when I run the reasoning this give me  over 1000 instances inferred...can
Topbraid export more than 1000 instances in a spreadsheet?

2013/3/4 aLe C <[email protected]>

> Thank you, very much!
>
> This was usefull for me.
>
> You know? if Hermit reasoning can run in topbraid Composer..
>
>
>
> 2013/3/1 Scott Henninger <[email protected]>
>
>>  ALe; This is more of a general reasoning question, so I'd suggest
>> looking for forums that get into more general aspects of inference.
>> What I can tell you is that the kind of reasoning you are attempting is
>> not possible in OWL because of Open World Assumptions.  I.e. it's not
>> possible to infer that something is not a member of a class because some
>> other information may become available that infers or asserts that it is a
>> member of the class.
>>
>> OTOH, if you want to perform this kind of negation, I'd suggest using
>> SPARQL.  Try running the following in your SPARQL view (my query is
>> approximate as your spec is semi-formal):
>>
>> CONSTRUCT {?person a :Extranjero}
>> WHERE {
>>     ?person a :Person .
>>     FILTER NOT EXISTS  {?person :hasnacionalidad "Mexicano" .}
>> }
>>
>> This gives you an idea of how the inference can be made.  To place this
>> in a system of inferences I'd suggest using SPIN (
>> http://www.w3.org/Submission/2011/SUBM-spin-overview-20110222/).
>>
>> To do this in Composer, configure inferencing to use SPIN (Inference >
>> Configure Inferencing..., click the "Customize settings..." checkbox,
>> Add>> TopSPIN and remove other reasoners from the pipeline.
>>
>> Create a new file by choosing File > New >  RDF/OWL/SPIN File.  When
>> that file is open in Composer, open the Imports tab and import your data
>> file (drag/drop the data file into the imports tab is easiest).
>>
>> Then create the two rules you want.  To infer persons as members of :Mexican,
>> double-click :Person to open it in the Class Form.  Then choose Add empty
>> row... for the spin:rule property and enter the following:
>>
>> CONSTRUCT { ?this a :Mexican .}
>> WHERE
>> {    ?this :hasnacionalidad "Mexicano" .
>> }
>>
>> The value of the ?this variable will be bound to each member of :Person
>> before executing the query.  I.e. if you have 50 members of :Person, the
>> rule will be executed for each one, binding ?this to each member.
>>
>> For the Extranjero inference, do the same as above - in the :Person 
>> classchoose Add empty row... for the spin:rule property and enter the 
>> following:
>>
>> CONSTRUCT {?this a :Extranjero .}
>> WHERE
>> {    FILTER NOT EXISTS  {?this :hasnacionalidad "Mexicano" .} .
>> }
>>
>> Note this is a slight variant of the first query i did above.
>>
>> Run inferences as you have done before and the inferences you are
>> looking for will be made.
>>
>> Generally speaking SPARQL and SPIN are a lot more direct about the
>> computations one wants to make and is therefore much easier to work
>> with.  In addition, it is far less constrained than OWL, so you can
>> practically run any computation or data transformation.  It is well wortha 
>> try, and I'd suggest
>> taking a look at a couple of resources:
>>   - http://www.topquadrant.com/products/SPIN.html
>>   - in Composer Help, go to TopBraid Composer > Application Development
>> Tools > SPIN
>>
>> -- Scott
>>
>>  On 2/28/2013 6:04 PM, aLe C wrote:
>>
>> Hi
>> I want to infer when a person isn't mexican, in other words,
>> I have instances (of a table imported by topbraid) with mexican persons
>> and foreign persons "Nicaraguan, Paraguayan, Puerto Rican, so on"
>> The ontology have two classes, "Mexican" y "Foreign"  with the property
>> "hasnacionalidad"
>>
>> Mexicano hasnacionalidad value "Mexicano"
>> Extranjero (not Mexicano)
>>
>> then,at the moment when run inferencing
>> it isn't infer nothing in "Extranjero" when the list have a lot of
>> "Extranjeros"
>>
>> I used not, disjoint and complementOf
>>
>> Regards!
>>
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