Metamath Proof Explorer


Theorem eqvrelsymb

Description: An equivalence relation is symmetric. (Contributed by NM, 30-Jul-1995) (Revised by Mario Carneiro, 12-Aug-2015) (Revised and distinct variable conditions removed by Peter Mazsa, 2-Jun-2019.)

Ref Expression
Hypothesis eqvrelsymb.1
|- ( ph -> EqvRel R )
Assertion eqvrelsymb
|- ( ph -> ( A R B <-> B R A ) )

Proof

Step Hyp Ref Expression
1 eqvrelsymb.1
 |-  ( ph -> EqvRel R )
2 1 adantr
 |-  ( ( ph /\ A R B ) -> EqvRel R )
3 simpr
 |-  ( ( ph /\ A R B ) -> A R B )
4 2 3 eqvrelsym
 |-  ( ( ph /\ A R B ) -> B R A )
5 1 adantr
 |-  ( ( ph /\ B R A ) -> EqvRel R )
6 simpr
 |-  ( ( ph /\ B R A ) -> B R A )
7 5 6 eqvrelsym
 |-  ( ( ph /\ B R A ) -> A R B )
8 4 7 impbida
 |-  ( ph -> ( A R B <-> B R A ) )