Re: constraints in algebra instead of calculus

On Jun 15, 9:23 am, Vadim Tropashko <vadimtro_inva..._at_yahoo.com> wrote:
> Remarkably, it is shaped very similarly to FK constraint
>
> A \/ X B \/ X
>
> that was mentioned earlier in the thread.
>
> This is fine, but what are the implications? Can it be demonstrated
> that Armstrong axioms easily follow from this formulation?

BTW, transitivity of FK is easy. Consider 3 relations

A(x,u), B(x,y,v), C(x,y,w)

where x,y,u,v,w individual attributes or sets of attributes. Again, let X(x) be the empty relation with attribute x, and Y(x,y) be the the empty relation with attributes x and y. Then,

table B foreign key (x) references A(x)

(what a syntax!) is formally an inequality

B \/ X A \/ X

(As a reminder, the partial order " " that is "less (or equal) than" in the relational lattice has the two special cases: 1. when both relations have the same set of attributes and the first relation has a superset of rows of the second one, and 2 The empty relation has sbset of attributes of the other empty relation
The partial order, therefore, generalizes the set theory subset relation).

Likewise,

C \/ Y B \/ Y

or in conventional language

table C foreign key (x,y) references B(x,y)

Note that also X Y as the set of attributes of the empty relation X is a subset of that of Y.

It follows that

C \/ X B \/ X

The proof is somewhat tedious (and I wonder if a quicker method or just a reference to LT would suffice):

By inequality definition we have:
B \/ Y \/ C \/ Y = C \/ Y
X \/ Y = X
Prove:
B \/ X \/ C \/ X = C \/ X

Proof:
B \/ X \/ C \/ X = B \/ X \/ C = B \/ X \/ Y \/ C = = B \/ X \/ Y \/ C = X \/ B \/ Y \/ C = X \/ C \/ Y = = X \/ C

Now that we proved C \/ X B \/ X we derive by transitivity

C \/ X A \/ X

That was quite a long way to establish the obvious:

table C foreign key (x) references A(x)

! Received on Fri Jun 15 2007 - 19:24:59 CEST