Re: What to call this operator?
From: Jon Heggland <heggland_at_idi.ntnu.no>
Date: Fri, 1 Jul 2005 11:11:07 +0200
Message-ID: <MPG.1d2f27139bce9f1f9896c1_at_news.ntnu.no>
Date: Fri, 1 Jul 2005 11:11:07 +0200
Message-ID: <MPG.1d2f27139bce9f1f9896c1_at_news.ntnu.no>
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....
--
Jon
Received on Fri Jul 01 2005 - 11:11:07 CEST