= Solution
\b[The literal semantic hypothesis is insufficient:] <a semantic Gödel fixed point need not be canonical>. A sentence can be false and provable in a <consistent first-order theory>[consistent], unsound <first-order theory>[arithmetic theory], in which case both sides of the stipulated semantic biconditional can be false.
For a concrete counterexample, assume PA is <consistent first-order theory>[consistent] and <set> $S=\mathrm{PA}+\neg\operatorname{Con}_{\mathrm{PA}}$. <Gödel second incompleteness theorem>[Gödel's second incompleteness theorem] makes this extension <consistent first-order theory>[consistent]: otherwise PA would prove its own <consistent first-order theory>[consistency]. It is still <bounded completeness of an arithmetic theory>[bounded-complete]. Use ordinary numeralwise proof and refutation enumerators for this recursively axiomatized <first-order theory>[theory], and take $G=\neg\operatorname{Con}_{\mathrm{PA}}$. This $G$ is false in the standard <natural numbers> but is an <axiom> of $S$. Since it is provable, $\exists yP(\ulcorner G\urcorner,y)$ is true; its <negation> is false. Thus the semantic <logical fixed point>[fixed-point] biconditional required in this part is true, with both sides false.
Choose $R$ by the usual canonical Rosser construction for these proof matrices. Its undecidability follows from the preceding parts, and it is a true $\Pi_1$ sentence: if it were false, its true $\Sigma_1$ <negation> would be provable in this <bounded completeness of an arithmetic theory>[bounded-complete] <first-order theory>[theory], contradicting undecidability. Hence $G$ is false and $R$ true, so their proposed semantic equivalence is false. This counterexample meets the literal assumptions and does not exploit a missing source page.
Under the intended canonical $\Pi_1$ Gödel/Rosser convention, the expected answer does hold. <consistent first-order theory>[Consistency] plus <bounded completeness of an arithmetic theory>[bounded completeness] implies $\Pi_1$ soundness: a false $\Pi_1$ theorem has a standard bounded counterexample whose <negation> the <first-order theory>[theory] proves. If canonical $G$ were false, its truth <logical fixed point>[fixed point] would make it provable, contradicting this soundness; thus $G$ is true. Canonical $R$ is true by the preceding $\Sigma_1$-negation argument, so \b[both canonical sentences are true and their equivalence is true]. This last conclusion needs that canonical construction or an equivalent additional restriction; truth of a biconditional alone does not supply it.
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