Metamath Proof Explorer


Theorem psseq1

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

Ref Expression
Assertion psseq1 ( 𝐴 = 𝐵 → ( 𝐴𝐶𝐵𝐶 ) )

Proof

Step Hyp Ref Expression
1 sseq1 ( 𝐴 = 𝐵 → ( 𝐴𝐶𝐵𝐶 ) )
2 neeq1 ( 𝐴 = 𝐵 → ( 𝐴𝐶𝐵𝐶 ) )
3 1 2 anbi12d ( 𝐴 = 𝐵 → ( ( 𝐴𝐶𝐴𝐶 ) ↔ ( 𝐵𝐶𝐵𝐶 ) ) )
4 df-pss ( 𝐴𝐶 ↔ ( 𝐴𝐶𝐴𝐶 ) )
5 df-pss ( 𝐵𝐶 ↔ ( 𝐵𝐶𝐵𝐶 ) )
6 3 4 5 3bitr4g ( 𝐴 = 𝐵 → ( 𝐴𝐶𝐵𝐶 ) )