I've been tasked with showing that given a regular language $L$ over $\Sigma={0}$, prove that the language $Minus(L)= \{ 0^x1^y | 0^{x-y}\in L \}$ is not regular.
I've tried to use $L=0^*$, which means that $\forall{x,y}\in{N}(0^{x-y}\in L)$, but I can't quite show that it's not formal.
I'd appreciate any insight, thank you!
I'm assuming that $\textrm{Minus}(L)$ is defined as $\{0^x 1^y \mid x\ge y, \,0^{x-y} \in L\}.$
The (non) regularity of $\textrm{Minus}(L)$ depends on the choice of $L$.
An example of language $L$ such that $\textrm{Minus}(L)$ is not regular is $L=\{0\}^*$. To see this suppose towards a contradiction that $\textrm{Minus}(L)$ is regular and let $p$ be its pumping length. Since $0^p1^p \in \textrm{Minus}(L)$, by the pumping lemma there must be some $1 \le i \le p$ such that $0^{p-i} 1^p \in \textrm{Minus}(L)$, however no such $i$ exists.
An example of a language $L$ such that $\textrm{Minus}(L)$ is regular is $\emptyset$. Indeed, $\textrm{Minus}(\emptyset) = \emptyset$.