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 . x5 ∈ x0 ⟶ ∀ x6 . x6 ∈ x0 ⟶ ∀ x7 . x7 ∈ x0 ⟶ x3 x5 x7 = x3 x6 x7 ⟶ x5 = x6.
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: ∀ x5 . x5 ∈ x0 ⟶ ∀ x6 . x6 ∈ x0 ⟶ x4 x5 x6 = x4 x6 x5.
Assume H10: x2 ∈ x0.
Assume H11: x2 = x1 ⟶ ∀ x5 : ο . x5.
Assume H12: ∀ x5 . x5 ∈ x0 ⟶ x4 x2 x5 = x5.
Assume H13:
∀ x5 . x5 ∈ x0 ⟶ (x5 = x1 ⟶ ∀ x6 : ο . x6) ⟶ ∃ x6 . and (x6 ∈ x0) (x4 x5 x6 = x2).
Assume H14: ∀ x5 . x5 ∈ x0 ⟶ ∀ x6 . x6 ∈ x0 ⟶ ∀ x7 . x7 ∈ x0 ⟶ x4 x5 (x3 x6 x7) = x3 (x4 x5 x6) (x4 x5 x7).
Let x5 of type ι be given.
Assume H15: x5 ∈ x0.
Let x6 of type ι be given.
Assume H16: x6 ∈ x0.
Let x7 of type ι be given.
Assume H17: x7 ∈ x0.
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,
λ x8 x9 . x9 = x3 x7 x6 leaving 3 subgoals.
The subproof is completed by applying H17.
The subproof is completed by applying H15.
Apply H18 with
λ x8 x9 . x9 = x3 x7 x6.
Apply H3 with
x6,
x7 leaving 2 subgoals.
The subproof is completed by applying H16.
The subproof is completed by applying H17.