Re: completeness of the relational lattice

On Jun 26, 2:31 pm, Vadim Tropashko <vadimtro_inva..._at_yahoo.com> wrote:
> On Jun 26, 1:48 pm, Jan Hidders <hidd..._at_gmail.com> wrote:
>
> > (R(x,y) * []) + (R(x,y) * W)
> > (22) = [x,y] + (R(x,y) * W)
> > (9) = ([x,y] + R(x,y)) * ([x,y] + W)
> > (10) = R(x,y) * ([x,y] + W)
> > (11) = R(x,y) * (([x] * [y]) + W)
> > (9) = R(x,y) * (([x] + W) * ([y] + W))
> > (32b) = R(x,y) * <x> * <y>
> > (28) = R(x,y)
>
> I don't think the axiom #9 (as written in one of your previous posts)
> is correct. First, it should be more restrictive than #8 and it looks
> less restrictive. Anyway, the first application of distributivity is
> correct. I doubt the second one. In the distributive law:
>
> A + (B*C) = (A+B) * (A+C)
>
> the attributes of B are required to be a superset of A, not subset.

Speaking of axom #9, I only now nocied that you applied earlier when proving the equality property:

• ((R(x,y) * <yz>) + [xy] (12)
• (R(x,y) + [xy]) * (<yz> + [xy]) (9)

In general

((R(x,y) * Q(y,z)) + S(x,y) != (R(x,y) + S(x,y)) * (Q(y,z) + S(x,y))

I admit being wrong about removing the S \/ H != 00 condition. Also I'm not quite positive that operating equalities of the kind <xyz>*<ab> is problem free as well... Received on Wed Jun 27 2007 - 00:17:16 CEST