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 ) ) |
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 ) ) |