$L_0=\{\langle M,w,0\rangle\mid M \text{ halts on } w\}$
$L_1=\{⟨M,w,1⟩\mid M \text{ does not halts on } w\}$
Here $\langle M,w,i \rangle$ is a triplet, whose first component $M$ is an encoding of a Turing Machine, second component $w$ is a string, and third component $i$ is a bit.
Let $L=L_0 \cup L_1$. Is $L$ non R.E ?
After seeing the question I was able to figure out that there are some strings that do not belong to $L_0$ as well as $L_1$.
Like, lets take the case for $L_0$ and a string $001\dots10−01−1$, ("$-$" shown for notation purpose only) where the first component describes a TM $M$ followed by input "$w=01$" and last bit "$1$". Now suppose M halts on "$01$". Still the given input is not in $L_0$ as the last bit is "$1$" and not "$0$" as required by $L_0$. So, this input must be in $L_1$. But since $M$ halts on $w$, this input is not in $L_1$ either. So there are infinite strings like these. I am not able to prove that these infinite set of strings are not r.e.
Is there any method to solve these kind of problems? I am not able to master these kind of problems.
Please someone help.