Turing Incomparability in Scott Sets

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Abstract

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For every Scott set F and every nonrecursive set X in F, there is a Y in F such that X and Y are Turing incomparable.

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Logic, 03D28

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http://salesiana.dossiersoluciones.com/handle/123456789/109039

Consulte el texto completo en el siguiente enlace:

https://arxiv.org/abs/math/0602439
 http://arxiv.org/abs/math/0602439

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arXiv