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[protege-discussion] is-a relationship

Nada Bajnaid nbajnaid at yahoo.com
Wed Nov 11 07:30:58 PST 2009


then can I inffer about the subclass relationship for ex: subclass (?x,?c) if x is-a subclass of c?

--- On Wed, 11/11/09, Timothy Redmond <tredmond at stanford.edu> wrote:


From: Timothy Redmond <tredmond at stanford.edu>
Subject: Re: [protege-discussion] is-a relationship
To: "User support for Core Protege and the Protege-Frames editor" <protege-discussion at lists.stanford.edu>
Date: Wednesday, November 11, 2009, 9:24 AM



On Nov 10, 2009, at 11:00 PM, Nada Bajnaid wrote:

> Thanks Auhood
>     Then can I have some SWRL rules with is-a and inffer about it?

Yes this is part of the SWRL language.  For instance to say that the SWRL variable ?x is a member of the Pizza class you can write

    Pizza(?x)

-Timothy


> 
> Regards
> Nada
> 
> --- On Tue, 11/10/09, A A <auhoodf at gmail.com> wrote:
> 
> From: A A <auhoodf at gmail.com>
> Subject: Re: [protege-discussion] is-a relationship
> To: "User support for Core Protege and the Protege-Frames editor" <protege-discussion at lists.stanford.edu>
> Date: Tuesday, November 10, 2009, 8:48 AM
> 
> 
> 
> I think the is-a relation is defined as a subclass relation implicitly. so you don't need to define it again.
> you can see the results when reasoning, e.g. if you write a query that finds all individuals of type persons  you should be able to get all females as well as all individuals defined as persons.
> 
> Auhood
> 
> 
> On 10 Nov 2009, at 12:26, Nada Bajnaid wrote:
> 
> > where could I read more about is-a relationship. if I have female subclass of person then we have female is-a person. Do I need to define object property and name it is-a or its already defined by the subclass relationship? and how I could use it in the reasoning?
> >
> > Thanks
> > Nada
> >
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