Re: Question about Date & Darwen <OR> operator

vc wrote:
> Mikito Harakiri wrote:
> >
> > Let A be a relation with attributes x and y. Let B has attributes y and
> > z. Then,
> >
> > A <OR> (A <AND> B) != A
> >
> > , since the header of A <AND> B has attributes x,y,z. There is no way
> > the subsequent <OR> operation to reduce it to x,y.
>
> You are confused by mixing up the language and an interpretation. If
> you consider A <OR> (A <AND> B) as a predicate formula, then the
> identity holds. One can also claim that the identity holds for the
> interpretation as well if one drops the column(s) representing an
> entire attribute domain(s) from the result, very much in the same
> fashion as when one added them to the original relations (your own
> transformation).

I don't believe A <OR> (A <AND> B) will have any columns representing an entire domain, unless A does already.

(BTW, I find it scales better if we speak of A as having attribute sets a and ab, and B having attribute sets b and ab. Once we get into three relations, using x, y, z breaks down, because you have seven sets of attributes, and you'd never keep them straight if it was x, y, z, w, v, u, t.)

> > Likewise,
> >
> > A <AND> (A <OR> B) = A
>
> See above.

This result *will* have columns representing an entire domain; those from B that are not in A, which in my scheme is labeled b.

Marshall Received on Tue Sep 06 2005 - 16:34:34 CEST