Theorem ab0 4315

Description: The class of sets verifying a property is the empty class if and only if that property is a contradiction. See also abn0 4320 (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 2936). (Contributed by BJ, 19-Mar-2021.) Avoid df-clel 2815, ax-8 2121. (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 2809	. 2 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ ⊥} ↔ ∀𝑥(𝜑 ↔ ⊥))
2		dfnul4 4270	. . 3 ⊢ ∅ = {𝑥 ∣ ⊥}
3	2	eqeq2i 2753	. 2 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ {𝑥 ∣ 𝜑} = {𝑥 ∣ ⊥})
4		nbfal 1562	. . 3 ⊢ (¬ 𝜑 ↔ (𝜑 ↔ ⊥))
5	4	albii 1826	. 2 ⊢ (∀𝑥 ¬ 𝜑 ↔ ∀𝑥(𝜑 ↔ ⊥))
6	1, 3, 5	3bitr4i 304	1 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ ∀𝑥 ¬ 𝜑)

Syntax hints:  ¬ wn 3   ↔ wb 207  ∀wal 1545   = wceq 1547  ⊥wfal 1559  {cab 2718  ∅c0 4268

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015  ax-9 2129  ax-10 2152  ax-11 2168  ax-12 2189  ax-ext 2712

This theorem depends on definitions:  df-bi 208  df-an 397  df-or 854  df-tru 1550  df-fal 1560  df-ex 1787  df-nf 1791  df-sb 2074  df-clab 2719  df-cleq 2732  df-dif 3893  df-nul 4269

This theorem is referenced by:  dfnf5  4317  abn0  4320  rab0  4321  rabeq0  4323  sticksstones22  42660