5
$\begingroup$

Are there languages $L_1$, $L_2$ such that such that $$L_1 \cup L_2\leq_m\, L_1\cap L_2,$$ and two other languages such that $$L_1 \cap L_2 \leq_m\, L_1 \cup L_2?$$ And if so, what are they? How can i go about finding these?

$\endgroup$
3
$\begingroup$

How about taking $L_1=L_2$? Then we will have both $L_1 \cup L_2\leq_m\, L_1\cap L_2$ and $L_1 \cap L_2 \leq_m\, L_1 \cup L_2$.

If you want to have $L_1\not= L_2$, you just try $L_2=L_1\cup\{w\}$, where $w$ is a string not in $L_1$.

If you are determined to have infinity elements in both $L_1\setminus L_2$ and $L_2\setminus L_1$, try $L_1=L\cup aL$ and $L_2=L\cup bL$, where $L$ is an infinite language over $\Sigma$ and $a,b$ are two new symbols not in $\Sigma$.

By now, you should be able to see that there are lots of way to construct examples that satisfy the requirements.

$\endgroup$
  • $\begingroup$ My bad, it should be $$L_1 \cup L_2\leq_m\, L_1\cap L_2,$$ not $$L_1 \cup L_2\leq_m\, L_2\cap L_2,$$ but i like your solution, any solution for the edited one? $\endgroup$ – Karlberg Jun 10 at 16:50
  • $\begingroup$ The same solution works. In fact, I did not notice it was $L_2\cap L_2$ when I wrote the answer. $\endgroup$ – Apass.Jack Jun 10 at 16:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.