Metamath Proof Explorer


Theorem sseq12d

Description: An equality deduction for the subclass relationship. (Contributed by NM, 31-May-1999)

Ref Expression
Hypotheses sseq1d.1 ⊢ φ → A = B
sseq12d.2 ⊢ φ → C = D
Assertion sseq12d ⊢ φ → A ⊆ C ↔ B ⊆ D

Proof

Step Hyp Ref Expression
1 sseq1d.1 ⊢ φ → A = B
2 sseq12d.2 ⊢ φ → C = D
3 1 sseq1d ⊢ φ → A ⊆ C ↔ B ⊆ C
4 2 sseq2d ⊢ φ → B ⊆ C ↔ B ⊆ D
5 3 4 bitrd ⊢ φ → A ⊆ C ↔ B ⊆ D