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 ( 𝜑 → ( 𝐴𝐶𝐵𝐶 ) )

Proof

Step Hyp Ref Expression
1 psseq1d.1 ( 𝜑𝐴 = 𝐵 )
2 psseq1 ( 𝐴 = 𝐵 → ( 𝐴𝐶𝐵𝐶 ) )
3 1 2 syl ( 𝜑 → ( 𝐴𝐶𝐵𝐶 ) )