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Re: minor typos in Test doc

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>


[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 GMT

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