Re: LAhE and quantifiers

xorxes:
> Some considerations about LAhE and quantifiers.
> 
> 1- lu'a an individual/member/component of
> 
> CLL does not say anything (that I could find) about default
> quantifiers for LAhEs. I won't assume any for now.
> 
> I take lu'a to work thus:
> 
> ro lu'a ko'a - each member of ko'a
> ro lu'a ko'a e ko'e - each member of both ko'a and ko'e
> ro lu'a ko'a a ko'e - each member of either ko'a or ko'e or both
> 
> {ro lu'a ko'a e ko'e} is equivalent to {ro lu'a ko'a ku'a ko'e},
> each member of the intersection of ko'a and ko'e, and {ro lu'a
> ko'a a ko'e} is equivalent to {ro lu'a ko'a jo'e ko'e}, each member
> of the union of ko'a and ko'e.
> 
> {ro lu'a ko'a enai ko'e} is each member of ko'a that is not a 
> member of ko'e, etc.
> 
> Similarly {ro lu'a ro lo broda} is each member of every broda
> (i.e. each member of the intersection of all broda) and 
> {ro lu'a su'o lo broda} is each member of at least one broda,
> i.e. each member of the union of all broda. For these to make
> sense, brodas have to be things with members.
> 
> {su'o da zo'u ro lu'a da} is different from {ro lu'a su'o da}.
> The first gives, for some X, each member of X. The second gives
> each member of at least some X (i.e. everything).
> 
> {ro da zo'u ro lu'a da} is different from {ro lu'a ro da}.
> The first is for each X, each member of X. The second
> one is each member of every X, (i.e. nothing).
> 
> We can expand {PA1 lu'a PA2 da} as {PA1 de poi PA2 da zo'u de cmima da} 

All this has to be the default story, since it follows inevitably from
the left-to-right scope rule & we would need a very very compelling
reason to go against it.

> 2- lu'i a set formed from
> 
> Here I'm not so sure. Is {lu'i ko'a} some set that has ko'a
> as one of its members, or is it the set that has ko'a as its only
> member, or something else? 

The something else might be the membership qua Group.

I think the *intent* of lu'i is clear -- it is somehow supposed to
magic a distributive into a set. I.e. lu'i ko'a e ko'e is intended
to be equivalent to ko'a ce ko'e. Getting that to work is quite a
tall order, though. Here's how it might be done:

(tu'o) lu'i ko'a e ko'e
= the set that includes ko'a e ko'e and that excludes ro da poi 
ko'a e ko'e na du ke'a
= the set {ko'a, ko'e}

> If we take it to be the reverse of lu'a, we would expand 
> {PA1 lu'i PA2 da} as {PA1 de poi PA2 da zo'u de se cmima da}.
> 
> So {ro lu'i ko'a} would be each set that has ko'a as a member,
> not necessarily as its only member. I have no idea whether this 
> is a useful notion or not. {su'o lu'i ko'a} is some set that
> has ko'a as one of its members. {ro lu'i ko'a enai ko'e} would 
> be each set that has ko'a as a member and does not have ko'e as 
> member, etc. {ro lu'i ro lo broda} is each set that contains all
> broda, {ro lu'i su'o lo broda} is each set that contains at least 
> some broda.

This is (a) correct & (b) pretty useless.

However, redefining {lu'i} as "set that excludes everything except"
gives more useful results.

> If we take the complement to be the full list of the members,
> we run into trouble with things like {lu'i ko'a enai ko'e}.
> How is that different from {lu'i ko'a}? 

On my suggestion: {su'o lu'i ko'a enai ko'e} 
= a set that includes ko'a and excludes ko'e

> And what is {lu'i ko'a a ko'e}?

{su'o lu'i ko'a a ko'e}
= a set that includes ko'a or ko'e and excludes everything thst is
either not ko'a or not ko'e. IOW, something that is the set {ko'a} 
or the set {ko'e}.

> A third possibility is that ko'a itself has members and then
> that {lu'i ko'a} is the set whose members are the members of
> ko'a. This view has been offered in the past, but I don't 
> think it is worth defending.

Agreed.

> 3- lu'o a mass formed from
> 
> This one has possibilities analogous to those of lu'i.
> 
> So, while {lu'a} seems to be fairly clear, the functioning of 
> {lu'i} and {lu'o} is not at all clear to me.

I think my suggestion for lu'i, which would apply also to lu'o,
is perhaps the best compromise between usefulness and consistency.

However, whereas {lu'a} ought happily to take either a set or a
mass as its argument, lu'i and lu'o would take only 'individuals'
(i.e. members/constituents). So {lu'i lu'o} and {lu'o lu'i} would
not convert a mass into a set and vice versa.

--And.