For $L_1,L_2 \in RE - R $ , I want to prove or disprove if the following can occur:

  1. $L_1 \cap L_2 \in R$
  2. $L_1 \cup L_2 \in R$
  3. $L_1 \cap L_2 \in R$ and $L_1 \cup L_2 \in R$

What I did:

  1. I think any two disjoint languages suffice, since the empty set is decidable.

  2. I think something along the lines of a language and its complement but I'm struggling to think of an example.

  3. It seems impossible but I have no idea how to prove it.

Any help/further insight would be welcomed!

  • $\begingroup$ Note that for 1, it's not enough to show that some particular $L_1$ and $L_2$ have an intersection in $R$. I'm guessing they want you to show it's true for all RE languages i.e. that RE is closed under intersection. If you're disproving closure, one counter-example is enough though. $\endgroup$ Commented Apr 15, 2014 at 20:37
  • $\begingroup$ @jmite It's absolutely enough to show that some particular $L_1$, $L_2$ have an intersection in $R$! The question asks, is it possible? So an answer to any of the parts would be either a particular $L_1$ and $L_2$ with the required property, a proof that such a pair exists, or a proof that no $L_1$, $L_2$ have that property. $\endgroup$ Commented Apr 15, 2014 at 20:41
  • $\begingroup$ For your second point, recall that if a language and its complement both are recursively enumerable, then they are recursive as well. $\endgroup$
    – Pål GD
    Commented Apr 15, 2014 at 21:01

2 Answers 2

  1. Correct. We can have $L_1\cap L_2=\emptyset\in R$. A good answer would give an example of such a pair $L_1$, $L_2$ so you should figure that out on your own.

  2. Correct but for the wrong reason. If $L_1$ and its complement are both $RE$, then both are recursive and the question says that neither is recursive. You should prove this yourself, as you'll need it for the next part.

    • Hint for this proof: show how to use machines that accept $L_1$ and $\overline{L_1}$ to give a machine that decides $L_1$.)
    • Hint for the question: take a recursives language $K_1$ and $K_2$ and set $L_1=K_1 \cup \text{something}$ and $L_2=K_2\cup\text{something else}$ in a way that gives $L_1\cup L_2=K_1\cup K_2$.
  3. Correct. We know that $L_1\in RE$. Use $L_1\cap L_2\in R$ and $L_1\cup L_2\in R$ to show that $\overline{L_1}\in RE$. By the extra proof I suggested you do in the previous part, that means that $L_1\in R$, contradicting the requirement that $L_1\in RE\setminus R$.


Hint for 2: Consider $L_1 = \{ 0 w : w \in L \} + \{ 1 w : w \in \{0,1\}^* \}$ and let $L_2$ be a complementary language.

Hint for 3: $L_2 \setminus (L_1 \cap L_2) = \overline{L_1} \cap L_2$ and $\overline{L_1 \cup L_2} = \overline{L_1} \cap \overline{L_2}$.

  • $\begingroup$ What is the explanation for $L_2 \setminus (L_1 \cap L_2) = \overline{L_1} \cap L_2 \in R$? $\endgroup$ Commented May 25, 2015 at 12:41
  • $\begingroup$ @Elimination It's in RE rather than in R. You're missing the point. Keep thinking. $\endgroup$ Commented May 25, 2015 at 13:42
  • $\begingroup$ I actually figured it out and thought I removed my comment. Thank you for responding though :) $\endgroup$ Commented May 25, 2015 at 14:09

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