Theorem dfnf5 4335

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 1806	. 2 ⊢ (Ⅎ𝑥𝜑 ↔ (∀𝑥𝜑 ∨ ∀𝑥 ¬ 𝜑))
2		abv 3466	. . 3 ⊢ ({𝑥 ∣ 𝜑} = V ↔ ∀𝑥𝜑)
3		ab0 4333	. . 3 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ ∀𝑥 ¬ 𝜑)
4	2, 3	orbi12i 925	. 2 ⊢ (({𝑥 ∣ 𝜑} = V ∨ {𝑥 ∣ 𝜑} = ∅) ↔ (∀𝑥𝜑 ∨ ∀𝑥 ¬ 𝜑))
5	1, 4	bitr4i 280	1 ⊢ (Ⅎ𝑥𝜑 ↔ ({𝑥 ∣ 𝜑} = V ∨ {𝑥 ∣ 𝜑} = ∅))

Syntax hints:  ¬ wn 3   ↔ wb 208   ∨ wo 858  ∀wal 1558   = wceq 1560  Ⅎwnf 1803  {cab 2740  Vcvv 3454  ∅c0 4285

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1815  ax-4 1829  ax-5 1930  ax-6 1987  ax-7 2028  ax-9 2152  ax-10 2175  ax-11 2191  ax-12 2212  ax-ext 2734

This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-tru 1563  df-fal 1573  df-ex 1800  df-nf 1804  df-sb 2091  df-clab 2741  df-cleq 2754  df-v 3456  df-dif 3907  df-nul 4286

This theorem is referenced by:  ab0orvALT  4337