Metamath Proof Explorer


Theorem ecopover

Description: Assuming that operation F is commutative (second hypothesis), closed (third hypothesis), associative (fourth hypothesis), and has the cancellation property (fifth hypothesis), show that the relation .~ , specified by the first hypothesis, is an equivalence relation. (Contributed by NM, 16-Feb-1996) (Revised by Mario Carneiro, 12-Aug-2015) (Proof shortened by AV, 1-May-2021)

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
ecopopr.cl ⊢ x ∈ S ∧ y ∈ S → x + ˙ y ∈ S
ecopopr.ass ⊢ x + ˙ y + ˙ z = x + ˙ y + ˙ z
ecopopr.can ⊢ x ∈ S ∧ y ∈ S → x + ˙ y = x + ˙ z → y = z
Assertion ecopover ⊢ ∼ ˙ Er S × S

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 ecopopr.cl ⊢ x ∈ S ∧ y ∈ S → x + ˙ y ∈ S
4 ecopopr.ass ⊢ x + ˙ y + ˙ z = x + ˙ y + ˙ z
5 ecopopr.can ⊢ x ∈ S ∧ y ∈ S → x + ˙ y = x + ˙ z → y = z
6 1 relopabiv ⊢ Rel ⁡ ∼ ˙
7 1 2 ecopovsym ⊢ f ∼ ˙ g → g ∼ ˙ f
8 1 2 3 4 5 ecopovtrn ⊢ f ∼ ˙ g ∧ g ∼ ˙ h → f ∼ ˙ h
9 vex ⊢ g ∈ V
10 vex ⊢ h ∈ V
11 9 10 2 caovcom ⊢ g + ˙ h = h + ˙ g
12 1 ecopoveq ⊢ g ∈ S ∧ h ∈ S ∧ g ∈ S ∧ h ∈ S → g h ∼ ˙ g h ↔ g + ˙ h = h + ˙ g
13 11 12 mpbiri ⊢ g ∈ S ∧ h ∈ S ∧ g ∈ S ∧ h ∈ S → g h ∼ ˙ g h
14 13 anidms ⊢ g ∈ S ∧ h ∈ S → g h ∼ ˙ g h
15 14 rgen2 ⊢ ∀ g ∈ S ∀ h ∈ S g h ∼ ˙ g h
16 breq12 ⊢ f = g h ∧ f = g h → f ∼ ˙ f ↔ g h ∼ ˙ g h
17 16 anidms ⊢ f = g h → f ∼ ˙ f ↔ g h ∼ ˙ g h
18 17 ralxp ⊢ ∀ f ∈ S × S f ∼ ˙ f ↔ ∀ g ∈ S ∀ h ∈ S g h ∼ ˙ g h
19 15 18 mpbir ⊢ ∀ f ∈ S × S f ∼ ˙ f
20 19 rspec ⊢ f ∈ S × S → f ∼ ˙ f
21 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
22 1 21 eqsstri ⊢ ∼ ˙ ⊆ S × S × S × S
23 22 ssbri ⊢ f ∼ ˙ f → f S × S × S × S f
24 brxp ⊢ f S × S × S × S f ↔ f ∈ S × S ∧ f ∈ S × S
25 24 simplbi ⊢ f S × S × S × S f → f ∈ S × S
26 23 25 syl ⊢ f ∼ ˙ f → f ∈ S × S
27 20 26 impbii ⊢ f ∈ S × S ↔ f ∼ ˙ f
28 6 7 8 27 iseri ⊢ ∼ ˙ Er S × S