Metamath Proof Explorer


Theorem elintrabg

Description: Membership in the intersection of a class abstraction. (Contributed by NM, 17-Feb-2007)

Ref Expression
Assertion elintrabg ( 𝐴 ∈ 𝑉 → ( 𝐴 ∈ ∩ { 𝑥 ∈ 𝐵 ∣ 𝜑 } ↔ ∀ 𝑥 ∈ 𝐵 ( 𝜑 → 𝐴 ∈ 𝑥 ) ) )

Proof

Step Hyp Ref Expression
1 eleq1 ⊢ ( 𝑦 = 𝐴 → ( 𝑦 ∈ ∩ { 𝑥 ∈ 𝐵 ∣ 𝜑 } ↔ 𝐴 ∈ ∩ { 𝑥 ∈ 𝐵 ∣ 𝜑 } ) )
2 eleq1 ⊢ ( 𝑦 = 𝐴 → ( 𝑦 ∈ 𝑥 ↔ 𝐴 ∈ 𝑥 ) )
3 2 imbi2d ⊢ ( 𝑦 = 𝐴 → ( ( 𝜑 → 𝑦 ∈ 𝑥 ) ↔ ( 𝜑 → 𝐴 ∈ 𝑥 ) ) )
4 3 ralbidv ⊢ ( 𝑦 = 𝐴 → ( ∀ 𝑥 ∈ 𝐵 ( 𝜑 → 𝑦 ∈ 𝑥 ) ↔ ∀ 𝑥 ∈ 𝐵 ( 𝜑 → 𝐴 ∈ 𝑥 ) ) )
5 vex ⊢ 𝑦 ∈ V
6 5 elintrab ⊢ ( 𝑦 ∈ ∩ { 𝑥 ∈ 𝐵 ∣ 𝜑 } ↔ ∀ 𝑥 ∈ 𝐵 ( 𝜑 → 𝑦 ∈ 𝑥 ) )
7 1 4 6 vtoclbg ⊢ ( 𝐴 ∈ 𝑉 → ( 𝐴 ∈ ∩ { 𝑥 ∈ 𝐵 ∣ 𝜑 } ↔ ∀ 𝑥 ∈ 𝐵 ( 𝜑 → 𝐴 ∈ 𝑥 ) ) )