Re: Relational symmetric difference is well defined

On Jun 19, 1:37 pm, "V.J. Kumar" <vjkm..._at_gmail.com> wrote:
> Jan Hidders <hidd..._at_gmail.com> wrote > > Really? For x="1", z="a" the formula says "forall y : A(1,y) <->
> > B(y,a)". For y=2 and y=3 the propositions A(1,y) and B(y,a) are both
> > true. For all other values for y both are false. So I would think the
> > formula holds for x=1, z=a.
>
> I was wrong by being pessimistic about the formula, but you are not right
> either. You formula is overly optimistic. Assuming quantification
> domains X = 1..100 and Z = 'a'..'z' and according to your predicate, the
> following pairs would be legit:
>
> (3, c), (4, d), (3, d),... etc., i.e. every pair such that A(x,y) and B
> (y,z) evaluate to false for each y. If you go ahead and try to set
> domain boundaries, you'd still need to deal with gaps.

If domain dependence is really a concern, the fix is trivial:

{(x,z)| (forall y : A(x,y) <-> B(y,z) & exists y: A(x,y) & exists y: B(y,z) }

or

{(x,z)| (exists y: A(x,y) & exists y: B(y,z) -> forall y : A(x,y) <-> B(y,z) } Received on Tue Jun 19 2007 - 23:14:20 CEST