I'm confused about a solution I saw about the following language not being regular:\begin{equation*}  L=\{0^n ~1 ~2^n : n >0\}\end{equation*} The example solution said that $L$ was the same as:\begin{equation*} L'=\{0^n1^n : n > 0 \}\end{equation*} so it wasn't regular. I understand why $L'$ isn't regular, but I don't understand how $L$ is the same as $L'$. How can you eliminate that middle symbol?
Removing constant symbols from language to prove irregularity
regular languages
The language $L'$ is the same as $L$ (in fact, it's an asymmetric relation) in two steps. If $L$ were regular, then so would be $L'' = \{0^n 2^n : n > 0 \}$ (take a DFA for $L$ and replace $1$ edges with $\epsilon$ edges). In its turn, $L''$ is the same as $L'$, up to change of symbols.