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Re: Cardinality Reasoning

From: Enrico Franconi <franconi@inf.unibz.it>
Date: Thu, 17 Aug 2006 17:18:25 +0200
Message-Id: <63E1B319-C8D8-4EE7-81C4-1D176520362C@inf.unibz.it>
Cc: public-owl-dev@w3.org
To: "Luke Steller" <Luke.Steller@infotech.monash.edu.au>

On 17 Aug 2006, at 15:21, Luke Steller wrote:

> I changed it slightly - theres one class called 'HasAtLeastOneColor'
> which printer1 is an instance of, then i have created one called
> 'DoesNotHaveAtLeastOneColor' which is its complement. Then
> 'Everything' is the union of the two.
> If 'Everything' contains every instance,

yes, since in any model any individual is instance of it,

> and ''HasAtLeastOneColor' contains only printer1,


> then why doesn't 'DoesNotHaveAtLeastOneColor'
> contain printer2 and all the colour instances?

because it is not true that in each single model printer2 (and the  
colour instances) don't have any colour (since you did not assert  
that about them). On the other hand, it is true that in each single  
model printer2 is either in HasAtLeastOneColor or in  
DoesNotHaveAtLeastOneColor, but not in both.

> Shouldn't complement
> specify every individual that is not in the set specified by the
> complement class?

Suppose you are in a black and white universe: each object is either  
white or blanck but not both. The union of black and white covers all  
the objects of the universe. However, if you have an object U of an  
unknown colour, then you can not deduce that U is either black or  
white for certain. The fact that you don't derive it, does not mean  
that you can derive the opposite: in fact, you will not derive that U  
is not white nor that it is not black.

> This is probably actually a question about complements themselves,
> because even when I take out the cardinality restrictions, if i make
> printer1 an instance of HasAtLeastOneColour, its complement class
> doesnt not contain all other instances. Why is this?

I hope I clarified.

Received on Thursday, 17 August 2006 15:18:53 UTC

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