[EM] Demonstration of equivalence of Schwartz definitions

[MIKE OSSIPOFF]

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






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