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Apr 19, 2015 at 2:10 comment added Turbo Since $\mathsf{3SAT}$ reduces to $\Pi$ which reduces to $\hat{\Pi}$, doesn't $\hat{\Pi}$ problem remain $\mathsf{NP}-\mathsf{complete}$? By definition $\hat{\Pi}\in\mathsf{NP}$, while since $\mathsf{3SAT}$ reduces to $\hat{\Pi}$ every problem in $\mathsf{NP}$ reduces to $\mathsf{3SAT}$ $\implies$ every problem in $\mathsf{NP}$ reduces to $\hat{\Pi}$.
Apr 18, 2015 at 22:05 history answered sdcvvc CC BY-SA 3.0