Theorem ab0 4402

Description: The class of sets verifying a property is the empty class if and only if that property is a contradiction. See also abn0 4408 (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 2947). (Contributed by BJ, 19-Mar-2021.) Avoid df-clel 2819, ax-8 2110. (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 2814	. 2 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ ⊥} ↔ ∀𝑥(𝜑 ↔ ⊥))
2		dfnul4 4354	. . 3 ⊢ ∅ = {𝑥 ∣ ⊥}
3	2	eqeq2i 2753	. 2 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ {𝑥 ∣ 𝜑} = {𝑥 ∣ ⊥})
4		nbfal 1552	. . 3 ⊢ (¬ 𝜑 ↔ (𝜑 ↔ ⊥))
5	4	albii 1817	. 2 ⊢ (∀𝑥 ¬ 𝜑 ↔ ∀𝑥(𝜑 ↔ ⊥))
6	1, 3, 5	3bitr4i 303	1 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ ∀𝑥 ¬ 𝜑)

Syntax hints:  ¬ wn 3   ↔ wb 206  ∀wal 1535   = wceq 1537  ⊥wfal 1549  {cab 2717  ∅c0 4352

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-9 2118  ax-10 2141  ax-11 2158  ax-12 2178  ax-ext 2711

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 847  df-tru 1540  df-fal 1550  df-ex 1778  df-nf 1782  df-sb 2065  df-clab 2718  df-cleq 2732  df-dif 3979  df-nul 4353

This theorem is referenced by:  dfnf5  4405  abn0  4408  rab0  4409  rabeq0  4411  sticksstones22  42125