# 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.