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 φA=B
Assertion psseq1d φACBC

Proof

Step Hyp Ref Expression
1 psseq1d.1 φA=B
2 psseq1 A=BACBC
3 1 2 syl φACBC