Theorem ab0 4260

Description: The class of sets verifying a property is the empty class if and only if that property is a contradiction. See also abn0 4266 (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 2935). (Contributed by BJ, 19-Mar-2021.) Avoid df-clel 2811, ax-8 2115. (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 2805	. 2 ⊢ (∀𝑥(𝜑 ↔ ⊥) ↔ {𝑥 ∣ 𝜑} = {𝑥 ∣ ⊥})
2		nbfal 1557	. . 3 ⊢ (¬ 𝜑 ↔ (𝜑 ↔ ⊥))
3	2	albii 1826	. 2 ⊢ (∀𝑥 ¬ 𝜑 ↔ ∀𝑥(𝜑 ↔ ⊥))
4		dfnul4 4211	. . 3 ⊢ ∅ = {𝑥 ∣ ⊥}
5	4	eqeq2i 2751	. 2 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ {𝑥 ∣ 𝜑} = {𝑥 ∣ ⊥})
6	1, 3, 5	3bitr4ri 307	1 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ ∀𝑥 ¬ 𝜑)

Syntax hints:  ¬ wn 3   ↔ wb 209  ∀wal 1540   = wceq 1542  ⊥wfal 1554  {cab 2716  ∅c0 4209

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1916  ax-6 1974  ax-7 2019  ax-9 2123  ax-10 2144  ax-11 2161  ax-12 2178  ax-ext 2710

This theorem depends on definitions:  df-bi 210  df-an 400  df-or 847  df-tru 1545  df-fal 1555  df-ex 1787  df-nf 1791  df-sb 2074  df-clab 2717  df-cleq 2730  df-dif 3844  df-nul 4210

This theorem is referenced by:  dfnf5  4263  abn0  4266  rab0  4268  rabeq0  4270  abfOLD  4289  0mpo0  7245  fsetdmprc0  8458