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This is true because every language in $NP$ is decidable and therefore HALTS but how do I formally show this?

bob
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  • What do you mean by mapping reducible? If you mean a many-one computable reduction, then since the language is in NP, the reduction itself can determine whether the instance is a Yes or a No. – Yuval Filmus Aug 10 '19 at 07:08

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It depends what type of reduction you're talking about. But if you mean a "many-one reduction with no time constraints" (a Turing machine that transforms instances of problem $L$ to instances of problem HALT such that the answers remain the same), then yes.

The algorithm goes something like this:

  • Take an input $x$
  • Use brute force to decide whether $x \in L$ (in potentially exponential but still finite time)
  • If the answer is "yes", let $T$ be a Turing machine that always accepts
  • If the answer is "no", let $T$ be a Turing machine that always diverges
  • Return $\langle T \rangle$

Now, when $\langle T \rangle$ is passed to a HALT oracle, the oracle will accept if $x \in L$, and reject if $x \not \in L$.

Draconis
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