L = { < M > | M is a turing machine and enter image description here }

Obviously, the language which L(M) is polynomially reducible to, is context free and hence recursive, so it is a decidable language .

Now, L(M) is reducible to decidable language.

I think this language is not decidable, as this language is non-trivial in the sense of rice theorem, as no recognizable language can be a part of L(M) and even this context free language belongs to L(M).

Is my understanding right ?

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    $\begingroup$ By the way, these "am I correct?" questions are not a good fit for StackExchange. There is no answer to that except "yes", which is not very informative. $\endgroup$ – chi Oct 4 '16 at 17:20

Yes, your understanding is right.

The first part (about context-free languages which are decidable) is unneeded. To apply Rice, you only have to show that the property at hand only depends on $L(M)$ (which it does, trivially) and prove that the set is nontrivial. So the whole exercise bogs down to: exhibit $M$ satisfying the reduction property, and $M′$ not satisfying it.

To prove/disprove reduction you may have to use the fact that the language $0^p1^{2p}$ is decidable, as you argued.

  • $\begingroup$ Can you please elaborate the first line more ? I am having a bit misunderstanding with that . $\endgroup$ – Garrick Oct 4 '16 at 17:25

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