Metamath Proof Explorer


Theorem ecovcom

Description: Lemma used to transfer a commutative law via an equivalence relation. (Contributed by NM, 29-Aug-1995) (Revised by David Abernethy, 4-Jun-2013)

Ref Expression
Hypotheses ecovcom.1 ⊢ C = S × S / ∼ ˙
ecovcom.2 ⊢ x ∈ S ∧ y ∈ S ∧ z ∈ S ∧ w ∈ S → x y ∼ ˙ + ˙ z w ∼ ˙ = D G ∼ ˙
ecovcom.3 ⊢ z ∈ S ∧ w ∈ S ∧ x ∈ S ∧ y ∈ S → z w ∼ ˙ + ˙ x y ∼ ˙ = H J ∼ ˙
ecovcom.4 ⊢ D = H
ecovcom.5 ⊢ G = J
Assertion ecovcom ⊢ A ∈ C ∧ B ∈ C → A + ˙ B = B + ˙ A

Proof

Step Hyp Ref Expression
1 ecovcom.1 ⊢ C = S × S / ∼ ˙
2 ecovcom.2 ⊢ x ∈ S ∧ y ∈ S ∧ z ∈ S ∧ w ∈ S → x y ∼ ˙ + ˙ z w ∼ ˙ = D G ∼ ˙
3 ecovcom.3 ⊢ z ∈ S ∧ w ∈ S ∧ x ∈ S ∧ y ∈ S → z w ∼ ˙ + ˙ x y ∼ ˙ = H J ∼ ˙
4 ecovcom.4 ⊢ D = H
5 ecovcom.5 ⊢ G = J
6 oveq1 ⊢ x y ∼ ˙ = A → x y ∼ ˙ + ˙ z w ∼ ˙ = A + ˙ z w ∼ ˙
7 oveq2 ⊢ x y ∼ ˙ = A → z w ∼ ˙ + ˙ x y ∼ ˙ = z w ∼ ˙ + ˙ A
8 6 7 eqeq12d ⊢ x y ∼ ˙ = A → x y ∼ ˙ + ˙ z w ∼ ˙ = z w ∼ ˙ + ˙ x y ∼ ˙ ↔ A + ˙ z w ∼ ˙ = z w ∼ ˙ + ˙ A
9 oveq2 ⊢ z w ∼ ˙ = B → A + ˙ z w ∼ ˙ = A + ˙ B
10 oveq1 ⊢ z w ∼ ˙ = B → z w ∼ ˙ + ˙ A = B + ˙ A
11 9 10 eqeq12d ⊢ z w ∼ ˙ = B → A + ˙ z w ∼ ˙ = z w ∼ ˙ + ˙ A ↔ A + ˙ B = B + ˙ A
12 opeq12 ⊢ D = H ∧ G = J → D G = H J
13 12 eceq1d ⊢ D = H ∧ G = J → D G ∼ ˙ = H J ∼ ˙
14 4 5 13 mp2an ⊢ D G ∼ ˙ = H J ∼ ˙
15 3 ancoms ⊢ x ∈ S ∧ y ∈ S ∧ z ∈ S ∧ w ∈ S → z w ∼ ˙ + ˙ x y ∼ ˙ = H J ∼ ˙
16 14 2 15 3eqtr4a ⊢ x ∈ S ∧ y ∈ S ∧ z ∈ S ∧ w ∈ S → x y ∼ ˙ + ˙ z w ∼ ˙ = z w ∼ ˙ + ˙ x y ∼ ˙
17 1 8 11 16 2ecoptocl ⊢ A ∈ C ∧ B ∈ C → A + ˙ B = B + ˙ A