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Re: completeness of the relational lattice

From: Jan Hidders <hidders_at_gmail.com>
Date: Tue, 26 Jun 2007 21:07:11 -0000
Message-ID: <1182892031.923869.212910@q75g2000hsh.googlegroups.com>


On 26 jun, 22:29, Vadim Tropashko <vadimtro_inva..._at_yahoo.com> wrote:
> On Jun 26, 10:55 am, Vadim Tropashko <vadimtro_inva..._at_yahoo.com>
> wrote:
>
> > 13b: S /\ 00 = S and H /\ 00 = H and S \/ H != 00 and S /\ H = H'
> > imply
> > ((R /\ <S \/ 1E>) \/ H) /\ <S \/ 1E> = (R /\ <S \/ 1E>) \/ H'
>
> Hmm, the S \/ H != 00 condition doesn't seem to be required. So
> rewritten a little the axiom
>
> S /\ 00 = S and H /\ 00 = H imply
>
> ((R /\ <S \/ 1E>) \/ H) /\ <S \/ 1E> = (R /\ <S \/ 1E>) \/ (H /\ <S \/ 1E>)

But this is equivalent to :

 ((R /\ <S \/ 1E>) \/ H) /\ <S \/ 1E> = (R /\ <S \/ 1E>) \/ (H /\ S)

which cannot be correct if H and S have no common columns. If you are projecting all the columns away that are involved in the equalities and afterwards add them back then that is not the same as not removing them.

And, yes, having a negative premisse is a very big problem.

Received on Tue Jun 26 2007 - 16:07:11 CDT

Original text of this message

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