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From:  Jan Hidders <hidders@gmail.com>
Newsgroups: comp.databases.theory
Subject: Re: Relational symmetric difference is well defined
Date: Wed, 20 Jun 2007 12:26:44 -0000
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On 20 jun, 13:57, "V.J. Kumar" <vjkm...@gmail.com> wrote:
> Jan Hidders <hidd...@gmail.com> wrote innews:1182338460.046186.165360@g4g2000hsf.googlegroups.com:
>
> > I think it is quite clear what I said: you take the formula "(forall
> > y : A(x, y) <-> B(y, z))" and you add "and (exists y : A(x,y)) and
> > (exists y : B(y, z))". That of course gives you "(forall y : A(x, y)
> > <-
> >> B(y, z)) and (exists y : A(x, y)) and (exists y : B(y, z))" which is
> > indeed equivalent with "(forall y : A(x,y) <-> B(y,z)) and (exists
> > y1 : A(x, y1)) and (exists y2 : B(y2, z))".
>
> Sounds good!  It should work now.

I'm so glad you agree. :-) Now, could it be that I've seen you
somewhere on a database conference?

-- Jan Hidders