Re: Possible tweak to datatype semantics

I'm with Peter on this.
I don't see a real benefit with this proposal.


AZ


Le 13/09/2013 08:03, Peter F. Patel-Schneider a écrit :
> What good would this change from the 2004 situation do?
>
> Even if inertia was strongly indicating that this change should not be
> made, I would vote against it.
>
> If you make this change, you have the situation that if x:dt is not a
> recognized datatype, the empty graph does not RDFS entail
>    x:dt rdf:type rdfs:Class .
> but
>    :a :p "foo"^^x:dt .
> does.
>
> I believe that your argument falls apart when you look closer at it.
> You are saying, in effect, that if x:dt is a recognized datatype then
> any well-typed literal with it as the datatype belongs to it, and the
> appearance of an ill-typed literal causes a contradiction, and thus
> entails any graph, including the graph that states that the ill-typed
> literal belongs to the x:dt, so why not make this hold even if x:dt is
> not a recognized datatype. However, when x:dt is *not* a recognized
> datatype this reasoning doesn't hold water, so there is no reason to
> modify the semantics to make it valid.
>
> You might just as well argue that if x:dt is a recognized datatype then
> it is a subclass of rdfs:Literal therefore anything should be a subclass
> of rdfs:Literal.
>
> peter
>
> On 09/12/2013 10:25 PM, Pat Hayes wrote:
>> I know its very late to even be talking about this, but Antoine's test
>> cases made me notice an oddity which the current semantics for
>> datatyped literals produces, and which would be easy to fix. So I'm
>> outlining it here in case the WG feels it would be worth doing.
>>
>> We distinguish 'recognized' datatype IRIs from the others, and right
>> now, if you see a literal with an unrecognized datatype IRI in it, say
>> x:dt, then you know nothing at all about what that literal means.
>> Absolutely nothing. So this inference:
>>
>> :a :p "foo"^^x:dt .
>>
>> |=
>>
>> :a :p _:x .
>> _:x rdf:type x:dt .
>>
>> is not a valid entailment. But if x:dt were recognized, it would be:
>> and moreover, you know this without knowing anything about x:dt. This
>> entailment is valid for ANY recognized datatype, and ANY string "foo".
>> So why isn't it valid for any datatype, recognized or not?  This is
>> clearly slightly irrational. A rational way to reason would be: I know
>> now, even without recognizing that datatype, that this inference will
>> be valid when I do recognize it; and I won't need to know anything
>> more about the datatype in order to make that inference; so why not
>> just pretend that I recognize the datatype and make the inference now,
>> to save time?
>>
>> We could fix this with the following changes.
>>
>> In section 7.1, add the condition (to the table, it would be the third
>> line out of three):
>>
>> For any literal "sss"^^aaa, if IL("sss"^^aaa) is defined then
>> <IL("sss"^^aaa), I(aaa)> is in IEXT(I(rdf:type))
>>
>> and add the explanatory text immediately below:
>> "The third condition applies to all datatyped literals, whether the
>> datatype IRI is recognized or not."
>>
>> And in section 7.2.1, at the beginning of the text, add the entailment
>> pattern (moved from section 8.1.1, and with "for ddd in D" removed):
>>
>> rdfD1  <if S contains>  xxx aaa "sss"^^ddd  <then S D-entails> xxx aaa
>> _:nnn .      _:nnn rdf:type ddd .
>>
>> together with its explanatory text from 8.1.1.
>>
>>
>> The advantage to RDF engines is that this is one less case where they
>> have to check whether or not a datatype is "recognized", and it means
>> that the interpolation lemma is more useful when there are datatyped
>> literals around.
>>
>> Any comments? Is this worth doing? Is this legally possible to do at
>> this LC stage? I would be willing to declare the current version an
>> error if that is what it takes :-)
>>
>> Pat
>>
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>>
>>
>>
>>
>>
>>
>
>


-- 
Antoine Zimmermann
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Received on Friday, 13 September 2013 12:44:47 UTC