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Answer :
Final answer:
The statement is true. If L1 is reducible to L2 in polynomial time and L2 is in NP, then L1 is also in NP.
Explanation:
The statement 'If L1 reduces to L2 in polynomial time, and if L2 is NP, then L1 must be NP' is True. This is based on the fact that if a language L1 can be reduced to another language L2 in polynomial time, and if L2 is in NP, then L1 is also in NP. The concept of polynomial time reduction is a crucial element in the theory of computational complexity within computer science. It allows us to categorize problems according to their intrinsic hardness. If you can reduce problem L1 to problem L2 in polynomial time, any solution to L2 can be used to solve L1. Therefore, if L2 is NP (Nondeterministic Polynomial time), which means we can verify a solution quickly, then L1 is also NP because we can solve L1 with our solution to L2.
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