Theorem ab0 4332

Description: The class of sets verifying a property is the empty class if and only if that property is a contradiction. See also abn0 4337 (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 2960). (Contributed by BJ, 19-Mar-2021.) Avoid df-clel 2839, ax-8 2149. (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 4283	. . 3 ⊢ ∅ = {𝑥 ∣ ⊥}
3	2	eqeq2i 2776	. 2 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ {𝑥 ∣ 𝜑} = {𝑥 ∣ ⊥})
4		nbfal 1585	. . 3 ⊢ (¬ 𝜑 ↔ (𝜑 ↔ ⊥))
5	4	albii 1852	. 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 1568   = wceq 1570  ⊥wfal 1582  {cab 2741  ∅c0 4281

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-9 2157  ax-10 2180  ax-11 2197  ax-12 2216  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-fal 1583  df-ex 1813  df-nf 1817  df-sb 2102  df-clab 2742  df-cleq 2755  df-dif 4050  df-nul 4282

This theorem is used by:  dfnf5  4334  abn0  4337  rab0OLD  4339  rabeq0  4341  kardval  35933  sticksstones22  43421