Theorem ab0 3905

Description: The class of sets verifying a property is the empty class if and only if that property is a contradiction. See also abn0 3908 (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 2782). (Contributed by BJ, 19-Mar-2021.)

Assertion

Ref	Expression
ab0	⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ ∀𝑥 ¬ 𝜑)

Proof of Theorem ab0

Step	Hyp	Ref	Expression
1		nfab1 2753	. . 3 ⊢ Ⅎ𝑥{𝑥 ∣ 𝜑}
2	1	eq0f 3884	. 2 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ ∀𝑥 ¬ 𝑥 ∈ {𝑥 ∣ 𝜑})
3		abid 2598	. . . 4 ⊢ (𝑥 ∈ {𝑥 ∣ 𝜑} ↔ 𝜑)
4	3	notbii 309	. . 3 ⊢ (¬ 𝑥 ∈ {𝑥 ∣ 𝜑} ↔ ¬ 𝜑)
5	4	albii 1737	. 2 ⊢ (∀𝑥 ¬ 𝑥 ∈ {𝑥 ∣ 𝜑} ↔ ∀𝑥 ¬ 𝜑)
6	2, 5	bitri 263	1 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ ∀𝑥 ¬ 𝜑)

Syntax hints:  ¬ wn 3   ↔ wb 195  ∀wal 1473   = wceq 1475   ∈ wcel 1977  {cab 2596  ∅c0 3874

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590

This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-clab 2597  df-cleq 2603  df-clel 2606  df-nfc 2740  df-v 3175  df-dif 3543  df-nul 3875

This theorem is referenced by:  dfnf5  3906  rab0  3909  rabeq0  3911  abf  3930