Re: Checking a relation in BCNF for MVDs

On 6 Jan 2003 21:20:25 +0100, you wrote in comp.databases.theory:

Thank you, Jan for your answer. I still have a heap of questions, though!

> Correct, provided that you mean with an all-key relation a relation
> that has only one candidate key and this candidate key contains all
> attributes.

That is what I meant.

> The trick is usually not to try all
> these possibilities but to read the description of the table and its
> contents and try to recognize formulations that indicate
> multi-valued dependencies.

If I understood it right, if I want to prove that a relation is in 4NF, I have to show there are no multi-valued dependencies in it (again, please correct me if I'm wrong; I'm only beginning to grasp those concepts). How can I prove it, though, if I do not take all possible combinations into account? Or is there a more elegant way to prove it?

And another question. Say I have the relations

r1(attr1, attr2)
r2(attr1, attr2, attr3)

The attributes [att1, attr2] form the primary key of r1 and they are a foreign key in r2, pointing to r1. r2 is an all-key-relation (as described above). Now, I have a FEELING that, since [attr1, attr2] is a foreign key, I can treat it as if it was ONE attribute in r2, and consequently do not have to check for MVDs in that relation any more. How can I formally justify this? Received on Wed Jan 08 2003 - 00:42:53 CET