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Two formal grammars over the same terminal alphabet are isomorphic when there is a bijection $\varphi :V_0\to V_1$ between their nonterminals such that $\varphi (S_0)=S_1$ andwhere $\varphi$ is extended to words by fixing each terminal and acting symbol by symbol.

The first grammar generatesbecause $A\to bAb$ adds two $b$'s and $A\to b$ terminates the derivation.

In the second grammar, $Z$ is a nonproductive nonterminal: every right-hand side of a $Z$-production still contains a $Z$. Consequently the branch $Y\to ZZ$ never yields a terminal word. Every terminal derivation from $Y$ instead uses $Y\to bYb$ repeatedly and ends with $Y\to b$, soThe grammars are therefore equivalent, although they are not isomorphic.

The grammars are isomorphic. The bijectionmaps every production of $P_0$ to the corresponding production of $P_1$. They are therefore equivalent. Notice that $B$ in the first grammar and $C$ in the second are unreachable; the isomorphism also preserves this irrelevant component.