Does consistency of Peano arithmetic follow from arithmetical completeness of modal logic GL?

Question

Solovay's Arithmetical Completeness theorem states that if $A$ is sentence of modal logic, then if every realization $A^*$ of $A$ is proved by $\mathsf{PA}$, then $\mathsf{GL}\vdash A$. By contraposition, we know that if $A$ isn't a theorem of $\mathsf{GL}$, then $A$ isn't a theorem of Peano arithmetic. So since there are sentences that aren't valid in $\mathsf{GL}$, by semantic completeness of $\mathsf{GL}$ it follows that they aren't theorems of $\mathsf{GL}$, so by arithmetical completeness it follows that they aren't provable in Peano arithmetic. Since it's not the case that Peano arithmetic can prove anything, it's not inconsistent. Is it wrong to jump to such a conclusion, and if so, why?

Answer 1

It appears that my question is a duplicate of this question. The answer is negative, because Solovay's proof of arithmetical completeness presupposes the consistency of Peano arithmetic.