For problems X and Y, Y is NP-complete and X reduces to Y in polynomial time.…

Answer: X is in NP, but not necessarily NP-complete.

• Why X is in NP: Y is NP-complete, so Y is in NP. There is a polynomial-time reduction from X to Y. On input x, compute the reduction f(x) in polynomial time and then run the nondeterministic polynomial-time verifier (or decider) for Y on f(x). This composes to a nondeterministic polynomial-time procedure for X, so X is in NP.

• Why X is not necessarily NP-hard or NP-complete: NP-hardness of X would require that every problem in NP reduces to X (equivalently, an NP-complete problem reduces to X). A reduction from X to Y does not provide reductions from all NP problems to X, so X may fail to be NP-hard and hence may not be NP-complete.

• Why the other statements are false:

  • The claim that a polynomial-time algorithm for X implies a polynomial-time algorithm for Y is false because the reduction goes from X to Y. A poly-time algorithm for Y implies a poly-time algorithm for X, not vice versa.

  • The claim that X is NP-complete is false because NP-completeness additionally requires NP-hardness, which is not established by a reduction from X to Y.

  • The claim that X is NP-hard is false because NP-hardness would require a reduction from an NP-complete problem to X (or from every NP problem to X), not a reduction from X to an NP-complete problem.