SQUALL to SPARQL Translator

Your SQUALL sentence:

For every person X that shares a parent with a person Y, return that X has sibling Y and Y has sibling X.

CONSTRUCT {

?x1 :sibling ?x3 .
?x3 :sibling ?x1 .
} WHERE { ?x1 a :person .
?x1 :parent ?x2 .
?x3 a :person .
FILTER (?x1 != ?x3) .
?x3 :parent ?x2 .

}

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).