Metamath Proof Explorer


Theorem sseld

Description: Membership deduction from subclass relationship. (Contributed by NM, 15-Nov-1995)

Ref Expression
Hypothesis sseld.1 ⊢ ( 𝜑 → 𝐴 ⊆ 𝐵 )
Assertion sseld ( 𝜑 → ( 𝐶 ∈ 𝐴 → 𝐶 ∈ 𝐵 ) )

Proof

Step Hyp Ref Expression
1 sseld.1 ⊢ ( 𝜑 → 𝐴 ⊆ 𝐵 )
2 ssel ⊢ ( 𝐴 ⊆ 𝐵 → ( 𝐶 ∈ 𝐴 → 𝐶 ∈ 𝐵 ) )
3 1 2 syl ⊢ ( 𝜑 → ( 𝐶 ∈ 𝐴 → 𝐶 ∈ 𝐵 ) )