Is it a sufficient condition to be in NP?

Suppose the following situation.

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

Can you conclude that $Din NP$?

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.