Re: A different definition of MINUS, Part 3

vadimtro@xxxxxxxxx wrote:
On Dec 19, 10:29 am, vadim...@xxxxxxxxx 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.
.

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