

Sun, 05 Mar 2000 12:42:31 PST
X-Originating-IP: [200.41.247.43]
To: [email-masked]
Subject: Re: [lojban] Final clubs finally in Lojban
Date: Sun, 05 Mar 2000 12:42:31 PST
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X-eGroups-From: "Jorge Llambias" <[email-masked]>
From: "Jorge Llambias" <[email-masked]>

la stivn cusku di'e

>1. Call a set, s, of clubs preclusive if being a member of any one
>of the clubs in s precludes being a member of any other club in s.
>
>2. Call a set, m, of clubs maximally preclusive if it is preclusive
>and every proper superset of of m is not preclusive.

Nice! More compact than my definitions.

Using {girzu} for "club", {klesi} for "set",
{vasru selkle} for "proper superset", and an
interesting interpretation of {natfe} for "preclude":

i pamai ro da poi klesi le'i girzu zo'u ca'e
da natfe klesi ijo ro de poi cmima da ku'o
ro di poi cmima da gi'enai du de zo'u
le ka ce'u cmima de cu natfe le ka ce'u cmima di

i remai ro da poi klesi le'i girzu zo'u ca'e
da nafrai klesi ijo da natfe klesi ijebo
ro de poi vasru selkle da na natfe klesi

>3. Call a set, f, of clubs the set of final clubs if f is the largest
>set of maximally preclusive clubs.

But f can't always be determined by that. In some situations
there is no single largest maximally preclusive set. For example,
we could have the sets {A,B}, {B,C} and {D,E} all being maximally
preclusive. (And I insist that this "largest" condition was not
a premise of the problem as posed.)

>4. Call a club, c, a final club iff it is a member of f.

Yes, I forgot that last step.

Now that we have the definitions, let me prove a theorem:

Theorem: Every club belongs to at least one maximally
preclusive set.

Proof: By construction: start with the club in question,
and examine every other club in any order. If membership
in the club under examination precludes membership in
the initial club and in every other already accepted club,
then accept it, else reject it. The set of accepted clubs
plus the initial club is by construction a maximally
preclusive set. QED.

Corollary: If only one maximally preclusive set exists,
then it must be the set of all clubs.

Proof: Since every club belongs to at least one m.p.s.,
then when there is only one m.p.s. all clubs must belong
to it.

co'o mi'e xorxes

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