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    If a problem is NP-complete and in P, then P = NP — Carmelics
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    Supports→The Boolean Halting Problem (BHP) is in P only if P equals NP

    If a problem is NP-complete and in P, then P = NP

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    Related propositions within the same area of thought.
    BHP is NP-completeNP is closed under polynomial-time reductionsThe Boolean Halting Problem (BHP) is in P only if P equals NP

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    If any NP-complete problem is in P, then every problem in NP reduces t...91%If NP ≠ coNP, then NP-complete problems cannot be in coNP, because an ...89%If P ⊊ NP, then NP-complete problems are not in P.88%If NP ≠ coNP, then problems in NP ∩ coNP are not NP-complete.87%

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    Since \(\sc{BHP}\) is \(\textbf{NP}\)-complete, it follows from the closure of \(\textbf{NP}\) under \(\leq_P\) that this problem is in \(\textbf{P}\) only if \(\textbf{P} = \textbf{NP}\). Since this is widely thought not to be the case, this provides some evidence that \(\sc{BHP}\) is an intrinsically difficult computational problem. But since \(\sc{BHP}\) is closely related to the model of computation \(\mathfrak{N}\) itself this may appear to be of little practical significance. It is thus of

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