Theorem ab0ALT 4330

Description: Alternate proof of ab0 4329, 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 2925	. . 3 ⊢ Ⅎ𝑥{𝑥 ∣ 𝜑}
2	1	eq0f 4294	. 2 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ ∀𝑥 ¬ 𝑥 ∈ {𝑥 ∣ 𝜑})
3		abid 2743	. . . 4 ⊢ (𝑥 ∈ {𝑥 ∣ 𝜑} ↔ 𝜑)
4	3	notbii 323	. . 3 ⊢ (¬ 𝑥 ∈ {𝑥 ∣ 𝜑} ↔ ¬ 𝜑)
5	4	albii 1852	. 2 ⊢ (∀𝑥 ¬ 𝑥 ∈ {𝑥 ∣ 𝜑} ↔ ∀𝑥 ¬ 𝜑)
6	2, 5	bitri 278	1 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ ∀𝑥 ¬ 𝜑)

This proof depends on syntax axioms:  ¬ wn 3   ↔ wb 209  ∀wal 1568   = wceq 1570   ∈ wcel 2145  {cab 2739  ∅c0 4279

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-10 2178  ax-11 2194  ax-12 2213  ax-ext 2733

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-fal 1583  df-ex 1813  df-nf 1817  df-sb 2100  df-clab 2740  df-cleq 2753  df-clel 2836  df-nfc 2910  df-dif 3902  df-nul 4280

This theorem is used by: (None)