Metamath Proof Explorer


Theorem ecopovsym

Description: Assuming the operation F is commutative, show that the relation .~ , specified by the first hypothesis, is symmetric. (Contributed by NM, 27-Aug-1995) (Revised by Mario Carneiro, 26-Apr-2015)

Ref Expression
Hypotheses ecopopr.1 ⊢ ∼ ˙ = x y | x ∈ S × S ∧ y ∈ S × S ∧ ∃ z ∃ w ∃ v ∃ u x = z w ∧ y = v u ∧ z + ˙ u = w + ˙ v
ecopopr.com ⊢ x + ˙ y = y + ˙ x
Assertion ecopovsym ⊢ A ∼ ˙ B → B ∼ ˙ A

Proof

Step Hyp Ref Expression
1 ecopopr.1 ⊢ ∼ ˙ = x y | x ∈ S × S ∧ y ∈ S × S ∧ ∃ z ∃ w ∃ v ∃ u x = z w ∧ y = v u ∧ z + ˙ u = w + ˙ v
2 ecopopr.com ⊢ x + ˙ y = y + ˙ x
3 opabssxp ⊢ x y | x ∈ S × S ∧ y ∈ S × S ∧ ∃ z ∃ w ∃ v ∃ u x = z w ∧ y = v u ∧ z + ˙ u = w + ˙ v ⊆ S × S × S × S
4 1 3 eqsstri ⊢ ∼ ˙ ⊆ S × S × S × S
5 4 brel ⊢ A ∼ ˙ B → A ∈ S × S ∧ B ∈ S × S
6 eqid ⊢ S × S = S × S
7 breq1 ⊢ f g = A → f g ∼ ˙ h t ↔ A ∼ ˙ h t
8 breq2 ⊢ f g = A → h t ∼ ˙ f g ↔ h t ∼ ˙ A
9 7 8 bibi12d ⊢ f g = A → f g ∼ ˙ h t ↔ h t ∼ ˙ f g ↔ A ∼ ˙ h t ↔ h t ∼ ˙ A
10 breq2 ⊢ h t = B → A ∼ ˙ h t ↔ A ∼ ˙ B
11 breq1 ⊢ h t = B → h t ∼ ˙ A ↔ B ∼ ˙ A
12 10 11 bibi12d ⊢ h t = B → A ∼ ˙ h t ↔ h t ∼ ˙ A ↔ A ∼ ˙ B ↔ B ∼ ˙ A
13 1 ecopoveq ⊢ f ∈ S ∧ g ∈ S ∧ h ∈ S ∧ t ∈ S → f g ∼ ˙ h t ↔ f + ˙ t = g + ˙ h
14 vex ⊢ f ∈ V
15 vex ⊢ t ∈ V
16 14 15 2 caovcom ⊢ f + ˙ t = t + ˙ f
17 vex ⊢ g ∈ V
18 vex ⊢ h ∈ V
19 17 18 2 caovcom ⊢ g + ˙ h = h + ˙ g
20 16 19 eqeq12i ⊢ f + ˙ t = g + ˙ h ↔ t + ˙ f = h + ˙ g
21 eqcom ⊢ t + ˙ f = h + ˙ g ↔ h + ˙ g = t + ˙ f
22 20 21 bitri ⊢ f + ˙ t = g + ˙ h ↔ h + ˙ g = t + ˙ f
23 13 22 bitrdi ⊢ f ∈ S ∧ g ∈ S ∧ h ∈ S ∧ t ∈ S → f g ∼ ˙ h t ↔ h + ˙ g = t + ˙ f
24 1 ecopoveq ⊢ h ∈ S ∧ t ∈ S ∧ f ∈ S ∧ g ∈ S → h t ∼ ˙ f g ↔ h + ˙ g = t + ˙ f
25 24 ancoms ⊢ f ∈ S ∧ g ∈ S ∧ h ∈ S ∧ t ∈ S → h t ∼ ˙ f g ↔ h + ˙ g = t + ˙ f
26 23 25 bitr4d ⊢ f ∈ S ∧ g ∈ S ∧ h ∈ S ∧ t ∈ S → f g ∼ ˙ h t ↔ h t ∼ ˙ f g
27 6 9 12 26 2optocl ⊢ A ∈ S × S ∧ B ∈ S × S → A ∼ ˙ B ↔ B ∼ ˙ A
28 5 27 syl ⊢ A ∼ ˙ B → A ∼ ˙ B ↔ B ∼ ˙ A
29 28 ibi ⊢ A ∼ ˙ B → B ∼ ˙ A