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 φ A = B
Assertion psseq1d φ A C B C

Proof

Step Hyp Ref Expression
1 psseq1d.1 φ A = B
2 psseq1 A = B A C B C
3 1 2 syl φ A C B C