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