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Re: Question on DL negation

From: Bijan Parsia <bparsia@cs.man.ac.uk>
Date: Mon, 5 Mar 2007 15:28:31 +0000
Message-Id: <10E9886A-DBDA-49F0-8DE4-6AE94BCC9B1C@cs.man.ac.uk>
Cc: Semantic Web <semantic-web@w3.org>
To: Matt Williams <matthew.williams@cancer.org.uk>

On 5 Mar 2007, at 15:13, Matt Williams wrote:

> Dear Bijan,
>
> Thanks a lot - very helpful, as usual.
>
> The approach with nominals is interesting - I'll have a play and  
> see what happens.
>
> I guess what I missed from my first question is that if:
>
> \exists hasRole.\top
>
> is a valid class expression

It is.

> (which I think it is) then:
>
> (\exists hasRole.\top)
> should be valid.

It is too. But this says the same thing as:
	\forall hasRole \neg\top
or
	\forall hasRole \bot

Which means if I am an instance of this class I hasRole to *nothing*,  
that is, I have no such roles at all.

This is a bit different that "bob doesn't love mary". I.e., in order  
to say "bob doesn't love mary" you have to say that bob doesn't love  
*anyone*. Poor bob!

> But since adding \top to the formula doesn't seem to add anything,  
> could one write (\exists hasRole) as a shorthand? I think the  
> answer is no, but I'm not clear why.

The shorthand is fine if it implies the above. Indeed, think of  
unqualified number restrictions which could be equivalently written  
via qualified number restrictions and top.

It might help to do the standard translation into FOL which will show  
you exactly how the variables are working.

Cheers,
Bijan.
Received on Monday, 5 March 2007 15:28:30 UTC

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