Theorem ab0 4373

Description: The class of sets verifying a property is the empty class if and only if that property is a contradiction. See also abn0 4379 (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 2941). (Contributed by BJ, 19-Mar-2021.) Avoid df-clel 2810, ax-8 2108. (Revised by Gino Giotto, 30-Aug-2024.) (Proof shortened by SN, 8-Sep-2024.)

Assertion

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

Proof of Theorem ab0

Step	Hyp	Ref	Expression
1		abbib 2804	. 2 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ ⊥} ↔ ∀𝑥(𝜑 ↔ ⊥))
2		dfnul4 4323	. . 3 ⊢ ∅ = {𝑥 ∣ ⊥}
3	2	eqeq2i 2745	. 2 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ {𝑥 ∣ 𝜑} = {𝑥 ∣ ⊥})
4		nbfal 1556	. . 3 ⊢ (¬ 𝜑 ↔ (𝜑 ↔ ⊥))
5	4	albii 1821	. 2 ⊢ (∀𝑥 ¬ 𝜑 ↔ ∀𝑥(𝜑 ↔ ⊥))
6	1, 3, 5	3bitr4i 302	1 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ ∀𝑥 ¬ 𝜑)

Syntax hints:  ¬ wn 3   ↔ wb 205  ∀wal 1539   = wceq 1541  ⊥wfal 1553  {cab 2709  ∅c0 4321

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1913  ax-6 1971  ax-7 2011  ax-9 2116  ax-10 2137  ax-11 2154  ax-12 2171  ax-ext 2703

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 846  df-tru 1544  df-fal 1554  df-ex 1782  df-nf 1786  df-sb 2068  df-clab 2710  df-cleq 2724  df-dif 3950  df-nul 4322

This theorem is referenced by:  dfnf5  4376  abn0  4379  rab0  4381  rabeq0  4383  abfOLD  4402  sticksstones22  40972