Metamath Proof Explorer

Theorem sseq2d

Description: An equality deduction for the subclass relationship. (Contributed by NM, 14-Aug-1994)

Ref Expression
Hypothesis sseq1d.1 
|- ( ph -> A = B )
Assertion sseq2d 
|- ( ph -> ( C C_ A <-> C C_ B ) )

Proof

Step Hyp Ref Expression
1 sseq1d.1 
 |-  ( ph -> A = B )
2 sseq2 
 |-  ( A = B -> ( C C_ A <-> C C_ B ) )
3 1 2 syl 
 |-  ( ph -> ( C C_ A <-> C C_ B ) )