Re: Question about Date & Darwen <OR> operator

Mikito Harakiri wrote:
> paul c wrote:
> > Mikito Harakiri wrote:
> > > followed by collapse of identical disjunction terms. Likewise, the
> > > <AND> operation could be defined. Therefore, the D&D algebra
> > > intuitively looks like boolean algebra, but it is certainly not.
> > > Absorption, doesn't hold: relation headers monothonically increase, so
> > > that there is no way for headers to match. Therefore, this nice boolean
> > > logic must break somewhere. Where?
> > > ...
> >
> > i may be missing the question (wouldn't be the first time) regarding 'no
> > way for headers to match', but i presume that's why they have <REMOVE>
> > in A-algebra (they also require projection in TutorialDee). i'm assuming
> > you mean something like a db with a couple of hundred relations each of
> > which had, say, ten attributes of which only one is common with some
> > other relation, so the equivalent single relation for that database
> > would have something like 9 X 200 = 1800 attributes (plus many,many,many
> > tuples! likely, no person could compose that expression, but maybe
> > there's a reason for a machine to do it. i don't know why, though. if
> > that's what you mean by 'break', i guess you're right in theory.
>
> 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).

>
> Likewise,
>
> A <AND> (A <OR> B) = A

See above. Received on Tue Sep 06 2005 - 14:50:33 CEST