Theorem ab0 4337

Description: The class of sets verifying a property is the empty class if and only if that property is a contradiction. See also abn0 4342 (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 2959). (Contributed by BJ, 19-Mar-2021.) Avoid df-clel 2838, ax-8 2145. (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 2832	. 2 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ ⊥} ↔ ∀𝑥(𝜑 ↔ ⊥))
2		dfnul4 4289	. . 3 ⊢ ∅ = {𝑥 ∣ ⊥}
3	2	eqeq2i 2776	. 2 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ {𝑥 ∣ 𝜑} = {𝑥 ∣ ⊥})
4		nbfal 1585	. . 3 ⊢ (¬ 𝜑 ↔ (𝜑 ↔ ⊥))
5	4	albii 1849	. 2 ⊢ (∀𝑥 ¬ 𝜑 ↔ ∀𝑥(𝜑 ↔ ⊥))
6	1, 3, 5	3bitr4i 306	1 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ ∀𝑥 ¬ 𝜑)

Syntax hints:  ¬ wn 3   ↔ wb 209  ∀wal 1568   = wceq 1570  ⊥wfal 1582  {cab 2741  ∅c0 4287

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-9 2153  ax-10 2176  ax-11 2192  ax-12 2213  ax-ext 2735

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1573  df-fal 1583  df-ex 1810  df-nf 1814  df-sb 2097  df-clab 2742  df-cleq 2755  df-dif 3909  df-nul 4288

This theorem is referenced by:  dfnf5  4339  abn0  4342  rab0OLD  4344  rabeq0  4346  kardval  35546  sticksstones22  42916