Theorem disj3

Description: Two ways of saying that two classes are disjoint. (Contributed by NM, 19-May-1998)

	Ref	Expression
Assertion	disj3	|- ( ( A i^i B ) = (/) <-> A = ( A \ B ) )

Proof

Step	Hyp	Ref	Expression
1		pm4.71	|- ( ( x e. A -> -. x e. B ) <-> ( x e. A <-> ( x e. A /\ -. x e. B ) ) )
2		eldif	|- ( x e. ( A \ B ) <-> ( x e. A /\ -. x e. B ) )
3	2	bibi2i	|- ( ( x e. A <-> x e. ( A \ B ) ) <-> ( x e. A <-> ( x e. A /\ -. x e. B ) ) )
4	1 3	bitr4i	|- ( ( x e. A -> -. x e. B ) <-> ( x e. A <-> x e. ( A \ B ) ) )
5	4	albii	|- ( A. x ( x e. A -> -. x e. B ) <-> A. x ( x e. A <-> x e. ( A \ B ) ) )
6		disj1	|- ( ( A i^i B ) = (/) <-> A. x ( x e. A -> -. x e. B ) )
7		dfcleq	|- ( A = ( A \ B ) <-> A. x ( x e. A <-> x e. ( A \ B ) ) )
8	5 6 7	3bitr4i	|- ( ( A i^i B ) = (/) <-> A = ( A \ B ) )