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>

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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