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Jun 13, 2019 at 6:56 comment added dkaeae 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.)
Jun 12, 2019 at 18:46 vote accept MLStudent
Jun 12, 2019 at 18:46
Jun 12, 2019 at 18:46 comment added MLStudent 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?
Jun 12, 2019 at 12:24 history answered dkaeae CC BY-SA 4.0