While learning about the pumping lemma, I came across the following question:

Given the language L is $ a^n(0|1)^* $ with $ c_0 \cdot c_1 = n $, where $ c_0 $ indicates the amount of zeros present, use the pumping lemma to prove that this language is not regular. Examples of valid words in L are 0, 1, a01, aa001, etc...

In regular english: a word with leading as, followed by any amount of either 0 or 1 characters, where the amount of zeros multiplied by the amount of ones needs to match the amount of as.

My initial attempt was the following:

  • Pick w as $ a^p0^p1 $. This obviously holds since $ p \cdot 1 = p $.
  • Split w into xyz as $ x = \epsilon $, $ y = a^p $, $ z = 0^p1 $.
  • Show that $ xy^2z = a^{2p}0^p1 $ does not hold, since $ p \cdot 1 \ne 2p $.

However, the answer key instead opts to introduce two new variables $ m \geq 0 $, $ n \gt 0 $, then splits on $ x = a^m $, $ y = a^n $, $ z = a^{p-n-m}0^p1 $ (splitting the sequence of a into three parts). Then, they too use $ i = 2 $ to show that $ xy^2z = a^{m}a^{2n}a^{p-n-m}0^p1 = a^{p+n}0^p1 $, which is not a member of the language (since n was more than 0).

As far as I can see, my attempt adheres to the $ |xy| \leq p $ condition of the pumping lemma: x is empty and $ |y| = p $. As such, I was under the impression that my answer was correct.

However, the huge increase in complexity of the answer in the answer key leads me to believe that this is not a valid approach. I have the sneaking suspicion this is because my answer actually does violate the $ |xy| \leq p $ condition.

Is my attempt a correct way to prove that the language is not regular? If not, what misconception/mistake did I make along the way?

  • $\begingroup$ I'm not sure what you mean by your title. The condition $|xy|\leq p$ doesn't depend on $i$: it's a fact about $x$, $y$ and $p$ only. $\endgroup$ May 20, 2019 at 16:06

1 Answer 1


Your mistake is in the following step:

  • Split $w$ into $xyz$ as $x = \epsilon$, $y = a^p$, $z = 0^p 1$.

You're not allowed to choose how $w$ splits into $xyz$. You should show that for any way of splitting $w$ into $xyz$ such that $|xy| \leq p$ and $y \neq \epsilon$, there exists $i$ such that $xy^iz \notin L$.

Here is a similar "proof" that the language $01^*$ isn't regular:

  • Pick $w$ as $01^p$.
  • Split $w$ into $xyz$ as $x = \epsilon$, $y = 0$, $z = 1^p$.
  • Show that $xy^2z = 0^21^p$ is not in the language.

The proof of the pumping lemma will choose a different splitting, for example $x = 0$, $y = 1$, $z = 1^{p-1}$.

  • $\begingroup$ Oh duh! Can't believe I forgot what is essentially the essence of the entire pumping lemma (showing that there's no way at all to satisfy it). Much love. $\endgroup$
    – Peter
    May 20, 2019 at 15:36

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