Re: Relational symmetric difference is well defined

From: Vadim Tropashko <>
Date: Fri, 15 Jun 2007 14:53:16 -0700
Message-ID: <>

On May 31, 6:40 pm, "V.J. Kumar" <> wrote:
> Vadim Tropashko <> wrote
> > On May 30, 8:52 pm, Marshall <> 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. Received on Fri Jun 15 2007 - 23:53:16 CEST

Original text of this message