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,
∀ x5 . x5 ∈ x0 ⟶ explicit_Ring_minus x0 x1 x3 x4 x5 = x4 x5 (explicit_Ring_minus x0 x1 x3 x4 x2).
Assume H1: ∀ x5 . x5 ∈ x0 ⟶ ∀ x6 . x6 ∈ x0 ⟶ x3 x5 x6 ∈ x0.
Assume H2: ∀ x5 . x5 ∈ x0 ⟶ ∀ x6 . x6 ∈ x0 ⟶ ∀ x7 . x7 ∈ x0 ⟶ x3 x5 (x3 x6 x7) = x3 (x3 x5 x6) x7.
Assume H3: ∀ x5 . x5 ∈ x0 ⟶ ∀ x6 . x6 ∈ x0 ⟶ x3 x5 x6 = x3 x6 x5.
Assume H4: x1 ∈ x0.
Assume H5: ∀ x5 . x5 ∈ x0 ⟶ x3 x1 x5 = x5.
Assume H6:
∀ x5 . x5 ∈ x0 ⟶ ∃ x6 . and (x6 ∈ x0) (x3 x5 x6 = x1).
Assume H7: ∀ x5 . x5 ∈ x0 ⟶ ∀ x6 . x6 ∈ x0 ⟶ x4 x5 x6 ∈ x0.
Assume H8: ∀ x5 . x5 ∈ x0 ⟶ ∀ x6 . x6 ∈ x0 ⟶ ∀ x7 . x7 ∈ x0 ⟶ x4 x5 (x4 x6 x7) = x4 (x4 x5 x6) x7.
Assume H9: x2 ∈ x0.
Assume H10: x2 = x1 ⟶ ∀ x5 : ο . x5.
Assume H11: ∀ x5 . x5 ∈ x0 ⟶ x4 x2 x5 = x5.
Assume H12: ∀ x5 . x5 ∈ x0 ⟶ x4 x5 x2 = x5.
Assume H13: ∀ x5 . x5 ∈ x0 ⟶ ∀ x6 . x6 ∈ x0 ⟶ ∀ x7 . x7 ∈ x0 ⟶ x4 x5 (x3 x6 x7) = x3 (x4 x5 x6) (x4 x5 x7).
Assume H14: ∀ x5 . x5 ∈ x0 ⟶ ∀ x6 . x6 ∈ x0 ⟶ ∀ x7 . x7 ∈ x0 ⟶ x4 (x3 x5 x6) x7 = x3 (x4 x5 x7) (x4 x6 x7).
Let x5 of type ι be given.
Assume H15: x5 ∈ x0.
Apply explicit_Ring_with_id_plus_cancelL with
x0,
x1,
x2,
x3,
x4,
x5,
explicit_Ring_minus x0 x1 x3 x4 x5,
x4 x5 (explicit_Ring_minus x0 x1 x3 x4 x2) leaving 5 subgoals.
The subproof is completed by applying H0.
The subproof is completed by applying H15.
The subproof is completed by applying L16.
The subproof is completed by applying L17.
Apply explicit_Ring_with_id_minus_R with
x0,
x1,
x2,
x3,
x4,
x5,
λ x6 x7 . x7 = x3 x5 (x4 x5 (explicit_Ring_minus x0 x1 x3 x4 x2)) leaving 3 subgoals.
The subproof is completed by applying H0.
The subproof is completed by applying H15.
Let x6 of type ι → ι → ο be given.
Apply H12 with
x5,
λ x7 x8 . x3 x7 ... = ...,
... leaving 2 subgoals.