Re: A different definition of MINUS, Part 3
From: <vadimtro_at_gmail.com>
Date: Fri, 19 Dec 2008 11:56:49 -0800 (PST)
Message-ID: <b03dd5c7-2b0d-4833-95a4-b759ff9af538_at_w39g2000prb.googlegroups.com>
Date: Fri, 19 Dec 2008 11:56:49 -0800 (PST)
Message-ID: <b03dd5c7-2b0d-4833-95a4-b759ff9af538_at_w39g2000prb.googlegroups.com>
On Dec 19, 10:29 am, vadim..._at_gmail.com wrote:
> x ^ R00 = dx ^ R00 & % x and dx have the same headers
> y ^ R00 = dy ^ R00 & % y and dy have the same headers
> x ^ R00 = y ^ R00 & % x and y have the same headers
> x ^ R00 = dz ^ R00 & % x and dz have the same headers
> (dz ^ (x ^ y)) v R00 = R00 & % dz disjoint with x ^ y
> (x v dx) ^ (x v dy) = (x ^ y) v dz % application of increments on
> the base relations
> % is the same as increment on
> join view
> -> x v dx = x v dz.