Metamath Proof Explorer

Theorem nssss

Description: Negation of subclass relationship. Compare nss . (Contributed by NM, 30-Jun-2004) (Proof shortened by Andrew Salmon, 25-Jul-2011)

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

Proof

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