Re: Checking a relation in BCNF for MVDs

Sebastian wrote:
>Hi,
>
>> A relation is in 4NF if for all non-trivial MVDs X->>Y the set X is a
>> superkey.
>
>This definition makes perfect sense to me now that I read a lot about
>4NF and compared various definitions. Anyway, our coursework is
>finished by now, and I think we've done a good job (IN SPITE of
>Conolly!).
>
>> and as a small reminder:
>>
>> An MVD X->>Y is trivial if and only if X + Y contains all attributes of
>> the relation.
>
>Hmm, don't want to be a smartass, but I remember reading somewhere
>that when X = Y, it is trivial as well!

Oops, yes, you are right of course. In fact the definition I gave is inconsistent because if X->>Y is trivial then its dual X->>(H-Y) with H the set of all attributes in the relation, should also be trivial. But if X = Y the my definition says that X->>X is not trivial but X->>(H-X) is trivial (because X + (H - X) = H), and that is a contradiction.

• Jan Hidders