Re: Can relvars be dissymetrically decomposed? (vadim and x insight demanded on that subject)
Date: 19 Jul 2006 15:55:00 -0700
Message-ID: <1153349700.395423.279200_at_p79g2000cwp.googlegroups.com>
Butting in again - I'm still reading the paper Aloha posted a reference to last night, but wanted to check on something in this post.
Here goes :
Cimode wrote:
> No. I mean a relation can be evaluated just as a function can be
> evaluated mathematically speaking. When a relvar hold a specific
> relvalue in time, then and only the such evaluation represents an
> relvalue N tuple set. The definition of the relation itself can not be
> equated to either a specific relvalue nor to a relvar. Besides the
> same relation can be expressed using several relvars: it still is the
> same relation.
>
> For instance F(x) is expressed as 2 * x + 2 if you express another
> function T(y) as =
> 2 * y + 2 then F = T
>
> When the variable x in F or the variable y for T holds the value 2 both
> F and T = 2 * 2 + 2 = 6 . By analogy F and T are equivalent to
> relations R1 and R2, just as *x* and *y* are relvars, just as *6* is a
> specific relvalue to which relation R1 is evaluated when *x* and *y*
> have a specific relvalue. An inverse reasoning allow to establish that
> a relvalue is necessarily a relation as well.
>
So, would this mean something like :
relation := what we would otherwise think of as the relation header
(plus a mapping ?)
relvar := what we would otherwise think of as an attribute
relvalue := the result of 'evaluating' a 'relation' with the 'relvars'
filled in with appropriate values
R : A -> B -> C -> ???
R : A -> B -> C -> Boolean
But what are the implications of that ? (Presumably, the map of the function could only return True, otherwise could you directly assert a falsehood ?) Alternatively, what else could be the result type of an evaluation like this ?
[ snippage ]
Back to reading ... Received on Thu Jul 20 2006 - 00:55:00 CEST