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As it is explained in Sipser's book, the following language is undecidable and he proves this using the computation history method.

$\qquad E = \{\langle M \rangle \mid M\ \mathrm{LBA}, L(M)=\emptyset\}$

I wanted to see if we could use the same method used for proving undecidability of the following language in the case of LBAs too (instead of using computation history method).

$\qquad E' = \{\langle M \rangle \mid M\ \mathrm{TM}, L(M)=\emptyset\}$

Is it possible?

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  • $\begingroup$ You have been asking many similar questions in short succession. What have you tried? Also, I don't quite get what you are asking. Are you looking for a new undecidability proof for $E$ or $E'$? $\endgroup$ – Raphael Apr 2 '13 at 7:44
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Consider the reduction that shows that LBA-emptiness is undecidable: you reduce from $\overline{A_{TM}}=\{<M,w>: M $ does not accept $w\}$.

So the reduction takes as input $<M,w>$, and outputs an LBA $T$ such that $L(T)\neq \emptyset$ iff $M$ accepts $w$.

Now, recall that an LBA is in particular a TM. So this same reduction actually outputs a TM, and actually shows that $E_{TM}$ is undecidable as well.

The reason this works is that the relationship between $E_{LBA}$ and $E_{TM}$ is such that if $M$ is an LBA, then $M\in E_{LBA}$ iff $M\in E_{TM}$. Since the reduction above always outputs an LBA, then you can just plug it in.

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