Theorem disj1 1734

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

Assertion

Ref	Expression
disj1	⊢ ((A ∩ B) = ∅ ↔ ∀x(x ∈ A → ¬ x ∈ B))

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

Proof of Theorem disj1

Step	Hyp	Ref	Expression
1		disj 1733	. 2 ⊢ ((A ∩ B) = ∅ ↔ ∀x ∈ A ¬ x ∈ B)
2		df-ral 1205	. . 3 ⊢ (∀x ∈ A ¬ x ∈ B ↔ ∀x(x ∈ A → ¬ x ∈ B))
3	2	bicomi 150	. 2 ⊢ (∀x(x ∈ A → ¬ x ∈ B) ↔ ∀x ∈ A ¬ x ∈ B)
4	1, 3	bitr4 154	1 ⊢ ((A ∩ B) = ∅ ↔ ∀x(x ∈ A → ¬ x ∈ B))

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

This theorem is referenced by:  disj2 1735  disj3 1736  undif4 1744  disjsn 1836  funun 2700  erdisj 3223  aceq5lem4 3561

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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