An ambiguous expression grammar with an unambiguous grammar for the same language

The first grammar is ambiguous because the word $a+a\ast a$ has both the parse $a+(a\ast a)$ and the parse $(a+a)\ast a$.

The second grammar generates the same terminal strings as the first. One inclusion follows by simultaneous induction on its $E$-, $T$-, and $F$-derivations, since every displayed production can be simulated by the first grammar. For the reverse inclusion, scan a generated word at parenthesis depth zero: split at its rightmost top-level + if one exists, otherwise at its rightmost top-level * if one exists, and otherwise recognize the forced factor $a$, $b$, or $(E)$. Induction on word length gives a derivation in the second grammar. The same decomposition is unique: the $E$-rule forces the rightmost top-level +, the $T$-rule forces the rightmost top-level *, and the outer symbols force the $F$-rule. Hence the second grammar is unambiguous and, in particular, $a+a\ast a$ has only the parse $a+(a\ast a)$.

By [L1], a language with an unambiguous grammar is not inherently ambiguous. Therefore the first grammar is a counterexample to [A1].