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Theorem ab0 3925
Description: The class of sets verifying a property is the empty class if and only if that property is a contradiction. See also abn0 3928 (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 2791). (Contributed by BJ, 19-Mar-2021.)
Assertion
Ref Expression
ab0 ({𝑥𝜑} = ∅ ↔ ∀𝑥 ¬ 𝜑)

Proof of Theorem ab0
StepHypRef Expression
1 nfab1 2763 . . 3 𝑥{𝑥𝜑}
21eq0f 3901 . 2 ({𝑥𝜑} = ∅ ↔ ∀𝑥 ¬ 𝑥 ∈ {𝑥𝜑})
3 abid 2609 . . . 4 (𝑥 ∈ {𝑥𝜑} ↔ 𝜑)
43notbii 310 . . 3 𝑥 ∈ {𝑥𝜑} ↔ ¬ 𝜑)
54albii 1744 . 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 1478   = wceq 1480  wcel 1987  {cab 2607  c0 3891
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1719  ax-4 1734  ax-5 1836  ax-6 1885  ax-7 1932  ax-9 1996  ax-10 2016  ax-11 2031  ax-12 2044  ax-13 2245  ax-ext 2601
This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1483  df-ex 1702  df-nf 1707  df-sb 1878  df-clab 2608  df-cleq 2614  df-clel 2617  df-nfc 2750  df-v 3188  df-dif 3558  df-nul 3892
This theorem is referenced by:  dfnf5  3926  rab0  3929  rabeq0  3931  abf  3950
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