Re: pc answers

[jorge@PHYAST.PITT.EDU]

la djer cusku di'e

> My predicate calculus formula:
>
> E^!3(x) (remna (x)  E^!9(y)(gerku(y)  & pencu(x,y)))
>
> declares that there are exactly 3 humans and exactly 9 dogs, for the
> scope of the entire sentence.

That's what I thought (and it seems that I convinced you) but now that
I see it again, I think I was wrong. Assuming there is another "&"
between "remna(x)" and "E^!9(y)", then you are not claiming that only
three humans exist. Only that remna(x) and that other complicated
claim about x are both true of only three objects. Each separately
may be true of more.

In any case, the E^!9(y) is within the scope of the other, so it
doesn't in any way say that the nine y's are the same for every x,
only that for each x there are nine y's that fit that relationship.
So your formula admits that up to 27 dogs are being touched in all.

Now I think that your formula (with an additional "&") is a good
representation of {ci remna cu pencu so gerku}. Sorry about the
confusion. Your formula was right, just the interpretation wasn't.
The dogs touched are not limited to nine for the whole sentence.
Only for each of the x.

> ro lo ci remna ku ro lo ci gerku zo'u ra pencu ri

That would be, in your notation:

( E^!3(x) remna(x) ) & ( E^!9(x)(gerku(x) ) &
(x)(y) ( (remna(x) & gerku(y)) -> pencu(x,y) )

There are three and only three things that are human &
there are nine and only nine things that are dogs &
for every x that is human and every y that is a dog, x touches y.

Just out of curiousity, what are the "^" and "!" for? I assume
it is some standard notation, but I don't know it.

Jorge