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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.

A formula provably equivalent to a boxed formula satisfies $F\to \square F$ in GL. The displayed fixed point then gives both $\square \square F$ and $\square \neg \square F$ under $F$, hence $\square \perp$. Boxed explosion proves the reverse implication. Under arithmetic realization this is a formal inconsistency statement; Sigma-1 soundness excludes its provability, while second incompleteness excludes provability of its negation.

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.

Bounded completeness of an arithmetic theory means that every bounded arithmetic first-order sentence true in the standard natural numbers is provable in $S$. If $S$ is first-order consistent and proves a false bounded arithmetic sentence $B$, bounded completeness of an arithmetic theory also proves its true bounded negation, a contradiction. Conversely, if every bounded theorem is true, $S$ cannot be inconsistent, because inconsistency would prove the false bounded arithmetic sentence $0=1$. ThusIf $S$ is one-consistent and proves $\exists xB(x)$ with bounded matrix, that existential sentence cannot be false: otherwise every $\neg B(\bar{n})$ would be a true bounded arithmetic sentence and hence provable, contrary to one-consistency. Thus $S$ has Sigma-1 soundness.

Conversely, Sigma-1 soundness excludes inconsistency. If $S⊢\exists xB(x)$, it supplies a standard witness $n$ with $B(n)$ true. Bounded completeness of an arithmetic theory proves $B(\bar{n})$, so first-order consistency forbids a formal proof of $\neg B(\bar{n})$. Hence there cannot simultaneously be refutations of all standard instances. ThereforeFinite blocks of existential logical variables are treated as finite tuples of witnesses, with the same argument.

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.

A logical formula $D(\vec{x},y)$ weakly defines a total numerical function $f$ in $S$ if, for every standard input tuple $\vec{n}$ and $m=f(\vec{n})$,Strong definability of a numerical function additionally requires the internal uniqueness statementTogether with the positive instance, this is equivalent to proving $\forall y(D(\overline{\stackrel{⃗}{n}},y)\leftrightarrow y=\bar{m})$ for each standard input. The universal quantifier is the extra strength: it covers nonstandard potential outputs in first-order models of $S$. These are the weak and strong function conventions used here. ^[ref]

Suppose the graph of the total function is Sigma-1. Its complement is also Sigma-1, sinceReplace the graph assertion on the right by its Sigma-1 definition; existential quantifiers can be collected into one block. Bounded completeness of an arithmetic theory supplies numeralwise stage enumerators for each of these two disjoint complementary sets. Apply (b) to them. Its separating logical formula is provable exactly at the required graph tuples in the positive direction and refutable at all the other standard tuples, proving weak definability of a numerical function.

Using the graph logical formula alone would only supply positive instances from true existential witnesses; the separation construction is what supplies the negative-instance formal proofs without assuming Sigma-1 soundness or completeness for false Sigma-1 statements.

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.

By part (c), one-consistency gives Sigma-1 soundness and hence ordinary consistency. First, $S$ cannot prove $\neg G$: if it did, the diagonal equivalence would give $S⊢\square G$. This is a ${\Sigma }_1$ sentence, so soundness would make it true. The stipulated standard interpretation of the arithmetical provability predicate would then supply an actual proof of $G$, contradicting consistency.

Now suppose $S⊢\square \neg G$. This too is a ${\Sigma }_1$ theorem. Soundness and the standard correctness of the predicate would imply that there is an actual $S$-proof of $\neg G$, which was just excluded. ThereforeThe two applications of soundness concern the provability assertions, not the truth of arbitrary arithmetic sentences.