Re: A test case

>>.....
>>
>># Rule 4 (this is new)
>>
>>{
>>    ?p rdfd:datatype ?d .
>>    ?s ?p ?l .
>>    ?l rdf:type rdfs:Literal
>>}
>>log:implies
>>{
>>    ?s ?p ?o.
>>    ?o rdfd:lex ?l
>>} .
>
>I don't think rule 4 is valid. That is, Im not sure quite what
>?l rdf:type rdfs:Literal .
>is intended to convey, but if its supposed to say that the object of
>the previous triple is a literal, then the rule is not valid. Here's
>a counterexample:
>
>Suppose IEXT(I(<ex:PPP>)) is the identity map. Then for example I satisfies
>
>_:x <ex:PPP> "10" .
><ex:PPP> rdfd:datatype <xsd:integer> .
>
>(map _:x to "10"; all literal strings are in the universe; "10" is in
>the lexical space of the datatype) but not
>
>
>_:x <ex:PPP> "10" .
><ex:PPP> rdfd:datatype <xsd:integer> .
>_:x <ex:PPP> _:y .
>_:y rdfd:lex "10" .
>
>since this requires _:y to denote an integer, and so requires 10 to
>equal "10" in that datatyped interpretation.

I follow you up to this point, but don't see how 10 is required
to equal "10". You have two sets of statements, which equate
to essentially

   "10" ex:PPP "10" .

and

   "10" ex:PPP 10 .
   10 rdfd:lex "10" .

I don't see where 10 must equal "10". No'one has declared that
ex:PPP is a daml:uniqueProperty. There is no reason why _:x
cannot have multiple, different values for the ex:PPP property.

[and, as an aside, you can't use daml:uniqueProperty and both
the inline and bnode idioms, ever, given the present MT which
has literals always denoting literals]

I'm probably missing something here (like several years of
formal math training ;-) but I don't see where the equality
of 10 and "10" is actually asserted or required.

Patrick

>Pat

--
               
Patrick Stickler              Phone: +358 50 483 9453
Senior Research Scientist     Fax:   +358 7180 35409
Nokia Research Center         Email: patrick.stickler@nokia.com

Received on Monday, 29 April 2002 02:01:40 UTC