In the last exercise of the last chapter Computability and Logic of Boolos et al we are asked to use $GL$'s arithmetical soundness theorem to find a weaker assumption than the $\omega$-consistency of $PA$ to show that the Gödel Sentence of PA, $G$, is not disprovable in $PA$.
Since $PA\vdash G\leftrightarrow \neg Prv(\ulcorner G\urcorner)$, we have $PA\vdash \neg G \leftrightarrow Prv(\ulcorner G\urcorner)$, whose corresponding modal sentence is $q\leftrightarrow\square\neg q$. I worked out the fixed point of this, and it results in $q\leftrightarrow \square \bot$, so by Arithmetical Soundness we have that $PA\vdash \neg G \leftrightarrow Prv(\ulcorner 0=1\urcorner)$.
So the disprovability of $G$ is equivalent to the affirmation of the existence a proof of inconsistency,... that is, is equivalent to the $\omega$-inconsistency of the existential predicate $Prv(\ulcorner 0=1\urcorner)$, given the consistency of $PA$.
Is there a way I am missing to strengthen this somewhat dissapointing result?