Metamath Proof Explorer


Theorem sseq2

Description: Equality theorem for the subclass relationship. (Contributed by NM, 25-Jun-1998)

Ref Expression
Assertion sseq2 ( 𝐴 = 𝐵 → ( 𝐶 ⊆ 𝐴 ↔ 𝐶 ⊆ 𝐵 ) )

Proof

Step Hyp Ref Expression
1 eqss ⊢ ( 𝐴 = 𝐵 ↔ ( 𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴 ) )
2 sstr2 ⊢ ( 𝐶 ⊆ 𝐴 → ( 𝐴 ⊆ 𝐵 → 𝐶 ⊆ 𝐵 ) )
3 2 com12 ⊢ ( 𝐴 ⊆ 𝐵 → ( 𝐶 ⊆ 𝐴 → 𝐶 ⊆ 𝐵 ) )
4 sstr2 ⊢ ( 𝐶 ⊆ 𝐵 → ( 𝐵 ⊆ 𝐴 → 𝐶 ⊆ 𝐴 ) )
5 4 com12 ⊢ ( 𝐵 ⊆ 𝐴 → ( 𝐶 ⊆ 𝐵 → 𝐶 ⊆ 𝐴 ) )
6 3 5 anbiim ⊢ ( ( 𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴 ) → ( 𝐶 ⊆ 𝐴 ↔ 𝐶 ⊆ 𝐵 ) )
7 1 6 sylbi ⊢ ( 𝐴 = 𝐵 → ( 𝐶 ⊆ 𝐴 ↔ 𝐶 ⊆ 𝐵 ) )