Theorem ab0 4308

Description: The class of sets verifying a property is the empty class if and only if that property is a contradiction. See also abn0 4314 (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 2944). (Contributed by BJ, 19-Mar-2021.) Avoid df-clel 2816, 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		abbi 2810	. 2 ⊢ (∀𝑥(𝜑 ↔ ⊥) ↔ {𝑥 ∣ 𝜑} = {𝑥 ∣ ⊥})
2		nbfal 1554	. . 3 ⊢ (¬ 𝜑 ↔ (𝜑 ↔ ⊥))
3	2	albii 1822	. 2 ⊢ (∀𝑥 ¬ 𝜑 ↔ ∀𝑥(𝜑 ↔ ⊥))
4		dfnul4 4258	. . 3 ⊢ ∅ = {𝑥 ∣ ⊥}
5	4	eqeq2i 2751	. 2 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ {𝑥 ∣ 𝜑} = {𝑥 ∣ ⊥})
6	1, 3, 5	3bitr4ri 304	1 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ ∀𝑥 ¬ 𝜑)

Syntax hints:  ¬ wn 3   ↔ wb 205  ∀wal 1537   = wceq 1539  ⊥wfal 1551  {cab 2715  ∅c0 4256

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-9 2116  ax-10 2137  ax-11 2154  ax-12 2171  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-tru 1542  df-fal 1552  df-ex 1783  df-nf 1787  df-sb 2068  df-clab 2716  df-cleq 2730  df-dif 3890  df-nul 4257

This theorem is referenced by:  dfnf5  4311  abn0  4314  rab0  4316  rabeq0  4318  abfOLD  4337  sticksstones22  40124