From jjllambias@hotmail.com Sun Mar 05 15:02:39 2000 Received: (qmail 15919 invoked from network); 5 Mar 2000 23:02:51 -0000 Received: from unknown (10.1.10.27) by m2.onelist.org with QMQP; 5 Mar 2000 23:02:51 -0000 Received: from unknown (HELO hotmail.com) (216.33.241.140) by mta2.onelist.org with SMTP; 5 Mar 2000 23:02:51 -0000 Received: (qmail 53415 invoked by uid 0); 5 Mar 2000 23:02:50 -0000 Message-ID: <20000305230250.53414.qmail@hotmail.com> Received: from 200.41.247.53 by www.hotmail.com with HTTP; Sun, 05 Mar 2000 15:02:50 PST X-Originating-IP: [200.41.247.53] To: lojban@onelist.com Subject: Re: [lojban] (unknown) Date: Sun, 05 Mar 2000 15:02:50 PST Mime-Version: 1.0 Content-Type: text/plain; format=flowed X-eGroups-From: "Jorge Llambias" From: "Jorge Llambias" la stivn cusku di'e >3. Let the set f be that set which contains all of the members of all >the possible sets m. No! That is the set of all clubs, because every club belongs to at least one maximally preclusive set. (BTW, that is called the union of all maximally preclusive sets.) But you cannot define f as more than one m.p.s. because then final clubs will not satisfy the circular definition. Only one m.p.s. can be the set of final clubs. >4. Call a club, c, a final club iff it is a member of f. > >Just tidying up. Does that do it? It is provably impossible to define f from the given premises, so nothing will do it. co'o mi'e xorxes ______________________________________________________ Get Your Private, Free Email at http://www.hotmail.com