Using Pumping Lemma to prove language $L = \{(01)^m 2^m \mid m \ge0\}$ is not regular

I'm trying to use pumping lemma to prove that $L = \{(01)^m 2^m \mid m \ge0\}$ is not regular.

This is what I have so far: Assume $L$ is regular and let $p$ be the pumping length, so $w = (01)^p 2^p$. Consider any pumping decomposition $w = xyz$ such that $|y| >0$ and $|xy| \le p$.

I'm not sure what to do next.

Am I on the right track? Or am I way off?

Hint: You need to consider what all the decompositions $w=xyz$ look like, so what are all the possible things $x$, $y$ and $z$ can be given that $xyz=(01)^p2^p$. Then you pump each one and see whether you get a contradiction, which will be a word not in your language. Each case needs to lead to a contradiction, which would then be a contradiction of the pumping lemma. Voila! The language would not be regular.

Of course, you need to work through the details and consider all the possible splittings.

StackExchange Computer Science Q#1027, answer score: 6