i'm trying to figure out whether this Union $\left [ L_1=\{a^lb^mc^m|l,m\ge 0\} \cup L(b^*c^*)\right]=K$ is regular or not, now since regular languages are closed under intersection, so i assume $K$ is regular then its intersection with $L(a^*b^*)$ should be regular which is $K\cap L(a^*b^*)=L(a^*)\cup L(b^*)$ right ?

1 - Does this implies that the Union is also regular ?

2 - i know that $L=\{a^lb^mc^m|l,m\ge1\}$ is CFL, but is $L_1$ also CFL ?, which is basically $L_1=L\cup \{\epsilon ,a,bc\}$

3 - is $K$ regular or not and how to prove it ?

Any Hints would be appreciated.

  • $\begingroup$ I can't comprehend your question. You are aware that $\cup$ is union, not intersection, right? And that the "left" half of $L_1$ is not regular? $\endgroup$
    – Raphael
    Dec 20 '17 at 11:13
  • $\begingroup$ @Raphael i actually meant the intersection of the Union with $L(a^*b^*)$ $\endgroup$
    – proless8
    Dec 20 '17 at 11:22
  • $\begingroup$ The fact that the intersection of $K$ with a regular language is a regular language, doesn't tell you anything about $K$. $L_1$ is indeed not regular - you can't pump $b^pc^p$, $p$ being the pumping length. $\endgroup$ Dec 20 '17 at 11:32
  • $\begingroup$ @potestasity That word can be pumped perfectly fine. Use $ab^pc^p$. That said, the OP seems to be looking for a proof using closure properties. $\endgroup$
    – Raphael
    Dec 20 '17 at 12:22
  • $\begingroup$ @Raphael I'm not sure I understand. $ab^pc^p$ can be pumped (by choosing $\epsilon$, $a$ and $b^pc^p$ as the three parts), but I fail to see how you can pump $b^pc^p$. $\endgroup$ Dec 20 '17 at 13:08

You are not starting with the right proof idea. It's this:

  1. I want to show $L$ is not regular.
  2. Assume towards a contradiction that $L$ is regular.
  3. Since REG is closed against $f$, I conclude that $L' = f(L)$ is also regular.
  4. But I already know that $L'$ is not regular (from elsewhere). Contradiction; $L$ can't be regular.

See here for complete examples.

  • $\begingroup$ thanks i understand the proof idea, so i will use the complement of $K$ to get i think this language $L=\{a^nb^nc^n\}$, but i can't get the complement of $L_1$ any idea ? $\endgroup$
    – proless8
    Dec 20 '17 at 14:15
  • 1
    $\begingroup$ @proless8 The complement of $L_1$ does not have a lot of structure, so that's unlikely to help. Hint: Try to isolate $b^mc^m$ which you should know to be non-regular (up to homomorphism). $\endgroup$
    – Raphael
    Dec 20 '17 at 15:53

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