Theorem ab0 4334

Description: The class of sets verifying a property is the empty class if and only if that property is a contradiction. See also abn0 4339 (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 2934). (Contributed by BJ, 19-Mar-2021.) Avoid df-clel 2812, ax-8 2116. (Revised by GG, 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 2806	. 2 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ ⊥} ↔ ∀𝑥(𝜑 ↔ ⊥))
2		dfnul4 4289	. . 3 ⊢ ∅ = {𝑥 ∣ ⊥}
3	2	eqeq2i 2750	. 2 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ {𝑥 ∣ 𝜑} = {𝑥 ∣ ⊥})
4		nbfal 1557	. . 3 ⊢ (¬ 𝜑 ↔ (𝜑 ↔ ⊥))
5	4	albii 1821	. 2 ⊢ (∀𝑥 ¬ 𝜑 ↔ ∀𝑥(𝜑 ↔ ⊥))
6	1, 3, 5	3bitr4i 303	1 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ ∀𝑥 ¬ 𝜑)

Syntax hints:  ¬ wn 3   ↔ wb 206  ∀wal 1540   = wceq 1542  ⊥wfal 1554  {cab 2715  ∅c0 4287

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 1912  ax-6 1969  ax-7 2010  ax-9 2124  ax-10 2147  ax-11 2163  ax-12 2185  ax-ext 2709

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-tru 1545  df-fal 1555  df-ex 1782  df-nf 1786  df-sb 2069  df-clab 2716  df-cleq 2729  df-dif 3906  df-nul 4288

This theorem is referenced by:  dfnf5  4336  abn0  4339  rab0  4340  rabeq0  4342  sticksstones22  42532