Metamath Proof Explorer

Theorem psseq1d

Description: An equality deduction for the proper subclass relationship. (Contributed by NM, 9-Jun-2004)

Ref Expression
Hypothesis psseq1d.1 
|- ( ph -> A = B )
Assertion psseq1d 
|- ( ph -> ( A C. C <-> B C. C ) )

Proof

Step Hyp Ref Expression
1 psseq1d.1 
 |-  ( ph -> A = B )
2 psseq1 
 |-  ( A = B -> ( A C. C <-> B C. C ) )
3 1 2 syl 
 |-  ( ph -> ( A C. C <-> B C. C ) )