Received: from mail-pd0-f192.google.com ([209.85.192.192]:55617) by stodi.digitalkingdom.org with esmtps (TLSv1:RC4-SHA:128) (Exim 4.80.1) (envelope-from ) id 1WHcX7-0001AN-12 for lojban-list-archive@lojban.org; Sun, 23 Feb 2014 09:08:11 -0800 Received: by mail-pd0-f192.google.com with SMTP id g10sf1556305pdj.9 for ; Sun, 23 Feb 2014 09:07:59 -0800 (PST) DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=googlegroups.com; s=20120806; h=date:from:to:message-id:in-reply-to:references:subject:mime-version :x-original-sender:reply-to:precedence:mailing-list:list-id :list-post:list-help:list-archive:sender:list-subscribe :list-unsubscribe:content-type; bh=m0jUp6egU6PnEtDf27QTHddx5BQ1A5AJu/Z2Dk6t3hI=; b=EZRt2to89OY9Fol9+wbaLJLoeZJ1strFjovWlEbPa5UOTu9ZLBEyMM7bTmUJmHYnMU ulVo/HuF0Ebueu5bISIYboxnptRuMHCqL1br4oThSWC6HG+iyZG/uILY6r9Ztr6NFTaV 9B/aYO0fBCJIIY/GKU0uKZqkoZt1UX17bQlvA2SeHjnxiV3Nth/mG2OedvKVvlyuY9kO GbUMBHVT6wPnc8ysbT543wXroFZOPtVR7x0L4HJ+GiOCgap6A7X9AAbHdUpl1HZ0o3pe TqeA8Fk50bhMUOhFUVBbZgYEPlseH0VQBhEkMiMxd2jLsSwePa9f39wgLgupxI2mqhEV AN6w== DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=gmail.com; s=20120113; h=date:from:to:message-id:in-reply-to:references:subject:mime-version :x-original-sender:reply-to:precedence:mailing-list:list-id :list-post:list-help:list-archive:sender:list-subscribe :list-unsubscribe:content-type; bh=m0jUp6egU6PnEtDf27QTHddx5BQ1A5AJu/Z2Dk6t3hI=; b=BxCkERjIPJyHqTMfcS/w60e+oCm0QDJPUjcZ0kz3HKOhjHxRp5pzeohv3rGCC9Bchj KYSbAIsGe/N1zohehKB/y4pFlSOQCMoS1aO9T9rFGZ49OCoXwWe+4lKTJDnzID32zyhP 6+aHPYXlpyiIyk5aYOu2740D1yT/DG9UKJp+/Bj3IyW4dyxdDCfhRBtmWMRw3wjDkWtX Pl92hGA+jisqOWj9zk0QpwzyTLPm/+ee43sOyNsRhw5Fb88tLTCevOfY7JTlk8eGXgNn rC1ZIwdCdrpOcqmrA0CF8xUbPiIb3u74vMKb8dotl9QujvsuWszfUXX36ufQOqnhag9J epaQ== X-Received: by 10.50.30.66 with SMTP id q2mr277888igh.1.1393175279040; Sun, 23 Feb 2014 09:07:59 -0800 (PST) X-BeenThere: lojban@googlegroups.com Received: by 10.51.18.68 with SMTP id gk4ls1572563igd.29.gmail; Sun, 23 Feb 2014 09:07:58 -0800 (PST) X-Received: by 10.50.70.103 with SMTP id l7mr276674igu.15.1393175278404; Sun, 23 Feb 2014 09:07:58 -0800 (PST) Date: Sun, 23 Feb 2014 09:07:56 -0800 (PST) From: guskant To: lojban@googlegroups.com Message-Id: <390f1b9f-6edd-42f2-8474-ad1f3610cca3@googlegroups.com> In-Reply-To: References: <52F26B9E.2090001@gmx.de> <5e023b9a-515c-432b-a389-8f9af4766b51@googlegroups.com> <52F29ED8.1050607@gmx.de> <372dd8f1-1920-4afa-8d11-aa55696982a0@googlegroups.com> <03555bbd-cc44-426f-94ee-65d557f2d301@googlegroups.com> <592497c0-5db5-420e-867f-8df1663eca27@googlegroups.com> <52F65A5C.90605@gmx.de> <348c23bf-6d9f-4a05-bfe7-69b141c03cb7@googlegroups.com> <52F776EE.6070406@gmx.de> <6ffd64d2-2e2c-4b83-8722-b7f262f5837a@googlegroups.com> <52F7A4D5.5070106@gmx.de> <56096dec-1969-420d-b4e5-b8539cbe0cc0@googlegroups.com> <52F8FAA2.9030009@gmx.de> <52FE053C.3000604@gmx.de> <1e6d5917-ad1e-4c5b-abb7-5deb92110b83@googlegroups.com> <68bacba4-a957-481c-ba00-211db2de8dc3@googlegroups.com> <2f4f0766-1f52-46f0-80af-b4de86d9b5bd@googlegroups.com> <618e6524-d7f0-46c9-8d0b-bbee2dd0cd41@googlegroups.com> <36c4c2b2-8f8c-4d44-ac8e-48c02d45a233@googlegroups.com> <4b6b2cb9-51e5-47f6-97a9-2dec16406864@googlegroups.com> Subject: Re: [lojban] Individuals and xorlo MIME-Version: 1.0 X-Original-Sender: gusni.kantu@gmail.com Reply-To: lojban@googlegroups.com Precedence: list Mailing-list: list lojban@googlegroups.com; contact lojban+owners@googlegroups.com List-ID: X-Google-Group-Id: 1004133512417 List-Post: , List-Help: , List-Archive: Sender: lojban@googlegroups.com List-Subscribe: , List-Unsubscribe: , Content-Type: multipart/alternative; boundary="----=_Part_346_31605270.1393175276610" X-Spam-Score: -0.1 (/) X-Spam_score: -0.1 X-Spam_score_int: 0 X-Spam_bar: / ------=_Part_346_31605270.1393175276610 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: quoted-printable Le dimanche 23 f=C3=A9vrier 2014 22:55:14 UTC+9, xorxes a =C3=A9crit : > > > > > On Sat, Feb 22, 2014 at 11:45 PM, guskant > > wrote: > >> >> {ro'oi da su'o pa mei} alone cannot be expanded to logical elements only= ,=20 >> (D1) (D2) neither, because a predicate {N mei} is not a logical element:= {N=20 >> mei} is a predicate that reflects natural number theory, not only predic= ate=20 >> logic. They are _distributively_ not tautology. >> > > I agree that before "su'o pa mei" is defined, "ro'oi da su'o pa mei" is= =20 > not a tautology. It is only a tautology once "su'o pa mei" has been=20 > introduced as a tautological predicate. In the context I brought it up, I= =20 > was in the process of defining the "PA mei" series of predicates, and I= =20 > started by defining "su'o pa mei" such that "ro'oi da su'o pa mei". I did= =20 > not explicitly write down any definition for "su'o pa mei", but the only= =20 > definition of "su'o pa mei" that makes "ro'oi da su'o pa mei" true is one= =20 > that defines it as a tautological predicate.=20 > > One thing I should have said, and which I took for granted, but I see you= =20 > didn't from something you say below, is that all the "PA mei" predicates= =20 > must be non-distributive. We don't want to infer from "ko'a jo'u ko'e re= =20 > mei" that "ko'a re mei" or "ko'e re mei". That would kill the very meanin= g=20 > of these predicates. > > You give {su'o pa mei} to all the referent that are individual(s) of a=20 universe of discourse, while I give {su'o pa mei} to certain members of it,= =20 including non-indiviidual members, not to all. Once {su'o pa mei} is given= =20 to a referent, it satisfies the predicate _non-distributively_. When "ko'a= =20 jo'u ko'e re mei" is true, "ko'a re mei" and "ko'e re mei" are false=20 according to (D1) and (D2). To smaller part of {ko'a} or {ko'e}, {su'o N=20 mei} is not applied. =20 > =20 > >> It seems that using "ko'a" as a place holder causes a problem. >> I use {ko'a} as a plural constant, not as a place holder.=20 >> For a place holder, {ke'a} and {ce'u} are suitable, because they are fre= e=20 >> variables: such usage is not described in CLL, but it is useful at least= in=20 >> the current discussion. >> >> When {ce'u} appears more than two times in a sequence of words, differen= t=20 >> sumti can be substituted for them, while only a common sumti can be=20 >> substituted for {ke'a}s. For the current purpose, using {ke'a} is better= . >> > > When using the language, yes. We don't need free variables for ordinary= =20 > use of the language. But when talking about the language, as we are doing= =20 > here, using ko'a, ko'e, ko'i, ... is more convenient. We may need to use= =20 > more than one free variable. (The next step is defining the restricted=20 > series of numerical predicates, with two places, "ko'a PA mei ko'e", and= =20 > using subscripts for the different places in addition to the numbers in t= he=20 > predicate just adds a lot of confusion.) Also, sometimes we need the fre= e=20 > variable to appear within a relative clause. I have always used ko'a, ko'= e,=20 > ... as the place holders when writing definitions for brivla. I haven't= =20 > found anything else more convenient. Some people prefer to write their=20 > definitions with "ka", "ce'u" and subscripts, but I find them unnecessari= ly=20 > cumbersome. > =20 > I agree. However, for the current discussion, distinction between plural=20 constant and free plural variable is necessary. For this purpose, I use=20 {ke'a} as a free plural variable as "zasni" here. =20 > > Using {ke'a}, our definitions are described as follows: >> (D1-7) ko'a su'o pa mei >> (D1) ke'a su'o N mei :=3D su'oi da poi me ke'a ku'o su'oi de poi me ke'a= =20 >> zo'u ge da su'o N-1 mei gi de na me da >> (D2) ke'a N mei :=3D ke'a su'o N mei gi'e nai su'o N+1 mei=20 >> (D3) lo PA broda :=3D zo'e noi ke'a PA mei gi'e broda >> >> When (D1) and (D2) are applied to a particular sumti, ke'a are replaced= =20 >> with it. As for (D3), ke'a is in noi-clause, and it is already fixed to= =20 >> zo'e, and is not replaced with another sumti, of course.=20 >> >> Because (D1-7) defines only for {ko'a}, (D1) (D2) (D3) are valid only fo= r=20 >> sumti that involves a referent of {ko'a} such as {ko'e noi ko'a me ke'a}= ,=20 >> {ko'i no'u ko'a jo'u ko'o} etc. (D1) (D2) (D3) are not used for other su= mti=20 >> unless (D1-7) is applied to one of the referents that is involved by the= =20 >> sumti. >> > > If D1-7 defines only for ko'a, then it is not necessarily valid for ro'oi= =20 > da poi me ko'a. You need "ro'oi da poi me ko'a cu su'o mei" if you want i= t=20 > to be valid for anything among ko'a. But that won't make it valid for ko'= a=20 > jo'u ko'o if something in ko'o is not in ko'a.=20 > No. When (D1-7) defines for {ko'a}, the referent of {ko'a} satisfies {su'o= =20 pa mei} _non-distributively_.=20 Any other referents that are {me ko'a} do not satisfy {su'o pa mei}. =20 > > I used only (D1) and logical axioms including transitivity of {me}. Any= =20 >> mention of {su'o pa mei} is not necessary for the proof.=20 >> > > Then there must be something wrong in the proof. You just cannot prove "= ganai=20 > ko'a su'o N mei gi ko'a su'o N-1 mei" for N=3D2 from just D1, because D1 = does=20 > not define "su'o pa mei". You may have forgotten the restriction on N=20 > somewhere in the proof. > > An order of all integers including zero and negative numbers is used for=20 the proof, not only for N>=3D3. (D1)+(D2) for N=3D1 produces contradiction,= but=20 (D1) alone does not produce contradiction for every integer, although it is= =20 meaningless. I used a larger set of numbers than what is required by the=20 proved proposition. It is just like using a higher dimensional space in a= =20 proof on a figure in a lower dimensional space. It is a valid procedure. =20 > =20 > >> For example, suppose that a speaker regards {lo nanba} is=20 >>> non-individual: >>> >>>> ro'oi da poi me lo nanba ku'o su'oi de poi me lo nanba zo'u de me da= =20 >>>> ijenai da me de >>>> >>>> That is, the speaker regards a half of {lo nanba} is also {me lo=20 >>>> nanba}.=20 >>>> >>> >>> Yes. >>> =20 >>> >>>> Even though there is no individual {lo nanba}, an expression {N mei} i= s=20 >>>> available with (D1-7) (D1) (D2) (D3). >>>> >>> >>> No: >>> >>> "lo nanba cu su'o pa mei" is true >>> "lo nanba cu su'o re mei" is true >>> "lo nanba cu su'o ci mei" is true >>> >> >> I call them {lo nanba xi re} and {lo nanba xi ci} respectively for=20 >> convenience. >> > > But it's the same "lo nanba"!=20 > > lo nanba cu su'o pa mei gi'e su'o re mei gi'e su'o ci mei gi'e ..." is=20 > true. > =20 > It cannot be true when=20 (D1-7} lo nanba cu su'o pa mei is defined to {lo nanba}.=20 In the definition=20 (D1) lo nanba cu su'o re mei :=3D su'oi da poi me lo nanba ku'o su'oi de po= i=20 me lo nanba zo'u ge da su'o pa mei gi de na me da {da su'o pa mei} is true only for the referent of {lo nanba} used in=20 (D1-7), that is, {lo nanba} itself, and it satisfies {su'o pa mei}=20 _non-distributively_. The other referents in the domain of {da poi me lo=20 nanba} do not satisfy {da su'o pa mei}. Therefore, there is no referent=20 that satisfies {ge da su'o pa mei gi de na me da}. =20 > If >> (D1-7) lo nanba xi pa cu su'o pa mei >> is defined, and if {naku ge lo nanba xi pa cu me lo nanba xi re/ci gi=20 >> naku lo nanba xi re/ci cu me lo nanba xi pa}, the first sentence is true= ,=20 >> and the second and the third are false. >> > > I don't see how that makes the second and third false. > =20 > As I discussed above, (D1-7) is defined only on a referent selected by a=20 speaker, and the referent satisfies {su'o pa mei} _non-distributively_.=20 When=20 (D1-7) lo nanba cu su'o pa mei is defined, any smaller part of {lo nanba} is not {su'o pa mei}. Therefore,= =20 the second and the third sentences cannot be produced for the same referent= . =20 > That is to say, if {(D1-7) lo nanba cu su'o pa mei} is defined, and if al= l=20 >> the appearances of {lo nanba} have a common referent, the first sentence= is=20 >> true, and the second and the third are false. >> > > No. Your starting point was that every part of lo nanba has a proper part= ,=20 > so for lo nanba, and for every one of its parts, "su'o N mei" is true for= =20 > every natural N, and for "lo nanba", and for every one of its parts, "N= =20 > mei" is false for every natural N. > . > There are infinite referents that are {me lo nanba}, but {su'o pa mei} was= =20 defined only on the referent of {lo nanba} itself _non-distributively_. Any= =20 other referents that are {me lo nanba} do not satisfy {su'o pa mei},=20 therefore {su'o N mei} neither. =20 --=20 You received this message because you are subscribed to the Google Groups "= lojban" group. To unsubscribe from this group and stop receiving emails from it, send an e= mail to lojban+unsubscribe@googlegroups.com. To post to this group, send email to lojban@googlegroups.com. Visit this group at http://groups.google.com/group/lojban. For more options, visit https://groups.google.com/groups/opt_out. ------=_Part_346_31605270.1393175276610 Content-Type: text/html; charset=UTF-8 Content-Transfer-Encoding: quoted-printable


Le dimanche 23 f=C3=A9vrier 2014 22:55:14 UTC+9, x= orxes a =C3=A9crit :



On Sat, Feb 22, 201= 4 at 11:45 PM, guskant <gusni...@gmail.com> wrote:

{ro'oi da su'o pa mei= } alone cannot be expanded to logical elements only, (D1) (D2) neither, bec= ause a predicate {N mei} is not a logical element: {N mei} is a predicate t= hat reflects natural number theory, not only predicate logic. They are _dis= tributively_ not tautology.

I agree that before "su'o pa mei" is= defined, "ro'oi da su'o pa mei" is not a tautology. It is only a tautology= once "su'o pa mei" has been introduced as a tautological predicate. In the= context I brought it up, I was in the process of defining the "PA mei" ser= ies of predicates, and I started by defining "su'o pa mei" such that "ro'oi= da su'o pa mei". I did not explicitly write down any definition for "su'o = pa mei", but the only definition of "su'o pa mei" that makes "ro'oi da su'o= pa mei" true is one that defines it as a tautological predicate. 

One thing I should have said, and which I took for gran= ted, but I see you didn't from something you say below, is that all the "PA= mei" predicates must be non-distributive. We don't want to infer from "ko'= a jo'u ko'e re mei" that "ko'a re mei" or "ko'e re mei". That would kill th= e very meaning of these predicates.



You give {su'o pa mei} to all the referent that are individual(s) of = a universe of discourse, while I give {su'o pa mei} to certain members of i= t, including non-indiviidual members, not to all. Once {su'o pa mei} is giv= en to a referent, it satisfies the predicate _non-distributively_. When "ko= 'a jo'u ko'e re mei" is true, "ko'a re mei" and "ko'e re mei" are false acc= ording to (D1) and (D2). To smaller part of {ko'a} or {ko'e}, {su'o N mei} = is not applied.


 
 
<= div>It seems that using "ko'a" as a place holder causes a problem.
I use {ko'a} as a plural constant, not as a place holder. 
<= div>For a place holder, {ke'a} and {ce'u} are suitable, because they are fr= ee variables: such usage is not described in CLL, but it is useful at least= in the current discussion.

When {ce'u} appears more than two times in a sequence o= f words, different sumti can be substituted for them, while only a common s= umti can be substituted for {ke'a}s. For the current purpose, using {ke'a} = is better.

When using the language, yes. We don= 't need free variables for ordinary use of the language. But when talking a= bout the language, as we are doing here, using ko'a, ko'e, ko'i, ... is mor= e convenient. We may need to use more than one free variable. (The next ste= p is defining the restricted series of numerical predicates, with two place= s,  "ko'a PA mei ko'e", and using subscripts for the different places = in addition to the numbers in the predicate just adds a lot of confusion.) =  Also, sometimes we need the free variable to appear within a relative= clause. I have always used ko'a, ko'e, ... as the place holders when writi= ng definitions for brivla. I haven't found anything else more convenient. S= ome people prefer to write their definitions with "ka", "ce'u" and subscrip= ts, but I find them unnecessarily cumbersome.
 


I agree. However, for the current discussion, distinction between p= lural constant and free plural variable is necessary. For this purpose, I u= se {ke'a} as a free plural variable as "zasni" here.


 

Using {ke'a}, our definitions are describ= ed as follows:
(D1-7) ko'a su'o pa mei
(D1) ke'a su'o N mei :=3D= su'oi da poi me ke'a ku'o su'oi de poi me ke'a zo'u ge da su'o N-1 mei gi = de na me da
(D2) ke'a N mei  :=3D ke'a su'o N mei gi'e nai su'o N+1 mei 
(D3) lo PA broda :=3D zo'e noi ke'a PA mei gi'e broda
=
When (D1) and (D2) are applied to a particular sumti, = ke'a are replaced with it. As for (D3), ke'a is in noi-clause, and it is al= ready fixed to zo'e, and is not replaced with another sumti, of course= . 

Because (D1-7) defines only for {ko'a}, (D1) (D2) (D3) = are valid only for sumti that involves a referent of {ko'a} such as {ko'e n= oi ko'a me ke'a}, {ko'i no'u ko'a jo'u ko'o} etc. (D1) (D2) (D3) are not us= ed for other sumti unless (D1-7) is applied to one of the referents that is= involved by the sumti.

If D1-7 defines only for ko'a, then = it is not necessarily valid for ro'oi da poi me ko'a. You need "ro'oi da po= i me ko'a cu su'o mei" if you want it to be valid for anything among ko'a. = But that won't make it valid for ko'a jo'u ko'o if something in ko'o is not= in ko'a. 

No. When (D1-7) defines for {ko'a}, the referent of {ko'a} sati= sfies {su'o pa mei} _non-distributively_. 
Any other referen= ts that are {me ko'a} do not satisfy {su'o pa mei}.


 
<= div dir=3D"ltr">

I used only (D1)= and logical axioms including transitivity of {me}. Any mention of {su'o pa= mei} is not necessary for the proof. 

Then there must be somet= hing wrong in the proof.  You just cannot prove "ganai ko'a su'o N mei gi ko'a su'o N= -1 mei" for N=3D2 from just D1, because D1 does not define "su'o pa mei". Y= ou may have forgotten the restriction on N somewhere in the proof.



An order of all integers including zero and negative numbers is used = for the proof, not only for N>=3D3. (D1)+(D2) for N=3D1 produces contrad= iction, but (D1) alone does not produce contradiction for every integer, al= though it is meaningless. I used a larger set of numbers than what is requi= red by the proved proposition. It is just like using a higher dimensional s= pace in a proof on a figure in a lower dimensional space. It is a valid pro= cedure.


 
 
  &nb= sp; For example, suppose that a speaker regards {lo nanba} is non-individua= l:
ro'oi da poi me lo nanba ku'o su'oi de poi me lo nanb= a zo'u de me da ijenai da me de

That is, the speak= er regards a half of {lo nanba} is also {me lo nanba}. 

Yes.
 
Even though there is no individual {lo nanba}, an exp= ression {N mei} is available with (D1-7) (D1) (D2) (D3).

No:

"lo nanba cu su'o pa= mei" is true
"lo nanba cu su'o re mei" is true
"lo nanba cu su'o ci mei" = is true

I cal= l them {lo nanba xi re} and {lo nanba xi ci} respectively for convenience.<= /div>

But it's the same "lo nanba"! <= /div>

lo nanba cu su'o pa mei gi'e su'o re mei gi'e su'o= ci mei gi'e ..." is true.
 


It cannot be true when 
(D1-7} lo nanba cu su'o pa mei
is defined to {lo nanba}. 

In the def= inition 

(D1) lo nanba cu su'o re mei :=3D su= 'oi da poi me lo nanba ku'o su'oi de poi me lo nanba zo'u ge da su'o pa mei= gi de na me da

{da su'o pa mei} is true only for = the referent of {lo nanba} used in (D1-7), that is, {lo nanba} itself, and = it satisfies {su'o pa mei} _non-distributively_. The other referents in the= domain of {da poi me lo nanba} do not satisfy {da su'o pa mei}. There= fore, there is no referent that satisfies {ge da su'o pa mei gi de na me da= }.


 
<= blockquote class=3D"gmail_quote" style=3D"margin:0px 0px 0px 0.8ex;border-l= eft-width:1px;border-left-color:rgb(204,204,204);border-left-style:solid;pa= dding-left:1ex">
If
(D1-7) lo nanba xi pa cu= su'o pa mei
is defined, and if {naku ge lo nanba xi pa cu me lo nanba xi re/ci gi = naku lo nanba xi re/ci cu me lo nanba xi pa}, the first sentence is true, a= nd the second and the third are false.

I don't see how that makes the second and third false.
 


<= div>As I discussed above, (D1-7) is defined only on a referent selected by = a speaker, and the referent satisfies {su'o pa mei} _non-distributively_. W= hen 
(D1-7) lo nanba cu su'o pa mei
is defined, an= y smaller part of {lo nanba} is not {su'o pa mei}. Therefore, the second an= d the third sentences cannot be produced for the same referent.
<= br>

 
That is to say, if {(D1-7) lo nanba cu su'= o pa mei} is defined, and if all the appearances of {lo nanba} have a commo= n referent, the first sentence is true, and the second and the third are fa= lse.

No. Your starting point was that eve= ry part of lo nanba has a proper part, so for lo nanba, and for every one o= f its parts, "su'o N mei" is true for every natural N, and for "lo nanba", = and for every one of its parts, "N mei" is false for every natural N.
.


There are infinite referents that are {me lo nanba}, but {su'o pa&nb= sp;mei} was defined only on the referent of {lo nanba} itself _non-distribu= tively_. Any other referents that are {me lo nanba} do not satisfy {su'o pa= mei}, therefore {su'o N mei} neither.


<= div> 

--
You received this message because you are subscribed to the Google Groups &= quot;lojban" group.
To unsubscribe from this group and stop receiving emails from it, send an e= mail to lojban+unsubscribe@googlegroups.com.
To post to this group, send email to lojban@googlegroups.com.
Visit this group at http:= //groups.google.com/group/lojban.
For more options, visit https://groups.google.com/groups/opt_out.
------=_Part_346_31605270.1393175276610--