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.
Apply explicit_Ring_E with
x0,
x1,
x2,
x3,
∀ x4 . prim1 x4 x0 ⟶ and (prim1 (explicit_Ring_minus x0 x1 x2 x3 x4) x0) (x2 x4 (explicit_Ring_minus x0 x1 x2 x3 x4) = x1).
Assume H2:
∀ x4 . prim1 x4 x0 ⟶ ∀ x5 . prim1 x5 x0 ⟶ ∀ x6 . prim1 x6 x0 ⟶ x2 x4 (x2 x5 x6) = x2 (x2 x4 x5) x6.
Assume H3:
∀ x4 . prim1 x4 x0 ⟶ ∀ x5 . prim1 x5 x0 ⟶ x2 x4 x5 = x2 x5 x4.
Assume H5:
∀ x4 . prim1 x4 x0 ⟶ x2 x1 x4 = x4.
Assume H6:
∀ x4 . prim1 x4 x0 ⟶ ∃ x5 . and (prim1 x5 x0) (x2 x4 x5 = x1).
Assume H8:
∀ x4 . prim1 x4 x0 ⟶ ∀ x5 . prim1 x5 x0 ⟶ ∀ x6 . prim1 x6 x0 ⟶ x3 x4 (x3 x5 x6) = x3 (x3 x4 x5) x6.
Assume H9:
∀ x4 . prim1 x4 x0 ⟶ ∀ x5 . prim1 x5 x0 ⟶ ∀ x6 . prim1 x6 x0 ⟶ x3 x4 (x2 x5 x6) = x2 (x3 x4 x5) (x3 x4 x6).
Assume H10:
∀ x4 . prim1 x4 x0 ⟶ ∀ x5 . prim1 x5 x0 ⟶ ∀ x6 . prim1 x6 x0 ⟶ x3 (x2 x4 x5) x6 = x2 (x3 x4 x6) (x3 x5 x6).
Let x4 of type ι be given.
Apply Eps_i_ex with
λ x5 . and (prim1 x5 x0) (x2 x4 x5 = x1).
Apply H6 with
x4.
The subproof is completed by applying H11.