Theorem reldisj

Description: Two ways of saying that two classes are disjoint, using the complement of B relative to a universe C . (Contributed by NM, 15-Feb-2007) (Proof shortened by Andrew Salmon, 26-Jun-2011)

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

Proof

Step	Hyp	Ref	Expression
1		dfss2	|- ( A C_ C <-> A. x ( x e. A -> x e. C ) )
2		pm5.44	|- ( ( x e. A -> x e. C ) -> ( ( x e. A -> -. x e. B ) <-> ( x e. A -> ( x e. C /\ -. x e. B ) ) ) )
3		eldif	|- ( x e. ( C \ B ) <-> ( x e. C /\ -. x e. B ) )
4	3	imbi2i	|- ( ( x e. A -> x e. ( C \ B ) ) <-> ( x e. A -> ( x e. C /\ -. x e. B ) ) )
5	2 4	syl6bbr	|- ( ( x e. A -> x e. C ) -> ( ( x e. A -> -. x e. B ) <-> ( x e. A -> x e. ( C \ B ) ) ) )
6	5	sps	|- ( A. x ( x e. A -> x e. C ) -> ( ( x e. A -> -. x e. B ) <-> ( x e. A -> x e. ( C \ B ) ) ) )
7	1 6	sylbi	|- ( A C_ C -> ( ( x e. A -> -. x e. B ) <-> ( x e. A -> x e. ( C \ B ) ) ) )
8	7	albidv	|- ( A C_ C -> ( A. x ( x e. A -> -. x e. B ) <-> A. x ( x e. A -> x e. ( C \ B ) ) ) )
9		disj1	|- ( ( A i^i B ) = (/) <-> A. x ( x e. A -> -. x e. B ) )
10		dfss2	|- ( A C_ ( C \ B ) <-> A. x ( x e. A -> x e. ( C \ B ) ) )
11	8 9 10	3bitr4g	|- ( A C_ C -> ( ( A i^i B ) = (/) <-> A C_ ( C \ B ) ) )