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Let $L = \{ab^ncd \mid n \geq 0\}$. If we take $p = 5$ and $w = abbcd$ and write $w_i = xy^iz$, where $x = abb$, $y=c$, $z=d$, then $w_2 = abbccd$ which is not in $L$. We conclude that $L$ is not regular but it obviously is!

What am I doing wrong?

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The pumping lemma says that there exists a decomposition of $w$ to $xyz$ such that $y$ can be pumped.

It doesn't say that for all decompositions of $w$ to $xyz$, $y$ can be pumped.

In your example, a possible decomposition is:

$x=a$

$y=bb$

$z=cd$

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  • $\begingroup$ Pumping lemma isn't an iff. It holds for some non regular languages, but always holds for a regular language. $\endgroup$ – jmite Feb 1 '15 at 13:18
  • $\begingroup$ @AswinAlagappan in your example, pumping $y$ even once will give: $aababb$, which is not in $L$. $\endgroup$ – Erel Segal-Halevi Feb 1 '15 at 13:48

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