From @uga.cc.uga.edu:lojban@cuvmb.bitnet Sun Jul 02 19:11:11 1995 Received: from punt3.demon.co.uk by stryx.demon.co.uk with SMTP id AA3712 ; Sun, 02 Jul 95 19:11:09 BST Received: from punt3.demon.co.uk via puntmail for ia@stryx.demon.co.uk; Sat, 01 Jul 95 01:27:10 GMT Received: from uga.cc.uga.edu by punt3.demon.co.uk id aa13455; 1 Jul 95 2:26 +0100 Received: from UGA.CC.UGA.EDU by uga.cc.uga.edu (IBM VM SMTP V2R2) with BSMTP id 8850; Fri, 30 Jun 95 21:24:41 EDT Received: from UGA.CC.UGA.EDU (NJE origin LISTSERV@UGA) by UGA.CC.UGA.EDU (LMail V1.2a/1.8a) with BSMTP id 0591; Fri, 30 Jun 1995 21:24:41 -0400 Date: Fri, 30 Jun 1995 18:23:17 -0700 Reply-To: "John E. Clifford" Sender: Lojban list From: "John E. Clifford" Subject: Re: pc answers X-To: lojban list To: Iain Alexander In-Reply-To: <199507010021.AA16047@mail.crl.com> Message-ID: <9507010226.aa13455@punt3.demon.co.uk> Status: R On Fri, 30 Jun 1995 jorge@PHYAST.PITT.EDU wrote: > la djer cusku di'e > > > My predicate calculus formula: > > > > E^!3(x) (remna (x) E^!9(y)(gerku(y) & pencu(x,y))) > > > > declares that there are exactly 3 humans and exactly 9 dogs, for the > > scope of the entire sentence. > > That's what I thought (and it seems that I convinced you) but now that > I see it again, I think I was wrong. Assuming there is another "&" > between "remna(x)" and "E^!9(y)", then you are not claiming that only > three humans exist. Only that remna(x) and that other complicated > claim about x are both true of only three objects. Each separately > may be true of more. Right. The problem here is close to the problem with the absence of referential expressions in Lojban: quantifiers pick up more than you want, so miss the real point. > In any case, the E^!9(y) is within the scope of the other, so it > doesn't in any way say that the nine y's are the same for every x, > only that for each x there are nine y's that fit that relationship. > So your formula admits that up to 27 dogs are being touched in all. No. Although the dogs are in the scope of the men, they are not interdependent; this form is equivalent (with the & as you note) to the form with the dog and man quantifiers reversed, 9x(dog x & 3y (man y & touch y x)) Think of the And form, "there is a cimei and there is a somei..." > Now I think that your formula (with an additional "&") is a good > representation of {ci remna cu pencu so gerku} No, for the reasons just given. That seems to require something like 3x(man x & Ay(man y => 9z (dog z & touch y z))) (the 3x man x can be taken out of the parenthesis with the rest, since unrelated to it). I am tentatively accepting the xorxes-and reading here. > > > ro lo ci remna ku ro lo ci gerku zo'u ra pencu ri > > That would be, in your notation: > > ( E^!3(x) remna(x) ) & ( E^!9(x)(gerku(x) ) & > (x)(y) ( (remna(x) & gerku(y)) -> pencu(x,y) ) > > There are three and only three things that are human & > there are nine and only nine things that are dogs & > for every x that is human and every y that is a dog, x touches y. That this is a symbolization of the Lojban is surely questionable, given the now quite muddied relationship between quantifiers and descriptors in Lojban and logic, but the logic at least does get the 3-men, 9-dogs all-touch-all story right (at a small price in reality and efficiency -- but it does pick out only one case in 512, so probably is about the right size, such things being logarithmic, I think). I think there are shorter forms. pc>|83