Re: Relational symmetric difference is well defined

On 20 jun, 13:57, "V.J. Kumar" <vjkm..._at_gmail.com> wrote:
> Jan Hidders <hidd..._at_gmail.com> wrote innews:1182338460.046186.165360_at_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