Theorem dfnf5 4329

Description: Characterization of nonfreeness in a formula in terms of its extension. (Contributed by BJ, 19-Mar-2021.)

Assertion

Ref	Expression
dfnf5	⊢ (Ⅎ𝑥𝜑 ↔ ({𝑥 ∣ 𝜑} = V ∨ {𝑥 ∣ 𝜑} = ∅))

Proof of Theorem dfnf5

Step	Hyp	Ref	Expression
1		nf3 1787	. 2 ⊢ (Ⅎ𝑥𝜑 ↔ (∀𝑥𝜑 ∨ ∀𝑥 ¬ 𝜑))
2		abv 3448	. . 3 ⊢ ({𝑥 ∣ 𝜑} = V ↔ ∀𝑥𝜑)
3		ab0 4327	. . 3 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ ∀𝑥 ¬ 𝜑)
4	2, 3	orbi12i 914	. 2 ⊢ (({𝑥 ∣ 𝜑} = V ∨ {𝑥 ∣ 𝜑} = ∅) ↔ (∀𝑥𝜑 ∨ ∀𝑥 ¬ 𝜑))
5	1, 4	bitr4i 278	1 ⊢ (Ⅎ𝑥𝜑 ↔ ({𝑥 ∣ 𝜑} = V ∨ {𝑥 ∣ 𝜑} = ∅))

Syntax hints:  ¬ wn 3   ↔ wb 206   ∨ wo 847  ∀wal 1539   = wceq 1541  Ⅎwnf 1784  {cab 2709  Vcvv 3436  ∅c0 4280

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 1968  ax-7 2009  ax-9 2121  ax-10 2144  ax-11 2160  ax-12 2180  ax-ext 2703

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1544  df-fal 1554  df-ex 1781  df-nf 1785  df-sb 2068  df-clab 2710  df-cleq 2723  df-v 3438  df-dif 3900  df-nul 4281

This theorem is referenced by:  ab0orvALT  4331