From pycyn@aol.com Tue Sep 24 06:56:00 2002 Return-Path: X-Sender: Pycyn@aol.com X-Apparently-To: lojban@yahoogroups.com Received: (EGP: mail-8_1_1_3); 24 Sep 2002 13:56:00 -0000 Received: (qmail 38411 invoked from network); 24 Sep 2002 13:55:59 -0000 Received: from unknown (66.218.66.216) by m15.grp.scd.yahoo.com with QMQP; 24 Sep 2002 13:55:59 -0000 Received: from unknown (HELO imo-m06.mx.aol.com) (64.12.136.161) by mta1.grp.scd.yahoo.com with SMTP; 24 Sep 2002 13:55:59 -0000 Received: from Pycyn@aol.com by imo-m06.mx.aol.com (mail_out_v34.10.) id r.174.f21596c (2612) for ; Tue, 24 Sep 2002 09:55:54 -0400 (EDT) Message-ID: <174.f21596c.2ac1c8e9@aol.com> Date: Tue, 24 Sep 2002 09:55:53 EDT Subject: Re: [lojban] tu'o usage To: lojban@yahoogroups.com MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="part1_174.f21596c.2ac1c8e9_boundary" X-Mailer: AOL 7.0 for Windows US sub 10509 From: pycyn@aol.com X-Yahoo-Group-Post: member; u=2455001 X-Yahoo-Profile: kaliputra --part1_174.f21596c.2ac1c8e9_boundary Content-Type: text/plain; charset="US-ASCII" Content-Transfer-Encoding: 7bit In a message dated 9/23/2002 9:59:05 PM Central Daylight Time, jjllambias@hotmail.com writes: << > >What won't work? > > Some of the negation relationships. Unless the simple forms are > assigned to (A-E-I+O+) (or (A+E+I-O-) but this one would be silly) > then some of those relationships don't work between the simple forms. >> Frinstance? And remember that the other assumption of all of this is that things talked about exist, that the Lojban axiom {ro da zasti} uses importing {ro} and works at the restricted level as well {ro broda cu zasti} (I know that you diagree with this -- at least that it is importing {ro} -- but non-importing {ro} makes precious little sense in the general case.) << I didn't label any system as yours. I understand you argued at some point for (A+E+I+O+) and at other times for (A+E-I+O-) for the simple forms. A+/A- is the one we always disagree about, since I want {ro broda cu brode} to be A- and you want it to be A+. >> And that it doesn't matter which, since we are dealing with non-empty sets. On the rare occasion when we want to deal with empty sets, we have to flag it anyhow, so what the basic system is is relatively insignificant. << << > >(the apparent exception being an aberration that ran briefly form > >about > >1858 to 1958). > >Are those the dates of some particular events? > >> >Boole's Laws of Thought to my first paper on the subject (class, not >published). Nobody can accuse you of being too modest! :) Is your epoch making paper available online? >> I didn't claim that my paper caused the end of an era, only that it marks for me the point when I knew the era was over. If I got away with it in a class paper at UCLA, the logic hub of the universe, then it the old doctrine must have been dead indeed. That paper is not only not available on-line (thank God) but is long ago (well, not all that long ago, since only my last wife makes me clean out my files) destroyed. << I'm glad Boole is on my side then >> Boole was, paradoxically, a lousy logician and was working almost entirely outside the tradition (or, rather, with a degenerate humanistic tradition that ran parallel to and against the logician tradition since the Renaissance). I have to admit sadly that Dodgson got sucked into the same tradition, with only minor rebellions (he used O- for a while). << I said that changing inner {ro} to {me'iro} was nonsense, not that the passage of a negation boundary did not affect the inner quantifier. If the inner quantifier is {ro}, then nothing is changed, because {ro} as inner quantifier in fact adds nothing, neither claim nor presupposition: {lo'i broda} always has ro members by definition. >> Let's see, negation boundaries do affect inner quantifiers except in the case of the most common one. That does seem to violate the notion that they are affected -- a rule is a rule after all and the effects of negation boundaries on the universal quantifier is one of the best established of such rules. As for {ro} adding nothing, it does at least exclude {no} (I know you disagree, but this is my turn) and, further, as the default, can be stuck in anywhere nothing is explicit (which is why I take it that nebgation does not affect it). What about {le broda}, where the default is {su'o} : does {naku le broda} go over to {ro le no broda naku}? If not, why not? << When the inner quantifier is something other than {ro}, then there is an additional claim or presuposition that {lo'i broda} has Q members. If it is a claim, then passing through {na} will affect that claim, but not by changing the inner quantifier into another inner quantifier. For example (asuming for the moment that the inner is claimed rather than presupposed): naku lo pa broda cu brode = naku ge lo broda cu brode gi pa da broda = ganai lo broda cu brode ginai pa da broda = ga ro lo broda naku brode ginai pa da broda And this cannot be written as {ro lo Q broda naku brode}. So if the inner quantifier is claimed, the manipulation rules are not at all simple, except when the inner is non-importing ro, which makes no claim or presupposition. Yet another argument in favour of non-importing ro. >> If INNER makes a claim, the presumably the modified claim can be made by another INNER. If it cannot, then INNER coes not make a claim. The contention now comes down to whether INNER is modified by negation passage. I don't see it happening nor do I find anything about it happening in CLL. >From this I infer that INNER does not make a claim at the assertion level. It does make a claim, but at another level and differently dealt with. I would say (if we have to do this) that the derivation goes naku lo pa broda cu brode = ro lo pa broda naku brode = ge ro lo broda naku gi pa da broda I don't, btw, think doing this is very enlightening, since it leaves out some interesting information. As for a non-importing {ro} in that position, the rule should still apply, making non-sense of the whole. But using importing {ro} also makes nonsense of the whole (well, a tautology -- or nearly so), so the result is that taking INNER as assertive is an error. On the other hand, taking INNER as presuppositional makes the same sense regardless of the nature of {ro}. &: << > Even > mathematicians and linguists pretty much get this right. The the confusion may be about what "this" is. >> That "all" has existential import. I guess I have to take back "linguists" -- but, gee, my people (Partee, Bill Bright, various Lakoffs) and McCawley had it right. << > I suppose you mean {ro broda cu brode} and {ro da poi broda cu > brode} entail {da broda} (or you mean "implicate" rather than > "entail"). I mean that {ro broda cu brode} and {ro da poi broda cu brode} DON'T ENTAIL {da broda}. (Caps for emphasis, not shouting.) That is, they are equivalent to {ro da ga na broda gi brode}. >> Yes, that is what you meant. Sorry to type ahead of my head. << In saying that, I'm just describing my habits of interpretation. > It is quite true that for many people much of the time > "All broda are brode" does not entail "There are broda," but by the > same token, {ro broda cu brode} or {ro da poi broda cu brode} are not > translations of that sentence (in that sense), Right. As I understand it, this is your position, legitimately backed up by an Argument from Authority, which I'm not confident I'm capable of understanding, while Jorge takes the contrary view. I am saying that I hope Jorge is right, so as to spare me having to unlearn my habits. Of course, if I'm thereby committing some horrible logical fallacy I would want to recant, but I don't (yet) see why {ro broda cu brode} and {ro da poi broda cu brode} can't be strictly equivalent to {ro da ga na broda gi brode}. >> They could be and, indeed, for some time were taken to be such by most people. When I first tried to find some space for A+, I was content to claim {ro da poi broda cu brode} as a plausible place. At which point, Cowan pointed out (correctly) that {ro} always implies {su'o} in Logic and that, therefore, I could claim the lot and reserve for A- the place that Logic gives it, namely the universal conditional. So, I took him up that, since it essentially what I really thought was the case anyhow. But, if your habit is really to read "All S is P" as {ro da ganai S da gi P da} (with an importing {ro}, notice), then you don't have to change that habit, only the one about collapsing that farther. And you can even collapse, if you are willing to concede that there are S's. [Calling citation -- or the threat of such -- Argument from Authority is prejudicial, even when modified by "legitimately": loading.] << My brand of English has "all" and "every" as nonimporting, and "each" as importing, but "each" quantifies over a definite class (i.e. it means "each of the"), so the importingness is probably an artefact of the definiteness. >> I'll take your word for it, even though I have found (as have more formal empiricial researchers on the issue) that people are not very clear about this and often display patterns incompatible with their conscious beliefs on the topic. In particular, though, people who allow both importing and non-importing meanings usually group "every" with "each" (as it is historically as well = "ever each"), so you constitute a group either new or too small to have been noted before. Your explanation for the position of "each" probably accounts for your case, which is basically a "no importing" one. << > The trick seems to be a metaconjuction that works at one level like > an ordinary conjunction but at another level is not attached until > all the other operations have been gone through (see some of the > stuff about interdefining the various types of quantifiers earlier this year). I take it that the 'operations' are 'gone through' from inside to outside, i.e. mainly right to left in a Lojban-style syntax? That is, if X has scope over Y, then Y is processed before X? In that case, yes. >> You lost me there somewhere -- see the example above on INNER for what I have in mind, loosely. But note that I would not really use {e} here, since if fails to flag the difference involved. With a different conjunction, denying such a compound with {naku} would be the equivalent of {na'i}. << If you have the logical formula: P and ASSERTED: Q how should that be expressed grammatically so that it comes out like Q PRESUPPOSED: and P >> I don't follow the formula, I think. Suppose that P presupposes Q. Then the whole situation is "P funny-and Q." Negating this would be "not P funny-and Q," {na'i}ing it would be either "not(P and Q)" ("and" not at all funny) or (better) "not Q whether P" ("whether" = Lojban {u}). lioNEL: << Indeed, I take the opposing views. As xorxes pointed it out, the whole issue seems to decide wether the INNER part is claimed or presupposed. IMO it is naturally claimed (the ro case being special, see below): I would find it very strange, to say the least, to consider something explicitly stated as something presupposed. >> Me too. But INNER is not stated, merely displayed and, thus, open to a variety of interpretations, of which "presupposed" is one. "Asserted" is another, but I can't find any cases of it actually working that way anywhere and many cases of the presupposing version, even without {ro}. --part1_174.f21596c.2ac1c8e9_boundary Content-Type: text/html; charset="US-ASCII" Content-Transfer-Encoding: 7bit In a message dated 9/23/2002 9:59:05 PM Central Daylight Time, jjllambias@hotmail.com writes:

<<
>What won't work?

Some of the negation relationships. Unless the simple forms are
assigned to (A-E-I+O+) (or (A+E+I-O-) but this one would be silly)
then some of those relationships don't work between the simple forms.

>>
Frinstance?  And remember that the other assumption of all of this is that things talked about exist, that the Lojban axiom {ro da zasti} uses importing {ro} and works at the restricted level as well {ro broda cu zasti}  (I know that you diagree with this -- at least that it is importing {ro} -- but non-importing {ro} makes precious little sense in the general case.)

<<
I didn't label any system as yours. I understand you argued at some
point for (A+E+I+O+) and at other times for (A+E-I+O-) for the simple
forms. A+/A- is the one we always disagree about, since I want
{ro broda cu brode} to be A- and you want it to be A+.
>>
And that it doesn't matter which, since we are dealing with non-empty sets.  On the rare occasion when we want to deal with empty sets, we have to flag it anyhow, so what the basic system is is relatively insignificant.

<<
<<
> >(the apparent exception being an aberration that ran briefly form
> >about
> >1858 to 1958).
>
>Are those the dates of some particular events?
> >>
>Boole's Laws of Thought to my first paper on the subject (class, not
>published).

Nobody can accuse you of being too modest! :) Is your epoch making
paper available online?
>>
I didn't claim that my paper caused the end of an era, only that it marks for me the point when I knew the era was over.  If I got away with it in a class paper at UCLA, the logic hub of the universe, then it the old doctrine must have been dead indeed. That paper is not only not available on-line (thank God) but is long ago (well, not all that long ago, since only my last wife makes me clean out my files) destroyed.

<<
I'm glad Boole is on my side then
>>
Boole was, paradoxically, a lousy logician and was working almost entirely outside the tradition (or, rather, with a degenerate humanistic tradition that ran parallel to and against the logician tradition since the Renaissance).  I have to admit sadly that Dodgson got sucked into the same tradition, with only minor rebellions (he used O- for a while).

<<
I said that changing inner {ro} to {me'iro} was nonsense, not
that the passage of a negation boundary did not affect the inner
quantifier. If the inner quantifier is {ro}, then nothing is changed,
because {ro} as inner quantifier in fact adds nothing, neither
claim nor presupposition: {lo'i broda} always has ro members
by definition.
>>
Let's see, negation boundaries do affect inner quantifiers except in the case of the most common one.  That does seem to violate the notion that they are affected -- a rule is a rule after all and the effects of negation boundaries on the universal quantifier is one of the best established of such rules.  As for {ro} adding nothing, it does at least exclude {no} (I know you disagree, but this is my turn) and, further, as the default, can be stuck in anywhere nothing is explicit (which is why I take it that nebgation does not affect it).  What about {le broda}, where the default is {su'o} : does {naku le broda} go over to {ro le no broda naku}?  If not, why not?

<<
When the inner quantifier is something other than {ro}, then
there is an additional claim or presuposition that {lo'i broda}
has Q members. If it is a claim, then passing through {na}
will affect that claim, but not by changing the inner quantifier
into another inner quantifier. For example (asuming for the
moment that the inner is claimed rather than presupposed):

naku lo pa broda cu brode
= naku ge lo broda cu brode gi pa da broda
= ganai lo broda cu brode ginai pa da broda
= ga ro lo broda naku brode ginai pa da broda

And this cannot be written as {ro lo Q broda naku brode}.
So if the inner quantifier is claimed, the manipulation rules are
not at all simple, except when the inner is non-importing ro,
which makes no claim or presupposition. Yet another argument
in favour of non-importing ro.
>>
If INNER makes a claim, the presumably the modified claim can be made by another INNER.  If it cannot, then INNER coes not make a claim.  The contention now comes down to whether INNER is modified by negation passage.  I don't see it happening nor do I find anything about it happening in CLL.  From this I infer that INNER does not make a claim at the assertion level.   It does make a claim, but at another level and differently dealt with.  I would say (if we have to do this) that the derivation goes
naku lo pa broda cu brode = ro lo pa broda naku brode
                                      =  ge ro lo broda naku gi pa da broda
I don't, btw, think doing this is very enlightening, since it leaves out some interesting information.
As for a non-importing {ro} in that position, the rule should still apply, making non-sense of the whole.  But using importing {ro} also makes nonsense of the whole (well, a tautology -- or nearly so), so the result is that taking INNER as assertive is an error.  On the other hand, taking INNER as presuppositional makes the same sense regardless of the nature of {ro}.

&:
<<
> Even
> mathematicians and linguists pretty much get this right.

The the confusion may be about what "this" is.
>>
That "all" has existential import.   I guess I have to take back "linguists" -- but, gee, my people (Partee, Bill Bright, various Lakoffs) and McCawley had it right.

<<
>   I suppose you mean {ro broda cu brode} and {ro da poi broda cu
> brode} entail {da broda} (or you mean "implicate" rather than
> "entail").

I mean that {ro broda cu brode} and {ro da poi broda cu brode}
DON'T ENTAIL {da broda}. (Caps for emphasis, not shouting.)
That is, they are equivalent to {ro da ga na broda gi brode}.
>>
Yes, that is what you meant.  Sorry to type ahead of my head.

<<
In saying that, I'm just describing my habits of interpretation.

> It is quite true that for many people much of the time
> "All broda are brode" does not entail "There are broda,"  but by the
> same token, {ro broda cu brode} or {ro da poi broda cu brode} are not
> translations of that sentence (in that sense),

Right. As I understand it, this is your position, legitimately backed
up by an Argument from Authority, which I'm not confident I'm capable
of understanding, while Jorge takes the contrary view.

I am saying that I hope Jorge is right, so as to spare me having to
unlearn my habits. Of course, if I'm thereby committing some horrible
logical fallacy I would want to recant, but I don't (yet) see why
{ro broda cu brode} and {ro da poi broda cu brode} can't be strictly
equivalent to {ro da ga na broda gi brode}.
>>
They could be and, indeed, for some time were taken to be such by most people.  When I first tried to find some space for A+, I was content to claim {ro da poi broda cu brode} as a plausible place.  At which point, Cowan pointed out (correctly) that {ro} always implies {su'o} in Logic and that, therefore, I could claim the lot and reserve for A- the place that Logic gives it, namely the universal conditional. So, I took him up that, since it essentially what I really thought was the case anyhow.  But, if your habit is really to read "All S is P" as {ro da ganai S da gi P da} (with an importing {ro}, notice), then you don't have to change that habit, only the one about collapsing that farther.  And you can even collapse, if you are willing to concede that there are S's.
[Calling citation -- or the threat of such -- Argument from Authority is prejudicial, even when modified by "legitimately": loading.]

<<
My brand of English has "all" and "every" as nonimporting, and
"each" as importing, but "each" quantifies over a definite class
(i.e. it means "each of the"), so the importingness is probably
an artefact of the definiteness.
>>
I'll take your word for it, even though I have found (as have more formal empiricial researchers on the issue) that people are not very clear about this and often display patterns incompatible with their conscious beliefs on the topic.  In particular, though, people who allow both importing and non-importing meanings usually group "every" with "each" (as it is historically as well = "ever each"), so you constitute a group either new or too small to have been noted before.  Your explanation for the position of "each" probably accounts for your case, which is basically a "no importing" one.

<<
> The trick seems to be a metaconjuction that works at one level like
> an ordinary conjunction but at another level is not attached until
> all the other operations have been gone through (see some of the
> stuff about interdefining the various types of quantifiers earlier this year).

I take it that the 'operations' are 'gone through' from inside to
outside, i.e. mainly right to left in a Lojban-style syntax? That is,
if X has scope over Y, then Y is processed before X? In that case,
yes.
>>
You lost me there somewhere -- see the example above on INNER for what I have in mind, loosely.  But note that I would not really use {e} here, since if fails to flag the difference involved.  With a different conjunction, denying such a compound with {naku} would be the equivalent of {na'i}.

<<
If you have the logical formula:

  P and ASSERTED: Q

how should that be expressed grammatically so that it comes out
like

  Q PRESUPPOSED: and P
>>
I don't follow the formula, I think.  Suppose that P presupposes Q.  Then the whole situation is  "P funny-and Q."  Negating this would be "not P funny-and Q," {na'i}ing it
would be either "not(P and Q)" ("and" not at all funny) or (better) "not Q whether P" ("whether" = Lojban {u}). 

lioNEL:
<<
Indeed, I take the opposing views. As xorxes pointed it out, the whole
issue seems to decide wether the INNER part is claimed or presupposed.
IMO it is naturally claimed (the ro case being special, see below):
I would find it very strange, to say the least, to consider something
explicitly stated as something presupposed.
>>
Me too.  But INNER is not stated, merely displayed and, thus, open to a variety of interpretations, of which "presupposed" is one.  "Asserted" is another, but I can't find any cases of it actually working that way anywhere and many cases of the presupposing version, even without {ro}. 
--part1_174.f21596c.2ac1c8e9_boundary--