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I need to prove that $\mathsf{NP}$ is a subset of the union of $\mathsf{DTIME}(2^{n^c})$ for all $c > 1$.

Let $L$ be a language/decision problem in $\mathsf{NP}$. Then $L$ can be decided given a polynomial-size certificate in polynomial time with a turing machine $M$. So then we enumerate all possible certificates of polynomial size. There are $2^l$ possible certificates for a certificate of length $l$. For a certificate of length up to $n^c$, there are $\sum_{l=0}^{n^c} 2^l = 2^{n^c + 1} - 1$ many certificates. Each certificate can be decided in polynomial time, so we get that each problem in $\mathsf{NP}$ can be done in $\mathsf{DTIME}(2^{n^c}n^c)$. What am I doing wrong?

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    $\begingroup$ Did you know that you can render your formulas using LaTeX? $\endgroup$ – Dave Clarke May 29 '12 at 6:34
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You are on the right track. To finish the proof you need to show that

$\qquad \displaystyle \mathsf{DTIME}(2^{n^k} n^k) \subseteq \mathsf{DTIME}(2^{n^c})$

for some constant $c$.

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  • $\begingroup$ so then c can be a function of k, but not n? $\endgroup$ – Michael Studebaker May 30 '12 at 5:39
  • $\begingroup$ Correct. As $k$ is constant for a specific $L$, so will be $c$ if selected by a function of $k$. $\endgroup$ – Mike B. May 30 '12 at 7:31

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