Theorem disj 1733

Description: Two ways of saying that two classes are disjoint (have no members in common).

Assertion

Ref	Expression
disj	⊢ ((A ∩ B) = ∅ ↔ ∀x ∈ A ¬ x ∈ B)

Distinct variable group(s):   x,A   x,B

Proof of Theorem disj

Step	Hyp	Ref	Expression
1		df-in 1491	. . . 4 ⊢ (A ∩ B) = {x∣(x ∈ A ∧ x ∈ B)}
2	1	cleq1i 1108	. . 3 ⊢ ((A ∩ B) = ∅ ↔ {x∣(x ∈ A ∧ x ∈ B)} = ∅)
3		cleqabr 1175	. . 3 ⊢ ({x∣(x ∈ A ∧ x ∈ B)} = ∅ ↔ ∀x((x ∈ A ∧ x ∈ B) ↔ x ∈ ∅))
4		imnan 207	. . . . 5 ⊢ ((x ∈ A → ¬ x ∈ B) ↔ ¬ (x ∈ A ∧ x ∈ B))
5		noel 1711	. . . . . 6 ⊢ ¬ x ∈ ∅
6	5	nbn 542	. . . . 5 ⊢ (¬ (x ∈ A ∧ x ∈ B) ↔ ((x ∈ A ∧ x ∈ B) ↔ x ∈ ∅))
7	4, 6	bitr2 152	. . . 4 ⊢ (((x ∈ A ∧ x ∈ B) ↔ x ∈ ∅) ↔ (x ∈ A → ¬ x ∈ B))
8	7	bial 695	. . 3 ⊢ (∀x((x ∈ A ∧ x ∈ B) ↔ x ∈ ∅) ↔ ∀x(x ∈ A → ¬ x ∈ B))
9	2, 3, 8	3bitr 155	. 2 ⊢ ((A ∩ B) = ∅ ↔ ∀x(x ∈ A → ¬ x ∈ B))
10		df-ral 1205	. 2 ⊢ (∀x ∈ A ¬ x ∈ B ↔ ∀x(x ∈ A → ¬ x ∈ B))
11	9, 10	bitr4 154	1 ⊢ ((A ∩ B) = ∅ ↔ ∀x ∈ A ¬ x ∈ B)

Syntax hints:  ¬ wn 1   → wi 2   ↔ wb 127   ∧ wa 196  ∀wal 672  {cab 1090   = wceq 1091   ∈ wcel 1092  ∀wral 1201   ∩ cin 1486  ∅c0 1707

This theorem is referenced by:  disj1 1734  dffr2 2171  onint 2261  onxpdisj 2476  tfrlem10 2958  zfreg 3447  zfreg2 3448  aceq5 3563  kmlem4 3583

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-ral 1205  df-v 1349  df-dif 1489  df-in 1491  df-nul 1708

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