Theorem ab0 4366

Description: The class of sets verifying a property is the empty class if and only if that property is a contradiction. See also abn0 4372 (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 2933). (Contributed by BJ, 19-Mar-2021.) Avoid df-clel 2802, ax-8 2100. (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 2796	. 2 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ ⊥} ↔ ∀𝑥(𝜑 ↔ ⊥))
2		dfnul4 4316	. . 3 ⊢ ∅ = {𝑥 ∣ ⊥}
3	2	eqeq2i 2737	. 2 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ {𝑥 ∣ 𝜑} = {𝑥 ∣ ⊥})
4		nbfal 1548	. . 3 ⊢ (¬ 𝜑 ↔ (𝜑 ↔ ⊥))
5	4	albii 1813	. 2 ⊢ (∀𝑥 ¬ 𝜑 ↔ ∀𝑥(𝜑 ↔ ⊥))
6	1, 3, 5	3bitr4i 303	1 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ ∀𝑥 ¬ 𝜑)

Syntax hints:  ¬ wn 3   ↔ wb 205  ∀wal 1531   = wceq 1533  ⊥wfal 1545  {cab 2701  ∅c0 4314

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-9 2108  ax-10 2129  ax-11 2146  ax-12 2163  ax-ext 2695

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 845  df-tru 1536  df-fal 1546  df-ex 1774  df-nf 1778  df-sb 2060  df-clab 2702  df-cleq 2716  df-dif 3943  df-nul 4315

This theorem is referenced by:  dfnf5  4369  abn0  4372  rab0  4374  rabeq0  4376  abfOLD  4395  sticksstones22  41477