# Re: Sixth normal form

Date: Fri, 10 Aug 2007 13:08:53 -0000

Message-ID: <1186751333.018671.305210_at_j4g2000prf.googlegroups.com>

On 9 aug, 04:15, "Brian Selzer" <br..._at_selzer-software.com> wrote:

*>
**>
*

> My use of the term "independent" is not the same as that defined by

*> Rissanen. I consider a projection over a set of attributes A to be
**> independent if and only if every dependent set of attributes determined by
**> any subset of A is also a subset of A. In other words, a projection over a
**> set of attributes A is independent if and only if for each functional
**> dependency X --> Y in the closure of the set of functional dependencies for
**> the relation schema containing A, if X is a subset of A then Y must also be
**> a subset of A.
*

Very good, finally a proper definition. So you apprently meant it as a unary predicate over sets of attributes and not as a binary relationship between sets of attributes. Let me see, if we have a relation R1(A,B,C) with an FD A->B then you call the following sets independent: {}, {B}, {C}, {A,B}, {B,C}, {A,B,C}. Correct? And if we have R2(A,B,C) with A->B and A->C then the following are independent: {}, {B}, {C}, {B,C}, {A,B,C}.

> Consider a simple relation schema, {A, B, C} such that A --> B and B --> C.

*> Rissanen would consider both of the projections {A, B} and {B, C}
**> independent. I don't.
*

Indeed. You only call {B,C} independent.

> The closure of the set of functional dependencies

*> includes A --> C, which can only be preserved by the inclusion dependency,
**> {A,B}[B] IN {B,C}[B].
*

Not necessarily. That depends on your definition of FDs over attributes in different relations. The usual definition in normalization theory is that they hold for a schema if they hold for the natural join of all relations in the schema. In that case the FD is preserved also without the inclusion dependency.

- Jan Hidders