Theorem disj4 1737

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

Assertion

Ref	Expression
disj4	⊢ ((A ∩ B) = ∅ ↔ ¬ (A ∖ B) ⊂ A)

Proof of Theorem disj4

Step	Hyp	Ref	Expression
1		disj3 1736	. 2 ⊢ ((A ∩ B) = ∅ ↔ A = (A ∖ B))
2		cleqcom 1103	. 2 ⊢ (A = (A ∖ B) ↔ (A ∖ B) = A)
3		dfpss2 1557	. . . 4 ⊢ ((A ∖ B) ⊂ A ↔ ((A ∖ B) ⊆ A ∧ ¬ (A ∖ B) = A))
4		difss 1596	. . . 4 ⊢ (A ∖ B) ⊆ A
5	3, 4	mpbiran 547	. . 3 ⊢ ((A ∖ B) ⊂ A ↔ ¬ (A ∖ B) = A)
6	5	bicon2i 194	. 2 ⊢ ((A ∖ B) = A ↔ ¬ (A ∖ B) ⊂ A)
7	1, 2, 6	3bitr 155	1 ⊢ ((A ∩ B) = ∅ ↔ ¬ (A ∖ B) ⊂ A)

Syntax hints:  ¬ wn 1   ↔ wb 127   = wceq 1091   ∖ cdif 1484   ∩ cin 1486   ⊆ wss 1487   ⊂ wpss 1488  ∅c0 1707

This theorem is referenced by:  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-ne 1192  df-ral 1205  df-v 1349  df-dif 1489  df-in 1491  df-ss 1492  df-pss 1494  df-nul 1708

metamath.org