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 ⊢ φ → A = B
psseq12d.2 ⊢ φ → C = D
Assertion psseq12d ⊢ φ → A ⊂ C ↔ B ⊂ D

Proof

Step Hyp Ref Expression
1 psseq1d.1 ⊢ φ → A = B
2 psseq12d.2 ⊢ φ → C = D
3 1 psseq1d ⊢ φ → A ⊂ C ↔ B ⊂ C
4 2 psseq2d ⊢ φ → B ⊂ C ↔ B ⊂ D
5 3 4 bitrd ⊢ φ → A ⊂ C ↔ B ⊂ D