Metamath Proof Explorer


Theorem zfausclOLD

Description: Obsolete version of sepgi as of 14-Jul-2026. (Contributed by NM, 21-Jun-1993) (Proof modification is discouraged.) (New usage is discouraged.)

Ref Expression
Hypothesis sepgi.1 ⊢ A ∈ V
Assertion zfausclOLD ⊢ ∃ y ∀ x x ∈ y ↔ x ∈ A ∧ φ

Proof

Step Hyp Ref Expression
1 sepgi.1 ⊢ A ∈ V
2 eleq2 ⊢ z = A → x ∈ z ↔ x ∈ A
3 2 anbi1d ⊢ z = A → x ∈ z ∧ φ ↔ x ∈ A ∧ φ
4 3 bibi2d ⊢ z = A → x ∈ y ↔ x ∈ z ∧ φ ↔ x ∈ y ↔ x ∈ A ∧ φ
5 4 albidv ⊢ z = A → ∀ x x ∈ y ↔ x ∈ z ∧ φ ↔ ∀ x x ∈ y ↔ x ∈ A ∧ φ
6 5 exbidv ⊢ z = A → ∃ y ∀ x x ∈ y ↔ x ∈ z ∧ φ ↔ ∃ y ∀ x x ∈ y ↔ x ∈ A ∧ φ
7 ax-sep ⊢ ∃ y ∀ x x ∈ y ↔ x ∈ z ∧ φ
8 1 6 7 vtocl ⊢ ∃ y ∀ x x ∈ y ↔ x ∈ A ∧ φ