Metamath Proof Explorer


Theorem ab0ALT

Description: Alternate proof of ab0 , shorter but using more axioms. (Contributed by BJ, 19-Mar-2021) (Proof modification is discouraged.) (New usage is discouraged.)

Ref Expression
Assertion ab0ALT ⊢ x | φ = ∅ ↔ ∀ x ¬ φ

Proof

Step Hyp Ref Expression
1 nfab1 ⊢ Ⅎ _ x x | φ
2 1 eq0f ⊢ x | φ = ∅ ↔ ∀ x ¬ x ∈ x | φ
3 abid ⊢ x ∈ x | φ ↔ φ
4 3 notbii ⊢ ¬ x ∈ x | φ ↔ ¬ φ
5 4 albii ⊢ ∀ x ¬ x ∈ x | φ ↔ ∀ x ¬ φ
6 2 5 bitri ⊢ x | φ = ∅ ↔ ∀ x ¬ φ