I'm going to show that the beatpath definition of the Schwartz set defines the same set as that defined by the unbeaten set definition of the Schwartz set, the definition by which I've defined Cloneproof SSD. Suppose A has an unreturned beatpath. That means that, for some B distinct from A, there's a beatpath from B to A, but not from A to B. {A} isn't an unbeaten set, because having a beatpath to A means that A is beaten. The set of B & the options having beatpaths to B is an unbeaten set, because if anything outside that set beat a member of that set, then it would have a beatpath to B. But if it did, it wouldn't be outside of that set. Because A doesn't have a beatpath to B, then A isn't a member of that unbeaten set. I'll call that unbeaten set S. In the beatpath from B to A, say X beats A. Any unbeaten set containing A must contain X. Likewise Y which beats X, etc., all the way back to B. Any unbeaten set containing A must also contain B, and everything that has a beatpath to B, by the same argument. So, A isn't in S, but any unbeaten set containing A must contain S. Since A isn't in S, then no unbeaten set containing A can contain only S. If S is a proper subset of some unbeaten set containing A, then, since S is an unbeaten set, any unbeaten set containing A can't be an innermost unbeaten set, because it contains a smaller unbeaten set, S. That means that A can't be in an unbeaten set that doesn't contain a smaller unbeaten set. By the unbeaten set definition of the Schwartz set, A isn't in the Schwartz set. Suppose A doesn't have an unreturned beatpath. The set of A and everything that has a beatpath to A is an unbeaten set, by the same argument used above for set S. The name "S" will now apply to the set of A & everything that has a beatpath to A. Since A doesn't have an unreturned beatpath, then A has a beatpath to everything else in S. For any B in S, since A has a beatpath to B, then A must be in any unbeaten set that B is in, by an argument previously used above. A is in an unbeaten set, S. And S doesn't contain any element that's in an unbeaten set that A isn't in. The only way for A to not be in the Schwartz set would be for S to contain a smaller unbeaten set that A isn't in. If every smaller unbeaten set that it contains also contains A, then A is in the innermost unbeaten set. If it contains no smaller unbeaten set, then likewise A is in the innermost unbeaten set. I've shown that an option with an unreturned beatpath isn't in an innermost unbeaten set, and that an option with no unreturned beatpath is in an innermost unbeaten set. Mike Ossipoff _________________________________________________________________ Get your FREE download of MSN Explorer at http://explorer.msn.com