Suppose the following situation.

  • You have a decision problem $D$.
  • You know that $SAT$ is $NP$-complete.
  • You know that $D\leq_p SAT$.

Can you conclude that $D\in NP$?

I think it's true because it means that $D$ is at least as hard as any other problem in $NP$.

As I know I am not able to post questions to answer with yes/no: If it's true, can you give a better explanation? If it's false, can you argue why is so?

PD: it's not homework.


1 Answer 1


You can obtain a non-deterministic polynomial-time Turing machine $T$ for $D$ as follows:

  • Since $D \le_p SAT$, there is a polynomial-time Turing machine $T'$ that maps any instance of $D$ into an instance of $SAT$. Simulate the $T'$ on $x$, where $x$ is the input of $T$.
  • Use any non-deterministic polynomial-time algorithm to decide SAT (e.g., guess the value of all the variables in the SAT formula and check if the formula is satisfied). By definition of Karp reduction, $T$ will accept if and only if $x \in D$.

This shows that $D \in \mathsf{NP}$.


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