Metamath Proof Explorer


Theorem abssi

Description: Inference of abstraction subclass from implication. (Contributed by NM, 20-Jan-2006)

Ref Expression
Hypothesis abssi.1 ⊢ ( 𝜑 → 𝑥 ∈ 𝐴 )
Assertion abssi { 𝑥 ∣ 𝜑 } ⊆ 𝐴

Proof

Step Hyp Ref Expression
1 abssi.1 ⊢ ( 𝜑 → 𝑥 ∈ 𝐴 )
2 1 ss2abi ⊢ { 𝑥 ∣ 𝜑 } ⊆ { 𝑥 ∣ 𝑥 ∈ 𝐴 }
3 abid2 ⊢ { 𝑥 ∣ 𝑥 ∈ 𝐴 } = 𝐴
4 2 3 sseqtri ⊢ { 𝑥 ∣ 𝜑 } ⊆ 𝐴