Theorem disj3 1736

Description: Two ways of saying that two classes are disjoint.

Assertion

Ref	Expression
disj3	⊢ ((A ∩ B) = ∅ ↔ A = (A ∖ B))

Proof of Theorem disj3

Step	Hyp	Ref	Expression
1		pm4.71 481	. . . 4 ⊢ ((x ∈ A → ¬ x ∈ B) ↔ (x ∈ A ↔ (x ∈ A ∧ ¬ x ∈ B)))
2		eldif 1496	. . . . 5 ⊢ (x ∈ (A ∖ B) ↔ (x ∈ A ∧ ¬ x ∈ B))
3	2	bibi2i 460	. . . 4 ⊢ ((x ∈ A ↔ x ∈ (A ∖ B)) ↔ (x ∈ A ↔ (x ∈ A ∧ ¬ x ∈ B)))
4	1, 3	bitr4 154	. . 3 ⊢ ((x ∈ A → ¬ x ∈ B) ↔ (x ∈ A ↔ x ∈ (A ∖ B)))
5	4	bial 695	. 2 ⊢ (∀x(x ∈ A → ¬ x ∈ B) ↔ ∀x(x ∈ A ↔ x ∈ (A ∖ B)))
6		disj1 1734	. 2 ⊢ ((A ∩ B) = ∅ ↔ ∀x(x ∈ A → ¬ x ∈ B))
7		dfcleq 1098	. 2 ⊢ (A = (A ∖ B) ↔ ∀x(x ∈ A ↔ x ∈ (A ∖ B)))
8	5, 6, 7	3bitr4 158	1 ⊢ ((A ∩ B) = ∅ ↔ A = (A ∖ B))

Syntax hints:  ¬ wn 1   → wi 2   ↔ wb 127   ∧ wa 196  ∀wal 672   = wceq 1091   ∈ wcel 1092   ∖ cdif 1484   ∩ cin 1486  ∅c0 1707

This theorem is referenced by:  disj4 1737  orddif 2326  php 3409  inf5 3472

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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