Using closure properties to show that $a^n b^m a^{n+m}$ is not regular

Question

I want to show that $L = \{a^n b^m a^{n+m} \mid n, m \geq 0\}$ is not regular.

Can I say that the complement of $L$ intersected with $a^*b^*$ equals $\{a^n b^n \mid n \geq 0\}$ and since I know that $\{a^n b^n \mid n \geq 0\}$ is not regular, then $L$ is not regular?

Or would I have to use the pumping lemma?

You can do whatever you want as long as it's logically valid. — Yuval Filmus, Oct 18 '15 at 21:51
@YuvalFilmus Does that make sense tho? Like I guess is that true? — user270494, Oct 18 '15 at 23:10
That's for you to answer. Try to be more confident in your knowledge. — Yuval Filmus, Oct 19 '15 at 07:05

score 2 · Accepted Answer · edited Apr 13 '17 at 12:48

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That line of reasoning can work in principle; you are trying to use closure properties of REG for a proof of contradiction. Make sure you understand that that's what you're doing.

Unfortunately, you do not get $\{a^nb^n\}$. For instance, $a^3 b^4 \in \overline{L} \cap a^*b^*$. Check out our reference question for more techniques you can try.

edited Apr 13 '17 at 12:48

Community

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answered Oct 19 '15 at 06:22

Raphael

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Actually can I just use the pumping lemma and state $a^0b^pa^p$ is in the language, then y = $a^k$ with 0 < k <= p, choose i = 0 so I get $a^{−k}b^pa^p$ which is not in the language? – user270494 Oct 22 '15 at 05:46
@user270494 No, it wouldn't work like that. I recommend you visit our reference questions. – Raphael Oct 22 '15 at 07:35

score 2 · Answer 2 · answered Oct 19 '15 at 14:25

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Hint: What do you get when you intersect $L$ with the language denoted by $b^*a^*$? It won't be quite what you need, but it's close. You can then get the $b^ma^m$ language by subtracting (i.e., set difference) some regular language from that.

answered Oct 19 '15 at 14:25

Rick Decker

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Using closure properties to show that $a^n b^m a^{n+m}$ is not regular

2 Answers2