Can someone help me prove that $L = \{ a^ib^jc^k | i < j \ and \ i+2j+3 < k \}$ is not a context free language?

I've tried applying the pumping lemma for CFGs and proving case by case (taking x=uvw, at first v is int $a^+$, then v is in $a^+b^+$ etc.but I'm failing to find a contradiction). Any help is welcomed.\

PS: The suggestion does not solve my problem. I know the technique of applying the pumping lemma for CFG to prome a language is not context free, but I don't know how to use it on this particular language.


Firstly I want to point out that what you want to prove that $L$ is a context free language or CFL not grammar. A CFL can be defined by a CFG, but they are very different things. Anyway, back to your question.

I'm going to give you somewhat informal reasoning, but I think you will be able to formalize it if necessary.

Let us assume that $L$ is in fact a CFL. Then the pumping lemma for CFL's holds for $L$. Let n be the number of the pumping lemma for L.

Consider the word $w = a^nb^{n+1}c^{3n+4} \in L$

From the pumping lemma we have that $w = uxyzv, |xyz| \leq n, |xz| \geq 1, (\forall i \geq 0)[ux^iyz^iv \in L$]

We have two inequalities we can try to break:

  1. $i < j$
  2. $i + 2j + 3 < k$

Assume $xz$ does not contain $c$. This means that $xz$ contains only $a$'s or $b$'s (or both). Since $xz \geq 1$ This means we can pump up the word as much as necessary to break inequality $(2)$ -The right side of the inequality will remain the same when we pump up, but the left side will increase by at least $1$ on each step.

So we have deduced that $xz$ must contain a $c$.

Assume that $xz$ contains only $c$'s. Since $xz \geq 1$ we can pump down to remove at least one $c$. This will again break inequality $(2)$ since the right side will become less than or equal to the left.

So we have deduced that $xz$ must contain both a $b$ and a $c$. But since $xz$ contains a $b$, this means that we can pump down and remove at least one $b$, thus breaking inequality $(1)$.

Thus we have shown that no matter how we break up $w$ we can always pump it out of the language. This is contradiction with out assumption that $L$ is a $CFL$ so it must mean that $L$ is not a $CFL$.

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