Metamath Proof Explorer


Theorem sepg

Description: Version of the axiom of separation where the "containing set" is a class variable and the sethood assumption is in the antecedent. (Contributed by NM, 21-Jun-1993) Put sepgi in closed form. (Revised by BJ, 2-Jul-2022)

Ref Expression
Assertion sepg ( 𝐴 ∈ 𝑉 → ∃ 𝑦 ∀ 𝑥 ( 𝑥 ∈ 𝑦 ↔ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ) )

Proof

Step Hyp Ref Expression
1 eleq2 ⊢ ( 𝑧 = 𝐴 → ( 𝑥 ∈ 𝑧 ↔ 𝑥 ∈ 𝐴 ) )
2 1 anbi1d ⊢ ( 𝑧 = 𝐴 → ( ( 𝑥 ∈ 𝑧 ∧ 𝜑 ) ↔ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ) )
3 2 bibi2d ⊢ ( 𝑧 = 𝐴 → ( ( 𝑥 ∈ 𝑦 ↔ ( 𝑥 ∈ 𝑧 ∧ 𝜑 ) ) ↔ ( 𝑥 ∈ 𝑦 ↔ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ) ) )
4 3 albidv ⊢ ( 𝑧 = 𝐴 → ( ∀ 𝑥 ( 𝑥 ∈ 𝑦 ↔ ( 𝑥 ∈ 𝑧 ∧ 𝜑 ) ) ↔ ∀ 𝑥 ( 𝑥 ∈ 𝑦 ↔ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ) ) )
5 4 exbidv ⊢ ( 𝑧 = 𝐴 → ( ∃ 𝑦 ∀ 𝑥 ( 𝑥 ∈ 𝑦 ↔ ( 𝑥 ∈ 𝑧 ∧ 𝜑 ) ) ↔ ∃ 𝑦 ∀ 𝑥 ( 𝑥 ∈ 𝑦 ↔ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ) ) )
6 ax-sep ⊢ ∃ 𝑦 ∀ 𝑥 ( 𝑥 ∈ 𝑦 ↔ ( 𝑥 ∈ 𝑧 ∧ 𝜑 ) )
7 5 6 vtoclg ⊢ ( 𝐴 ∈ 𝑉 → ∃ 𝑦 ∀ 𝑥 ( 𝑥 ∈ 𝑦 ↔ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ) )