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Re: reification test case

From: Pat Hayes <phayes@ai.uwf.edu>
Date: Mon, 4 Feb 2002 18:44:03 -0600
Message-Id: <p0510140eb884da691a25@[65.212.118.208]>
To: jos.deroo.jd@belgium.agfa.com
Cc: w3c-rdfcore-wg@w3.org
>[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)

Ah, no. Wait a minute. Those two triples *in isolation* describe 
identical graphs, but that doesn't mean you can arbitrarily 
substitute one for another in a larger graph. For example these are 
NOT the same graph:

_:x foo baz
_:x woog plaz

_:y foo baz
_:x woog plaz

The first has 3 nodes, the second has 4 nodes. And they don't have 
the same entailments, either. The first entails the second, but not 
vice versa.

Pat

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Received on Monday, 4 February 2002 19:43:33 EST

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