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
∀ x3 . x3 ∈ x0 ⟶ explicit_Group_inverse x0 x1 x3 = explicit_Group_inverse x0 x2 x3.
Assume H3:
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).
Assume H4:
∃ 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 H3 with
∀ x3 . x3 ∈ x0 ⟶ explicit_Group_inverse x0 x1 x3 = explicit_Group_inverse x0 x2 x3.
Assume H5: ∀ x3 . x3 ∈ x0 ⟶ ∀ x4 . x4 ∈ x0 ⟶ x1 x3 x4 ∈ x0.
Assume H6: ∀ x3 . x3 ∈ x0 ⟶ ∀ x4 . x4 ∈ x0 ⟶ ∀ x5 . x5 ∈ x0 ⟶ x1 x3 (x1 x4 x5) = x1 (x1 x3 x4) x5.
Let x3 of type ι be given.
Assume H10: x3 ∈ x0.