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 . x4 ∈ x0 ⟶ x2 (explicit_Ring_minus x0 x1 x2 x3 x4) x4 = x1.
Assume H1: ∀ x4 . x4 ∈ x0 ⟶ ∀ x5 . x5 ∈ x0 ⟶ x2 x4 x5 ∈ x0.
Assume H2: ∀ x4 . x4 ∈ x0 ⟶ ∀ x5 . x5 ∈ x0 ⟶ ∀ x6 . x6 ∈ x0 ⟶ x2 x4 (x2 x5 x6) = x2 (x2 x4 x5) x6.
Assume H3: ∀ x4 . x4 ∈ x0 ⟶ ∀ x5 . x5 ∈ x0 ⟶ x2 x4 x5 = x2 x5 x4.
Assume H4: x1 ∈ x0.
Assume H5: ∀ x4 . x4 ∈ x0 ⟶ x2 x1 x4 = x4.
Assume H6:
∀ x4 . x4 ∈ x0 ⟶ ∃ x5 . and (x5 ∈ x0) (x2 x4 x5 = x1).
Assume H7: ∀ x4 . x4 ∈ x0 ⟶ ∀ x5 . x5 ∈ x0 ⟶ x3 x4 x5 ∈ x0.
Assume H8: ∀ x4 . x4 ∈ x0 ⟶ ∀ x5 . x5 ∈ x0 ⟶ ∀ x6 . x6 ∈ x0 ⟶ x3 x4 (x3 x5 x6) = x3 (x3 x4 x5) x6.
Assume H9: ∀ x4 . x4 ∈ x0 ⟶ ∀ x5 . x5 ∈ x0 ⟶ ∀ x6 . x6 ∈ x0 ⟶ x3 x4 (x2 x5 x6) = x2 (x3 x4 x5) (x3 x4 x6).
Assume H10: ∀ x4 . x4 ∈ x0 ⟶ ∀ x5 . x5 ∈ x0 ⟶ ∀ x6 . x6 ∈ x0 ⟶ x3 (x2 x4 x5) x6 = x2 (x3 x4 x6) (x3 x5 x6).
Let x4 of type ι be given.
Assume H11: x4 ∈ x0.
Apply H3 with
explicit_Ring_minus x0 x1 x2 x3 x4,
x4,
λ x5 x6 . x6 = x1 leaving 3 subgoals.
Apply explicit_Ring_minus_clos with
x0,
x1,
x2,
x3,
x4 leaving 2 subgoals.
The subproof is completed by applying H0.
The subproof is completed by applying H11.
The subproof is completed by applying H11.
Apply explicit_Ring_minus_R with
x0,
x1,
x2,
x3,
x4 leaving 2 subgoals.
The subproof is completed by applying H0.
The subproof is completed by applying H11.