Re: What to call this operator?
Date: 1 Jul 2005 10:13:01 -0700
Jon Heggland wrote:
> In article <1120156799.149832.136240_at_o13g2000cwo.googlegroups.com>,
> mikharakiri_nospaum_at_yahoo.com says...
> > > x y <= A
> > >
> > > That's right, instead of bracket notation A(x,y) saying that relation A
> > > has attributes x and y, we can just write "A >= x y" implying that A is
> > > a superset of join of empty relations x and y.
> > On a symmetrical note, lets use capital letters X, Y etc to denote an
> > infinite relation each of which is a full domain. Then
> > A <= X Y
> Nice, but is it not also so that A >= x y z and A <= X Y Z for A(x,y)?
> Which makes that notation less useful....
No, A <= X Y Z is false:
A X Y Z != A
A X Y Z != X Y Z
A union X Y Z != A
A union X Y Z != X Y Z
(In the above formulas we assume that join takes precedence over union, which allows some bracket economy).
However, A <= X and A <= Y.
A >= x y z formula can be interpreted that the A(x,y) predicate can be in fact be written as A(x,y,z)! Received on Fri Jul 01 2005 - 19:13:01 CEST