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Theorem ab0 3984
Description: The class of sets verifying a property is the empty class if and only if that property is a contradiction. See also abn0 3987 (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 2824). (Contributed by BJ, 19-Mar-2021.)
Assertion
Ref Expression
ab0 ({𝑥𝜑} = ∅ ↔ ∀𝑥 ¬ 𝜑)

Proof of Theorem ab0
StepHypRef Expression
1 nfab1 2795 . . 3 𝑥{𝑥𝜑}
21eq0f 3958 . 2 ({𝑥𝜑} = ∅ ↔ ∀𝑥 ¬ 𝑥 ∈ {𝑥𝜑})
3 abid 2639 . . . 4 (𝑥 ∈ {𝑥𝜑} ↔ 𝜑)
43notbii 309 . . 3 𝑥 ∈ {𝑥𝜑} ↔ ¬ 𝜑)
54albii 1787 . 2 (∀𝑥 ¬ 𝑥 ∈ {𝑥𝜑} ↔ ∀𝑥 ¬ 𝜑)
62, 5bitri 264 1 ({𝑥𝜑} = ∅ ↔ ∀𝑥 ¬ 𝜑)
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wb 196  wal 1521   = wceq 1523  wcel 2030  {cab 2637  c0 3948
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1762  ax-4 1777  ax-5 1879  ax-6 1945  ax-7 1981  ax-9 2039  ax-10 2059  ax-11 2074  ax-12 2087  ax-13 2282  ax-ext 2631
This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-tru 1526  df-ex 1745  df-nf 1750  df-sb 1938  df-clab 2638  df-cleq 2644  df-clel 2647  df-nfc 2782  df-v 3233  df-dif 3610  df-nul 3949
This theorem is referenced by:  dfnf5  3985  rab0  3988  rabeq0  3990  abf  4011
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