Re: minor typos in Test doc

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/~seanb

Received on Wednesday, 13 August 2003 07:23:17 UTC