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Theorem ab0 4109
Description: The class of sets verifying a property is the empty class if and only if that property is a contradiction. See also abn0 4112 (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.)
Assertion
Ref Expression
ab0 ({𝑥𝜑} = ∅ ↔ ∀𝑥 ¬ 𝜑)

Proof of Theorem ab0
StepHypRef Expression
1 nfab1 2918 . . 3 𝑥{𝑥𝜑}
21eq0f 4084 . 2 ({𝑥𝜑} = ∅ ↔ ∀𝑥 ¬ 𝑥 ∈ {𝑥𝜑})
3 abid 2762 . . . 4 (𝑥 ∈ {𝑥𝜑} ↔ 𝜑)
43notbii 310 . . 3 𝑥 ∈ {𝑥𝜑} ↔ ¬ 𝜑)
54albii 1898 . 2 (∀𝑥 ¬ 𝑥 ∈ {𝑥𝜑} ↔ ∀𝑥 ¬ 𝜑)
62, 5bitri 265 1 ({𝑥𝜑} = ∅ ↔ ∀𝑥 ¬ 𝜑)
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wb 197  wal 1632   = wceq 1634  wcel 2148  {cab 2760  c0 4073
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1873  ax-4 1888  ax-5 1994  ax-6 2060  ax-7 2096  ax-9 2157  ax-10 2177  ax-11 2193  ax-12 2206  ax-13 2411  ax-ext 2754
This theorem depends on definitions:  df-bi 198  df-an 384  df-or 864  df-tru 1637  df-ex 1856  df-nf 1861  df-sb 2053  df-clab 2761  df-cleq 2767  df-clel 2770  df-nfc 2905  df-v 3357  df-dif 3732  df-nul 4074
This theorem is referenced by:  dfnf5  4110  rab0  4113  rabeq0  4114  abf  4133
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