From jjllambias@hotmail.com Sun Sep 29 13:58:20 2002 Return-Path: X-Sender: jjllambias@hotmail.com X-Apparently-To: lojban@yahoogroups.com Received: (EGP: mail-8_1_1_4); 29 Sep 2002 20:58:19 -0000 Received: (qmail 38746 invoked from network); 29 Sep 2002 20:58:19 -0000 Received: from unknown (66.218.66.218) by m10.grp.scd.yahoo.com with QMQP; 29 Sep 2002 20:58:19 -0000 Received: from unknown (HELO hotmail.com) (216.33.241.128) by mta3.grp.scd.yahoo.com with SMTP; 29 Sep 2002 20:58:19 -0000 Received: from mail pickup service by hotmail.com with Microsoft SMTPSVC; Sun, 29 Sep 2002 13:58:19 -0700 Received: from 200.69.6.40 by lw8fd.law8.hotmail.msn.com with HTTP; Sun, 29 Sep 2002 20:58:19 GMT To: lojban@yahoogroups.com Bcc: Subject: Re: [lojban] Re: paroi ro mentu Date: Sun, 29 Sep 2002 20:58:19 +0000 Mime-Version: 1.0 Content-Type: text/plain; format=flowed Message-ID: X-OriginalArrivalTime: 29 Sep 2002 20:58:19.0687 (UTC) FILETIME=[F0A8BB70:01C267FA] From: "Jorge Llambias" X-Originating-IP: [200.69.6.40] X-Yahoo-Group-Post: member; u=6071566 X-Yahoo-Profile: jjllambias2000 X-Yahoo-Message-Num: 16227 la djorden cusku di'e >First of all, what you're talking about here is totally different: >na behaves differently because it needs to export to the leftmost >end of the prenex (inverting any quantifiers) before being interpreted. I did not use {na}. I used {naku} both times, which exports in the order where it appears. {na} goes directly to the leftmost without inverting anything. >Next, though, is that all of the above interpretations work provided >that ko'a and ko'e either can do quantifier inversion automatically >(which I think makes sense) or that in this case they were bound >to single items so inversion is a no op: > naku ko'a .e ko'e broda == > naku zo'u ko'a .e ko'e broda == > naku zo'u ge ko'a broda gi ko'e broda >It is false that: ko'a and ko'e broda. Correct so far. >This is the truth function FFFT, Nope. It is the negation of TFFF, i.e. FTTT. It is the case that either ko'a is not broda or ko'e is not broda (or both). In other words: naku ko'a broda ija naku ko'e broda Just as passing a negation through {ro} changes it to {su'o}, passing a negation through {e} changes it to {a}. The rest is an expansion of {ko'a e ko'e naku broda}: >which you can get with > ko'a na.enai ko'e broda > ko'a .e ko'e na broda >or > ko'a na broda .ijenai ko'e broda > ko'a na broda .ije ko'e na broda >which means > naku ko'a broda .ije naku ko'e broda == > naku zo'u ko'a broda .ije naku zo'u ko'e broda >works fine. See the section starting on pg. 407. >I'm not even sure what the relation you're suggesting is anyway. >You have "ko'a .e ko'e" and can say "ro le re broda" meaning the >same thing... so what? You can always say the same thing in many >different ways, and the transformation loses information. It's deeper than that. You can think of a quantification with {ro} as a long string of conjunctions: ro broda = le broda e le broda e le broda e le broda e ... where each {le broda} picks one member of {lo'i broda}. See also http://nuzban.wiw.org/wiki/index.php?DeMorgan%27s%20Laws for more about this. mu'o mi'e xorxes _________________________________________________________________ Send and receive Hotmail on your mobile device: http://mobile.msn.com