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Subject: Re: [lojban] {zo'e} as close-scope existentially quantified plural variable
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Martin Bays, On 04/11/2011 23:37:
> Let me try to clarify the basic problem I see with kinds, which
> I understand xorxes' dialect of lojban to suffer from. Then we can ask
> whether yours does too.
>
> Since we don't know exactly what we mean by 'kinds' (I certainly don't
> claim to), let me call the kind of kind which suffers from the problem
> a 'malkind'; the following description of the problem should be taken
> as a definition of malkinds.
>
>
> Suppose we have a forall-exists statement. Any at all will do;
> let's consider
>      "every French person wears a beret"
> and render it in lojban as
>      (A) {ro faspre cu dasni su'o ransedyta'u}.
>
> Then with malkinds, if (A) is true it would also be true that
>      (B) {su'o ransedyta'u cu se dasni ro faspre},
> i.e. we can just swap the quantifiers over and get another true
> statement.
>
> How does that work? For (B) to be true, we need something which
> ransedyta'us and which every French person wears. With malkinds, (A)
> being true implies that there is such a thing - namely the malkind
> corresponding to berets.
>
> (Various terms have been used which I take to refer to such a malkind,
> amongst them "berets", "the kind 'berets'", "Beret", and "Mr. Beret")

Hmm. This is well set out.

As I understand things, (B) is a legitimate entailment of (A) only if there=
 is only one beret, i.e. that all frenchmen wear the same beret. I think th=
at's the crucial point.

As for whether all frenchmen do wear the same beret, that depends on beret =
differentiation criteria. By the usual beret differentiation criteria, they=
 don't wear the same beret. But given that it is possible to say that we al=
l admire (the same) Obama and that millions of children each play with (the=
 same) Barbie, I think that it would be possible to think of a Barbie-like =
Beret that pops up on the heads of many different frenchmen. I leave open w=
hether the Barbie-like Beretrequires a different predicate from the berets =
that each pop up on only one head.

I note that natural language appears to be rampantly malkindful:
"What's *the hat that every frenchman wears*? *It* is a beret. *It* is worn=
 by every frenchman."
 =20
If I may be permitted to impute thoughts to you, I think your basic objecti=
on is to being able to consider Barbie-like Beret to be a beret. You don't =
object to "Every American should vote for an Obama" entailing "An Obama sho=
uld be voted for by every American", since you are happy to accept that the=
re is only one Obama. But you don't accept that there is only one beret.

> So in the sense that if we would say that one is true then we'd also say
> that the other is true, (A) and (B) are equivalent in malkindful lojban.

Do you mean they are truth-conditionally equivalent, or simply that each, w=
hen supplemented by auxiliary assumptions, can be inferred from the other? =
If Barbie-like Beret is a malkind, then (B) is derivable from (A) only if i=
t is also the case that all frenchmen wear the same beret; if they all wear=
 different berets, you can't derive (B).

So it seems to me that either (A) doesn't entail (B) malkindfully or that x=
orxesianism is not malkindful.

> If xorxes says (B) - or {ro faspre cu dasni lo ransedyta'u}, which
> appears to be approximately equivalent - I don't know how, beyond my
> prior knowledge of which was more likely, to tell whether he really means
> to make the surprising statement that all french people share a single
> beret, or just the (za'a also false!) statement that every french person
> wears a beret.

I think (B), at least with {pa ransedyta'u} rather than {su'o ransedyta'u},=
 means they share a single beret. And I think {ro faspre cu dasni lo ransed=
yta'u} means almost the same thing, except that it does not exclude the pos=
sibility that there is more than oneberet worn by every frenchman.

> Generally, with malkinds, the order of quantifiers in a sentence gives
> *no* information, at least until you bring in informal things like
> emphasis and convention.
>
> It seems to me rather obvious that this should be considered a problem!
>
> Can we agree on that much?

As said above, either malkindfulness is a problem but is not xorxesian, or =
malkindfulness is xorxesian but is not a problem (because the order of quan=
tifiers does matter in xorxesianism).
 =20
>>>> It'd be the individuative cmavo. I guess the one you call "Lion" is
>>>> used where X is a lion and Y is a lion but you don't know (or don't
>>>> say) whether X =3D Y.
>>>
>>> Err. Maybe. I don't think I understand you there.
>>
>> Well, I was just groping towards trying to get an understanding of
>> what Lion is.
>>
>>> But if you mean to consider kinds as equivalence classes of mundanes
>>> ("imaginary elements", in mathematical logic jargon), I may be with you=
.
>>
>> Explain a bit further, and then I might be able to say whether this is
>> what I mean.
>>
>> I don't know what equivalence classes and imaginary elements are, but
>> if they're what you get in situations where broda(X) and broda(Y), but
>> you don't know or choose not to say whether X=3DY, then maybe I'm saying
>> that whenever you say broda(Z), Z is one of these equivalence classes
>> thingos.
>
> It's more that you take your things and divide them up into sections,
> and consider two things to be equivalent if they're in the same section.
>
> The 'imaginaries' bit implies that the sections shouldn't be chosen just
> arbitrarily, but should be following some rule.
>
> So we decide that even though we were considering all the things to be
> different, we no longer care about some of the differences, and consider
> certain things to be the-same-as-for-present-purposes other things.
>
> We can then go one step further and consider these sections
> ("equivalence classes") as things in themselves ("imaginary elements").
>
> So this may not be quite what you seem to be saying, but it may be
> close.

It strikes me as very close or else bang on. I would consider everything to=
 be an imaginary element.

>> Sure, we know what the difference between one lion and two lions is.
>> But there are these cases where you can't tell the difference. And
>> I think that these cases in which the speaker can't tell the
>> difference should be generalized into a case where for whatever reason
>> the speaker doesn't tell the difference.
>
> But do we really need to create a new entity to do that? In examples
> like the "lion(s) in your garden every day", we can just give a vague
> count - {su'o cinfo}, in that case.

Yes, but it looks like one lion, not like a group of one or more lions. Con=
trast "the candidate your friends are going to vote for", which means they =
each vote for only one candidate, albeit not necessarily the same one, with=
 "the candidates your friends are going to vote for", which allows that the=
y each vote for more than one candidate.
 =20
>>>> If the difference between (i), (iii) and (iv) is that in (i)
>>>> disambiguation is by tense, in (iii) disambiguation is by special
>>>> individuating cmavo, and in (iv) disambiguation is solely by glorking,
>>>> then I reject (i) because I don't see how it could work,
>>>
>>> Do you see that it couldn't work?
>>
>> Yes. If {ko'a broda ko'e} you'd want to disambiguate both the criteria
>> by which ko'a counts as a single broda (be ko'e) and the criteria by
>> which ko'e counts as a single se broda (be ko'a). I don't see how the
>> tense system could do that.
>
> i.e. it works only on unary predicates?

Yes.

> But we're talking about using it on a descriptor, which takes a unary
> predicate anyway. In {lo broda be ko'e} there's only one place to
> deal with; similarly for {lo se broda be ko'a}.

So the same problem doesn't arise with {pa da broda pa de}? What I mean is,=
 yes we talking about {lo}, but if you need to indicate differentiation cri=
teria on the unary predicate complement of {lo}, why wouldn't you also need=
 to indicate differentiation criteria on predicate places in general?

> Do you mean that there's then a bootstrapping issue - we need ko'a to
> get ko'e and need ko'e to get ko'a?

No.

> Given e.g. a binary predicate P(x,y), which let's say is to start with
> defined only when x is a foo and y is quux,
> (i) has us define what P(X,Y) means where X is a bunch of foos and Y is
> a bunch of quuxs (here a bunch of foos corresponds to a set of foos);
> meanwhile, (iii) (or something like it) has us define what P(x/~, y/~)
> means, where x/~ is an imaginary foo - i.e. one of the new things we get
> when we consider a new, coarser notion of equality of foos - and y/~ is
> an imaginary quux.
>
> So we need to consider the properties of these new beasties X and x/~.
> One possible, arguably natural, scheme for this in the case of x/~ leads
> to the quantifier-permuting ambiguities discussed at the top of this
> post.
>
> Why is X better? Actually, it isn't - they're pretty much dual. The
> difference is *just* that we're allowing {su'o da} and {ro broda} to
> pick up things like x/~, but not to pick up things like X.
>
> So one solution (similar to something I've suggested in different
> language before) might actually be to allow these imaginaries in
> addition to bunches, and allow that those e.g. deriving from lions do
> themselves cinfo, *but* require that (usual singular) quantifiers do not
> pick them up.
>
> {lo}, meanwhile, could be defined to be allowed to pick up any of them.
>
> We might also define/clarify other quantifiers and gadri to be allowed
> to pick up various combinations of bunches and imaginaries.

Since I think everything is an imaginary, in the sense of being a generaliz=
ation over potential subtypes, something that doesn't pick up imaginaries d=
oesn't pick up anything.

> Other than the fact that I doubt I've made sufficiently clear what
> I mean by an 'imaginary', and the issue that I'm not really sure that
> what I mean by it covers all the cases you and xorxes want to be
> covered, I suppose your main problem with this would be that it still
> singles out a particular "layer" of e.g. lions to be the things picked
> up by {su'o cinfo}.

Yes.

>But I don't see a way around that if we want to
> solve the quantifier-swapping issue (which I really think we do).

I don't think there is a quantifier-swapping issue. There's only disagreeme=
nt on beret-counting. Or Obama-counting: if I don't agree that there is onl=
y one Obama, then I'd object to you claiming that "ro prenu cu prami su'o O=
bama" and "su'o Obama cu se prami ro prenu" are equivalent.
--And.

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