Suppose X and Y are decision problems for which X≤ P Y, i.e., X is polynomial-time reducible to Y . If X is NP-complete and Y is in NP, why Y must also be NP-complete.
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NP-complete is a subset of NP not the other way around. – apokryfos Apr 24 '22 at 05:09
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1@Progman even though the question is not very well asked, I don't think it should be moved to the CS forum, because this question is about very basic notions of CS, which are also discussed in SO. – jthulhu Apr 24 '22 at 19:54
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If X is NP complete, in particular it is NP hard, that is, every NP problem Z is polynomial-time reducible to X, which in turn is polynomial-time reducible to Y, thus Y is also NP hard. Being both NP and NP hard means you're NP complete, therefore Y is NP complete.
jthulhu
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