Theorem ab0 4336

Description: The class of sets verifying a property is the empty class if and only if that property is a contradiction. See also abn0 4341 (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 2961). (Contributed by BJ, 19-Mar-2021.) Avoid df-clel 2840, ax-8 2147. (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 2834	. 2 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ ⊥} ↔ ∀𝑥(𝜑 ↔ ⊥))
2		dfnul4 4290	. . 3 ⊢ ∅ = {𝑥 ∣ ⊥}
3	2	eqeq2i 2778	. 2 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ {𝑥 ∣ 𝜑} = {𝑥 ∣ ⊥})
4		nbfal 1578	. . 3 ⊢ (¬ 𝜑 ↔ (𝜑 ↔ ⊥))
5	4	albii 1842	. 2 ⊢ (∀𝑥 ¬ 𝜑 ↔ ∀𝑥(𝜑 ↔ ⊥))
6	1, 3, 5	3bitr4i 306	1 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ ∀𝑥 ¬ 𝜑)

Syntax hints:  ¬ wn 3   ↔ wb 209  ∀wal 1561   = wceq 1563  ⊥wfal 1575  {cab 2743  ∅c0 4288

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1818  ax-4 1832  ax-5 1933  ax-6 1990  ax-7 2031  ax-9 2155  ax-10 2178  ax-11 2194  ax-12 2215  ax-ext 2737

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1566  df-fal 1576  df-ex 1803  df-nf 1807  df-sb 2094  df-clab 2744  df-cleq 2757  df-dif 3910  df-nul 4289

This theorem is referenced by:  dfnf5  4338  abn0  4341  rab0OLD  4343  rabeq0  4345  sticksstones22  42797