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RE: RDF and OWL rules

From: Geoff Chappell <geoff@sover.net>
Date: Mon, 7 Apr 2003 00:13:07 -0400
To: "'Jos De_Roo'" <jos.deroo@agfa.com>
Cc: <www-archive@w3.org>
Message-ID: <026d01c2fcbb$feaa4210$835ec6d1@GSCLAPTOP>

> -----Original Message-----
> From: Jos De_Roo [mailto:jos.deroo@agfa.com]
> Sent: Sunday, April 06, 2003 9:20 AM
> To: geoff@sover.net
> Cc: www-archive@w3.org
> Subject: RE: RDF and OWL rules
> So I think that
> {:i a [ owl:intersectionOf (:A :B :C)]} => {:i a :A, :B, :C}.
> and
> {:i a :A, :B, :C} => {:i a [ owl:intersectionOf (:A :B :C)]}.

I can see this variation on it - it just requires that a class that is
the subClassOf A, B, & C has the same class extension as a class that is
an intersection of A, B, & C. 

Actually, I guess we can't know that the class extensions are the same -
after all, we know that the class defined by an intersection is fully
described (since the rdf list is closed to new additions), but we don't
really know that a class that is the subClassOf A, B, & C is not also a
subClassOf D. So maybe the best we can say is that a class that is the
subClassOf A, B, & C is a subClassOf a class that is an intersection of
A, B, & C.

So this way would hold (because of normal subClassOf rules)

{:i a :A, :B, :C} => {:i a [ owl:intersectionOf (:A :B :C)]}.

And the other would hold because an intersection of A, B, & C will be
subClassOf A, B, & C by definition

{:i a [ owl:intersectionOf (:A :B :C)]} => {:i a :A, :B, :C}.

> but I'm not seeing how to do that for all cases...
> Anyhow, I still think that [1] is OK at least we get [R2]

I guess I still don't see how [1] will work. I'm beginning to wonder if
the test is in fact valid (though I'm still far more inclined to believe
my understanding is incomplete). I'm not doubting that we can create
rules that prove the test, but want to be sure that the rules will be
truly licensed by the semantics.

It just seems that the most we can know is that a restriction on
cardinality and an intersection of restrictions on min and max
cardinalities (all with same values) must have the same class extension.
But that only lets us say that the classes are equivalentClass(es), not
that they are indentical (which is what the test claims). 

In the end it may not really matter since we have to axiomatize this
anyway (i.e. we're not reasoning on integers and sets to decide that the
class extensions must be the same). Still it makes me question my

> > I hope I've managed to explain myself :-)
> I think so ;-)
> > Regards,
> >

Received on Monday, 7 April 2003 00:16:09 UTC

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