Theorem ab0ALT 4375

Description: Alternate proof of ab0 4373, shorter but using more axioms. (Contributed by BJ, 19-Mar-2021.) (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

Ref	Expression
ab0ALT	⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ ∀𝑥 ¬ 𝜑)

Proof of Theorem ab0ALT

Step	Hyp	Ref	Expression
1		nfab1 2905	. . 3 ⊢ Ⅎ𝑥{𝑥 ∣ 𝜑}
2	1	eq0f 4339	. 2 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ ∀𝑥 ¬ 𝑥 ∈ {𝑥 ∣ 𝜑})
3		abid 2713	. . . 4 ⊢ (𝑥 ∈ {𝑥 ∣ 𝜑} ↔ 𝜑)
4	3	notbii 319	. . 3 ⊢ (¬ 𝑥 ∈ {𝑥 ∣ 𝜑} ↔ ¬ 𝜑)
5	4	albii 1821	. 2 ⊢ (∀𝑥 ¬ 𝑥 ∈ {𝑥 ∣ 𝜑} ↔ ∀𝑥 ¬ 𝜑)
6	2, 5	bitri 274	1 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ ∀𝑥 ¬ 𝜑)

Syntax hints:  ¬ wn 3   ↔ wb 205  ∀wal 1539   = wceq 1541   ∈ wcel 2106  {cab 2709  ∅c0 4321

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-10 2137  ax-11 2154  ax-12 2171  ax-ext 2703

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 846  df-tru 1544  df-fal 1554  df-ex 1782  df-nf 1786  df-sb 2068  df-clab 2710  df-cleq 2724  df-clel 2810  df-nfc 2885  df-dif 3950  df-nul 4322

This theorem is referenced by: (None)