Metamath Proof Explorer


Theorem sseq12

Description: Equality theorem for the subclass relationship. (Contributed by NM, 31-May-1999)

Ref Expression
Assertion sseq12 ⊢ A = B ∧ C = D → A ⊆ C ↔ B ⊆ D

Proof

Step Hyp Ref Expression
1 sseq1 ⊢ A = B → A ⊆ C ↔ B ⊆ C
2 sseq2 ⊢ C = D → B ⊆ C ↔ B ⊆ D
3 1 2 sylan9bb ⊢ A = B ∧ C = D → A ⊆ C ↔ B ⊆ D