Let x0 of type ι be given.
Let x1 of type ι be given.
Let x2 of type ι be given.
Let x3 of type ι → ι → ι be given.
Let x4 of type ι → ι → ι be given.
Apply explicit_Ring_with_id_E with
x0,
x1,
x2,
x3,
x4,
prim1 (explicit_Ring_minus x0 x1 x3 x4 x2) x0.
Assume H2:
∀ x5 . prim1 x5 x0 ⟶ ∀ x6 . prim1 x6 x0 ⟶ ∀ x7 . prim1 x7 x0 ⟶ x3 x5 (x3 x6 x7) = x3 (x3 x5 x6) x7.
Assume H3:
∀ x5 . prim1 x5 x0 ⟶ ∀ x6 . prim1 x6 x0 ⟶ x3 x5 x6 = x3 x6 x5.
Assume H5:
∀ x5 . prim1 x5 x0 ⟶ x3 x1 x5 = x5.
Assume H6:
∀ x5 . prim1 x5 x0 ⟶ ∃ x6 . and (prim1 x6 x0) (x3 x5 x6 = x1).
Assume H8:
∀ x5 . prim1 x5 x0 ⟶ ∀ x6 . prim1 x6 x0 ⟶ ∀ x7 . prim1 x7 x0 ⟶ x4 x5 (x4 x6 x7) = x4 (x4 x5 x6) x7.
Assume H10: x2 = x1 ⟶ ∀ x5 : ο . x5.
Assume H11:
∀ x5 . prim1 x5 x0 ⟶ x4 x2 x5 = x5.
Assume H12:
∀ x5 . prim1 x5 x0 ⟶ x4 x5 x2 = x5.
Assume H13:
∀ x5 . prim1 x5 x0 ⟶ ∀ x6 . prim1 x6 x0 ⟶ ∀ x7 . prim1 x7 x0 ⟶ x4 x5 (x3 x6 x7) = x3 (x4 x5 x6) (x4 x5 x7).
Assume H14:
∀ x5 . prim1 x5 x0 ⟶ ∀ x6 . prim1 x6 x0 ⟶ ∀ x7 . prim1 x7 x0 ⟶ x4 (x3 x5 x6) x7 = x3 (x4 x5 x7) (x4 x6 x7).
Apply explicit_Ring_with_id_minus_clos with
x0,
x1,
x2,
x3,
x4,
x2 leaving 2 subgoals.
The subproof is completed by applying H0.
The subproof is completed by applying H9.