Re: [lojban] Individuals and xorlo

On Sat, Feb 22, 2014 at 11:45 PM, guskant <gusni.kantu@gmail.com> wrote:

{ro'oi da su'o pa mei} alone cannot be expanded to logical elements only, (D1) (D2) neither, because a predicate {N mei} is not a logical element: {N mei} is a predicate that reflects natural number theory, not only predicate logic. They are _distributively_ 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" series 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 granted, 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 the very meaning of these predicates.

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.
For a place holder, {ke'a} and {ce'u} are suitable, because they are free 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 of words, different sumti can be substituted for them, while only a common sumti 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 about the language, as we are doing here, using ko'a, ko'e, ko'i, ... is more convenient. We may need to use more than one free variable. (The next step is defining the restricted series of numerical predicates, with two places, "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 writing definitions for brivla. I haven't found anything else more convenient. Some people prefer to write their definitions with "ka", "ce'u" and subscripts, but I find them unnecessarily cumbersome.

Using {ke'a}, our definitions are described as follows:

(D1-7) ko'a su'o pa mei
(D1) ke'a su'o N mei := 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 := ke'a su'o N mei gi'e nai su'o N+1 mei
(D3) lo PA broda := 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 already 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 noi ko'a me ke'a}, {ko'i no'u ko'a jo'u ko'o} etc. (D1) (D2) (D3) are not used 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 poi 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.

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 something 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=2 from just D1, because D1 does not define "su'o pa mei". You may have forgotten the restriction on N somewhere in the proof.

For example, suppose that a speaker regards {lo nanba} is non-individual:

ro'oi da poi me lo nanba ku'o su'oi de poi me lo nanba zo'u de me da ijenai da me de

That is, the speaker regards a half of {lo nanba} is also {me lo nanba}.

Yes.

Even though there is no individual {lo nanba}, an _expression_ {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 call them {lo nanba xi re} and {lo nanba xi ci} respectively for convenience.

But it's the same "lo nanba"!

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

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, and the second and the third are false.

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

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 common referent, the first sentence is true, and the second and the third are false.

No. Your starting point was that every part of lo nanba has a proper part, so for lo nanba, and for every one of 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.

mu'o mi'e xorxes