If a problem X (decision problem) is known to be NP-Complete, and proven to be reduced to problem Y, can you then say problem Y is NP-Complete?

Question

If a problem X (decision problem) is known to be NP-Complete, and proven to be reduced to problem Y in polynomialtime, can you then say problem Y is NP-Complete?

My first thought was, no, problem Y needs to be shown that it is in NP. But after further thought, if X is reduced to Y, Y is already considered to be NP-Complete. Now I'm just confused...any help would be appreciated.

Answer 1

Argumentum per contrarium:

If X ∈ NP and X ⇔ Y and Y ∉ NP then X ∉ NP.

Answer 2

Problem X - Unsure
Problem Y - In NP

To prove X is in NP, you show that you can follow steps to reduce every problem in X to a problem in Y. Then you know that the X problem is at least as hard as the equivalent Y problem.

So no, you need to start with Y and then reduce to X.