Why adding transitivity breaks language filter in SPARQL Query?

sparql_query_inference

@prefix tpedia: <http://tatipedia.org/> .
@prefix owl: <http://www.w3.org/2002/07/owl#> .
@prefix rdfs: <http://www.w3.org/2000/01/rdf-schema#> .

tpedia:new_york a tpedia:Place ;
                rdfs:label "Nova Iorque"@pt ;
                rdfs:label "New York"@en .

tpedia:london a tpedia:Place ;
                rdfs:label "Londres"@pt ;
                rdfs:label "London"@en .

tpedia:name rdf:type owl:DatatypeProperty ;
            rdfs:subPropertyOf rdfs:label .

tpedia:munich a tpedia:Place ;
                tpedia:name "Munique"@pt ;
                tpedia:name "Munich"@en .

The issue was solved using inference.

First, we need to add the inference rule using the Virtuoso ISQL:

rdfs_rule_set ('http://tatipedia.org/property_ruleset', 'http://tatipedia.org');

After this, the query bellow will work as expected:

DEFINE input:inference <http://tpedia.org/property_ruleset>
SELECT DISTINCT ?subject ?label
WHERE {
    ?subject a <http://tatipedia.org/Place>;
             rdfs:label ?label  .
    FILTER (langMatches(lang(?label), "pt"))
}