I was reviewing All NP problems reduce to NP-complete problems: so how can NP problems not be NP-complete?

I understand that the general way we show a problem A is in NP is to show there exists a poly-time certifier for A. However, I am confused if say, we know nothing about A, and it reduces to an NP-Complete problem B, can we now say that A is in NP?

I understand that all problems in NP reduce to any NP-Complete problems, but I'm not sure if this will prove anything. I also understand to show a problem is NP-Complete, we first show it's in NP with a poly-time verifier/certificate and then reduce an NP-Complete problem to it. But confused about this other scenario:

Let A be an unknown problem. Let B be a known problem in NP-Complete. A reduces to B What does this tell us about A?


1 Answer 1


Yes (assuming you meant poly-time reductions).

Being in $NP-complete$ is equivalent to being in both classes:

  1. NP
  2. NP-hard

In particular, $B\in NP$. Now, $A\in NP$ since $A\le_p B$ and $NP$ is closed under poly-time reductions (you can easily see how to create a non-deterministic poly-time TM for $A$ given the reduction $f$ and a TM for $B$)


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