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 ) )