Metamath Proof Explorer


Theorem ab0

Description: The class of sets verifying a property is the empty class if and only if that property is a contradiction. See also abn0 (from which it could be proved using as many essential proof steps but one fewer syntactic step, at the cost of depending on df-ne ). (Contributed by BJ, 19-Mar-2021) Avoid df-clel , ax-8 . (Revised by Gino Giotto, 30-Aug-2024) (Proof shortened by SN, 8-Sep-2024)

Ref Expression
Assertion ab0 ( { 𝑥𝜑 } = ∅ ↔ ∀ 𝑥 ¬ 𝜑 )

Proof

Step Hyp Ref Expression
1 abbib ( { 𝑥𝜑 } = { 𝑥 ∣ ⊥ } ↔ ∀ 𝑥 ( 𝜑 ↔ ⊥ ) )
2 dfnul4 ∅ = { 𝑥 ∣ ⊥ }
3 2 eqeq2i ( { 𝑥𝜑 } = ∅ ↔ { 𝑥𝜑 } = { 𝑥 ∣ ⊥ } )
4 nbfal ( ¬ 𝜑 ↔ ( 𝜑 ↔ ⊥ ) )
5 4 albii ( ∀ 𝑥 ¬ 𝜑 ↔ ∀ 𝑥 ( 𝜑 ↔ ⊥ ) )
6 1 3 5 3bitr4i ( { 𝑥𝜑 } = ∅ ↔ ∀ 𝑥 ¬ 𝜑 )