Re: completeness of the relational lattice

From: Vadim Tropashko <vadimtro_invalid_at_yahoo.com>
Date: Fri, 22 Jun 2007 17:53:25 -0700
Message-ID: <1182560005.936433.46650_at_j4g2000prf.googlegroups.com>

On Jun 22, 5:38 pm, Jan Hidders <hidd..._at_gmail.com> wrote:
> On 23 jun, 01:05, Vadim Tropashko <vadimtro_inva..._at_yahoo.com> wrote:
> > A(R \/ B) = A(R) intersect A(B)
> > Note that function A is not a homomorhism, that is the dual identity
> > A(R /\ B) = A(R) union A(B)
> > doesn't hold.
>
> Really? It doesn't? Can you give a counterexample?

Oops, you are right. In my notation it corresponds to

00 /\ (R /\ B) = (00 /\ R) /\ (00 /\ B)

I confused it with the identity

00 \/ (R /\ B) = (00 \/ R) /\ (00 \/ B)

which doesn't hold.

So function A is a true lattice homomorphism to boolean algebra of header sets.

> > BTW, if you have finction A, then we probably don't need element 00,
> > right? (The element A(01) is just an empty set in the boolean algebra
> > of headers, not the lattice element 00)
>
> Not sure what you are asking. A is not really part of the algebra, but
> only used in describing the rules, so A(01) is not an expression in
> the algebra.

I cleared my confusion. What would be the result of join betwwen empty relation R and empty relation S with disjoint attributes? 00, of course. Received on Sat Jun 23 2007 - 02:53:25 CEST

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