Re: Relational symmetric difference is well defined

On 15 jun, 16:00, "V.J. Kumar" <vjkm..._at_gmail.com> wrote:
> Jan Hidders <hidd..._at_gmail.com> wrote innews:1181910061.495472.280190@c77g2000hse.googlegroups.com:
>
>
>
> > On 1 jun, 03:40, "V.J. Kumar" <vjkm..._at_gmail.com> wrote:
> >> Vadim Tropashko <vadimtro_inva..._at_yahoo.com> wrote
> >> innews:1180628927.976321.267880_at_a26g2000pre.googlegroups.com:
>
> >> > On May 30, 8:52 pm, Marshall <marshall.spi..._at_gmail.com> wrote:
> >> >> Can you clarify the difference between set containment join and
> >> >> set equality join? The inverse of join is much on my mind these
> >> >> days.
>
> >> > Set equality join
>
> >> > A(x,y)/=B(y,z) is {(x,z)| {y|A(x,y)}={y|A(y,z)} }
>
> >> > Set containment join
>
> >> > A(x,y)/=B(y,z) is {(x,z)| {y|A(x,y)}>{y|A(y,z)} }
>
> >> > where the ">" is "subset of".
>
> >> 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.
>
> > This was perhaps already clear, but it is the *formulation* of the
> > semantics which is not first-order. The semantics themselves are
> > clearly first order since they can be defined in first order logic or
> > the flat relational algebra.
>
> This is very intriguing !

Not really. It is pretty obvious that in the above formulation of the semantics of the joins you can replace the higher order expression with a first order formula.

• Jan Hidders