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If a language is defined such that
$L = (0+1)^{\ast}$ if $\mathsf{P} = \mathsf{NP}$ and $\emptyset$ otherwise

Then $L$ is a regular language if $\mathsf{P} = \mathsf{NP}$, otherwise it is the empty langauge. Therefore $\mathsf{P} = \mathsf{NP}$ , $L$ is recursive (being regular), but is $L$ still recursive if $\mathsf{P} \neq \mathsf{NP}$?

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  • $\begingroup$ Welcome to cstheory, a Q&A site for research-level questions in theoretical computer science (TCS). Your question does not appear to be a research-level question in TCS. Please see the FAQ for more information on what is meant by this. We have migrated your question to Computer Science which has a broader scope. $\endgroup$ – Kaveh Jan 10 '13 at 2:47
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    $\begingroup$ In both cases L is regular ans thus recursive. $\endgroup$ – Ran G. Jan 10 '13 at 4:22
  • $\begingroup$ @Ran G. Turn into an answer? $\endgroup$ – Yuval Filmus Jan 10 '13 at 11:35
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The language $L = \emptyset$ is indeed a recursive set.

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