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Theorem ab0 4370
Description: The class of sets verifying a property is the empty class if and only if that property is a contradiction. See also abn0 4376 (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 2805, ax-8 2101. (Revised by Gino Giotto, 30-Aug-2024.) (Proof shortened by SN, 8-Sep-2024.)
Assertion
Ref Expression
ab0 ({𝑥𝜑} = ∅ ↔ ∀𝑥 ¬ 𝜑)

Proof of Theorem ab0
StepHypRef Expression
1 abbib 2799 . 2 ({𝑥𝜑} = {𝑥 ∣ ⊥} ↔ ∀𝑥(𝜑 ↔ ⊥))
2 dfnul4 4320 . . 3 ∅ = {𝑥 ∣ ⊥}
32eqeq2i 2740 . 2 ({𝑥𝜑} = ∅ ↔ {𝑥𝜑} = {𝑥 ∣ ⊥})
4 nbfal 1549 . . 3 𝜑 ↔ (𝜑 ↔ ⊥))
54albii 1814 . 2 (∀𝑥 ¬ 𝜑 ↔ ∀𝑥(𝜑 ↔ ⊥))
61, 3, 53bitr4i 303 1 ({𝑥𝜑} = ∅ ↔ ∀𝑥 ¬ 𝜑)
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wb 205  wal 1532   = wceq 1534  wfal 1546  {cab 2704  c0 4318
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1906  ax-6 1964  ax-7 2004  ax-9 2109  ax-10 2130  ax-11 2147  ax-12 2164  ax-ext 2698
This theorem depends on definitions:  df-bi 206  df-an 396  df-or 847  df-tru 1537  df-fal 1547  df-ex 1775  df-nf 1779  df-sb 2061  df-clab 2705  df-cleq 2719  df-dif 3947  df-nul 4319
This theorem is referenced by:  dfnf5  4373  abn0  4376  rab0  4378  rabeq0  4380  abfOLD  4399  sticksstones22  41624
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