Theorem dfnul3OLD 4358

Description: Obsolete version of dfnul4 4354 as of 23-Sep-2024. (Contributed by NM, 25-Mar-2004.) (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

Ref	Expression
dfnul3OLD	⊢ ∅ = {𝑥 ∈ 𝐴 ∣ ¬ 𝑥 ∈ 𝐴}

Proof of Theorem dfnul3OLD

Step	Hyp	Ref	Expression
1		pm3.24 402	. . . . 5 ⊢ ¬ (𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐴)
2		equid 2011	. . . . 5 ⊢ 𝑥 = 𝑥
3	1, 2	2th 264	. . . 4 ⊢ (¬ (𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐴) ↔ 𝑥 = 𝑥)
4	3	con1bii 356	. . 3 ⊢ (¬ 𝑥 = 𝑥 ↔ (𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐴))
5	4	abbii 2812	. 2 ⊢ {𝑥 ∣ ¬ 𝑥 = 𝑥} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐴)}
6		dfnul2 4355	. 2 ⊢ ∅ = {𝑥 ∣ ¬ 𝑥 = 𝑥}
7		df-rab 3444	. 2 ⊢ {𝑥 ∈ 𝐴 ∣ ¬ 𝑥 ∈ 𝐴} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐴)}
8	5, 6, 7	3eqtr4i 2778	1 ⊢ ∅ = {𝑥 ∈ 𝐴 ∣ ¬ 𝑥 ∈ 𝐴}

Syntax hints:  ¬ wn 3   ∧ wa 395   = wceq 1537   ∈ wcel 2108  {cab 2717  {crab 3443  ∅c0 4352

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-9 2118  ax-ext 2711

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1540  df-fal 1550  df-ex 1778  df-sb 2065  df-clab 2718  df-cleq 2732  df-rab 3444  df-dif 3979  df-nul 4353

This theorem is referenced by: (None)