Re: Re: Higher-arity to RDF binary

On Thu, May 20, 2010 at 15:40, Chris Dollin <chris.dollin@epimorphics.com>wrote:

> On Thursday 20 May 2010 02:34:10 pm Jitao Yang wrote:
> > 2010/5/20 Chris Dollin <chris.dollin@epimorphics.com>
> >
> > > On Thursday 20 May 2010 02:25:03 pm Jitao Yang wrote:
> > >
> > > > > @prefix eh: <http://rdf.epimorphics.com/chris/examples#>.
> > > > > @prefix some: <http://example.com/rdf#>.
> > > > >
> > > > > some:p eh:assertion
> > > > >    [eh:on some:d; eh:value some:o]
> > > > >    , [eh:on some:d1; eh:value some:o1]
> > > > >    .
> > > > >
> > > > > if I understand correctly? The above representation could be
> translated
> > > :
> > > >
> > > > some:p               eh:assertion             _blanknode1 ;
> > > > _blanknode1      eh:on                       some:d  ;
> > > > _blanknode1      eh:value                   some:o  ;
> > > > some:p               eh:assertion             _blanknode2 ;
> > > > _blanknode2      eh:on                        some:d1 ;
> > > > _blanknode2      eh:value                    some:o1 .
> > >
>

What I mean is based on the above representation,



>  > > Yes.
> > >
> > > > but it seems lost the connections between b, p and o?
>
 > >
> > > What's b?
> > >
> >
> > sorry, should be d
>
> The connections are there:
>
>    P assertion [on D; value O]
>
> encodes the formula
>
>    DescPr(D, P, O).
>
> Always assuming I haven't completely misunderstood what's
> going on ...
>
>
we can not reason that the value of p is o.
The above representation is a RDF reification, do think so?


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