Re: A different definition of MINUS, Part 3
From: paul c <toledobythesea_at_oohay.ac>
Date: Fri, 19 Dec 2008 12:23:06 -0800
Message-ID: <PnT2l.28787$R43.4295_at_newsfe08.iad>
>
> There were typos, some weak, and some wrong assumptions! Here is
> corrected "intersection view update" statement:
>
> x ^ R00 = dx ^ R00 & % x and dx have the same headers
> y ^ R00 = dy ^ R00 & % y and dy have the same headers
> x ^ R00 = y ^ R00 & % x and y have the same headers
> (x ^ y) ^ R00 = dz ^ R00 & % x ^ y and dz have the same headers
> (dx ^ x) v R00 = R00 & % dx disjoint with x
> (dy ^ y) v R00 = R00 & % dy disjoint with y
> (dz ^ (x ^ y)) v R00 = R00 & % dz disjoint with x ^ y
> dx = dz v (x ^ R00) & % dx is a projection of dz
> dy = dz v (y ^ R00) % dy is a projection of dz
> -> (x v dx) ^ (y v dy) = (x ^ y) v dz. % Then, application of
> increments
> % on the base relations
> % is the same as increment
> % on join view
> .
> It can be proved as follows. Since dz, x and dx all have the same
> header, then dx = dz. The statement conclusion is nothing more than
> distributivity assertion under condition of all relations having the
> same header.
> ...
Date: Fri, 19 Dec 2008 12:23:06 -0800
Message-ID: <PnT2l.28787$R43.4295_at_newsfe08.iad>
vadimtro_at_gmail.com wrote:
> 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.
>
> There were typos, some weak, and some wrong assumptions! Here is
> corrected "intersection view update" statement:
>
> 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 ^ y) ^ R00 = dz ^ R00 & % x ^ y and dz have the same headers
> (dx ^ x) v R00 = R00 & % dx disjoint with x
> (dy ^ y) v R00 = R00 & % dy disjoint with y
> (dz ^ (x ^ y)) v R00 = R00 & % dz disjoint with x ^ y
> dx = dz v (x ^ R00) & % dx is a projection of dz
> dy = dz v (y ^ R00) % dy is a projection of dz
> -> (x v dx) ^ (y v dy) = (x ^ y) v dz. % Then, application of
> increments
> % on the base relations
> % is the same as increment
> % on join view
> .
> It can be proved as follows. Since dz, x and dx all have the same
> header, then dx = dz. The statement conclusion is nothing more than
> distributivity assertion under condition of all relations having the
> same header.
> ...
Thanks, that is more reassurance, that the problem has nothing to do with the algebra but rather either one of these must change to avoid the problem: 1) the definitions used by various languages including Tutorial D OR, 2) the typical dbms assumption that a single named 'table' with a constant heading always results. Received on Fri Dec 19 2008 - 21:23:06 CET