Metamath Proof Explorer


Theorem psseq2i

Description: An equality inference for the proper subclass relationship. (Contributed by NM, 9-Jun-2004)

Ref Expression
Hypothesis psseq1i.1 𝐴 = 𝐵
Assertion psseq2i ( 𝐶𝐴𝐶𝐵 )

Proof

Step Hyp Ref Expression
1 psseq1i.1 𝐴 = 𝐵
2 psseq2 ( 𝐴 = 𝐵 → ( 𝐶𝐴𝐶𝐵 ) )
3 1 2 ax-mp ( 𝐶𝐴𝐶𝐵 )