ x0 ⟶ ∀ x5 . x5 ∈ x0 ⟶ ∀ x6 . x6 ∈ x0 ⟶ x2 x4 (x2 x5 x6) = x2 (x2 x4 x5) x6)) (∀ x4 . x4 ∈ x0 ⟶ ∀ x5 . x5 ∈ x0 ⟶ x2 x4 x5 = x2 x5 x4)) (x1 ∈ x0)) (∀ x4 . x4 ∈ x0 ⟶ x2 x1 x4 = x4)) (∀ x4 . x4 ∈ x0 ⟶ ∃ x5 . and (x5 ∈ x0) (x2 x4 x5 = x1))) (∀ x4 . x4 ∈ x0 ⟶ ∀ x5 . x5 ∈ x0 ⟶ x3 x4 x5 ∈ x0),
∀ x4 . ... ⟶ ∀ x5 . ... ⟶ ∀ x6 . ... ⟶ x3 x4 (x3 x5 x6) = x3 (x3 ... ...) ...,
...,
... leaving 4 subgoals.