Theorem ab0 4328

Description: The class of sets verifying a property is the empty class if and only if that property is a contradiction. See also abn0 4333 (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 2956). (Contributed by BJ, 19-Mar-2021.) Avoid df-clel 2835, ax-8 2146. (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 2829	. 2 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ ⊥} ↔ ∀𝑥(𝜑 ↔ ⊥))
2		dfnul4 4280	. . 3 ⊢ ∅ = {𝑥 ∣ ⊥}
3	2	eqeq2i 2773	. 2 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ {𝑥 ∣ 𝜑} = {𝑥 ∣ ⊥})
4		nbfal 1584	. . 3 ⊢ (¬ 𝜑 ↔ (𝜑 ↔ ⊥))
5	4	albii 1851	. 2 ⊢ (∀𝑥 ¬ 𝜑 ↔ ∀𝑥(𝜑 ↔ ⊥))
6	1, 3, 5	3bitr4i 306	1 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ ∀𝑥 ¬ 𝜑)

This proof depends on syntax axioms:  ¬ wn 3   ↔ wb 209  ∀wal 1567   = wceq 1569  ⊥wfal 1581  {cab 2738  ∅c0 4278

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-9 2154  ax-10 2177  ax-11 2193  ax-12 2212  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1572  df-fal 1582  df-ex 1812  df-nf 1816  df-sb 2099  df-clab 2739  df-cleq 2752  df-dif 4047  df-nul 4279

This theorem is used by:  dfnf5  4330  abn0  4333  rab0OLD  4335  rabeq0  4337  kardval  35876  sticksstones22  43311