 Solution

Write $BA=\exists y\mathrm{Prov}(┌A┐,y)$ and fix a refutable first-order sentence $\perp$, so $S⊢\neg \perp$ and ${\mathrm{Con}}_S=\neg B\perp$. First prove the internal equivalenceBecause $\perp \to G$ is provable, the derivability conditions give $B\perp \to BG$. The fixed-point equivalence $G\leftrightarrow \neg BG$ therefore yields $G\to \neg B\perp$.

For the reverse implication, necessitating $G\to \neg BG$ and applying distribution gives $BG\to B(\neg BG)$. positive introspection gives $BG\to B(BG)$. Finally the provable implication $BG\to (\neg BG\to \perp )$, necessitated and distributed, givesConsequently $S⊢BG\to B\perp$, so $\neg B\perp \to \neg BG\to G$. The same argument works with any chosen refutable first-order sentence in place of $\perp$; their formal consistency statements are provably equivalent.

By the classical deduction theorem for propositional logic applied to adjoining a first-order sentence, $S+\{G\}$ is consistent exactly when $S⊬\neg G$. The displayed equivalence makes $S+\{G\}$ and $S+\{{\mathrm{Con}}_S\}$ deductively equivalent. ThereforeNo soundness assumption on that extension is being substituted for consistency.