Theorem nss

Description: Negation of subclass relationship. Exercise 13 of TakeutiZaring p. 18. (Contributed by NM, 25-Feb-1996) (Proof shortened by Andrew Salmon, 21-Jun-2011)

	Ref	Expression
Assertion	nss	|- ( -. A C_ B <-> E. x ( x e. A /\ -. x e. B ) )

Proof

Step	Hyp	Ref	Expression
1		exanali	|- ( E. x ( x e. A /\ -. x e. B ) <-> -. A. x ( x e. A -> x e. B ) )
2		df-ss	|- ( A C_ B <-> A. x ( x e. A -> x e. B ) )
3	1 2	xchbinxr	|- ( E. x ( x e. A /\ -. x e. B ) <-> -. A C_ B )
4	3	bicomi	|- ( -. A C_ B <-> E. x ( x e. A /\ -. x e. B ) )