Metamath Proof Explorer


Theorem psseq2

Description: Equality theorem for proper subclass. (Contributed by NM, 7-Feb-1996)

Ref Expression
Assertion psseq2 ( 𝐴 = 𝐵 → ( 𝐶𝐴𝐶𝐵 ) )

Proof

Step Hyp Ref Expression
1 sseq2 ( 𝐴 = 𝐵 → ( 𝐶𝐴𝐶𝐵 ) )
2 neeq2 ( 𝐴 = 𝐵 → ( 𝐶𝐴𝐶𝐵 ) )
3 1 2 anbi12d ( 𝐴 = 𝐵 → ( ( 𝐶𝐴𝐶𝐴 ) ↔ ( 𝐶𝐵𝐶𝐵 ) ) )
4 df-pss ( 𝐶𝐴 ↔ ( 𝐶𝐴𝐶𝐴 ) )
5 df-pss ( 𝐶𝐵 ↔ ( 𝐶𝐵𝐶𝐵 ) )
6 3 4 5 3bitr4g ( 𝐴 = 𝐵 → ( 𝐶𝐴𝐶𝐵 ) )