Metamath Proof Explorer


Theorem breqan12rd

Description: Equality deduction for a binary relation. (Contributed by NM, 8-Feb-1996)

Ref Expression
Hypotheses breq1d.1
|- ( ph -> A = B )
breqan12i.2
|- ( ps -> C = D )
Assertion breqan12rd
|- ( ( ps /\ ph ) -> ( A R C <-> B R D ) )

Proof

Step Hyp Ref Expression
1 breq1d.1
 |-  ( ph -> A = B )
2 breqan12i.2
 |-  ( ps -> C = D )
3 1 2 breqan12d
 |-  ( ( ph /\ ps ) -> ( A R C <-> B R D ) )
4 3 ancoms
 |-  ( ( ps /\ ph ) -> ( A R C <-> B R D ) )