Metamath Proof Explorer


Theorem releqd

Description: Equality deduction for the relation predicate. (Contributed by NM, 8-Mar-2014)

Ref Expression
Hypothesis releqd.1 φA=B
Assertion releqd φRelARelB

Proof

Step Hyp Ref Expression
1 releqd.1 φA=B
2 releq A=BRelARelB
3 1 2 syl φRelARelB