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Theorem dfnf5 4119
Description: Characterization of non-freeness in a formula in terms of its extension. (Contributed by BJ, 19-Mar-2021.)
Assertion
Ref Expression
dfnf5 (Ⅎ𝑥𝜑 ↔ ({𝑥𝜑} = V ∨ {𝑥𝜑} = ∅))

Proof of Theorem dfnf5
StepHypRef Expression
1 nf3 1881 . 2 (Ⅎ𝑥𝜑 ↔ (∀𝑥𝜑 ∨ ∀𝑥 ¬ 𝜑))
2 abv 3359 . . 3 ({𝑥𝜑} = V ↔ ∀𝑥𝜑)
3 ab0 4118 . . 3 ({𝑥𝜑} = ∅ ↔ ∀𝑥 ¬ 𝜑)
42, 3orbi12i 938 . 2 (({𝑥𝜑} = V ∨ {𝑥𝜑} = ∅) ↔ (∀𝑥𝜑 ∨ ∀𝑥 ¬ 𝜑))
51, 4bitr4i 269 1 (Ⅎ𝑥𝜑 ↔ ({𝑥𝜑} = V ∨ {𝑥𝜑} = ∅))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wb 197  wo 873  wal 1650   = wceq 1652  wnf 1878  {cab 2751  Vcvv 3350  c0 4081
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1890  ax-4 1904  ax-5 2005  ax-6 2070  ax-7 2105  ax-9 2164  ax-10 2183  ax-11 2198  ax-12 2211  ax-13 2352  ax-ext 2743
This theorem depends on definitions:  df-bi 198  df-an 385  df-or 874  df-tru 1656  df-ex 1875  df-nf 1879  df-sb 2063  df-clab 2752  df-cleq 2758  df-clel 2761  df-nfc 2896  df-v 3352  df-dif 3737  df-nul 4082
This theorem is referenced by:  ab0orv  4120
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