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Re: mergeg in current Semantics ED

From: Antoine Zimmermann <antoine.zimmermann@emse.fr>
Date: Sat, 09 Mar 2013 00:56:50 +0100
Message-ID: <513A7AC2.7010603@emse.fr>
To: "Peter F. Patel-Schneider" <pfpschneider@gmail.com>
CC: Pat Hayes <phayes@ihmc.us>, RDF WG <public-rdf-wg@w3.org>
Le 08/03/2013 15:29, Peter F. Patel-Schneider a écrit :
>
> 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.

No, you said:

 >>> 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)

{G1,G2} *is* Ex p1(s1,x) ^ Ex p2(s2,x)

(unless you do not use the letter 'E' to mean the existential quantifier?)


AZ

>
> 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/
Received on Friday, 8 March 2013 23:57:24 GMT

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