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

Date: Wed, 13 Aug 2003 12:20:12 +0100 (GMT Daylight Time)

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

cc: connolly@w3.org, <jjc@hpl.hp.com>, <www-webont-wg@w3.org>

Message-ID: <Pine.WNT.4.44.0308131205040.1724-100000@potato>

Date: Wed, 13 Aug 2003 12:20:12 +0100 (GMT Daylight Time)

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

cc: connolly@w3.org, <jjc@hpl.hp.com>, <www-webont-wg@w3.org>

Message-ID: <Pine.WNT.4.44.0308131205040.1724-100000@potato>

On Wed, 13 Aug 2003, Jos De_Roo wrote: > > [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? Similar, yes. If we have: 1. all R.(C and D) 2. (all R.C) and (all R.D) 3. all R.(C or D) 4. (all R.C) or (all R.D) Then we get: 1. == 2. 1. => 4. 2. => 4. 4. => 3. > What about complement? Is > complement some R.C = all R.(complement C)? complement (some R.C) == all R.(complement C) complement (all R.C) == some R.(complement C) Cheers, Sean -- Sean Bechhofer seanb@cs.man.ac.uk http://www.cs.man.ac.uk/~seanbReceived on Wednesday, 13 August 2003 07:23:17 UTC

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