Theorem dfnul3 1710

Description: Alternate definition of the empty set..

Assertion

Ref	Expression
dfnul3	⊢ ∅ = {x ∈ A∣ ¬ x ∈ A}

Proof of Theorem dfnul3

Step	Hyp	Ref	Expression
1		pm3.24 496	. . . . 5 ⊢ ¬ (x ∈ A ∧ ¬ x ∈ A)
2		cleqid 1102	. . . . 5 ⊢ x = x
3	1, 2	2th 540	. . . 4 ⊢ (¬ (x ∈ A ∧ ¬ x ∈ A) ↔ x = x)
4	3	bicon1i 193	. . 3 ⊢ (¬ x = x ↔ (x ∈ A ∧ ¬ x ∈ A))
5	4	biabi 1181	. 2 ⊢ {x∣ ¬ x = x} = {x∣(x ∈ A ∧ ¬ x ∈ A)}
6		dfnul2 1709	. 2 ⊢ ∅ = {x∣ ¬ x = x}
7		df-rab 1208	. 2 ⊢ {x ∈ A∣ ¬ x ∈ A} = {x∣(x ∈ A ∧ ¬ x ∈ A)}
8	5, 6, 7	3eqtr4 1126	1 ⊢ ∅ = {x ∈ A∣ ¬ x ∈ A}

Syntax hints:  ¬ wn 1   ∧ wa 196   = weq 797  {cab 1090   = wceq 1091   ∈ wcel 1092  {crab 1204  ∅c0 1707

This theorem is referenced by:  kmlem3 3582

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-rab 1208  df-v 1349  df-dif 1489  df-nul 1708

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