Epsilon-productions can be eliminated except possibly at the start symbol

Let $N$ be the nullable variables of $G$. Form a grammar $G_{\epsilon }$ by deleting every production $A\to \epsilon$ and, for each remaining production $A\to x_1\cdots x_k$, adding every nonempty right-hand side obtained by deleting any chosen subset of nullable occurrences among the $x_i$.

If $\epsilon \mathrm{\notin }L(G)$, clause 1 already proves the theorem. Assume instead that $\epsilon \in L(G)$. Introduce a fresh start variable $S_0$ and add the two productions $S_0\to S$ and $S_0\to \epsilon$, leaving all productions of $G_{\epsilon }$ unchanged.

Any derivation of a nonempty terminal word in $G$ can be shortened to one in $G_{\epsilon }$ by collapsing each maximal family of nullable subderivations into a single use of one of the added productions. Conversely, every added production in $G_{\epsilon }$ can be expanded back in $G$ by re-inserting derivations of the deleted nullable variables to $\epsilon$. Therefore [L2] gives $L(G_{\epsilon })=L(G)\setminus \{\epsilon \}$.

The rule $S_0\to \epsilon$ restores the missing empty word, while step 2.1 already preserved every nonempty terminal word. Because $S_0$ is fresh, it appears on no right-hand side, so this is the only $\epsilon$-production of the new grammar $G^{\mathrm{\prime }}$. Hence $L(G^{\mathrm{\prime }})=L(G)$.