Past exam of the model theory and Gödel's incompleteness theorems course of the University of Oxford / 2007 / A / 1 / d

Start with the supplied bounded-complete, one-consistent first-order theory $T$ proving a false first-order sentence $F$. Replace it by its deductive closure, which preserves these properties and contains $F$. Enumerate the arithmetic first-order sentences as $X_0,X_1,\dots$. Recursively extend the current first-order theory by either $X_n$ or $\neg X_n$, choosing an extension which remains one-consistent as permitted by the supplied lemma, and then take its deductive closure. Call the resulting increasing sequence $T_n$, with $T_0$ the initial first-order theory, and put $S={\bigcup }_n{T}_n$.

Every stage is still bounded-complete. By (b), every stage has Sigma-1 soundness. If $S$ proved a false Sigma-1 first-order sentence, the finite formal proof would use assumptions from one stage $T_n$, contradicting its Sigma-1 soundness. Thus $S$ has Sigma-1 soundness and is bounded-complete, so (b) makes it one-consistent.

This soundness of an arithmetic theory argument is essential: the definition of one-consistency involves infinitely many standard refutations, so it would be insufficient merely to say that a violation has a finite formal proof. What has a finite formal proof is the false existential sentence, and bounded completeness of an arithmetic theory is what turns one-consistency into the needed soundness of an arithmetic theory condition.

The union is deductively closed because each formal proof has finitely many assumptions. It contains either $X$ or $\neg X$ for every first-order sentence, by the numeralwise enumeration in arithmetic, and contains the false first-order sentence $F$. If it were omega-consistent, (c) would make it correct on all first-order sentences, contradicting $F$. We have therefore constructedOmega-consistency always implies one-consistency by restricting its matrices. This counterexample proves that the converse is strictly false.