semantics of "most"

[cbmvax!uunet!gnu.ai.mit.edu!grackle!bob]

   I don't know if I've explained this terribly clearly. I haven't
   found any reference which states the problem with "most" (and some
   other quantifiers) in an intuitive way...

It appears to me that the mathematics you were trying to express are
based on the metaphor of a container, with your only permitted test
being whether something is in or is not in the container.  Using this
metaphor, it is easy to demonstrate what you are trying to show; it is
also easy to show how to handle "most".

By `container', I mean `box', `room', `body' or sand castle; the sorts
of things that children learn have an inside and an outside.

In this case, we have more than one container, and entities are in
more than one container at the same time.  (I am in a chair now, and
in a room, at the same time; however, the one container does not have
to be inside the other.)

   we can translate "Two
   men walk" as "two x.(man(x) & walk(x))", i.e.  "Ex.Ey.(x /= y &
   man(x) & man(y) & walk(x) & walk(y))".

There exist x and y 
such that x and y are not identical 
and x is located in the container `man' 
and y is located in the container `man'
and x is located in the container `walk' 
and y is located in the container `walk'.

It is easy to imagine walking as an attribute of `man' so you can
visualize this metaphor as two dolls in a box marked `walk' which is
in a bigger box marked `man'. 

   Secondly, I should make it clear what I mean by "most". A
   paraphrase is "more than half of", i.e. "Most men walk" is
   equivalent to (1) "more than half of the individuals who are men
   are individuals who walk."

   Now, look at how we translate sentences into first order logic and
   interpret the resulting formulae: English	A man walks.
   FOL	Ex.(man(x) & walk(x)) Interp	Examine the universe U, and
   see if you can find at least one individual i from U for which
   man(i) and walk(i) both hold.

Examine the boxes: do you find at least one doll in both the `walk'
box and the `man' box?

   English	Every man walk.
   FOL	Ax.(man(x) => walk(x)) Interp	Examine the universe U, and
   check that for every individual i from U, if man(i) holds, then
   walk(i) holds. (And if man(i) does not hold, we don't care if
   walk(i) does or not.)

Check the boxes:  is every doll that is in a `man' box also in a
`walk' box?

Now to a problem:

   English	Two men walk.
   FOL	2x.(man(x) & walk(x)) Interp	Examine the universe U, and
   see if you can find at least two different individuals i from U for
   which man(i) and walk(i) both hold.

This sneaks in a new idea, which is counting; you cannot `find at
least two different individuals' unless you can count.  (You can find
that you have different individuals because they are not identical.
But you don't have numbers, which means you won't know that the
existance of non-identical entities implies at least two entities.)

My understanding is that the problem with `most' occurs if you are
restricted to determining membership and equality.  If you can count
and do other related operations, you can handle `most'.

   And if we try it for "most":
   English	Most men walk.  
   ??FOL	Most x.(man(x) c walk(x)),
   where c is a logical connective.  Interp	Examine the universe
   U, and see if more than half the individuals i in U are such that
   the fact of i being a man is related by c to the fact of walking.

You cannot "Examine the universe U, and see if more than half"  unless
you have a test for determining halfness.  If your only tests are
whether a doll is in a box and whether it is different from another
doll, you cannot determine halfness.  

   One solution which is widely used is to adopt what are called
   "generalized quantifiers". Now, instead of translating a sentence
   into a quantifier over a variable, and a formula which uses that
   variable, we express the quantifier as a relation between two sets.

   Thus English	A man walks.  GQ	The set of men has a non-empty
   intersection with the set of walkers.

New container image!  Sitting on the beach you construct sand
enclosures:  walls of sand around a small area.  You can see whether
someone has put dolls into the courtyard of your sandcastle.

In the `non-intersection' case, the `man' castle and the `walk' castle
enclose different areas of the beach.  In the `intersection' case, one
castle is inside the other, or, at least, some of the territory is
shared by both.

Thus:

   English	Every man walks.  GQ	The set of men is a subset of
   the set of walkers.

And:

   English	Two men walk.  GQ	The intersection of the set of
   men and the set of walkers contains at least two individuals.


But now we have trouble again:

   English	Most men walk.
   GQ	The intersection of the set of men and the set of walkers
   contains at least half as many members as the set of men.

How do you say "at least half" unless you count?


But now, suppose you are able to count...

Two ways to determine more than half.  

Without counting , you can pair off the dolls, creating two equal sets
and possibly a third set of one doll left over.  Then you can pair off
the set of dolls that are both in the set of men and the set of
walkers and see whether you have left over dolls.  If so, you have got
`at least half'.  (This is the usual technique for comparing
infinities.)

Alternatively, you can count the dolls in each set and do arithmetic.
`At least half' becomes the name of a test procedure involving
counting, adding, division, and comparing:

  `At least half' is true if the number of dolls that are both in the
  set of men and the set of walkers is greater than the number of
  dolls in the set of all the dolls divided by two.

Looking at it this way, it appears to me rather obvious that you
cannot handle "most" with a logic in which you cannot count (or pair
and compare). 

Of course, a mathematician could argue that if you can determine that
you have different individuals because they are not identical, this
makes introducing the notion of `more than one' so simple that you can
use the idea in your thinking; and if you can go this far, you
can compare sets pair by pair.....which suggests that you can handle
`most'....  (I think that this was all done a century ago.)

Anyhow, Lojban does have numbers and counting predicates.  What this
tells me is that if you want a lojban computer to handle most, you
will need to program the computer to compute (or at least, give it the
ability to pair and compare).

The usual statement about the logical languages is that their grammar
is "based on logic", not that it is limited to a small set of logical
operations.

    Robert J. Chassell               bob@gnu.ai.mit.edu
    Rattlesnake Mountain Road        (413) 298-4725 or (617) 253-8568 or
    Stockbridge, MA 01262-0693 USA   (617) 876-3296 (for messages)