From: Jos De_Roo <jos.deroo@agfa.com>

Date: Wed, 13 Aug 2003 12:54:59 +0200

To: "Sean Bechhofer <seanb" <seanb@cs.man.ac.uk>

Cc: connolly@w3.org, jjc@hpl.hp.com, www-webont-wg@w3.org

Message-ID: <OF8F94BA62.A68C58E7-ONC1256D81.003AEA60-C1256D81.003BF78E@agfa.be>

Date: Wed, 13 Aug 2003 12:54:59 +0200

To: "Sean Bechhofer <seanb" <seanb@cs.man.ac.uk>

Cc: connolly@w3.org, jjc@hpl.hp.com, www-webont-wg@w3.org

Message-ID: <OF8F94BA62.A68C58E7-ONC1256D81.003AEA60-C1256D81.003BF78E@agfa.be>

[Sean's further explanation of http://www.w3.org/2002/03owlt/description-logic/inconsistent107.rdf] > Well, in general, the intersection of > > restriction(a:r someValuesFrom a:d) > restriction(a:r someValuesFrom a:c) > > means that there must be a d related via r and a c related via r, but they > don't *have* to be the same individual -- there could be two of them. > > In this case of this test, however, the extra constraint on the > cardinality of r (due to the negation of the minCardinality) means > that there can *only* be one, hence the contradiction. > > The various different ways of combining restrictions using intersection > and union are related as follows: > > 1. some R.(C and D) > 2. (some R.C) and (some R.D) > 3. some R.(C or D) > 4. (some R.C) or (some R.D) > > Given no other contraints, we know that it's always the case that: > > 1. => 2. > 2. => 3. > 2. => 4. > 3. == 4. > > Apologies if this is teaching grandma to suck eggs (another marvellous > phrase :-). No apologies, this is extremely useful teaching! Above implications are exactly what I'm missing in my understanding and I believe it will simplify our clumsy implementation that we have right now. Is all R.(C and D) similar? What about complement? Is complement some R.C = all R.(complement C)? Is the some/all converse also the case? Thanks in advance ;-) -- Jos De Roo, AGFA http://www.agfa.com/w3c/jdroo/Received on Wednesday, 13 August 2003 06:55:42 UTC

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