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I am currently studying turing computability and related problems such as the halting problem with a background in formal languages.

I know that the class of recursive (decidable) languages is a closure under union, intersection and complement, and that recursively enumerable languages (semi-decidable) are a closure under union and intersection, but what about non-recursive (undecidable) languages?

To be more specific, I am trying to prove that a specific language of the form $$L = \{ w \in \Sigma^* : w \in A \lor w \in B \}$$ is non-recursive. I managed to prove that neither $A$ nor $B$ are recursive by reducing them to the universal halting problem, but I am not sure what that implies for $L$.

The exact language is $$L = \left\{ w\#u \in \left\{ 0, 1, \# \right\}^* | \\ M_w \text{ will halt with input } u \text{ or } M_u \text{ will halt with input } w \right\}$$ where $M_b$ is the turing machine whose transition function $\delta$ is represented using Gödel numbering as a binary string $b$.

Is the class of non-recursive languages a closure under union? If it is, how could I prove it myself or where could I find an existing proof?

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  • $\begingroup$ What are you actually asking? The question in your title is rather different from the two questions in the body. $\endgroup$ – David Richerby Feb 25 '17 at 16:33
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Non-recursive languages are closed under complement, but not under union or intersection.

Indeed, a decider for $A$ exists iff there is a decider for $A^c$. Hence the closure under complement.

However, let $B$ be any non-recursive language. $B^c$ is also non-recursive. But $B \cap B^c = \emptyset$ and $B \cup B^c = \Sigma^*$ are both recursive. Hence, non-recursive languages are not closed under union or intersection.

In the example you mention, you can not conclude anything about $L$ from $A,B$ being non recursive.

I will provide a hint for $L$, only. When you have a language of the form $L=\{f(x,y) | x,y\mbox{ s.t. } p(x,y)\}$, it is sometimes convenient to choose a particular value $a$ and study the related language $L_a = \{ y | y\mbox{ s.t. } p(a,y) \}$ first. Maybe one can choose $a$ to make $p(a,y)$ simple enough so that other proof techniques apply.

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  • $\begingroup$ Well, this is not the anwer I hoped for, but your explanation is surprisingly simple and appears to be correct. Thank you! $\endgroup$ – just.kidding Feb 25 '17 at 12:02
  • $\begingroup$ @just.kidding I don't wish to spoil the exercise, but I added a hint. $\endgroup$ – chi Feb 25 '17 at 12:08
  • $\begingroup$ Thanks! I guess if $L_a$ is undecidable, then $L$ is as well, because $L$ must be decidable for all tuples $(x, y)$ in order to be decidable? $\endgroup$ – just.kidding Feb 25 '17 at 12:15
  • $\begingroup$ @just.kidding Yes. The point is now to find $a$ such that $L_a$ is undecidable, yet it reduces to $L$. $\endgroup$ – chi Feb 25 '17 at 15:24

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