Metamath Proof Explorer


Theorem psseq12d

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

Ref Expression
Hypotheses psseq1d.1
|- ( ph -> A = B )
psseq12d.2
|- ( ph -> C = D )
Assertion psseq12d
|- ( ph -> ( A C. C <-> B C. D ) )

Proof

Step Hyp Ref Expression
1 psseq1d.1
 |-  ( ph -> A = B )
2 psseq12d.2
 |-  ( ph -> C = D )
3 1 psseq1d
 |-  ( ph -> ( A C. C <-> B C. C ) )
4 2 psseq2d
 |-  ( ph -> ( B C. C <-> B C. D ) )
5 3 4 bitrd
 |-  ( ph -> ( A C. C <-> B C. D ) )