Metamath Proof Explorer


Theorem sseq12d

Description: An equality deduction for the subclass relationship. (Contributed by NM, 31-May-1999)

Ref Expression
Hypotheses sseq1d.1
|- ( ph -> A = B )
sseq12d.2
|- ( ph -> C = D )
Assertion sseq12d
|- ( ph -> ( A C_ C <-> B C_ D ) )

Proof

Step Hyp Ref Expression
1 sseq1d.1
 |-  ( ph -> A = B )
2 sseq12d.2
 |-  ( ph -> C = D )
3 1 sseq1d
 |-  ( ph -> ( A C_ C <-> B C_ C ) )
4 2 sseq2d
 |-  ( ph -> ( B C_ C <-> B C_ D ) )
5 3 4 bitrd
 |-  ( ph -> ( A C_ C <-> B C_ D ) )