Message-Id: <m0sRrgh-0000YjC@xiron.pc.helsinki.fi>
Date:         Fri, 30 Jun 1995 21:48:34 EDT
Reply-To:     jorge@PHYAST.PITT.EDU
Sender:       Lojban list <LOJBAN@CUVMB.BITNET>
From:         jorge@PHYAST.PITT.EDU
Subject:      Re: pc answers
To:           Veijo Vilva <veion@XIRON.PC.HELSINKI.FI>
Content-Length: 1271
Lines: 36

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