Re: Can relvars be dissymetrically decomposed? (vadim and x insight demanded on that subject)
From: Bob Badour <bbadour_at_pei.sympatico.ca>
Date: Tue, 18 Jul 2006 00:11:36 GMT
Message-ID: <YGVug.11802$pu3.271497_at_ursa-nb00s0.nbnet.nb.ca>
>
> Well, the INSERT does an assignment after evaluating a relational
> expression - so doing an INSERT to a relvar R is the same as assigning
> to R the result of a relational expression involving R's current value.
Date: Tue, 18 Jul 2006 00:11:36 GMT
Message-ID: <YGVug.11802$pu3.271497_at_ursa-nb00s0.nbnet.nb.ca>
erk wrote:
> Cimode wrote:
>
>>[snip] >> >>>>[snip] >>>>For instance, if you consider the function F(X) = A(X) and you add >>>>the value B to F(X), you are basically doing a translation making a new >>>>function T(X) = F(X) + B = A(X) + B In this case, the result of adding >>>>B to the function can be expressed (characterized) as a mathematical >>>>translation (jumping from F(X) to T(X)). It says a lot about the >>>>function F(X) behavior and may help describe it better over time. >>> >>>But F isn't changing. You're creating a new function, not changing an >>>existing one. Assuming the expression F(X)+B makes sense, T has the >> >>Now that's finally getting interesting... If I follow your reasoning >>that would mean that once an INSERT is done, there would be necessarily >>a new relation resulting from the INSERT operation performed?
>
> Well, the INSERT does an assignment after evaluating a relational
> expression - so doing an INSERT to a relvar R is the same as assigning
> to R the result of a relational expression involving R's current value.
A simpler answer would have been 'Yes'. A relation value is a relation, and one doesn't usually qualify the term. One qualifies a relvar as a relation variable because a relvar is not a relation. The value the variable identifies is a relation.
[snip] Received on Tue Jul 18 2006 - 02:11:36 CEST