Hi Joshua,
I wasn't very clear in my example. I have my rule r1 which defines ?s p
?o. However, I can't define that the property p of ?s is ?o with the
traditional mechanisms of inference of OWL or OWL2. Thus, I have a rule
that defines:
(?s p ?o) <- c1, ... .
However, I need to produce a reified triple that has an associated
certainty factor:
(?f p ?e),
(?f p1 ?p),
(?e p2 ?p),
makeTemp(?bn) ->
(?bn rdf:subject ?f),
(?bn rdf:predicate p3),
(?bn rdf:object ?e),
(?bn exa:certainty "0.9"^^xsd:decimal).
I'm not having a good understanding what rules are for?
Taking your example, why not:
(?s rdf:type ?superclass),
c1 ->
t1, t2.
(and the previous example doesn't work as a rule, only works as OWL
inference mechanisms).
Or why not:
(?f rdf:type ?superclass),
(?f p1 ?p),
(?e p2 ?p),
makeTemp(?bn) ->
(?bn rdf:subject ?f),
(?bn rdf:predicate p3),
(?bn rdf:object ?e),
(?bn exa:certainty "0.9"^^xsd:decimal).
Miguel
On 15/03/14 01:45, "Joshua TAYLOR" <[email protected]> wrote:
>On Fri, Mar 14, 2014 at 4:23 PM, Miguel Bento Alves
><[email protected]> wrote:
>> Hi Joshua,
>>
>> r1 is a backward rule [r1: Š <- Š]. In my understanding, backward rules
>> are things that relate two resources, overcoming the limitations of OWL
>> and OWL2.
>
>A backward rule doesn't relate two resources. A backward rule
>provides a pattern for a triple as the head, and some conditions under
>which the triple can be inferred. E.g.,
>
>[r: (?s rdf:type ?superclass) <-
> (?s rdf:type ?subclass)
> (?subclass rdfs:subClassOf ?superclass)]
>
>When the reasoner tries to infer a triple of the form (?s rdf:type
>?superclass), it will try to use this rule by checking whether matches
>for the body can be found. It doesn't make sense to say that the rule
>r relates two resources; that's not what a rule is. A rule is a
>basis for for inferring triples from existing data and other
>conditions.
>
>> Thus, I think that the example that I gave make sense:
>> (r1 is a backward rule)
>>
>> (?f r1 ?e),
>> (?f p1 ?p),
>> (?e p2 ?p),
>> makeTemp(?bn) ->
>> (?bn rdf:subject ?f),
>> (?bn rdf:predicate p3),
>> (?bn rdf:object ?e),
>> (?bn exa:certainty "0.9"^^xsd:decimal).
>
>Again, I'm not sure what (?f r1 ?e) is supposed to mean. In the rule
>syntax, it's just a triple with variables as the subject and object,
>and a URI r1 as the predicate. In the example rule r that I gave,
>what would you expect (?s r ?o) to mean?
>
>> Moreover, this rule works well if I materialize r1.
>> Construct {
>> (?f r1 ?e)
>> }
>> Where {
>> (?f r1 ?e)
>> }
>> (of course, if I materialize a rule I remove from ruleset).
>
>I'm not sure what this is supposed to mean. If you construct the data
>that matched a triple pattern, then you'll get the same data that
>matched the triple pattern. I'm not sure how this is "a
>materialization of r1".
>
>> Do you think that this understanding of what is a (backward) rule is not
>> correct?
>
>I think it's not correct, but I'm still not sure what you think it
>would mean to have a backward rule [r: ... <- ...] and then write a
>triple pattern (?s r ?o). Rules aren't predicates, so it doesn't
>make sense to me to try to use them as predicates.
>
>//JT
>--
>Joshua Taylor, http://www.cs.rpi.edu/~tayloj/
