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Is the following statement true?

If a problem P1 is in NP and polynomial time reducible to P2, where P2 is NP-complete, then P1 is also NP-complete.

Intuitively I think the answer is No because I need to prove that it is NP-hard as well for P1 to be NP-complete. But I cannot get the exact proof.

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  • $\begingroup$ I'm not sure what you mean by "I need to prove that all NP Hard as well for p1 to be NPC". Informally, saying that a problem is NP-hard is saying "If I could solve this problem, I could solve everything in NP with only a little more work." Does the scenario of this question allow you to do that? Note also that the two possible answers to the question are "Yes, P1 is NP-complete: here's a proof" and "No, his doesn't prove that P1 is NP-complete: here's an explanation of why not." In the negative case, you can't prove that P1 isn't NP-complete because every NPC problem is also reducible to P2. $\endgroup$ – David Richerby Apr 22 '15 at 11:24
  • $\begingroup$ that was a typo I meant ''it is NP hard as well for p1 to be NPC''(refering to the two properties for a problem to be NPC) @DavidRicherby. $\endgroup$ – yuugen Apr 22 '15 at 13:10
  • $\begingroup$ OK -- now I understand what you're asking. So, you (correctly) think we're in the "no" case of my comment. You just need to explain why the P1 being polytime reducible to an NP-complete problem isn't a proof that P1 is NP-complete. $\endgroup$ – David Richerby Apr 22 '15 at 13:15
  • $\begingroup$ Yes..I can't put my finger on that :(..could you please elaborate $\endgroup$ – yuugen Apr 22 '15 at 13:24
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    $\begingroup$ Hint: give a counter-example. $\endgroup$ – Raphael Apr 22 '15 at 14:04
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Hint: Every problem in NP is reducible to $P_2$. In particular, if $P_1$ is in NP but not NP-complete then it reduces to $P_2$ without being NP-complete. So the question is:

Are there problems in NP which are not NP-complete?

Perhaps you can answer this question somehow. Try to think of the simplest possible problems.

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  • $\begingroup$ I suggest rephrasing the question as follows: "Are there problems in NP (but not in P) that are not NP-hard?" $\endgroup$ – André Souza Lemos Apr 22 '15 at 21:56
  • $\begingroup$ Well, problems in P can't be NP-hard. Unless P = NP, of course... $\endgroup$ – André Souza Lemos Apr 22 '15 at 21:59
  • $\begingroup$ I am not contesting the exercise, on the contrary. Just suggesting what I think would be a better version of the question, that's all. As it reads now, the answer would be trivially "yes", unless you assume implicitly what I have added. $\endgroup$ – André Souza Lemos Apr 22 '15 at 22:07
  • $\begingroup$ Not to be byzantine, but assuming explicitly that the problem is not in P is, in my opinion, an informative hint that would enhance the educational value of the exercise. $\endgroup$ – André Souza Lemos Apr 22 '15 at 22:24
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – André Souza Lemos Apr 22 '15 at 22:27

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