Prove that L = { wu; wu in {a,b}* and |w| = |u| } is not regular

Question

Prove that $L = \{wu, wu \in \{a,b\}^* \ \land |w| = |u| \}$ is not regular.

I'm trying to approach this problem with the pumping lemma.

So I know I have to pick a string s, which can be split into xyz components in terms of the pumping length, p.

But I cannot seem to find an s.

For example using $ s = a^p b^p$ or $s = a^p a b^p b $, since $|xy| \le p \land |y| > 0$, then y must contain some a's. So we can try to pump $x y^i z$ to see the new string is still in L.

The issue is that if y contains an even number of a's it can always be pumped because the resulting string will have even length so w can be the first half and u be the second half.

Answer 1

As Rick Decker said $L$ is a regular language. For example: $$L= \mathscr{L}(\,(aa | ab | ba | bb)^*\,)$$

If your teacher asked you to prove that $L$ is not regular in an exam then he is a big troll. Otherwise remember that the pumping lemma is only usefull to prove that a language is not regular and if you fail to find the contradiction in the lemma then that doesn't prove anything. $L$ can be regular or irregular at that point.

In this answer I covered some common mistakes when students use the pumping lemma: Common Mistakes