And:
>I note also that if you want to say "b has exactly 1 friend", with my
>method you'd do it thus:
>
> set
> V
> |
> e--m-->d<--friend--b
> |
> #
> |
> V
> 1
>
## Yes, that makes sense.
>but this seems not to work for your method, because it doesn't rule out:
>
> set
> V
> |
> e--m-->d<--friend--b--friend-->f
> |
> #
> |
> V
> 1
## Hmmmm. I see the problem, but I don't think it can be ruled out simply
by the formal logic of the system. After all, 'P and not P' is
syntactically well formed, but has to be ruled out.
>
>Right. So is it the case that Friend1 Isa the Friend relation that
>points to d, and that this relation in turn Isa Friend (with no specified
>arguments)?
## Yes, if d is present.
>
>> >It would seem that by inheritance from d, a and c are friend of b.
>> >So where do 'friend1' and 'friend2' fit in?
>> ## No, in my analysis of "a and c are friends of b" there's no d. That's
>> only needed for the generalisation that all b's friends are women.
>
>OK, but what's the answer for the case where all b's friends are
>women?
## Then d is present, as the node corresponding to "d's typical friend".
Richard (= Dick) Hudson
Phonetics and Linguistics, University College London,
Gower Street, London WC1E 6BT.
+44(0)20 7679 3152; fax +44(0)20 7383 4108;
http://www.phon.ucl.ac.uk/home/dick/home.htm
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|