- From: Antoine Zimmermann <antoine.zimmermann@emse.fr>
- Date: Fri, 08 Mar 2013 08:02:39 +0100
- To: Peter Patel-Schneider <pfpschneider@gmail.com>
- CC: Pat Hayes <phayes@ihmc.us>, RDF WG <public-rdf-wg@w3.org>
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 > > > > > > ------------------------------------------------------------ > IHMC (850)434 8903 <tel:%28850%29434%208903> or (650)494 3973 > <tel:%28650%29494%203973> > 40 South Alcaniz St. (850)202 4416 <tel:%28850%29202%204416> office > Pensacola (850)202 4440 <tel:%28850%29202%204440> fax > FL 32502 (850)291 0667 <tel:%28850%29291%200667> mobile > phayesAT-SIGNihmc.us http://www.ihmc.us/users/phayes > > > > > > -- 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 07:03:05 UTC