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    NP is closed under polynomial-time many-one reductions — Carmelics
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    Supports→BHP is in P only if P equals NP

    NP is closed under polynomial-time many-one reductions

    Philosophy of LanguageTruth & Knowledge
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    BHP is NP-completeBHP is in P only if P equals NPIf any NP-complete problem is in P, then every problem in NP reduces to it and i...

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    NP is closed under polynomial-time reductions94%NP is closed under polynomial-time many-one reducibility92%NP is closed under polynomial time many-one reducibility, meaning if Y...87%f is a polynomial-time many-one reduction of Y to BHP83%

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    Similarly, parts i) and ii) respectively implies that \(\textbf{P} \subsetneq \textbf{EXP}\) and \(\textbf{NP} \subsetneq \textbf{NEXP}\). And it similarly follows from part iii) that \(\textbf{L} \subsetneq \textbf{PSPACE}\). Note that since every deterministic Turing machine is, by definition, a non-deterministic machine, we clearly have \(\textbf{P} \subseteq \textbf{NP}\) and \(\textbf{PSPACE} \subseteq \textbf{NPSPACE}\). 2 Suppose that \(f(n)\) is both time and space constructible. Then

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