Re: constraints in algebra instead of calculus

From: Aloha Kakuikanu <aloha.kakuikanu_at_yahoo.com>
Date: Fri, 15 Jun 2007 17:58:39 -0000
Message-ID: <1181930319.559285.148200@e9g2000prf.googlegroups.com>

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 - 12:58:39 CDT