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@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 Mon Jul 17 2006 - 19:11:36 CDT