# Re: Date and McGoveran comments on view updating 'problem'

From: <vadimtro_at_gmail.com>
Date: Thu, 11 Dec 2008 13:48:37 -0800 (PST)

On Dec 11, 1:28 pm, vadim..._at_gmail.com wrote:
> On Dec 8, 4:13 pm, vadim..._at_gmail.com wrote:
>
> > As for deletion here is my worksheet
> ....
> > Now, we are proving the following decomposition
>
> > (SP v D')^(S v D') = (SP ^ S) v D'.
>
> This assertion is wrong. If D is the set of tuples that we delete from
> SP ^ S, then the condition is:
>
> (SP ^ (D' v (SP ^ R00))) ^ (S ^ (D' v (S ^ R00))) = (SP ^ S) ^ D'
>
> Informally, on the right side we have a relation which is the result
> of deleting D tuples from SP ^ S. On the left side is a relation that
> is a join of two base relations SP and S, each trimmed with some
> projection of D.
>
> I fail to derive it with any assumptions, except the obvious, like
> demanding that D acted only on SP (or S). Therefore, there might be
> some truth to people questioning the validity of deletion from join
> view...

Ugh, wrong again! Here is the deletion case solved:

1. (SP ^ S) ^ R00 = D ^ R00 "Tuples D are deleted from SP ^ S, therefore they have the same header"
2. (D ^ (SP ^ S)') ^ R00 = R00 "D is subset of SP ^ S"

The reason why I was confused is that set containment can be written in "more intuitive" fashion as

D ^ (SP ^ S) = SP ^ S

and this appeared to be a weaker condition! Received on Thu Dec 11 2008 - 22:48:37 CET

Original text of this message