= Solution
Use the two formal <consistent first-order theory>[consistency] equivalences expressly supplied for this part. If $S\vdash G\leftrightarrow R$, the <derivability conditions> give $S\vdash\Box\neg G\leftrightarrow\Box\neg R$. Taking <negations> and applying those supplied equivalences yields
$$
S\vdash\operatorname{Con}_{S+\operatorname{Con}_S}\leftrightarrow\operatorname{Con}_S.
$$
<One-consistency> implies $\Sigma_1$ soundness under <bounded completeness of an arithmetic theory>[bounded completeness]. Hence $S+\operatorname{Con}_S$ is <consistent first-order theory>[consistent]: if not, $S$ would prove $\neg\operatorname{Con}_S$, a false $\Sigma_1$ sentence. But in that <consistent first-order theory>[consistent] extension the last equivalence and its <consistent first-order theory>[consistency] <axiom> prove its own <consistent first-order theory>[consistency] $\operatorname{Con}_{S+\operatorname{Con}_S}$. <Gödel second incompleteness theorem>[Gödel's second incompleteness theorem] rules this out. Therefore
$$
\boxed{S\nvdash G\leftrightarrow R.}
$$
This establishes the requested unprovability using the explicit formal hypotheses, without inferring internal provability from the semantic equivalence discussed in the preceding part. The hypotheses for the second theorem are the standard effective arithmetic presentation and adequate <arithmetical provability predicate>[provability predicates] stipulated here.
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