[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]
Re: pc answers
djer's formula (plus an "&"):
> > > E^!3(x) ( remna(x) & E^!9(y)(gerku(y) & pencu(x,y)))
my comment:
> > 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.
pc:
> No. Although the dogs are in the scope of the men, they are not
> interdependent; this form is equivalent (with the & as you note) to the
> form with the dog and man quantifiers reversed, 9x(dog x & 3y (man y &
> touch y x)) Think of the And form, "there is a cimei and there is a
> somei..."
Wouldn't that be:
E^!3(x) E^!9(y) (remna(x) & gerku(y) & pencu(x,y))
Otherwise, how do you write the subordinate case in that notation?
Say a, b and c are the three men in question.
I want to claim:
( remna(a) & E^!9(y)(gerku(y) & pencu(a,y)) )
& ( remna(b) & E^!9(y)(gerku(y) & pencu(b,y)) )
& ( remna(c) & E^!9(y)(gerku(y) & pencu(c,y)) )
and furthemore, that a, b, and c are the only things that satisfy this.
Wouldn't the first formula say just that? Otherwise, why put the
E^!9(y) inside of the claim for x, it seems like a misleading notation.
Jorge