Re: Can relvars be dissymetrically decomposed? (vadim and x insight demanded on that subject)

From: Cimode <>
Date: 18 Jul 2006 03:45:05 -0700
Message-ID: <>

Bob Badour wrote:
> 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.

*Strong* caution must be put here before assuming relations and relvalue are synonyms. relvalues are not to be confused with relations in abstract thinking. The only thing that can be said is that a relvalue is *necessarily* the *evaluation* of a specific relation. An N tuples relvalue necessarily constitutes the *evaluation* a relation R1 but does not equate to R1.

Question is how a relation can be defined or specified otherwise than by the relvalues it may evaluate to? When a function F(x) is defined as 2 * x + 2 such definition is independent from the values to which the function is evaluated. The value F(2) is an evaluation of function F when the variable x = 2. Stating that F(x) = 2 changes the nature of the function.

The*usual* vulgarization intended (shorthand) meaning or *usage* of the term relation does not constitute a proof.

> [snip]
Received on Tue Jul 18 2006 - 12:45:05 CEST

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