0

Given three languages $L_1, L_2, L_3$ with $L_1$ and $L_3$ being semi-decidable and $L_1 \subseteq L_2, L_2 \subseteq L_3$. Can I deduce from these properties, that $L_2$ is also semi-decidable and how would I proof this?

Intuitionally it seems obvious that $L_2$ is also semi-decidable, but I didn't find a way to prove (or falsify) it.

So far I only found information about the closure of semi-decidability under $\cup$ and $\cap$. Can I use these properties in any way?

duelle
  • 11
  • 4
  • 3
    This question has been answered many times on this site. – Sam Jones Aug 05 '15 at 15:44
  • Thanks for your hint. Could you point me to them? As I couldn't find any fitting questions when I searched beforehand. Perhaps I used the wrong keywords (not a native speaker -,-). – duelle Aug 05 '15 at 15:47
  • 2
    There isn't an explicit answer to your question on this question, but, you can apply the same principal to your question: http://cs.stackexchange.com/questions/2528/are-supersets-of-non-regular-languages-also-non-regular/ – Sam Jones Aug 05 '15 at 15:53
  • See also http://cs.stackexchange.com/q/17966/755, http://cs.stackexchange.com/q/22839/755, http://cs.stackexchange.com/q/22039/755, http://cs.stackexchange.com/q/12527/755, http://cs.stackexchange.com/q/17860/755, and http://cs.stackexchange.com/q/38486/755 where you will find answers to many similar questions -- those answers can be readily adjusted to this case as well. – D.W. Aug 05 '15 at 19:33

1 Answers1

0

No, because every language $L$ satisfies $$ \emptyset \subseteq L \subseteq \Sigma^* $$ and $\emptyset$ and $\Sigma^*$ are both semi-decidable.

David Richerby
  • 81,689
  • 26
  • 141
  • 235
Yuval Filmus
  • 276,994
  • 27
  • 311
  • 503