Metamath Proof Explorer


Theorem sselda

Description: Membership deduction from subclass relationship. (Contributed by NM, 26-Jun-2014)

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

Proof

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