Jurriaan  --

Here's how to do non-subsumption in datalog+negation. 

You can run this example by pointing Netscape 7 or Mozilla to www.reengineeingllc.com
Click on Internet Business Logic, then on GO, then pick NonSubsume1 . 

If you then ask for a table headed
 
      some-classA does not subsume some-classB

you will get that Class 1 does not subsume Class 2 , and also that Class 2 does not
subsume Class 1.   You can also ask for a step-by-step explanation.

Hope this helps.                 -- Adrian


|  Showing that one class does not subsume another.


some-item is in some-classB but not in some-classA
--------------------------------------------------
that-classA does not subsume that-classB


some-classA and some-classB are named differently
some-item is a member of that-classA
not : that-item is a member of that-classB
--------------------------------------------------
that-item is in that-classA but not in that-classB


some-item is a member of some-class
------------------------------------
there is a class named that-class


there is a class named some-classA-name
there is a class named some-classB-name
that-classA-name is not equal that-classB-name
----------------------------------------------------------
that-classA-name and that-classB-name are named differently


this-item is a member of this-class
===================================
 D                       Class 1
 E                       Class 1
 E                       Class 2
 F                       Class 2
 D                       Class 3





                           INTERNET BUSINESS LOGIC
 
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Adrian Walker
Reengineering LLC
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At 01:37 PM 8/15/03 +0200, you wrote:
Hi everybody,
 
I'd like to express in OWL (or in description logic) that two classes are in a non-subsumption relation with each other. This is not as strong as disjointness, as there may be individuals that are members of both classes. In the case of "A non-subsumes B", it says that there exists an individual which is member of B and not of A. Does anyone know if this is possible?
 
Thanks in advance,
 
Jurriaan