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Theorem dfnf5 4336
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 1787 . 2 (Ⅎ𝑥𝜑 ↔ (∀𝑥𝜑 ∨ ∀𝑥 ¬ 𝜑))
2 abv 3506 . . 3 ({𝑥𝜑} = V ↔ ∀𝑥𝜑)
3 ab0 4335 . . 3 ({𝑥𝜑} = ∅ ↔ ∀𝑥 ¬ 𝜑)
42, 3orbi12i 911 . 2 (({𝑥𝜑} = V ∨ {𝑥𝜑} = ∅) ↔ (∀𝑥𝜑 ∨ ∀𝑥 ¬ 𝜑))
51, 4bitr4i 280 1 (Ⅎ𝑥𝜑 ↔ ({𝑥𝜑} = V ∨ {𝑥𝜑} = ∅))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wb 208  wo 843  wal 1535   = wceq 1537  wnf 1784  {cab 2801  Vcvv 3496  c0 4293
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2116  ax-9 2124  ax-10 2145  ax-11 2161  ax-12 2177  ax-ext 2795
This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-tru 1540  df-ex 1781  df-nf 1785  df-sb 2070  df-clab 2802  df-cleq 2816  df-clel 2895  df-nfc 2965  df-v 3498  df-dif 3941  df-nul 4294
This theorem is referenced by:  ab0orv  4337
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