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From:  Vadim Tropashko <vadimtro_invalid@yahoo.com>
Newsgroups: comp.databases.theory
Subject: Re: Relational symmetric difference is well defined
Date: Fri, 15 Jun 2007 14:53:16 -0700
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On May 31, 6:40 pm, "V.J. Kumar" <vjkm...@gmail.com> wrote:
> Vadim Tropashko <vadimtro_inva...@yahoo.com> wrote innews:1180628927.976321.267880@a26g2000pre.googlegroups.com:
>
> > On May 30, 8:52 pm, Marshall <marshall.spi...@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.

Well, natural join is the second order expression too:

A(x,y)/\B(y,z)  is  {(x,z)| {y|A(x,y)}intersect{y|A(y,z)} != {} }

So having the second order definition is not necessarily that bad.

The situation may be analogous to the fundamental theorem of algebra,
which states that the complex numbers field is algebraically closed.
The most common proof of the theorem (by Gauss) depends on the
analytic structure of C and uses second order arguments. However, one
can give a first order, purely algebraic proof of more general result
that if R is any real closed field, then C=R[i] is algebraically
closed.