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Re: Adoption of entailment in SPARQL

From: <jos.deroo@agfa.com>
Date: Tue, 20 Sep 2005 11:51:09 +0200
To: franconi@inf.unibz.it
Cc: Pat Hayes <phayes@ihmc.us>, RDF Data Access Working Group <public-rdf-dawg@w3.org>, public-rdf-dawg-request@w3.org
Message-ID: <OF6DB511F2.79049A11-ONC1257082.0034CC06-C1257082.003611C5@agfa.com>

[...]

> (Referring to <http://lists.w3.org/Archives/Public/www-archive/
> 2005Sep/0009>, the query and its proof doesn't capture the main idea
> behind the "worker example". In fact, the graph in the message
> doesn't implies that :Andrea is an instance of :EMPLOYEE, so the
> tuple (:Paul :Andrea :Caroline) cannot be in the answer of the given
> query. Most likely, the theorem prover didn't consider the certain
> answers (those true in every model) but possible answers (as those
> true in at least one model). The only variable in the SELECT should
> be ?X, then the query returns :Paul because of reasoning by case on
> all the possible models of the graph.)

True; those were indeed not certain answers :)
I now understand (and tested my understanding with proof engines)
that

{ :Paul a :WORKER.
  :Paul :has-friend ?Y.
  ?Y a :EMPLOYEE.
  ?Y :has-friend ?Z.
  ?Z a :MANAGER }

is entailed
whereas

{ :Paul a :WORKER.
  :Paul :has-friend :Andrea.
  :Andrea a :EMPLOYEE.
  :Andrea :has-friend ?Z.
  ?Z a :MANAGER }

is *not* entailed.



-- 
Jos De Roo, AGFA http://www.agfa.com/w3c/jdroo/
Received on Tuesday, 20 September 2005 09:52:02 GMT

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