Re: constraints in algebra instead of calculus

On Jun 15, 9:24 am, Vadim Tropashko <vadimtro_inva..._at_yahoo.com> wrote:
> 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):

Actually, there is much easier derivation of the

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

We have:

B \/ X < A \/ X
C \/ Y < B \/ Y

Join the first inequality with Y and the second inequality with X:

B \/ X \/ Y< A \/ X \/ Y
C \/ Y \/ X < B \/ Y \/ X

By transitivity we have

C \/ Y \/ X < A \/ X \/ Y

Now that Y \/ X = X it reduces to:

C \/ X < A \/ X Received on Fri Jun 15 2007 - 19:58:39 CEST