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Re: reification test case

From: <jos.deroo.jd@belgium.agfa.com>
Date: Mon, 4 Feb 2002 10:03:53 +0100
To: patrick.stickler@nokia.com
Cc: w3c-rdfcore-wg@w3.org
Message-Id: <OFD94BA0E7.21A10B33-ONC1256B56.00309B63@bayer-ag.com>
[sorry for the delay in answer, but I still have to scrape this from the
rdfcore archive because my mail is still not coming through]

> > actually I see that already
> >
> > _:s1 <property>      "property" .
> >
> > entails
> >
> > _:s2 <property>      "property" .
> >
> > so I don't see the point of reification
> >
> > --
> > Jos
> >
> >
>
> I'm quite curious how you come to this result, since
> bNodes are distinct and there is no definition by
> RDF, that I'm aware of, that two bNodes of
> type rdf:Statement which have an intersection of
> the same S, P, and O triples are the same "thing".
>
> The two bNodes reify the same triple, but are
> distinct reifications in their own right. No?
> Why wouldn't we treat them as distinct resources?
>
> What am I missing here (honestly)?

Patrick, this is plain MT
At the end of chapter 2. Simple entailment between RDF graphs.
you may find

[[[
It might be thought that the operation of changing a bound variable
would be an example of an inference which was valid but not covered
by the interpolation lemma, e.g. the inference of

_:x foo baz

from

_:y foo baz

Recall however that by our conventions, these two expressions describe
identical RDF graphs.
]]]

after all, bNodes are blank (circles with nothing in)

--
Jos
Received on Monday, 4 February 2002 05:07:35 EST

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