Metamath Proof Explorer

Theorem eqvreleqd

Description: Equality theorem for equivalence relation, deduction version. (Contributed by Peter Mazsa, 23-Sep-2021)

Ref Expression
Hypothesis eqvreleqd.1 
|- ( ph -> R = S )
Assertion eqvreleqd 
|- ( ph -> ( EqvRel R <-> EqvRel S ) )

Proof

Step Hyp Ref Expression
1 eqvreleqd.1 
 |-  ( ph -> R = S )
2 eqvreleq 
 |-  ( R = S -> ( EqvRel R <-> EqvRel S ) )
3 1 2 syl 
 |-  ( ph -> ( EqvRel R <-> EqvRel S ) )