One-consistency ($1\text{-consistency}$)

An arithmetic theory is one-consistent if it does not prove $\exists x\delta (x)$ while proving every $\neg \delta (\overline{n})$, for a bounded arithmetic formula $\delta$ and all standard natural numbers $n$. If the theory proves all true bounded sentences, this is equivalent to Sigma-1 soundness: a false existential theorem has every standard instance refuted by bounded completeness; conversely a true witness and its provable negation contradict the assumption of a consistent first-order theory.

Given a bounded-complete one-consistent arithmetic first-order theory with a false theorem, enumerate first-order sentences and successively choose one-consistent one-sentence extensions. Every stage has Sigma-1 soundness, and a false Sigma-1 theorem in the union would already have a formal proof at one stage. The union is therefore one-consistent and complete while retaining the false theorem. It cannot be omega-consistent, since a complete bounded-complete omega-consistent first-order theory is sound.