- From: Pat Hayes <phayes@ihmc.us>
- Date: Fri, 8 Mar 2013 14:14:33 -0600
- To: Antoine Zimmermann <antoine.zimmermann@emse.fr>
- Cc: Peter Patel-Schneider <pfpschneider@gmail.com>, RDF WG <public-rdf-wg@w3.org>
On Mar 8, 2013, at 1:13 AM, Antoine Zimmermann wrote:
> Le 07/03/2013 23:56, Peter Patel-Schneider a écrit :
>>
>>
>> On Thu, Mar 7, 2013 at 2:41 PM, Pat Hayes <phayes@ihmc.us
>> <mailto:phayes@ihmc.us>> wrote:
>>
>>
>> On Mar 7, 2013, at 3:12 PM, Peter Patel-Schneider wrote:
>>
>> > I think that the current document makes the entailment not work.
>> >
>> > G1 is Ex p1(s1,x)
>> > G2 is Ex p2(s2,x)
>>
>> No, its not. If they share a blank node, they must be in the same
>> scope; and the existential is defined at the scope, not at the
>> graph, level. So under the conditions given by Antoine, the Ex is
>> outside the conjunction of G1 and G2.
>>
>>
>> I see exactly the opposite of this in the current document.
>>
>> "If E is an RDF graph then I(E) = true if [I+A](E) = true for some
>> mapping A from the set of blank nodes in the scope of E to IR, otherwise
>> I(E)= false."
>>
>> So, consider I with I(s1) = s1, I(s2) = s2, I(p1) = p1, I(p2) = p2,
>> IEXT(p1) = {<s1,s1>}, IEXT(p2) = {<s2,s2>}
>> (add the other stuff to minimally turn this into a simple interpretation).
>>
>> Then I(G1) = true, from the mapping A1:x->s1
>> and I(G2) =true, from the mapping A2:x->s2
>> but there is no mapping for x that makes I({G1,G2}) = true.
>
> I({G1,G2}) is not defined in the document, and the existential depends on a graph. You have an exists quantifier each time you consider a different graph.
No, you don't. The existential is attached to the scope of the bnode, not to the graph boundary. Thats the whole point of having bnode scopes. (They often coincide, but the point remains.) And the existential binds a bnode identifier, not the bnode itself. (You can't bind something that has no lexical form to bind.)
Pat
> One could write things like this:
>
> forall E (E is a graph => (I(E) = true <=> exists A, [I+A](E) = true))
>
> or even:
>
> forall E (E is a graph => (I(E) = true <=> exists A(E), [I+A(E)](E) = true))
>
>
> AZ
>
>
>>
>> So, yes, the existential in {G1,G2} is global in the current document,
>> but that is precisely what makes the difference.
>>
>> peter
>>
>>
>
>
> --
> Antoine Zimmermann
> ISCOD / LSTI - Institut Henri Fayol
> École Nationale Supérieure des Mines de Saint-Étienne
> 158 cours Fauriel
> 42023 Saint-Étienne Cedex 2
> France
> Tél:+33(0)4 77 42 66 03
> Fax:+33(0)4 77 42 66 66
> http://zimmer.aprilfoolsreview.com/
>
>
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Received on Friday, 8 March 2013 20:15:07 UTC