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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    $\begingroup$ 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

The language $L = \emptyset$ is indeed a recursive set.


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