Solution
= Solution
Let $\Sigma=\{a,b\}$, $V=\{S\}$, and
$$
P=\{S\to aSa, S\to b\}.
$$
Then $\mathcal L(G)=\{a^nba^n:n\geq0\}$. Adding $S\to\varepsilon$ also produces $a^{2n}$; for example $aa$ is generated but lies in neither $\mathcal L(G)$ nor $\{\varepsilon\}$. Thus the proposed equality can fail because the new erasing rule may be used on a noninitial occurrence of $S$.