Re: [xdi] Minutes: XDI TC Telecon Thursday 1-2PM PT 2009-02-12

[Giovanni Bartolomeo <giovanni.bartolomeo@uniroma2.it>]

Dear Drummond,

Thanks for these detailed minutes. Let me check if I've understood correctly:
In this example, let us assume that +ball/+color/+ball+color is a 
valid dictionary entry (class definition).
Let suppose +aBall/$is$a/+ball (i.e. +aBall is an instance of +ball).
Since +aBall inherits properties from +Ball, I can say 
+aBall/+color/+red, which is an "instance level" predicate.
But I cannot say +aBall/+color/+ball+color (right?)
Main question: would +aBall/+color/+aBall+color make any sense?
Second question: how to know that +aBall is "an instance"? $is$a 
might also mean "subclassOf", right?

Thanks (and sorry for being so obsessive on this topic, countless 
nights spent on this :-(,

Giovanni



At 06.34 13/02/2009, Drummond Reed wrote:
>We next discussed another concern of Giovanni's with some "strange"
>statements that can be made with $has. This is related to this email:
>
>         http://lists.oasis-open.org/archives/xdi/200901/msg00054.html
>
>It is the example about balls and their colors, where +x is +ball and +y is
>+color. This produces the XDI RDF statement +ball/+color/+ball+color. But in
>the same document we have also +ball/+color/+red. So you have two different
>objects of the same predicate, which is fine in RDF, but one (=red) is an
>instance of color and the other (+ball+color) is a subclass of color.
>
>After some discussion, we realized that what was deceiving about these
>examples is that they confuse the dictionary/class space (Bill would say
>"T-box") from the individual/instance space ("A-box"). For example, in some
>email threads we had talked about having +ball/+color/+ball+color and
>+ball/+color/+red in the same graph. While this is technically possible, it
>doesn't make sense because the first statement is appropriate to a
>dictionary and the second would be more appropriate to an instance, except
>that +ball is not a typical instance. More typical would be
>=jbradley+ball/+color/+red.