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Re: how to make ill-typed literals inconsistent - ISSUE 109

From: Antoine Zimmermann <antoine.zimmermann@emse.fr>
Date: Thu, 17 Jan 2013 10:36:59 +0100
Message-ID: <50F7C63B.8090600@emse.fr>
To: public-rdf-wg@w3.org
This solution is fine for me.


AZ

Le 16/01/2013 20:44, Pat Hayes a écrit :
>
> On Jan 16, 2013, at 10:59 AM, 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 is already required. What is needed is to change the current
> condition for when the lexical form is not in the lexical space of
> the datatype map. The current semantics says that in this case, the
> literal is required to denote something ouside LV. I propose to
> change this so that ill-typed literals simply do not denote at all.
> (This is more natural in any case: the current not-in-LV condition is
> artificial.) The main semantic conditions for RDF then make any
> triple containing such a literal false. The only serious change to
> the basic semantics is to allow the IL map to be partial. This looks
> like it might make RDF into a free logic, and it would if there were
> any quantifiers, but as there aren't, it seems safe.
>
> Pat
>
>
>> 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
>>
>
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-- 
Antoine Zimmermann
ISCOD / LSTI - Institut Henri Fayol
École Nationale Supérieure des Mines de Saint-Étienne
158 cours Fauriel
42023 Saint-Étienne Cedex 2
France
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Received on Thursday, 17 January 2013 09:37:25 GMT

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