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

From: [email-masked]
Date: Sat, 4 Aug 2001 18:32:24 EDT

In a message dated 8/4/2001 4:14:14 PM Central Daylight Time, 
[email-masked] 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.