I was studying Turing languages for an exam and I came up with this problem for wich I haven't found a solution online. This is my question:

Let's say we have $L_1, L_2 \subseteq\{0,1\}^*$. $L_1$ is Turing-recognisable and $L_2$ is Turing-decidable.

Is $L_u = L_1 \cup L_2$ decidable? Is it recognisable?

My guess is that $L_u$ is just recognisable. Here is my attemp on proving it.

A language $L \subseteq \{0,1\}^*$ is decidable if and only if both $L$ and $L^c$ are recognisable (where $L^c = \{0,1\}^* \setminus L$). My idea was to prove that $L_u$ is recognisable and $L_u^c$ is not.

In our case $L_u$ is recognisable (trivial) and $L_u^c = (L_1 \cup L_2)^c = (L_1^c \cap L_2^c)$. We know that $L_1^c$ is recognisable and $L_2^c$ is not-recognisable (or else, $L_2$ would have been decidable). At this point I got stuck. I'm pretty sure that $L_u^c$ is not-recognisable, but I found this example:

Let $L_x = \{stuff\} \cup L_1$ be a not-recognisable language and $L_1$ the language we defined above. The intersection $L_x \cap L_1 = L_1$ wich by definition is recognisable.

Follow up question:

Can we sometimes recognise the intersection between a not-recognisable language and a recognisable one? In other words, is my proof wrong somehow?

  • 1
    $\begingroup$ Please ask only one question per post. Thank you! $\endgroup$
    – D.W.
    Jun 18 at 17:04

2 Answers 2


The answer to the question "Is $L_u = L_1 \cup L_2$ decidable?" is "sometimes".

For a positive example, let both $L_1$ and $L_2$ be the empty language.
For a negative example, let $L_1$ be Turing-recognisable but not Turing-co-recognisable such as the language of halting Turing machines, and let $L_2$ be the empty language.

The answer to the question "Is $L_u = L_1 \cup L_2$ Turing-recognisable?" is "Yes". In fact, we have a stronger statement below.

Proposition: the union of two Turing-recognisable languages are Turing-recognisable.
Proof: Suppose $L_1$ and $L_2$ are recognised by Turing machine $T_1$ and $T_2$ respectively. Let $T_3$ be the Turing machine that upon input $s$ simulates $T_1$ and $T_2$ simultaneously. If and only if either $T_1$ or $T_2$ accepts, $T_3$ accepts. Then $T_3$ accepts the union of $L_1$ and $L_2$.

The answer to "Can we sometimes recognise the intersection between a not-recognisable language and a recognisable one?" is "yes".

For example, let the recognisable one be the empty language.

  • $\begingroup$ Thanks for the answer. I edited my question to make more clear my point. Basically if $L_u$ is "sometimes" decidable, is it always recognisable? $\endgroup$
    – Pietro
    Jun 18 at 17:03
  • $\begingroup$ Yes. In fact, Turing-recognisability are closed under both union and intersection. So is decidability. $\endgroup$
    – John L.
    Jun 18 at 17:30

A few things,

  • It's hard to find what your proof attempt is trying to do. I know you're stuck, but you should at least have a strategy of what you want to do. In your proof, a good idea is to use AFSOC and look for a contradiction.
  • The online theorem you found makes no sense, if we have $L_x = S \cup L_1$ but we don't know anything about $L_1$, I'm not sure why there is a claim about recognizability.

So we want to know if $L_1 \cup L_2$ is decidable, right? Intuitively, this is probably not true, because if $L_1$ is just recognizable, adding a decidable language to it should probably not in general make it decidable all of a sudden (it can though, but remember if we have no assumptions about $L_1, L_2$, then this has to hold in all cases).

So normally when we have an intuition that something doesn't work, we use contradiction or a counterexample.

You can have 2 options here, so you can either

  1. AFSOC $L_u = L_1 \cup L_2$ is decidable, and try to show a contradiction, most likely that $L_1$ is decidable.
  2. Find a counterexample

I think 2 is much easier to find. You can just take any recognizable (but not decidable) language $L_1$, and then take a trivial subset that is decidable, let's say a single element, to make $L_2$. Then $L_2$ is decidable, but $L_1 \cup L_2 = L_1$ which is not decidable. (note that we can take some element, because if $L_1$ is empty, then it cannot be undecidable).

  • $\begingroup$ You're right, I'm gonna edit my question to make my strategy more clear. About my "online theorem", $L_1$ is the language I already defined (recognisable). I'll make this more clear aswell. Also, is my "proof" wrong? Can we prove that $L_u$ is recognisable? $\endgroup$
    – Pietro
    Jun 18 at 16:57
  • $\begingroup$ Proving recognisability is trivial, because we know $L_u = L_1 \cup L_2$, and $L_1, L_2$ are both recognisable. So we can easily build a TM $T_u$ defined as $T_u \, : \, \text{Run } T_1 \text{ and } T_2 \text{ and return either if they accept}$. Notice that this $T_u$ defined is not necessarily decidable, since we don't know if $T_1$ terminates. $\endgroup$
    – mikinty
    Jun 18 at 17:27

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