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Theorem ab0 4159
Description: The class of sets verifying a property is the empty class if and only if that property is a contradiction. See also abn0 4162 (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 2986). (Contributed by BJ, 19-Mar-2021.)
Assertion
Ref Expression
ab0 ({𝑥𝜑} = ∅ ↔ ∀𝑥 ¬ 𝜑)

Proof of Theorem ab0
StepHypRef Expression
1 nfab1 2957 . . 3 𝑥{𝑥𝜑}
21eq0f 4134 . 2 ({𝑥𝜑} = ∅ ↔ ∀𝑥 ¬ 𝑥 ∈ {𝑥𝜑})
3 abid 2801 . . . 4 (𝑥 ∈ {𝑥𝜑} ↔ 𝜑)
43notbii 311 . . 3 𝑥 ∈ {𝑥𝜑} ↔ ¬ 𝜑)
54albii 1904 . 2 (∀𝑥 ¬ 𝑥 ∈ {𝑥𝜑} ↔ ∀𝑥 ¬ 𝜑)
62, 5bitri 266 1 ({𝑥𝜑} = ∅ ↔ ∀𝑥 ¬ 𝜑)
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wb 197  wal 1635   = wceq 1637  wcel 2157  {cab 2799  c0 4123
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1877  ax-4 1894  ax-5 2001  ax-6 2069  ax-7 2105  ax-9 2166  ax-10 2186  ax-11 2202  ax-12 2215  ax-13 2422  ax-ext 2791
This theorem depends on definitions:  df-bi 198  df-an 385  df-or 866  df-tru 1641  df-ex 1860  df-nf 1864  df-sb 2062  df-clab 2800  df-cleq 2806  df-clel 2809  df-nfc 2944  df-v 3400  df-dif 3779  df-nul 4124
This theorem is referenced by:  dfnf5  4160  rab0  4163  rabeq0  4164  abf  4183
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