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 2933). (Contributed by BJ, 19-Mar-2021.) Avoid df-clel 2811, ax-8 2115. (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 2805	. 2 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ ⊥} ↔ ∀𝑥(𝜑 ↔ ⊥))
2		dfnul4 4287	. . 3 ⊢ ∅ = {𝑥 ∣ ⊥}
3	2	eqeq2i 2749	. 2 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ {𝑥 ∣ 𝜑} = {𝑥 ∣ ⊥})
4		nbfal 1556	. . 3 ⊢ (¬ 𝜑 ↔ (𝜑 ↔ ⊥))
5	4	albii 1820	. 2 ⊢ (∀𝑥 ¬ 𝜑 ↔ ∀𝑥(𝜑 ↔ ⊥))
6	1, 3, 5	3bitr4i 303	1 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ ∀𝑥 ¬ 𝜑)

Syntax hints:  ¬ wn 3   ↔ wb 206  ∀wal 1539   = wceq 1541  ⊥wfal 1553  {cab 2714  ∅c0 4285

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-9 2123  ax-10 2146  ax-11 2162  ax-12 2184  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1544  df-fal 1554  df-ex 1781  df-nf 1785  df-sb 2068  df-clab 2715  df-cleq 2728  df-dif 3904  df-nul 4286

This theorem is referenced by:  dfnf5  4334  abn0  4337  rab0  4338  rabeq0  4340  sticksstones22  42418