# If NP $\neq$ Co-NP then is P $\neq$ NP

Does the proof of the widely believed result P $\neq$ NP depend on the proof of NP $\neq$ Co-NP ?

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## migrated from cstheory.stackexchange.comJan 10 '13 at 2:48

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Which proofs are you referring to? What do you mean by depending? –  Kaveh Jan 10 '13 at 2:44

Only in one direction. As $\mathsf{P}=\text{co-}\mathsf{P}$, if $\mathsf{NP}\neq\text{co-}\mathsf{NP}$ then we would know that $\mathsf{P}\neq\mathsf{NP}$. However the reverse implication doesn't hold. If $\mathsf{P}\neq\mathsf{NP}$ then it's possible that either $\mathsf{NP}\neq\text{co-}\mathsf{NP}$ or $\mathsf{NP}=\text{co-}\mathsf{NP}$.