Substituting a variable's productions into a leading variable occurrence preserves the language

Let $xAy{\Rightarrow }_{G}xuBvy$ be one use of the deleted production in a derivation of a terminal word, and follow the distinguished occurrence of $B$ created by this step. It is eventually rewritten by some $B\to w_i$, since the displayed rules are all the productions of $B$. Every intervening step rewrites a different variable occurrence. Such a step commutes with the rewrite of the distinguished $B$: applying the two context-free productions in either order gives the same sentential form. We may therefore move $B\to w_i$ leftward past all intervening steps and replace the resulting adjacent fragment $xAy{\Rightarrow }_{G}xuBvy{\Rightarrow }_{G}xu{w}_{i}vy$ by the single $H$-step $xAy{\Rightarrow }_{H}xu{w}_{i}vy$. Restoring the commuted surrounding-context steps gives the same final terminal word.

Conversely, every use in $H$ of a substituted production $A\to u{w}_{i}v$ can be expanded in $G$ to the two steps $xAy{\Rightarrow }_{G}xuBvy{\Rightarrow }_{G}xu{w}_{i}vy\mathrm{.}$ Replacing each substituted step in a derivation of a terminal word in $H$ by this two-step fragment yields a derivation in $G$ with the same yield. Hence [L2] gives $L(H)\subseteq L(G)$.

If a terminal derivation in $G$ never uses the deleted production $A\to uBv$, then it is also a derivation in $H$. Otherwise apply step 1.1 to one use of $A\to uBv$. This decreases by one the number of uses of the deleted production in the finite derivation without changing its terminal yield. Repeating finitely many times produces an $H$-derivation of the same word, so [L2] gives $L(G)\subseteq L(H)$.

The two inclusions from steps 2.1 and 1.2 prove $L(H)=L(G)$.