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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