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Re: mergeg in current Semantics ED

From: Peter F. Patel-Schneider <pfpschneider@gmail.com>
Date: Fri, 08 Mar 2013 06:29:26 -0800
Message-ID: <5139F5C6.4040200@gmail.com>
To: Antoine Zimmermann <antoine.zimmermann@emse.fr>
CC: Pat Hayes <phayes@ihmc.us>, RDF WG <public-rdf-wg@w3.org>

On 03/07/2013 11:02 PM, Antoine Zimmermann wrote:
> 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
Umm, I said at the beginning of the message that the entailment does not follows.

peter
Received on Friday, 8 March 2013 14:29:59 GMT

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