Let x0 of type ι be given.

Let x1 of type ι be given.

Let x2 of type ι be given.

Assume H0: Field_Hom x0 x1 x2.

Let x3 of type ο be given.

Assume H1: Field x0 ⟶ Field x1 ⟶ x2 ∈ setexp (field0 x1) (field0 x0) ⟶ ap x2 (field3 x0) = field3 x1 ⟶ ap x2 (field4 x0) = field4 x1 ⟶ (∀ x4 . x4 ∈ field0 x0 ⟶ ∀ x5 . x5 ∈ field0 x0 ⟶ ap x2 (field1b x0 x4 x5) = field1b x1 (ap x2 x4) (ap x2 x5)) ⟶ (∀ x4 . x4 ∈ field0 x0 ⟶ ∀ x5 . x5 ∈ field0 x0 ⟶ ap x2 (field2b x0 x4 x5) = field2b x1 (ap x2 x4) (ap x2 x5)) ⟶ (∀ x4 . x4 ∈ field0 x0 ⟶ ap x2 (Field_minus x0 x4) = Field_minus x1 (ap x2 x4)) ⟶ (∀ x4 . x4 ∈ field0 x0 ⟶ ap x2 x4 = field3 x1 ⟶ x4 = field3 x0) ⟶ (∀ x4 . x4 ∈ field0 x0 ⟶ ∀ x5 . x5 ∈ field0 x0 ⟶ ap x2 x4 = ap x2 x5 ⟶ x4 = x5) ⟶ (∀ x4 . x4 ∈ field0 x0 ⟶ ∀ x5 . x5 ∈ omega ⟶ ap x2 (CRing_with_id_omega_exp x0 x4 x5) = CRing_with_id_omega_exp x1 (ap x2 x4) x5) ⟶ x3.

Apply and7E with Field x0, Field x1, x2 ∈ setexp (field0 x1) (field0 x0), ap x2 (field3 x0) = field3 x1, ap x2 (field4 x0) = field4 x1, ∀ x4 . x4 ∈ field0 x0 ⟶ ∀ x5 . x5 ∈ field0 x0 ⟶ ap x2 (field1b x0 x4 x5) = field1b x1 (ap x2 x4) (ap x2 x5), ∀ x4 . x4 ∈ field0 x0 ⟶ ∀ x5 . x5 ∈ field0 x0 ⟶ ap x2 (field2b x0 x4 x5) = field2b x1 (ap x2 x4) (ap x2 x5), x3 leaving 2 subgoals.

The subproof is completed by applying H0.

Assume H2: Field x0.

Assume H3: Field x1.

Assume H4: x2 ∈ setexp (field0 x1) (field0 x0).

Assume H5: ap x2 (field3 x0) = field3 x1.

Assume H6: ap x2 (field4 x0) = field4 x1.

Assume H7: ∀ x4 . ... ⟶ ∀ x5 . ... ⟶ ap x2 (field1b x0 x4 x5) = field1b x1 ... ....

...

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Proofgold Proof