Re: [lojban] du'u & ka (was: ce'u (was: vliju'a

[pycyn@aol.com]

In a message dated 8/4/2001 4:14:14 PM Central Daylight Time,
xod@sixgirls.org writes:

> I *think* I recall a weak consensus that du'u = ce'u-less ka,
> which implies that ka must contain an implicit or explicit
> ce'u.




I recall Cowan saying something like this too. I would love to hear more
about this!

Whether Cowan said it or not, it makes a good deal of logical sense.  On one
popular view, a proposition is a function from worlds into truth values. On
that view, a n-place property is a function from worlds into the set of sets
of n-tuples in the domains of the worlds.  A property defined without a
{ce'u} is 0-placed and so, in each world it gives the set of 0-tuples that
satisfies it.  Since there is only 1 0-tuple (0), there are only two such
sets, 0 and {0}.  The result is thus isomorphic to a proposition at least.  
Of course, there are other ways of doing things that give quite different
results.