[protege-discussion] is-a relationship

[Timothy Redmond]

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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