Metamath Proof Explorer

Theorem breqd

Description: Equality deduction for a binary relation. (Contributed by NM, 29-Oct-2011)

Ref Expression
Hypothesis breq1d.1 
|- ( ph -> A = B )
Assertion breqd 
|- ( ph -> ( C A D <-> C B D ) )

Proof

Step Hyp Ref Expression
1 breq1d.1 
 |-  ( ph -> A = B )
2 breq 
 |-  ( A = B -> ( C A D <-> C B D ) )
3 1 2 syl 
 |-  ( ph -> ( C A D <-> C B D ) )