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 2957). (Contributed by BJ, 19-Mar-2021.) Avoid df-clel 2836, ax-8 2143. (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 2830	. 2 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ ⊥} ↔ ∀𝑥(𝜑 ↔ ⊥))
2		dfnul4 4287	. . 3 ⊢ ∅ = {𝑥 ∣ ⊥}
3	2	eqeq2i 2774	. 2 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ {𝑥 ∣ 𝜑} = {𝑥 ∣ ⊥})
4		nbfal 1574	. . 3 ⊢ (¬ 𝜑 ↔ (𝜑 ↔ ⊥))
5	4	albii 1838	. 2 ⊢ (∀𝑥 ¬ 𝜑 ↔ ∀𝑥(𝜑 ↔ ⊥))
6	1, 3, 5	3bitr4i 305	1 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ ∀𝑥 ¬ 𝜑)

Syntax hints:  ¬ wn 3   ↔ wb 208  ∀wal 1557   = wceq 1559  ⊥wfal 1571  {cab 2739  ∅c0 4285

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1814  ax-4 1828  ax-5 1929  ax-6 1986  ax-7 2027  ax-9 2151  ax-10 2174  ax-11 2190  ax-12 2211  ax-ext 2733

This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-tru 1562  df-fal 1572  df-ex 1799  df-nf 1803  df-sb 2090  df-clab 2740  df-cleq 2753  df-dif 3907  df-nul 4286

This theorem is referenced by:  dfnf5  4334  abn0  4337  rab0OLD  4339  rabeq0  4341  sticksstones22  42749