Theorem dfnf5 4370

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 1780	. 2 ⊢ (Ⅎ𝑥𝜑 ↔ (∀𝑥𝜑 ∨ ∀𝑥 ¬ 𝜑))
2		abv 3477	. . 3 ⊢ ({𝑥 ∣ 𝜑} = V ↔ ∀𝑥𝜑)
3		ab0 4367	. . 3 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ ∀𝑥 ¬ 𝜑)
4	2, 3	orbi12i 911	. 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 205   ∨ wo 844  ∀wal 1531   = wceq 1533  Ⅎwnf 1777  {cab 2701  Vcvv 3466  ∅c0 4315

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-9 2108  ax-10 2129  ax-11 2146  ax-12 2163  ax-ext 2695

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 845  df-tru 1536  df-fal 1546  df-ex 1774  df-nf 1778  df-sb 2060  df-clab 2702  df-cleq 2716  df-v 3468  df-dif 3944  df-nul 4316

This theorem is referenced by:  ab0orvALT  4372