Theorem ab0orvALT 4366

Description: Alternate proof of ab0orv 4365, shorter but using more axioms. (Contributed by Mario Carneiro, 29-Aug-2013.) (Revised by BJ, 22-Mar-2020.) (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

Ref	Expression
ab0orvALT	⊢ ({𝑥 ∣ 𝜑} = V ∨ {𝑥 ∣ 𝜑} = ∅)

Distinct variable group:   𝜑,𝑥

Proof of Theorem ab0orvALT

Step	Hyp	Ref	Expression
1		nfv 1913	. 2 ⊢ Ⅎ𝑥𝜑
2		dfnf5 4364	. 2 ⊢ (Ⅎ𝑥𝜑 ↔ ({𝑥 ∣ 𝜑} = V ∨ {𝑥 ∣ 𝜑} = ∅))
3	1, 2	mpbi 230	1 ⊢ ({𝑥 ∣ 𝜑} = V ∨ {𝑥 ∣ 𝜑} = ∅)

Syntax hints:   ∨ wo 847   = wceq 1539  Ⅎwnf 1782  {cab 2712  Vcvv 3464  ∅c0 4315

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1794  ax-4 1808  ax-5 1909  ax-6 1966  ax-7 2006  ax-9 2117  ax-10 2140  ax-11 2156  ax-12 2176  ax-ext 2706

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1542  df-fal 1552  df-ex 1779  df-nf 1783  df-sb 2064  df-clab 2713  df-cleq 2726  df-v 3466  df-dif 3936  df-nul 4316

This theorem is referenced by: (None)