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Re: test case A revisited

From: Patrick Stickler <patrick.stickler@nokia.com>
Date: Mon, 01 Jul 2002 16:12:18 +0300
To: ext Brian McBride <bwm@hplb.hpl.hp.com>, RDF Core <w3c-rdfcore-wg@w3.org>
Message-ID: <B9462FE2.17AC6%patrick.stickler@nokia.com>

On 2002-07-01 14:39, "ext Brian McBride" <bwm@hplb.hpl.hp.com> wrote:

> 
> Friday's telecon reminded me that I had left test case A in for a
> reason.  There was more I had to say about it, and writing that message the
> following occurred to me.
> 
> Test case A says:
> 
>  <s1> <p> "lit" .
>  <s2> <p> "lit" .
> 
> can we conclude that value of both properties are the same.
> 
> Consider
> 
>  _:b1 rdf:type rdf:Seq .
>  _:b1 rdf:_1   "10" .
>  _:b2 rdf:type rdf:Seq .
>  _:b2 rdf:_1   "10" .
> 
> This would require that the first member of each sequence is the same.
> 
> Given that we also have a common superproperty of the rdf:_xxx properties,
> this essentially means that all literals which are members of any container
> must all have the same dataype, i.e. all literals in containers must be tidy.
> 
> I suggest this is incompatible with the untidy literals and a yes to test
> case A above.

I agree.

Thus, I stand (or actually sit ;-) corrected about test A being
irrelevant. It is not irrelevant, but joins with B and C such
that a yes answer is a vote for tidyness.

Which makes me wonder whether the inquiry is now a tad bit
imbalanced...  'yes' to (A and B and C) or D ... hmmmmm....

Anyway, just send it and let's see what we get.

I retract my earlier objection.

Patrick
 
--
               
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, 1 July 2002 09:12:19 EDT

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