Re: mergeg in current Semantics ED

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. 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
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Received on Friday, 8 March 2013 07:14:03 UTC