Theorem dfnf5 4405

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 1784	. 2 ⊢ (Ⅎ𝑥𝜑 ↔ (∀𝑥𝜑 ∨ ∀𝑥 ¬ 𝜑))
2		abv 3500	. . 3 ⊢ ({𝑥 ∣ 𝜑} = V ↔ ∀𝑥𝜑)
3		ab0 4402	. . 3 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ ∀𝑥 ¬ 𝜑)
4	2, 3	orbi12i 913	. 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 846  ∀wal 1535   = wceq 1537  Ⅎwnf 1781  {cab 2717  Vcvv 3488  ∅c0 4352

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-9 2118  ax-10 2141  ax-11 2158  ax-12 2178  ax-ext 2711

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 847  df-tru 1540  df-fal 1550  df-ex 1778  df-nf 1782  df-sb 2065  df-clab 2718  df-cleq 2732  df-v 3490  df-dif 3979  df-nul 4353

This theorem is referenced by:  ab0orvALT  4407