# Reduction from deciders of the universal to deciders of the empty language

$ALL_{TM} = \{\langle M\rangle | \; M$ 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?

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.