{
  "http://es.wikipedia.org/wiki/Divide_and_Conquer_(Stargate_SG-1)" : { "http://xmlns.com/foaf/0.1/primaryTopic" : [ { "type" : "uri", "value" : "http://es.dbpedia.org/resource/Divide_and_Conquer_(Stargate_SG-1)" } ] } ,
  "http://es.dbpedia.org/resource/Divide_and_Conquer_(Stargate_SG_1)" : { "http://dbpedia.org/ontology/wikiPageRedirects" : [ { "type" : "uri", "value" : "http://es.dbpedia.org/resource/Divide_and_Conquer_(Stargate_SG-1)" } ] } ,
  "http://es.dbpedia.org/resource/Crossroads_(Stargate_SG-1)" : { "http://dbpedia.org/ontology/subsequentWork" : [ { "type" : "uri", "value" : "http://es.dbpedia.org/resource/Divide_and_Conquer_(Stargate_SG-1)" } ] ,
    "http://es.dbpedia.org/property/siguiente" : [ { "type" : "uri", "value" : "http://es.dbpedia.org/resource/Divide_and_Conquer_(Stargate_SG-1)" } ] } ,
  "http://es.dbpedia.org/resource/Window_of_Opportunity_(Stargate_SG-1)" : { "http://dbpedia.org/ontology/previousWork" : [ { "type" : "uri", "value" : "http://es.dbpedia.org/resource/Divide_and_Conquer_(Stargate_SG-1)" } ] ,
    "http://es.dbpedia.org/property/previo" : [ { "type" : "uri", "value" : "http://es.dbpedia.org/resource/Divide_and_Conquer_(Stargate_SG-1)" } ] } ,
  "http://es.dbpedia.org/resource/Divide_and_Conquer_(Stargate_SG-1)" : { "http://www.w3.org/1999/02/22-rdf-syntax-ns#type" : [ { "type" : "uri", "value" : "http://dbpedia.org/ontology/TelevisionEpisode" } ,
      { "type" : "uri", "value" : "http://www.w3.org/2002/07/owl#Thing" } ,
      { "type" : "uri", "value" : "http://schema.org/CreativeWork" } ,
      { "type" : "uri", "value" : "http://dbpedia.org/ontology/Work" } ,
      { "type" : "uri", "value" : "http://www.wikidata.org/entity/Q386724" } ,
      { "type" : "uri", "value" : "http://schema.org/TVEpisode" } ] ,
    "http://www.w3.org/2000/01/rdf-schema#label" : [ { "type" : "literal", "value" : "Divide and Conquer (Stargate SG-1)" , "lang" : "es" } ] ,
    "http://www.w3.org/2000/01/rdf-schema#comment" : [ { "type" : "literal", "value" : "Divide and Conquer (Divide y Vencer\u00E1s) es el quinto episodio de la cuarta temporada de la serie de televisi\u00F3n de ciencia ficci\u00F3n Stargate SG-1, y el septuag\u00E9simo primer cap\u00EDtulo de toda la serie." , "lang" : "es" } ] ,
    "http://www.w3.org/2002/07/owl#sameAs" : [ { "type" : "uri", "value" : "http://es.dbpedia.org/resource/Divide_and_Conquer_(Stargate_SG-1)" } ] ,
    "http://xmlns.com/foaf/0.1/name" : [ { "type" : "literal", "value" : "Divide and Conquer" , "lang" : "es" } ] ,
    "http://purl.org/dc/terms/subject" : [ { "type" : "uri", "value" : "http://es.dbpedia.org/resource/Categor\u00EDa:Episodios_de_Stargate_SG-1" } ,
      { "type" : "uri", "value" : "http://es.dbpedia.org/resource/Categor\u00EDa:Episodios_de_televisi\u00F3n_de_2000" } ] ,
    "http://xmlns.com/foaf/0.1/isPrimaryTopicOf" : [ { "type" : "uri", "value" : "http://es.wikipedia.org/wiki/Divide_and_Conquer_(Stargate_SG-1)" } ] ,
    "http://dbpedia.org/ontology/wikiPageID" : [ { "type" : "literal", "value" : 1716625 , "datatype" : "http://www.w3.org/2001/XMLSchema#integer" } ] ,
    "http://dbpedia.org/ontology/wikiPageRevisionID" : [ { "type" : "literal", "value" : 117065644 , "datatype" : "http://www.w3.org/2001/XMLSchema#integer" } ] ,
    "http://dbpedia.org/ontology/wikiPageExternalLink" : [ { "type" : "uri", "value" : "http://www.imdb.com/title/tt0709073/" } ,
      { "type" : "uri", "value" : "http://www.stargate-sg1-solutions.com/wiki/4.05_%22Divide_And_Conquer%22_Episode_Guide" } ,
      { "type" : "uri", "value" : "http://www.scifi.com/stargate/episodes/season04/0405/" } ,
      { "type" : "uri", "value" : "http://www.gateworld.net/sg1/s4/405.shtml" } ] ,
    "http://dbpedia.org/ontology/wikiPageLength" : [ { "type" : "literal", "value" : "6455" , "datatype" : "http://www.w3.org/2001/XMLSchema#nonNegativeInteger" } ] ,
    "http://es.dbpedia.org/property/t\u00EDtulo" : [ { "type" : "literal", "value" : "Divide and Conquer" , "lang" : "es" } ] ,
    "http://es.dbpedia.org/property/estrellas" : [ { "type" : "uri", "value" : "http://es.dbpedia.org/resource/J.R._Bourne" } ,
      { "type" : "uri", "value" : "http://es.dbpedia.org/resource/Andrew_Jackson_(Actor)" } ,
      { "type" : "uri", "value" : "http://es.dbpedia.org/resource/Kirsten_Robek" } ,
      { "type" : "uri", "value" : "http://es.dbpedia.org/resource/Phillip_Mitchell" } ,
      { "type" : "uri", "value" : "http://es.dbpedia.org/resource/Vanessa_Angel" } ,
      { "type" : "uri", "value" : "http://es.dbpedia.org/resource/Teryl_Rothery" } ] ,
    "http://es.dbpedia.org/property/serie" : [ { "type" : "uri", "value" : "http://es.dbpedia.org/resource/Stargate_SG-1" } ] ,
    "http://es.dbpedia.org/property/producci\u00F3n" : [ { "type" : "literal", "value" : 405 , "datatype" : "http://www.w3.org/2001/XMLSchema#integer" } ] ,
    "http://dbpedia.org/ontology/episodeNumber" : [ { "type" : "literal", "value" : "5" , "datatype" : "http://www.w3.org/2001/XMLSchema#nonNegativeInteger" } ] ,
    "http://dbpedia.org/ontology/title" : [ { "type" : "literal", "value" : "Divide and Conquer" , "lang" : "es" } ,
      { "type" : "literal", "value" : "Divide y Vencer\u00E1s" , "lang" : "es" } ] ,
    "http://dbpedia.org/ontology/seasonNumber" : [ { "type" : "literal", "value" : "4" , "datatype" : "http://www.w3.org/2001/XMLSchema#nonNegativeInteger" } ] ,
    "http://dbpedia.org/ontology/director" : [ { "type" : "uri", "value" : "http://es.dbpedia.org/resource/Martin_Wood" } ] ,
    "http://dbpedia.org/ontology/guest" : [ { "type" : "uri", "value" : "http://es.dbpedia.org/resource/Vanessa_Angel" } ,
      { "type" : "uri", "value" : "http://es.dbpedia.org/resource/J.R._Bourne" } ,
      { "type" : "uri", "value" : "http://es.dbpedia.org/resource/Kirsten_Robek" } ,
      { "type" : "uri", "value" : "http://es.dbpedia.org/resource/Phillip_Mitchell" } ,
      { "type" : "uri", "value" : "http://es.dbpedia.org/resource/Teryl_Rothery" } ,
      { "type" : "uri", "value" : "http://es.dbpedia.org/resource/Andrew_Jackson_(Actor)" } ] ,
    "http://dbpedia.org/ontology/previousWork" : [ { "type" : "uri", "value" : "http://es.dbpedia.org/resource/Crossroads_(Stargate_SG-1)" } ] ,
    "http://dbpedia.org/ontology/series" : [ { "type" : "uri", "value" : "http://es.dbpedia.org/resource/Stargate_SG-1" } ] ,
    "http://dbpedia.org/ontology/subsequentWork" : [ { "type" : "uri", "value" : "http://es.dbpedia.org/resource/Window_of_Opportunity_(Stargate_SG-1)" } ] ,
    "http://dbpedia.org/ontology/writer" : [ { "type" : "uri", "value" : "http://es.dbpedia.org/resource/Tor_Alexander_Valenza" } ] ,
    "http://www.w3.org/ns/prov#wasDerivedFrom" : [ { "type" : "uri", "value" : "http://es.wikipedia.org/wiki/Divide_and_Conquer_(Stargate_SG-1)?oldid=117065644&ns=0" } ] ,
    "http://dbpedia.org/ontology/abstract" : [ { "type" : "literal", "value" : "Divide and Conquer (Divide y Vencer\u00E1s) es el quinto episodio de la cuarta temporada de la serie de televisi\u00F3n de ciencia ficci\u00F3n Stargate SG-1, y el septuag\u00E9simo primer cap\u00EDtulo de toda la serie." , "lang" : "es" } ] ,
    "http://es.dbpedia.org/property/temporada" : [ { "type" : "literal", "value" : 4 , "datatype" : "http://www.w3.org/2001/XMLSchema#integer" } ] ,
    "http://es.dbpedia.org/property/escritor" : [ { "type" : "uri", "value" : "http://es.dbpedia.org/resource/Tor_Alexander_Valenza" } ] ,
    "http://es.dbpedia.org/property/director" : [ { "type" : "uri", "value" : "http://es.dbpedia.org/resource/Martin_Wood" } ] ,
    "http://es.dbpedia.org/property/siguiente" : [ { "type" : "uri", "value" : "http://es.dbpedia.org/resource/Window_of_Opportunity_(Stargate_SG-1)" } ] ,
    "http://es.dbpedia.org/property/episodios" : [ { "type" : "uri", "value" : "http://es.dbpedia.org/resource/Anexo:Episodios_de_Stargate_SG-1" } ] ,
    "http://es.dbpedia.org/property/previo" : [ { "type" : "uri", "value" : "http://es.dbpedia.org/resource/Crossroads_(Stargate_SG-1)" } ] ,
    "http://es.dbpedia.org/property/emisi\u00F3n" : [ { "type" : "literal", "value" : 28 , "datatype" : "http://www.w3.org/2001/XMLSchema#integer" } ] ,
    "http://es.dbpedia.org/property/t\u00EDtuloTrad" : [ { "type" : "literal", "value" : "Divide y Vencer\u00E1s" , "lang" : "es" } ] ,
    "http://es.dbpedia.org/property/episodio" : [ { "type" : "literal", "value" : 5 , "datatype" : "http://www.w3.org/2001/XMLSchema#integer" } ] }
}
