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According to Wikipedia https://en.wikipedia.org/wiki/RP_(complexity), $P \ne NP$ implies that $RP$ is a strict subset of $NP$. Does anybody have a reference? Furthermore, am I correct that if this indeed the case, then $NP-COMPLETE \cap RP = \emptyset$ since we can use $NP$ completeness to solve all other $NP$ problems?

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    $\begingroup$ Wiki states that we need to assume this in addition to P=BPP, which makes it trivial. This implication is not known unconditionally. $\endgroup$ – Ariel Jun 26 at 17:05
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You have not accurately summarized the statement in Wikipedia. The statement in Wikipedia also needs the extra assumption that $P=BPP$, which is widely conjectured but has not been proven to be true.

With that clarification, no reference is needed -- the reasoning is straightforward and already described on Wikipedia. If $P=BPP$, then $P=RP=co-RP$ (since $BPP=co-BPP$ and $RP \subseteq BPP$ and $co-RP \subseteq BPP$). If additionally $P \ne NP$, then it follows that $RP \ne NP$.

Yes, if $RP \ne NP$, then $NPC \cap RP = \emptyset$.

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