Theorem ab0ALT 4338

Description: Alternate proof of ab0 4337, 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 2927	. . 3 ⊢ Ⅎ𝑥{𝑥 ∣ 𝜑}
2	1	eq0f 4302	. 2 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ ∀𝑥 ¬ 𝑥 ∈ {𝑥 ∣ 𝜑})
3		abid 2745	. . . 4 ⊢ (𝑥 ∈ {𝑥 ∣ 𝜑} ↔ 𝜑)
4	3	notbii 323	. . 3 ⊢ (¬ 𝑥 ∈ {𝑥 ∣ 𝜑} ↔ ¬ 𝜑)
5	4	albii 1849	. 2 ⊢ (∀𝑥 ¬ 𝑥 ∈ {𝑥 ∣ 𝜑} ↔ ∀𝑥 ¬ 𝜑)
6	2, 5	bitri 278	1 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ ∀𝑥 ¬ 𝜑)

Syntax hints:  ¬ wn 3   ↔ wb 209  ∀wal 1568   = wceq 1570   ∈ wcel 2143  {cab 2741  ∅c0 4287

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-8 2145  ax-9 2153  ax-10 2176  ax-11 2192  ax-12 2213  ax-ext 2735

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1573  df-fal 1583  df-ex 1810  df-nf 1814  df-sb 2097  df-clab 2742  df-cleq 2755  df-clel 2838  df-nfc 2912  df-dif 3909  df-nul 4288

This theorem is referenced by: (None)