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${\sf UP}$ is defined in terms of unambiguous-SAT which asks if there exits at most one solution or no solution. On the other hand, ${\sf US}$ is defined in terms of unique-SAT which asks if there exists exactly one solution. I would like to know whether it is known that they are Turing equivalent, ${\sf P^{UP}}={\sf P^{US}}$. Any references would be appreciated.

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$US$ contains $coNP$, so $P^{US}$ is at least as powerful as $P^{coNP}=P^{NP}$. The (mostly) common wisdom suggests that $P^{UP}$ is strictly contained in $P^{NP} \subseteq P^{US}$. – Joshua Grochow Nov 17 '12 at 2:57

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up vote 10 down vote accepted

[Since it's been a while and no one has added any further answers, I'm reposting my comment as an answer.]

${\sf US}$ contains ${\sf coNP}$, so ${\sf P^{US}}$ is at least as powerful as ${\sf P^{coNP}} = {\sf P^{NP}}$. The (mostly) common wisdom suggests that ${\sf P^{UP}}$ is strictly contained in ${\sf P^{NP}} \subseteq {\sf P^{US}}$.

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