Metamath Proof Explorer


Theorem ersym

Description: An equivalence relation is symmetric. (Contributed by NM, 4-Jun-1995) (Revised by Mario Carneiro, 12-Aug-2015)

Ref Expression
Hypotheses ersym.1 φ R Er X
ersym.2 φ A R B
Assertion ersym φ B R A

Proof

Step Hyp Ref Expression
1 ersym.1 φ R Er X
2 ersym.2 φ A R B
3 errel R Er X Rel R
4 1 3 syl φ Rel R
5 brrelex12 Rel R A R B A V B V
6 4 2 5 syl2anc φ A V B V
7 brcnvg B V A V B R -1 A A R B
8 7 ancoms A V B V B R -1 A A R B
9 6 8 syl φ B R -1 A A R B
10 2 9 mpbird φ B R -1 A
11 df-er R Er X Rel R dom R = X R -1 R R R
12 11 simp3bi R Er X R -1 R R R
13 1 12 syl φ R -1 R R R
14 13 unssad φ R -1 R
15 14 ssbrd φ B R -1 A B R A
16 10 15 mpd φ B R A