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I want to prove that $ L = {a^n b^m c^{ \lfloor \frac{n}{m} \rfloor } } $

isn't context free language, so I choose N - constant from lemma

so the word is $ w = a^N b^N c $ and $ w = uvxyz $

1 case

v and y contains only a $ v,y \in a^{*} $

and $ w_2 = a^{N+B} b^{N} c $ where $ B > 0 $ is it correct to say that it doesn't belong to language, because $ N + B \neq N $ ?

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or second version

$ v,y \in a^{*} $ so $w_0 = uv^0xy^0z = uxz $ and $ w_0 = a^{N-B}b^Nc $ which doesn't belong to language because $ \lfloor \frac{N-B}{N} \rfloor = 0 \neq 1 $

and what with case where $ v \in a^+ $ and $ y \in b^+ $

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    $\begingroup$ I trust you realize that you left off quite a few other cases. For instance, it might be the case that $vxy = a^tb^s$. Also, in the case you noted, it might be the case that $\lfloor (N+B)/N\rfloor$ might still be 1 (though that can be fixed by more pumping). $\endgroup$ – Rick Decker Jun 12 '14 at 17:27
  • $\begingroup$ Yes, I know that there are more cases, but I was not sure if this one is correct, but if this one is incorrect, how to prove that ? I added second version $\endgroup$ – user19334 Jun 12 '14 at 17:35
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    $\begingroup$ Strings of that form are a context free language, so you won't be able to get a contradiction. $\endgroup$ – vonbrand Jun 12 '14 at 20:03
  • $\begingroup$ maybe on $ w = a^{N}b^{N}c $ pumping lemma not work , but language L isn't context free. I proved it to word $ w = a^{N^2} b^N c^N $ $\endgroup$ – user19334 Jun 12 '14 at 20:32