Non context free language that is pumpable?

Question

For my homework assignment, I have to come up with a non-cfl that is pumpable. I came up with the following: $$ C = \{a^n b^n c^n d^m \mid n \ge 1 \text{ and } m \ge 1 \} $$

I'm not sure whether this works. For the pumping lemma, let $p$ being the pumping length. If I generate a string with $p$ $a$'s, $b$'s, and $c$'s, and only one $d$, my only choice for $vxy$ would be $d$. Pumping this down to $v^0xy^0$ gives $a^p b^p c^p$, thus escaping the language. However, if I let $m$ be greater than or equal to zero, if I choose a string with no $d$'s, then I am forced to put either $a$'s, $b$'s or $c$'s in my $vxy$ string.

See Example of a non-context free language that nonetheless CAN be pumped? for some examples. — Gilles 'SO- stop being evil', Oct 12 '13 at 22:36
I don't get the examples there; they seem a little too complex lol. this is a homework problem, and i have to proof that it's not a cfl by arguing my way through, so i'd like to keep the language simple. — Yuen Hsi, Oct 12 '13 at 23:50
@Raphael If that is the spirit of this site, then you should close the question. — Hendrik Jan, Oct 13 '13 at 20:20
@Raphael So did you get in University of Kaiserslautern and become a PhD student without ever asking for help for any homework assignments? — Yuen Hsi, Oct 13 '13 at 20:28
@HendrikJan No, I can't, all that is covered by policy is upholding the normal standards. Insofar, my above comment is more advice than moderator's note. — Raphael, Oct 14 '13 at 07:43
@YuenHsi I don't see how that is relevant here. Note, though, that "asking for help" and "asking for a solution" are different things; your question is somewhere in the middle. Homework is only there to help you understand things (it produces only work for the teachers!) and this is faciliated by doing the work, not knowing the answer. Many students misunderstand and fail, angry, "but I looked at all the answers!". So please, no offense intended but only advice. — Raphael, Oct 14 '13 at 07:46

score 0 · Answer 1 · answered Oct 13 '13 at 11:23

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As you say "thus escaping the language" which means your example language is not pumpable: for $m=1$ we cannot pump $d$ and stay in the language. Instead consider $D= C \cup \{a,b,c\}^*$. Now we can pump $d$ whenever it is present, and we can pump any string that does not contain $d$.

answered Oct 13 '13 at 11:23

Hendrik Jan

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Hey, thanks for your response; I thought of doing that but I didn't know how to proof that D is not context free. Even though C isn't a CFL, I thought CFL languages are not closed under union? {a, b, c}* is regular, isn't it? – Yuen Hsi Oct 13 '13 at 20:29

Non context free language that is pumpable?

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