Metamath Proof Explorer


Theorem psseq12d

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

Ref Expression
Hypotheses psseq1d.1 ( 𝜑𝐴 = 𝐵 )
psseq12d.2 ( 𝜑𝐶 = 𝐷 )
Assertion psseq12d ( 𝜑 → ( 𝐴𝐶𝐵𝐷 ) )

Proof

Step Hyp Ref Expression
1 psseq1d.1 ( 𝜑𝐴 = 𝐵 )
2 psseq12d.2 ( 𝜑𝐶 = 𝐷 )
3 1 psseq1d ( 𝜑 → ( 𝐴𝐶𝐵𝐶 ) )
4 2 psseq2d ( 𝜑 → ( 𝐵𝐶𝐵𝐷 ) )
5 3 4 bitrd ( 𝜑 → ( 𝐴𝐶𝐵𝐷 ) )