- From: Richard Cyganiak <richard@cyganiak.de>
- Date: Wed, 16 Jan 2013 18:05:38 +0000
- To: "Patel-Schneider, Peter" <Peter.Patel-Schneider@nuance.com>, Pat Hayes <phayes@ihmc.us>
- Cc: RDF Working Group WG <public-rdf-wg@w3.org>
Peter, Pat,
I see -- the ill-typed literal is required to denote something that both is in LV and is not in LV, leading to a contradiction.
Going through the ISSUE-109 thread again, I think Antoine proposed a similar mechanism here:
http://lists.w3.org/Archives/Public/public-rdf-wg/2012Nov/0194.html
Pat didn't like Antoine's approach:
http://lists.w3.org/Archives/Public/public-rdf-wg/2012Nov/0196.html
Pat proposed what appears to be a different way of achieving the same result:
http://lists.w3.org/Archives/Public/public-rdf-wg/2012Nov/0195.html
Anyway, I trust that there is *some* way of making this work in the Semantics. Therefore I've re-assigned ISSUE-109 against the RDF Semantics product, assuming that the best way of resolving the issue is to adopt one of the proposals above (plus a small change in Concepts to reflect that semantic change).
Best,
Richard
On 16 Jan 2013, at 16:59, Patel-Schneider, Peter wrote:
> During the call today there was some discussion of ill-typed literals.
>
> *IF* one wants ill-typed literals to be inconsistent then one has to tweak the semantics.
> The effect is (roughly) to require that the interpretations for literals whose datatype is in the datatype map belong to the value space for that datatype. This looks a lot like the situation where that literal is range-required to be in the datatype.
>
> As far as wording goes, RDF semantics would change something like:
>
> Current:
> if <aaa,x> is in D then for any typed literal "sss"^^ddd in V with I(ddd) = x ,
> if sss is in the lexical space of x then IL("sss"^^ddd) = L2V(x)(sss), otherwise IL("sss"^^ddd) is not in LV
>
> Revised:
> if <aaa,x> is in D then for any typed literal "sss"^^ddd in V with I(ddd) = x ,
> IL("sss"^^ddd) is in LV
> if <aaa,x> is in D then for any typed literal "sss"^^ddd in V with I(ddd) = x ,
> if sss is in the lexical space of x then IL("sss"^^ddd) = L2V(x)(sss), otherwise IL("sss"^^ddd) is not in LV
>
> This may look odd, but the net result is that there can be no models for ill-typed literals.
>
>
> Note: I'm not here an advocate for this change, just noting how it could be done.
>
> peter
>
Received on Wednesday, 16 January 2013 18:06:09 UTC