= Solution
Put $T=S\cup\{\neg G_S\}$. The first supplied equivalence, retained in $T$, combines with its new axiom to give
$$
T\vdash\neg\operatorname{Con}(S).
$$
The supplied implication $\operatorname{Con}(T)\to\operatorname{Con}(S)$ then yields, by contraposition,
$$
T\vdash\neg\operatorname{Con}(T).
$$
Finally the second supplied equivalence is $T\vdash G_T\leftrightarrow\operatorname{Con}(T)$, so propositional reasoning gives
$$
\boxed{T\vdash\neg G_T.}
$$
No extra <first-order consistency> assumption is needed for this formal derivation. If $S$ is <first-order consistent>, part (b) shows that adding $\neg G_S$ preserves <first-order consistency>, because inconsistency of the extension would imply $S\vdash G_S$. Such a <first-order consistent> extension therefore proves the <negation> of its own true <Gödel sentence>, illustrating why <first-order consistency> alone does not imply <one-consistency>.
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