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$ALL_{TM} = \{\langle M\rangle | \; M$ is $TM$ and $L(M)=\Sigma^*\}$
$E_{TM} = \{\langle M\rangle | \; M$ is $TM$ and $L(M)=\emptyset\}$

I can't find reduction of $ALL_{TM}$ to $E_{TM}$. But I can't proof that it does not exist. Is there is such a reduction?

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The point to note is if you can reduce $L_1$ to $L_2$ then the same reduction/mapping can be used to see that $\overline{L_1}$ is reducible to $\overline{L_2}$. The following is the proof, avoid if you just needed the hint.
Now we know $E_{TM}$ is not Turing recognizable but $\overline{E_{TM}}$ is. Secondly we know $\overline{ALL_{TM}}$ is not Turing recognizable . So say if $ALL_{TM}$ is reducible to $E_{TM}$, then it would mean $\overline{ALL_{TM}}$ is reducible to $\overline{E_{TM}}$, and as $\overline{E_{TM}}$ is Turing recognizable $\overline{ALL_{TM}}$ would become Turing recognizable which is a contradiction.

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  • $\begingroup$ Yes other moderators told me that. I was ready to remove my answer but it was decided that in future I should not post answers to questions directly asking for solutions. Now I take care of it. $\endgroup$
    – sashas
    Jan 26 '16 at 17:22
  • $\begingroup$ Okay, never mind then. :) (Please flag this for removel after reading. Thanks!) $\endgroup$
    – Raphael
    Jan 26 '16 at 23:51

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