I looked through many sources which give this as an example for cfl. It also makes sense according to this: Properties of CFL

But it fails the pumping lemma test. Let's take n=5. According to the Pumping Lemma, we can decompose s into uvxyz like: u=0, v=0, x=0,y=01, z=111

If we pump it, it clearly fails. So, is 0^n1^n context free or not?

  • $\begingroup$ You don't get to choose the pumping length and you don't get to choose the decomposition. $\endgroup$
    – Steven
    Nov 13, 2023 at 10:12

1 Answer 1


You have just shown that the given language does not meet the condition of the pumping lemma for $n=5$ and one particular decomposition.

To conclude that $L$ is not context free you need to show that the language does not meet the condition of the pumping lemma for some word of sufficiently large length $n$ and all decompositions of such word (within the constraints of the pumping lemma).

It turns out that the words of $L = \{0^n 1^n \mid n \ge 0\}$ admit a valid decomposition as soon as $n \ge 1$. Here is a decomposition of $0^5 1^5$ into $uvxyz$ that can be pumped: $u=0^4$, $v=0$, $x=\varepsilon$, $y=1$, $z=1^4$.

This is not surprising, since $L$ is indeed context-free as the argument in your question shows.

  • $\begingroup$ Thank you for the response. I was unaware of the fact that the language had to be proven against every decomposition for a length n. I was taught something different. Your answer has really cleared up things for me. $\endgroup$ Nov 13, 2023 at 16:20

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