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At first I want to mention that I am new in ontology and protege. I try to develop an ontology using protege 5.0. I defined some data properties with enumerated range which I defined in data range expression like this: {"Balanced" , "Clever" , "Energetic" , "Lazy" , "Talkative"}. Now I want to define an individual and want to choose a value from the range I defined above as filler for a datatype property. How can I do this?

thanks for your help in advance

Ignazio
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Sanam
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1 Answers1

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It is not possible to declare an individual from that range, because datatype ranges are composed of literals, not individuals.

If you want an individual labeled, for example, as "Balanced", create an individual with any IRI you think adequate and add an rdfs:label to it, as "Balanced"

Edit: as discussed in the comments, what is wanted is a data property assertion p(A, "literal")

I have done this with the following steps: - create a data property p (data property tab) - create an individual A (individual tab) - select A, a data property assertion tab appears among others - select the + button - select the desired property and insert the desired value

Note: In the question, a datatype enumeration is mentioned - I have not added it to the example, but the procedure is the same if the property p has a declared range.

This is shown in the picture below for reference.

enter image description here

Ignazio
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    "It is not possible to declare an individual from that range, because datatype ranges are composed of literals, not individuals." I think OP, who appears to be new to Protégé might have wanted to declare a new individual, and give it a value for the property that has the enumerated range. E.g., OP has declared property p with range {"x","y"} and now wants to declare individual a and assert that p(a,"y"). – Joshua Taylor Sep 05 '14 at 23:24
  • That's a possible interpretation. So, the last sentence of the question should be "Now I want to define an individual and assert a value for one of the properties from the range I defined above." – Ignazio Sep 06 '14 at 08:55
  • Thanks for your help. Yes as Joshua wrote I want to have an individual p(a, "y"). How can I do that? – Sanam Sep 08 '14 at 08:15