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For the proof of "Cook-Levin Theorem", for a Turing Machine $M$ that accepts a language $L \in NP$ and input $x \in \{0,1\}^*$, we can create a SAT Formula, that is satisfiable if and only if $M$ accepts $x$. Could we adopt this construction so that for any Turing Machine $M$ and input $x \in \{0,1\}^*$ we can create a SAT Formula $\phi$ that is satisfiable if and only if $M$ accepts $x$ (even if this SAT Formula has more than polynomial length of $|M|$)? Or would that contradicts Rice's Theorem?

Edit: As dkaeae correctly pointed out, defining a SAT formula that is satisfiable iff a TM $M$ accepts an input $x$ is indeed possible. What I meant to ask though is, whether a reduction in the sense of a computable function exists (albeit not being limited to running in polynomial time, but indeed being somehow limited in the running time).

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The proof of the Cook-Levin theorem shows that for any nondeterministic Turing machine $M$, any input $x$, and any (reasonable) time bound $t(n)$, we can construct (efficiently) a SAT formula of size proportional to $t(|x|)^2$ that is satisfiable iff $M$ accepts $x$ within $t(|x|)$ steps.

If you know that $M$ is supposed to halt within $t(n)$ steps, then the SAT formula captures the behavior of $M$ on $x$. But for a general Turing machine we don't possess any a priori bound on the running time. This is why you cannot solve the halting problem in this way.

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Certainly you can. Given $M$ and an input $w \in \{ 0,1 \}^\ast$, produce $x \lor x$ if $M$ accepts $x$ and $x \land \lnot x$ otherwise.

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  • $\begingroup$ Thanks! I will edit my question to add: Can we define a reduction, which has a running time somehow bounded by the input length (not necessarily by a polynomial though), but still outputs a sat formula that is satisfiable iff $M$ accepts $x$? I think in this case, Yuval's answer holds, correct? $\endgroup$ – MLStudent Jun 12 '19 at 18:46
  • $\begingroup$ Indeed. It seems the (only?) requirement for the reduction to work is that the time of $M$ is bounded by a computable function. (This implies, among other things, that $M$ always halts on every input.) $\endgroup$ – dkaeae Jun 13 '19 at 6:56

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