 Solution

An arithmetic theory is one-consistent if there is no bounded arithmetic formula $\delta (x)$ for which it proves $\exists x\delta (x)$ and also proves $\neg \delta (\overline{n})$ for every standard natural number $n$. A finite tuple of existential witnesses can be coded by one natural number. The associated Sigma-1 soundness property says that every ${\Sigma }_1$ sentence it proves is true in the standard natural numbers.

Suppose first that the theory is one-consistent, but proves a false ${\Sigma }_1$ sentence $\exists x\delta (x)$. For every standard $n$, $\neg \delta (\overline{n})$ is then a true bounded sentence. Bounded completeness makes each such sentence a theorem, violating one-consistency.

Conversely suppose the theory is ${\Sigma }_1$-sound. If it proved both the existential sentence and every negated standard instance, soundness would supply an actual witness $n$. Its positive instance $\delta (\overline{n})$ is a true bounded sentence and hence provable. Together with the assumed negative instance this would make the theory inconsistent; an inconsistent theory proves a false ${\Sigma }_1$ sentence, contrary to soundness. Thusunder the stated bounded-completeness assumption. In particular either condition implies that the theory is a consistent first-order theory.