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I would like to find out if this language is context free or not:

$\qquad L=\{a^{i}b^{j}c^{k} \mid i<j,i+2j+3<k\}$.

I've tried to apply the pumping lemma taking out $w=a^n b^{n+1}c^{3n+6}$ which is in the language, where $n$ is the natural number given by the lemma. I can write $w=uvwxy$ such that $|vx|≥1$ and $|vwx|≤n$. But I didn't get to any contradiction yet.

Any ideas?

Raphael
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KrasivaM
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    If you don't get any contradictions it means nothing. $L$ may or may not be CFL at this point. So you can choose another word $w$ to try to find a contradiction or suppose that $L$ is CFL and try to write the corresponding context-free grammar or draw the push-down automaton that denotes $L$. Here is explained in detail how to use pumping lemma correctly. (mod-edited to fit in a comment) – Renato Sanhueza Jun 16 '15 at 17:10
  • Thank you! Yes, I know that not getting a contradiction doesn't mean anything, but I don't know which word to choose in order to get one. Or should I intersect this language with a regular one first? – KrasivaM Jun 16 '15 at 19:37
  • This may not be possible with the Pumping lemma; you can always pump $c$s. Try Ogden's lemma as explained here. Hint: enforce that $a$ and/or $b$ has to be pumped, but $c$ can not. – Raphael Jun 17 '15 at 12:46
  • Raphael, in the case of the $c$'s, pumping down should work just fine. – Ran G. Jun 17 '15 at 23:12

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