Theorem ab0 4332

Description: The class of sets verifying a property is the empty class if and only if that property is a contradiction. See also abn0 4335 (from which it could be proved using as many essential proof steps but one fewer syntactic step, at the cost of depending on df-ne 3017). (Contributed by BJ, 19-Mar-2021.)

Assertion

Ref	Expression
ab0	⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ ∀𝑥 ¬ 𝜑)

Proof of Theorem ab0

Step	Hyp	Ref	Expression
1		nfab1 2979	. . 3 ⊢ Ⅎ𝑥{𝑥 ∣ 𝜑}
2	1	eq0f 4304	. 2 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ ∀𝑥 ¬ 𝑥 ∈ {𝑥 ∣ 𝜑})
3		abid 2803	. . . 4 ⊢ (𝑥 ∈ {𝑥 ∣ 𝜑} ↔ 𝜑)
4	3	notbii 322	. . 3 ⊢ (¬ 𝑥 ∈ {𝑥 ∣ 𝜑} ↔ ¬ 𝜑)
5	4	albii 1816	. 2 ⊢ (∀𝑥 ¬ 𝑥 ∈ {𝑥 ∣ 𝜑} ↔ ∀𝑥 ¬ 𝜑)
6	2, 5	bitri 277	1 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ ∀𝑥 ¬ 𝜑)

Syntax hints:  ¬ wn 3   ↔ wb 208  ∀wal 1531   = wceq 1533   ∈ wcel 2110  {cab 2799  ∅c0 4290

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1907  ax-6 1966  ax-7 2011  ax-8 2112  ax-9 2120  ax-10 2141  ax-11 2157  ax-12 2173  ax-ext 2793

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-tru 1536  df-ex 1777  df-nf 1781  df-sb 2066  df-clab 2800  df-cleq 2814  df-clel 2893  df-nfc 2963  df-dif 3938  df-nul 4291

This theorem is referenced by:  dfnf5  4333  rab0  4336  rabeq0  4337  abf  4355  0mpo0  7236