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Re: need to determine what RDF is

From: patrick hayes <phayes@ai.uwf.edu>
Date: Thu, 30 May 2002 17:21:36 -0500
Message-Id: <p05111a12b91c53483854@[65.217.30.61]>
To: Dan Connolly <connolly@w3.org>
Cc: danbri@w3.org, bwm@hplb.hpl.hp.com, www-rdf-comments@w3.org, em@w3.org, pfps@research.bell-labs.com

>On Thu, 2002-05-30 at 11:10, patrick hayes wrote:
>>  >On Thu, 2002-05-30 at 10:26, Peter F. Patel-Schneider wrote:
>>  >[...]
>>  >>  I'm only interested in relationships between RDF graphs.  Which such
>>  >>  relationships are RDF relationships?
>>  >>
>>  >>  My view is that the only such relationships are RDF entailment and RDFS
>>  >>  entailment.  Any agent that computes any other relationship between RDF
>>  >>  graphs is not doing RDF.
>>  >
>>  >Why is RDFS special? It's just the first of many RDF vocabularies,
>>  >no?
>>
>>  Its more than just an RDF vocabulary because it has some extra
>>  semantic conditions.
>
>
>That doesn't look special, to me; I expect each vocabulary
>to come with some extra semantic conditions.

Wait. Each vocabulary comes with some content that is expressed in 
the RDF graph(s) that use that vocabulary, sure. But that doesn't 
extent the model theory; it doesn't add any *extra* semantic 
conditions, it just leverages the semantic conditions that the 
language spec provides.

>  > For example, if RDFS is considered purely as an
>>  RDF vocabulary, then rdfs:subClassOf is not required to be
>>  transitive;
>
>???

Why are you puzzled? There isn't anything in the RDF spec (here 
considered as separate from the RDFS spec) which says that it is 
transitive, and one cannot express transitivity in an RDF graph, so 
how could it possible be required to be transitive?

>How does 'RDF vocabulary' get that meaning?
>Never mind; I'll stop using 'RDF vocabulary' if
>that's what it means to you.
>
>
>>  in fact, there is no way to express transitivity of a
>>  property in RDF.
>
>Right; we wrote it in semi-formal english in the spec

In the RDFS spec it is built into the model theory; in the RDF spec 
it simply isn't there.

>(and in rdf:comment's, I think).
>
>>  Considered as RDFS, however, it is required to be
>>  transitive. The distinction is made exact in the MT document by
>>  distinguishing between RDF-entailment and RDFS-entailment. In
>>  general, if a particular vocabulary has some extra semantic
>>  conditions attached to it, then there will be a new notion of
>>  entailment (I used the generic term 'vocabulary entailment' for this
>>  idea in the MT document:
>>  http://www.w3.org/TR/rdf-mt/#NameSpaceEntailment ).
>
>Yes, of course; and then there are all the notions of entailment
>we get by combining these vocab... er... extensions.
>
>That's what I meant by "lots of formalisms in the framework"
>over in rdf-logic.

But that makes 'the framework' meaningless (or so all-encompassing 
that it is ridiculous to talk about it.) Who knows what extensions 
will appear in the future, and what new semantic conventions will be 
described?

>  > Most RDF
>>  vocabularies, eg dublin core, RSS, have no extra semantics attached
>>  to them.
>
>Yes, I think they do.
>
>The semantics may be informal -- we might not be able
>to communicate them to a machine

Well then there is no point in discussing them in this context. We 
aren't trying to re-do the human web in RDF plus all of human 
language, we are trying to do the semantic web. If something can't be 
conveyed to a machine then it isn't on the semantic web.

>-- but there are certainly
>interpretations that are *not* consistent with the
>specification

Can that specification be used by a software agent to rule out those 
inconsistent interpretations? If not, it's merely decoration: it has 
no role in our discussions. Lets not delude ourselves into thinking 
that our programs are going to be able to understand everything. 
There's a well-known term in foundations of mathematics for this: 
nonstandard models. Almost any formal representation has nonstandard 
models, ie ones that weren't intended by the humans who designed the 
formalization. But it has them, whether they are nonstandard or not. 
Pretending that the nonstandard models aren't there is just 
delusional thinking.

Pat Hayes
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Received on Thursday, 30 May 2002 18:22:24 GMT

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