Prove that the language {|M2 accepts , and M1 doesn't accept } isn't in RE

Question

I was asked to prove that the language {(M1),(M2)|M2 accepts (M1) , and M1 doesn't accept (M2)} isn't in RE, and not in co-RE.

I tried reductions but didn't seem one that works, I also tried variations of the proof that Ld isn't in RE, but didn't managed that either.

Any one has an idea how to prove that?

Thanks in advance :)

Answer 1

We show reduction from $HP$ and $\overline{HP}$

$$ f(,x) = , $$

$M_{x}$ on input $w$ will act as follows:

1. run $M$ on $x$.
2. accept.

$M_{\phi}$ just rejects every input word right away (so it's language is empty).

Now observe that if $M$ halts on $x$, then $L(M_{x})=\Sigma^{*}$ and otherwise it accepts the empty language too, so correctness follows.

The reduction from $\overline{HP}$ would be: $$ f(,x) = , $$

$M_x$ as before, and $M_{\Sigma^{*}}$ accepts every input word right away.

correctness follows again from the same observation.