# Show that the Pumping Lemma for CFLs is not powerful enough to prove that the language L = {aibjck |i ≠j ≠ k ≠ i } is not context free

Show that the Pumping Lemma for CFLs is not powerful enough to prove that the language $L = \{a^ib^jc^k \mid i ≠j ≠ k ≠ i \}$ is not context free.

From my understanding, we want to prove that all 3 conditions of the pumping lemma hold. This will not guarantee that the language is context-free. However, I have found a case where it breaks

Let the string be $abbccc$

If $u = \lambda$, $v = a$, $x = aa$, $y = a$, $z = bbccc$.

Therefore, when we pump the string for $i = 2$, we get $aaaaaabbcccccc$, which is NOT in the language because we have the same number of $a$'s and $c$'s. How is this possible?

• Have you tried w = bcc. The professor hinted to find a w such that it is not pumpable. Mar 22 '18 at 0:23
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– Raphael
Mar 22 '18 at 9:29

The Pumping Lemma states: "... there exists a constant $p$ such that for each word $w\in L$ with $|w|\ge p$ there exists a decomposition $w = uvxyz$ such that ...", where $p$ is the pumping constant.