SQUALL to SPARQL Translator

Your SQUALL sentence:

Which rdf:Property relates a publication to a person?

SELECT DISTINCT ?x1 WHERE {

?x1 a rdf:Property .
?x2 a :publication .
?x3 a :person .
?x2 ?x1 ?x3 .

}

Run at DBpedia SPARQL endpoint (assuming prefixes res: for resources, : and dbo: for ontology, and dbp: for properties in addition to DBpedia namespace definitions).

Load in DBpedia SPARQL Explorer (assuming the same prefixes as above).