Intersection and union of a regular and a non-regular language Let $L_1$ be regular, $L_1 \cap L_2$ regular, $L_2$ not regular. Show that $L_1 \cup L_2$ is not regular or give a counterexample.


I tried this: Look at $L_1 \setminus (L_2 \cap L_1)$. This one is regular. I can construct a finite automaton for this: $L_1$ is regular, $L_2 \cap L_1$ is regular, so remove all the paths (finite amount) for $L_1 \cap L_2$ from the finite amount of paths for $L_1$. So there are a finite amount of paths left for this whole thing. This thing is disjoint from $L_2$, but how can I prove that the union of $L_1 \setminus (L_1 \cap L_2)$ (regular) and $L_2$ (not regular) is not regular?
We can prove this by contradiction. 
Lets define $\overline{L_1} = \Sigma^* \setminus L_1$. Then we can reformulate $L_2$:

$L_2 = ((L_1 \cup L_2) \setminus L_1) \cup (L_1 \cap L_2) = ((L_1 \cup L_2) \cap \overline{L_1}) \cup (L_1 \cap L_2)$

We know:


Regular Languages are closed under union, intersection and complement
$\overline{L_1}$ and $L_1 \cap L_2$ are regular
$L_2$ is not regular


Now assume $L_1 \cup L_2$ is regular: Then $((L_1 \cup L_2) \cap \overline{L_1}) \cup (L_1 \cap L_2)$ is regular (as it is only a union/intersection of regular languages), so $L_2$ would be regular. That is a contradiction, therefore our assumption is false, and $L_1 \cup L_2$ can not be regular.