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RE: SEM: All OWL reasoners will be incomplete?

From: Jeremy Carroll <jjc@hpl.hp.com>
Date: Fri, 06 Sep 2002 15:08:04 +0200
To: Ian Horrocks <horrocks@cs.man.ac.uk>
Cc: www-webont-wg@w3.org
Message-id: <BHEGLCKMOHGLGNOKPGHDKEALCAAA.jjc@hpl.hp.com>


> Given a sufficiently expressive language (i.e., one like OWL), this
> doesn't represent any restriction w.r.t. what you ask for below, i.e.,
> being able to determine if ontology O1 models ontology O2. This is
> trivially reducible to ontology consistency.

IIUC only if you can negate an Ontology.

not(prop rdf:type owl:TransitiveProperty )

is not in OWL.



>                 Similarly
> for a DL, the frame axioms might state that there is a subset of the
> set of all property names (the transitive properties) whose
> interpretations must be transitively closed. 

In terms of Pat's document then

If E is:
owl:TransitiveProperty
then x is in ICEXT(I(E)) iff:
<y, z> and <z, u> in IEXT(x) implies <y, u> in IEXT(x)

should be weakened to

If E is:
owl:TransitiveProperty
then [[ x is in ICEXT(I(E)) implies
{ <y, z> and <z, u> in IEXT(x) implies <y, u> in IEXT(x) }
]]



I still haven't understood which way you prefer to go on this.
Sorry. (I found your message quite difficult :( ).
e.g.
>Having said that, the expressive power of OWL means that entailment
>w.r.t. ontologies containing property inclusion and transitivity
>axioms is still trivially reducible to ontology consistency.

Could you please give an example:

e.g.

{ empty }
does not entail
{ foo rdf:type owl:TransitiveProperty }

how do I convert that into a consistency question?

Jeremy
Received on Friday, 6 September 2002 09:08:22 GMT

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