Re: Newbie question about db normalization theory: redundant keys OK?

"David Portas" <dportas_at_gmail.com> wrote in message news:98a2073c-612a-4f98-bf79-a0d441b99313_at_d21g2000prf.googlegroups.com...
> On Dec 14, 7:49 pm, "David Cressey" <cresse..._at_verizon.net> wrote:
> >
> >
> > Someone else questioned my categorization of DKNF as "final". What I
meant
> > was this: there are no normal forms more restrictive than DKNF. This
has
> > been proven, although I have not seen the proof, and maybe wouldn't be
> > able to follow it if I did. The other participant pointed out that
further
> > table decomposition is possible. That's true, but that doesn't mean
that
> > the resulting schema is of a higher normal form.
> >
>
> Define "more restrictive". My interpretation is that you think a
> schema in DKNF must also be in 6NF, 5NF, ... etc. This is false
> though. Example:
>
> R{a*,b,c}
>
> where a* is the only candidate key.
>
> Given the dependencies {a}->{b} and {a}->{c} relation R IS in DKNF but
> NOT in 6NF. Does this mean 6NF is "more restrictive"? It is true to
> say that 6NF is not a proper subset of DKNF because there are some 6NF
> schemas that are not in DKNF.
>

As I already said, 6NF baffles me. Why is the relation above not in 6NF?

> --
> David Portas
Received on Sat Dec 15 2007 - 12:21:32 CET