Re: model theory for RDF/S

>From: Pat Hayes <phayes@ai.uwf.edu>
>Subject: Re: model theory for RDF/S
>Date: Thu, 27 Sep 2001 13:04:15 -0500
>
>>  >>  >17/ Because of the complexity of RDFS, I won't believe the Schema Lemma
>>  >>  >     until I see a completely worked out proof.
>>  >>
>>  >>  Fair enough. I no longer believe it myself. What I am sure of is that
>>  >>  there is *some* closure table for which it is correct, however. Also,
>>  >>  it should be stated so as to explicitly rule out the rdfs:Literal
>>  >>  class.
>>  >
>>  >Probably.  However there might not be a finite schema-closure for full RDF!
>>
>>  There must be, since the Herbrand universes are always finite. Until
>>  one can build functional terms recursively, any set of generation or
>>  closure rules is just going to do combinatorics on the finite set of
>>  possible triples.
>>
>>  Pat
>
>Not necessarily so, at least not in the thinking of at least one RDF
>person.  I have heard comments to the effect that every statement in RDF
>has a reification (although, of course, not vice versa).  This would, I
>think, require an infinite Herbrand universe.

Phrases like 'every statement in RDF' are meaningless out of context. 
In the model theory we are always talking relative to a vocabulary, 
and if that vocabulary is finite then there are only finitely many 
triples. Of course for an infinite vocabulary, the closure may be 
infinite; the comment in section 5 says that the process will 
terminate for any *finite* graph.

For a mathematical audience I would have defined the closure in terms 
of the set-theoretic least upper bound, rather than in terms of a set 
of rules terminating, but that seemed inappropriate here.

Pat
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Received on Thursday, 27 September 2001 20:54:50 UTC