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_Field_E with
x0,
x1,
x2,
x3,
x4,
∀ x5 . prim1 x5 x0 ⟶ ∀ x6 . prim1 x6 x0 ⟶ ∀ x7 . prim1 x7 x0 ⟶ x3 x5 x7 = x3 x6 x7 ⟶ x5 = x6.
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 H9:
∀ x5 . prim1 x5 x0 ⟶ ∀ x6 . prim1 x6 x0 ⟶ x4 x5 x6 = x4 x6 x5.
Assume H11: x2 = x1 ⟶ ∀ x5 : ο . x5.
Assume H12:
∀ x5 . prim1 x5 x0 ⟶ x4 x2 x5 = x5.
Assume H13:
∀ x5 . prim1 x5 x0 ⟶ (x5 = x1 ⟶ ∀ x6 : ο . x6) ⟶ ∃ x6 . and (prim1 x6 x0) (x4 x5 x6 = x2).
Assume H14:
∀ x5 . prim1 x5 x0 ⟶ ∀ x6 . prim1 x6 x0 ⟶ ∀ x7 . prim1 x7 x0 ⟶ x4 x5 (x3 x6 x7) = x3 (x4 x5 x6) (x4 x5 x7).
Let x5 of type ι be given.
Let x6 of type ι be given.
Let x7 of type ι be given.
Assume H18: x3 x5 x7 = x3 x6 x7.
Apply explicit_Field_plus_cancelL with
x0,
x1,
x2,
x3,
x4,
x7,
x5,
x6 leaving 5 subgoals.
The subproof is completed by applying H0.
The subproof is completed by applying H17.
The subproof is completed by applying H15.
The subproof is completed by applying H16.
Apply H3 with
x7,
x5,
... leaving 3 subgoals.