Re: mergeg in current Semantics ED

Le 07/03/2013 23:25, Peter Patel-Schneider a écrit :
> I think that the current document makes the entailment not work.
>
> G1 is Ex p1(s1,x)
> G2 is Ex p2(s2,x)
>
> {G1,G2} is Ex p1(s1,x) ^ p2(s2,x)
>
> In particular, {G1,G2} is *not* Ex p1(s1,x) ^ Ex p2(s2,x)

What?

{G1,G2} entails G iff all interpretations that make G1 and G2 true also 
make G true. Let us consider interpretation I:

IR = {x,y,z,t}
IS = {(<s1>,x),(<s2>,y),(<p1>,z),(<p2>,t)}
IEXT(<p1>) = {(x,y)}
IEXT(<p2>) = {(y,z)}

Let us examine the truth of G1 and G2 under this interpretation:

"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."

Consider the mapping: A1 = {(b,y)} (possibly, the mapping contains other 
things if there are more bnodes "in the scope", whatever this means.)
[I+A1](G1) = true
So I satisfies G1

Consider the mapping: A2 = {(b,z)}
[I+A2](G2) = true
So I satisfies G2

Now, in order to satisfy G, there must exist a mapping A such that A(b) 
= y and A(b) = z. Assuming that y and z are different, the mapping 
cannot exist such that [I+1](G) = true, so G is not satisfied by I.

So, there exists an interpretation I that makes both G1 and G2 true but 
does not make G true, therefore, G is not entailed.

I am really surprised that I have to show you the proof explicitly.


AZ

>
> peter
>
>
> On Thu, Mar 7, 2013 at 12:51 PM, Pat Hayes <phayes@ihmc.us
> <mailto:phayes@ihmc.us>> wrote:
>
>
>     On Mar 7, 2013, at 12:30 PM, Peter Patel-Schneider wrote:
>
>      >
>      >
>      > On Thu, Mar 7, 2013 at 10:22 AM, Pat Hayes <phayes@ihmc.us
>     <mailto:phayes@ihmc.us>> wrote:
>      >
>      > On Mar 7, 2013, at 9:52 AM, Antoine Zimmermann wrote:
>      >
>      > > There is a problem with the definition of merge in the draft.
>      > >
>      > > I'm using math notations instead of a concrete serialisation
>     syntax because I want to show things at the abstract syntax level,
>     which is what RDF Semantics relies on.
>      > >
>      > > Let us take a blank node b from the set of blank nodes. Let us
>     consider the two graphs G1 = {(<s1>,<p1>,b)} and G2 = {(<s2>,<p2>,b)}.
>      >
>      > You have the same bnode in both graphs, so they must be in the
>     same scope, right? For example, both are subgraphs of a larger
>     graph, or both in the same dataset.
>      >
>      > >
>      > > Let us ask ourselves whether {G1,G2} entails:
>      > >
>      > > G = {(<s1>,<p1>,b),(<s2>,<p2>,b)}
>      > >
>      > > The answer is trivially NO wrt the current semantics of the ED.
>      >
>      > If those really are the same b, then the answer is YES, and I
>     claim that it should be.
>      >
>      >
>      > I don't know how you are going to get this to go through.
>
>     Do you mean, technically or politically? Technically, this is true
>     now (that G1 and G2 together entail G) and it also was in the 2004
>     semantics, if the G1 and G2 were for example subgraphs of G.
>     Politically, I think we have debated this to death and the new
>     account based on scopes is exactly what the WG wants. For example,
>     we have an explicit decision that all bnodes in (all graphs in) a
>     dataset shall share bnodes in common, so to standardize those bnodes
>     apart would be definitely a mistake.
>
>     Pat
>
>
>      >
>      > peter
>      >
>      >
>
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>


-- 
Antoine Zimmermann
ISCOD / LSTI - Institut Henri Fayol
École Nationale Supérieure des Mines de Saint-Étienne
158 cours Fauriel
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France
Tél:+33(0)4 77 42 66 03
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http://zimmer.aprilfoolsreview.com/

Received on Friday, 8 March 2013 07:03:05 UTC