Re: Relational symmetric difference is well defined

On Jun 1, 3:58 am, "V.J. Kumar" <vjkm..._at_gmail.com> wrote:
> Marshall <marshall.spi..._at_gmail.com> wrote >
> >> The above formulas obviously are no longer first-order expressions.
> >> Along with the increased expressive power (e.g. it's trivial to
> >> define a powerset), you will reap the usual drawbacks of the higher
> >> order logic.
>
> > Can you expand on that?
>
> > Although it's clear the above operator(s) are different in
> > kind from join and union, it's not obvious (to me anyway)
> > that this difference translates into higher-orderness.
>
> {y|A(x,y)} is a set variable that can range over arbitrary sets of domain
> elements rather than just the domain elements. Similarly, {r|r<=R}, a
> powerset of R, where R is some relation and <= is the set containment
> symbol, is also a second order formula.

Thanks.

Marshall Received on Fri Jun 01 2007 - 16:23:59 CEST