Metamath Proof Explorer


Theorem sseq1d

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

Ref Expression
Hypothesis sseq1d.1 φA=B
Assertion sseq1d φACBC

Proof

Step Hyp Ref Expression
1 sseq1d.1 φA=B
2 sseq1 A=BACBC
3 1 2 syl φACBC