Metamath Proof Explorer


Theorem sseq12i

Description: An equality inference for the subclass relationship. (Contributed by NM, 31-May-1999) (Proof shortened by Eric Schmidt, 26-Jan-2007)

Ref Expression
Hypotheses sseq1i.1 ⊢ A = B
sseq12i.2 ⊢ C = D
Assertion sseq12i ⊢ A ⊆ C ↔ B ⊆ D

Proof

Step Hyp Ref Expression
1 sseq1i.1 ⊢ A = B
2 sseq12i.2 ⊢ C = D
3 sseq12 ⊢ A = B ∧ C = D → A ⊆ C ↔ B ⊆ D
4 1 2 3 mp2an ⊢ A ⊆ C ↔ B ⊆ D