Theorem dfnf5 4388

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 1783	. 2 ⊢ (Ⅎ𝑥𝜑 ↔ (∀𝑥𝜑 ∨ ∀𝑥 ¬ 𝜑))
2		abv 3490	. . 3 ⊢ ({𝑥 ∣ 𝜑} = V ↔ ∀𝑥𝜑)
3		ab0 4386	. . 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 1535   = wceq 1537  Ⅎwnf 1780  {cab 2712  Vcvv 3478  ∅c0 4339

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-9 2116  ax-10 2139  ax-11 2155  ax-12 2175  ax-ext 2706

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1540  df-fal 1550  df-ex 1777  df-nf 1781  df-sb 2063  df-clab 2713  df-cleq 2727  df-v 3480  df-dif 3966  df-nul 4340

This theorem is referenced by:  ab0orvALT  4390