Prove or disprove the following statement: for arbitrary regular expressions $r_1$ and $r_2$ over an alphabet $\Sigma$ such that $\epsilon$ belongs to $L(r_1)$, there exists a regular expression $r$ over $\Sigma$ such that $r = r_1r + r_2$.
My Solution: Let us consider $\Sigma =\{a\}$. $L_1$ is even number of symbols so $r_1 =(aa)^*$ and $L_2$ is odd number of symbols so $r_2 = a(aa)^*$. $r$ is either $r_2$ which is possible or $rr_1$. Now if we keep replacing $r$ with $rr_1$ in the expression we will end up getting an infinitely large string because $r$ cannot be $\epsilon$ (thats what I think). So in my opinion this language is not regular. But I am not sure if I am thinking correctly.