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I am trying to proof the properties of the complement and concatenation of two non-context-free languages $L_1$ and $L_2$.
I believe that both of these languages are closed under complement and concatenation but can't seem to find a solid proof for it. I'm leaning towards them being closed because I can't find any counter examples.
I already proved that non-context-free languages are not closed under union and intersection, so I can use those properties (if applicable).

So in short what I want to proof, given $L_1, L_2$ are non-context-free languages:
$L_1L_2 \in S_{nonContextFree}$
$\overline{L_1} \in S_{nonContextFree}$

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    $\begingroup$ Languages cannot be closed/not closed under complementation, what you're referring to are classes of languages. As for your question, a small hint: if you find it hard to prove the claim, perhaps try to find a counterexample :) $\endgroup$ – Shaull Jan 5 at 10:21
  • $\begingroup$ @Shaull whoops, something go lost in translation there! Thanks for the counterexample tip, but as mentioned in my question, I couldn't find any. Therefor my suspicion that the. classes of non-context-free languages are closed under these operations. $\endgroup$ – Bram Vanbilsen Jan 5 at 10:29
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As for closure under complementation -- consider the following hint: context-free languages are not closed under complementation.

Regarding concatenation, this is slightly more tricky, but non-context free languages are not closed under complementation.

There are many ways of finding counterexamples for the latter, one is the following "trick":

Let $L$ be some non-context free language over some alphabet $\Sigma$, that does not contain $0,1$. Now define two languages: $A=(1\cdot \Sigma^*)\cup (0\cdot L)\cup \{\epsilon\}$ and $B=(0\cdot \Sigma^*)\cup (1\cdot L)\cup \{\epsilon\}$

It is not hard to prove that neither $A$ nor $B$ are context free, basically due to presence of $L$.

However, we have that $A\cdot B=\{0,1\}\cdot \Sigma^*\cup \{\epsilon\}$, which is regular (and in particular context-free).

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  • $\begingroup$ For complementation, does that automatically mean that it is also not closed under non-context free languages? I do not directly see how you would derive one from the other. as they are completely disjunct. $\endgroup$ – Bram Vanbilsen Jan 5 at 11:51
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    $\begingroup$ Consider a context-free language $L$, whose complement is not context free. What is the complement of the complement? $\endgroup$ – Shaull Jan 5 at 12:21
  • $\begingroup$ Wow, makes a whole lot of sense... Thanks for the help! $\endgroup$ – Bram Vanbilsen Jan 5 at 12:31

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