Dankness asked > Is there a model of ZFC inside which ZFC does not have a model?
Noah Schweber answered: > Yes. > Recall that by the Completeness Theorem, having a model and being > consistent are the same thing. Also, by Incompleteness, ZFC doesn't > prove its own consistency. Finally, ZFC can prove the Soundness > Theorem - that an inconsistent theory has no models! > So - assuming ZFC has a model - ZFC is consistent. If ZFC is > consistent, then ZFC can't prove "ZFC is consistent." By completeness, > this means there's a model of ZFC satisfying "ZFC is inconsistent." > Since ZFC proves the Soundness Theorem, this model must think that ZFC > has no model! Algém teria a bondade de esclarecer o que significam "modelos pensantes" e por quê o teorema de *completude* está sendo invocado para *ZFC*? -- Hermógenes Oliveira -- Você está recebendo esta mensagem porque se inscreveu no grupo "LOGICA-L" dos Grupos do Google. Para cancelar inscrição nesse grupo e parar de receber e-mails dele, envie um e-mail para logica-l+unsubscr...@dimap.ufrn.br. Para postar neste grupo, envie um e-mail para logica-l@dimap.ufrn.br. Visite este grupo em https://groups.google.com/a/dimap.ufrn.br/group/logica-l/. Para ver esta discussão na web, acesse https://groups.google.com/a/dimap.ufrn.br/d/msgid/logica-l/877fdp62nn.fsf%40camelot.oliveira.
