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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.

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

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