Theorem ab0 4305

Description: The class of sets verifying a property is the empty class if and only if that property is a contradiction. See also abn0 4311 (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 2943). (Contributed by BJ, 19-Mar-2021.) Avoid df-clel 2817, ax-8 2110. (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 2811	. 2 ⊢ (∀𝑥(𝜑 ↔ ⊥) ↔ {𝑥 ∣ 𝜑} = {𝑥 ∣ ⊥})
2		nbfal 1554	. . 3 ⊢ (¬ 𝜑 ↔ (𝜑 ↔ ⊥))
3	2	albii 1823	. 2 ⊢ (∀𝑥 ¬ 𝜑 ↔ ∀𝑥(𝜑 ↔ ⊥))
4		dfnul4 4255	. . 3 ⊢ ∅ = {𝑥 ∣ ⊥}
5	4	eqeq2i 2751	. 2 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ {𝑥 ∣ 𝜑} = {𝑥 ∣ ⊥})
6	1, 3, 5	3bitr4ri 303	1 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ ∀𝑥 ¬ 𝜑)

Syntax hints:  ¬ wn 3   ↔ wb 205  ∀wal 1537   = wceq 1539  ⊥wfal 1551  {cab 2715  ∅c0 4253

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-9 2118  ax-10 2139  ax-11 2156  ax-12 2173  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-tru 1542  df-fal 1552  df-ex 1784  df-nf 1788  df-sb 2069  df-clab 2716  df-cleq 2730  df-dif 3886  df-nul 4254

This theorem is referenced by:  dfnf5  4308  abn0  4311  rab0  4313  rabeq0  4315  abfOLD  4334  sticksstones22  40052