Thanks, Holger. My need in this case isn't to query inferred data using 
SPARQL or GraphQL, but rather to reuse the inferred data as an input for 
another EDG property shape with another sh:values rule in it. I think I'm 
hearing that that's not possible.

Can you point me to where I can read more about the materialization option 
and decide whether there's a way to make that work for this use case? 
Thanks.

On Tuesday, January 18, 2022 at 4:03:32 PM UTC-8 Holger Knublauch wrote:

>
> On 2022-01-19 9:44 am, cbur...@healthwise.org wrote:
>
> I have defined a property shape and am successfully inferring its objects 
> using a sh:values rule.
>
> Can you remind me whether it is possible to use this property and its 
> inferred data in the sh:values rule for a different property shape? If so, 
> is there a special way that it needs to be done -- like, does the data I 
> inferred in the first property shape need to be materialized in some way 
> before I can make use of it?  -- 
>
> In general, the RDF graphs that TopBraid operates on do not automatically 
> "see" the extra inferred triples, so they are not computed as RDF triples 
> on the fly. However, our higher level query languages GraphQL and 
> JavaScript/ADS do see them whenever they are requested. The user interface 
> goes through GraphQL and will therefore display these values as if they 
> were materialized.
>
> Now, within SHACL itself it depends on what you want to do. If you have 
> SHACL-AF inference rules including other sh:values rules you can safely use 
> path expressions 
> https://w3c.github.io/shacl/shacl-af/#node-expressions-path and the 
> values will be computed on the fly. For SHACL constraints there is no such 
> mechanism unless you are expressing the constraints in SPARQL. There you 
> can use the magic property 
>
>     (?focusNode ?predicate) tosh:values ?result 
>
> to query the inferred values.
>
> The other option, of course, is to materialize the inferences, e.g. using 
> the Inferences panel or the Transform tab.
>
> Does this help?
>
> Holger
>
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