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

Date: Thu, 11 Dec 2008 14:24:48 -0800 (PST)

On Dec 11, 1:48 pm, vadim..._at_gmail.com wrote:
> 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...
>
> 2.  (D ^ (SP ^ S)') ^ R00 = R00
> "D is subset of SP ^ S"

Join instead of union typo again! These pendulum swings become entertaining...

There is little lesson here. The set containment expressed via empty intersection is a theorem in RL:

x ^ R00 = y ^ R00 ->
((x ^ y') v R00 = R00
<-> x ^ y = x). Received on Thu Dec 11 2008 - 23:24:48 CET

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