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I have this language: $L = \{a^{n+2} b^m a^{2n} b^{3n}\mid n,m >=0 \}$ and I am trying to prove that it is not CFL. I assumed that my word is $a^{p+2} b^m a^{2p} b^{3p}$ (where $p$ is the pumpung length) and I applied the pumping lemma. I can find a contradiction for every case except for the case when the part that I am pumping is only $b$'s (from $b^m$). What would be the best choice of the word for pumping lemma?

I've already read some questions here that are similar to this one, but this kind of combination is diferent from those. First, third and fourth terms are in a relation here despide the examples where are other kind of relations.

Razvan
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  • Let me direct you towards our reference questions which cover your problem in detail. I suggest that you work through the related questions listed there and continue trying to solve your problem. We're happy to help you understand the concepts but just solving exercises for you is unlikely to achieve that. – D.W. Jun 12 '15 at 19:18
  • I already read that and much more, but I just can't figure out what word should I take to prove this. – Razvan Jun 12 '15 at 19:53
  • Keep trying. Keep trying other words -- you tried one word and found it doesn't lead to a contradiction, so maybe you need to try a different word. Alternatively, your next step is to do some more studying and practice on the pumping lemma. You might try working some easier exercises. But this isn't the place to outsource "solve my exercise for me". – D.W. Jun 12 '15 at 20:07
  • I asked for a hint, not to solve it. Ive already tried more than 10 words and noting worked for me. Also I tried some closure properties and there is no point in keep trying without a result. – Razvan Jun 12 '15 at 20:51
  • Hint: $b^m$ give you trouble, so check if you really, really have to have it. – Raphael Jun 12 '15 at 21:43
  • No, I don't need it and I tried to remove it, but at some intersections that I tried I removed also $b^{3n}$ and I need this. – Razvan Jun 12 '15 at 22:06

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