I'm trying to proof that the following language is not regular using pumping lemma. $L=\{w\bar{w}|w\in \{0,1\}^* and\ \bar{w}\ is\ one's\ complement\ of\ w\}$

I started by stating that:

$|w\bar{w}| = 2p = |xyz|$

Because of Pumping Lemma the following has to be true:

$1 \leq |y| \leq |xy| \leq p$

Because $|xy| \leq p$ and $|w|=|\bar{w}|$ have to be true, $\bar{w}$ has to be completly in $z$. I now tried somehow to manipulate the first half, so that it always evaluates to a contradiction, but I am stuck.

  • 1
    $\begingroup$ You need to add “for all” or “exists” here and there, or you can’t get anywhere. $\endgroup$
    – gnasher729
    Commented Feb 3, 2020 at 21:00

1 Answer 1


Just show that after processing $0^n$ and $0^{n’}$ with n != n’ you must have reached two different states. (Which is easy: In one state, processing $1^n$ leads to an accepting state, in the other state it doesn’t). Therefore there is no finite set of states.

Or you just take the pumping lemma, p arbitrary large, and w = $0^p 1^p$. Which makes y = $0^k$ for some k >= 1.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.