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Proving NP is a subset of the union of exponential DTIME
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dtimeprovingtheexponentialunionsubset
Problem
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?
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?
Solution
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 $c$.
$\qquad \displaystyle \mathsf{DTIME}(2^{n^k} n^k) \subseteq \mathsf{DTIME}(2^{n^c})$
for some constant $c$.
Context
StackExchange Computer Science Q#2151, answer score: 6
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