 Solution

The definition of $a$ makes the diagonal instance $E_a[\bar{a}]$ exactly $G_S$. The representing logical formula therefore has the standard interpretationIf $S⊢G_S$, choose an actual witness $m$ to the left-hand existential statement. The bounded arithmetic sentence $A(\bar{a},\bar{m})$ is true and hence provable by ${\Sigma }_0$-completeness. But $G_S$ also gives $S⊢\neg A(\bar{a},\bar{m})$, a contradiction. ThusNow assume one-consistency. It implies first-order consistency, so no formal proof of $G_S$ exists and $A(a,m)$ is false for each standard $m$. Each bounded arithmetic sentence $\neg A(\bar{a},\bar{m})$ is consequently true and provable. A formal proof of $\neg G_S$ would give a formal proof of $\exists v_{2}A(\bar{a},v_2)$ while every standard witness instance is refuted, contrary to one-consistency. HenceThe two hypotheses play different roles: ordinary first-order consistency excludes the positive Gödel sentence, while the stronger hypothesis also excludes its negation.