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An arithmetic theory is bounded-complete if it proves every true closed formula with only bounded quantifiers. It then proves every true Sigma-1 sentence by choosing a standard witness. If it is also consistent, it cannot prove a false Pi-1 sentence: a standard bounded counterexample supplies a provable contradiction to that universal theorem. This arithmetic condition is distinct from bounded completeness of a statistical model or family.

A two-variable formula enumerates a set $A$ in a theory $S$ when membership supplies a standard numeral stage whose instance is provable, while nonmembership makes every standard-stage instance refutable. The notion contains these proof obligations, not merely a syntactic complexity restriction on the formula. With bounded completeness, a bounded formula whose existential closure defines a computably enumerable set provides such an enumerator.^[ref]

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.

Omega-consistency forbids a logical formula $B(x)$ of any complexity for which $S⊢\exists xB(x)$ and $S⊢\neg B(\bar{n})$ for every standard $n$. The difference from one-consistency is that the matrix is unrestricted. It implies first-order consistency.

For a complete first-order consistent classical first-order theory, theorem membership obeys the ordinary Boolean truth tables. For example, it proves a conjunction exactly when it proves both conjuncts; completeness and first-order consistency also make theoremhood of a negation the complement of theoremhood of the first-order sentence. Omega-consistency additionally gives the witness propertyThe reverse implication is existential introduction. For the forward implication, if no standard instance were provable, completeness would give a formal proof of each negated instance, violating omega-consistency. This works also after replacing additional free variables by numerals.

Now induct on logical formulas with their remaining free variables replaced by standard numerals. At the atomic level, the equivalence between theoremhood and standard truth follows from bounded completeness of an arithmetic theory and bounded soundness of an arithmetic theory from (b). The Boolean steps use the truth-table facts just proved. The existential step uses the witness property and the induction hypothesis on its matrix; universal quantification follows by negation. Hencefor every first-order sentence $X$. Completeness ensures it decides each first-order sentence, and this equivalence ensures every decision is correct.

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 logical formula is ${\Sigma }_0$, or bounded, if all its quantifiers are bounded by arithmetical first-order terms. A ${\Sigma }_1$ logical formula has an initial block of existential quantifiers followed by a bounded matrix, up to the usual logical transformations. Soundness of an arithmetic theory at a given level means that every first-order sentence of that level proved by the first-order theory is true in the standard natural numbers. Here ${\Sigma }_0$-completeness means that every true bounded arithmetic sentence is provable.

A first-order theory is one-consistent if there is no bounded arithmetic formula $A(x)$ for which it proves $\exists xA(x)$ while proving $\neg A(\bar{n})$ for every standard natural number $n$. This is a restricted form of omega-consistency. In particular one-consistency implies first-order consistency, since an inconsistent first-order theory proves every first-order sentence and its negation.

Suppose a ${\Sigma }_0$-complete first-order theory $S$ is first-order consistent. If it proved a false bounded arithmetic sentence $\theta$, the true bounded arithmetic sentence $\neg \theta$ would also be provable by completeness, contradicting first-order consistency. Thus $S$ is ${\Sigma }_0$-sound. Conversely ${\Sigma }_0$-soundness excludes inconsistency because an inconsistent first-order theory proves the false bounded arithmetic sentence $0=1$. ThereforeIf $S$ is one-consistent but proved a false existential sentence $\exists xA(x)$ with bounded matrix, every standard instance $\neg A(\bar{n})$ would be true and hence provable. This violates one-consistency. Thus it is ${\Sigma }_1$-sound. Conversely, ${\Sigma }_1$-soundness implies first-order consistency by the previous argument, since bounded arithmetic sentences are included at the existential level. If $S$ proved $\exists xA(x)$, soundness of an arithmetic theory would give a standard witness $n$ with $A(n)$ true. Bounded completeness of an arithmetic theory would then give $S⊢A(\bar{n})$, so first-order consistency prevents simultaneous formal proofs of all the negated instances. HenceMultiple existential logical variables can be treated as a finite tuple of witnesses; the same reasoning applies.

If PA is inconsistent, every desired conclusion is already a theorem, so it suffices to analyze the first-order consistent case. Use $Q(u,v)$ for the bounded arithmetical proof relation and $\mathrm{neg}(u)$ for the code of the negation of the logical formula coded by $u$.

Suppose $PA⊢E_n$ and let $m$ be the code of an actual formal proof. By first-order consistency there is no formal proof of $\neg E_n$, in particular none with code at most $m$. Thusis a true bounded arithmetic sentence. Bounded completeness of an arithmetic theory proves it in PA, and existential introduction gives $PA⊢{\mathrm{Pr}}_{PA}^{\ast }(\bar{n})$.

Conversely suppose $PA⊢\neg E_n$ and choose an actual formal proof code $k$ of that negation. First-order consistency excludes every formal proof of $E_n$, so each of the finitely many statements $\neg Q(\bar{n},\bar{j})$ for $j<k$ is true and provable. The supplied finite-case theorem permits PA to combine them into the assertion that no formal proof code below $k$ proves $E_n$.

PA also proves $Q(\overline{\mathrm{neg}(n)},\bar{k})$. For any candidate formal proof code $v\ge \bar{k}$, this one code already refutes the universal no-earlier-negation condition in the Rosser predicate symbol. For $v<\bar{k}$, the finite verification refutes its positive formal proof condition. The supplied order dichotomy and finite-case expansion cover all $v$. ThereforeThe argument uses bounded verification of a finite initial segment and one fixed counter-witness beyond it; it does not infer that PA proves the absence of all ordinary formal proofs merely from first-order consistency.

The literal semantic hypothesis is insufficient: a semantic Gödel fixed point need not be canonical. A sentence can be false and provable in a consistent, unsound arithmetic theory, in which case both sides of the stipulated semantic biconditional can be false.

For a concrete counterexample, assume PA is consistent and set $S=\mathrm{PA}+\neg {\mathrm{Con}}_{\mathrm{PA}}$. Gödel's second incompleteness theorem makes this extension consistent: otherwise PA would prove its own consistency. It is still bounded-complete. Use ordinary numeralwise proof and refutation enumerators for this recursively axiomatized theory, and take $G=\neg {\mathrm{Con}}_{\mathrm{PA}}$. This $G$ is false in the standard natural numbers but is an axiom of $S$. Since it is provable, $\exists yP(┌G┐,y)$ is true; its negation is false. Thus the semantic fixed-point biconditional required in this part is true, with both sides false.

Choose $R$ by the usual canonical Rosser construction for these proof matrices. Its undecidability follows from the preceding parts, and it is a true ${\Pi }_1$ sentence: if it were false, its true ${\Sigma }_1$ negation would be provable in this bounded-complete theory, contradicting undecidability. Hence $G$ is false and $R$ true, so their proposed semantic equivalence is false. This counterexample meets the literal assumptions and does not exploit a missing source page.

Under the intended canonical ${\Pi }_1$ Gödel/Rosser convention, the expected answer does hold. Consistency plus bounded completeness implies ${\Pi }_1$ soundness: a false ${\Pi }_1$ theorem has a standard bounded counterexample whose negation the theory proves. If canonical $G$ were false, its truth fixed point would make it provable, contradicting this soundness; thus $G$ is true. Canonical $R$ is true by the preceding ${\Sigma }_1$-negation argument, so both canonical sentences are true and their equivalence is true. This last conclusion needs that canonical construction or an equivalent additional restriction; truth of a biconditional alone does not supply it.

Use the two formal consistency equivalences expressly supplied for this part. If $S⊢G\leftrightarrow R$, the derivability conditions give $S⊢\square \neg G\leftrightarrow \square \neg R$. Taking negations and applying those supplied equivalences yieldsOne-consistency implies ${\Sigma }_1$ soundness under bounded completeness. Hence $S+{\mathrm{Con}}_S$ is consistent: if not, $S$ would prove $\neg {\mathrm{Con}}_S$, a false ${\Sigma }_1$ sentence. But in that consistent extension the last equivalence and its consistency axiom prove its own consistency ${\mathrm{Con}}_{S+{\mathrm{Con}}_S}$. Gödel's second incompleteness theorem rules this out. ThereforeThis establishes the requested unprovability using the explicit formal hypotheses, without inferring internal provability from the semantic equivalence discussed in the preceding part. The hypotheses for the second theorem are the standard effective arithmetic presentation and adequate provability predicates stipulated here.

Pi-1 soundness means that every closed Pi-1 theorem is true in the standard natural numbers. Consistency plus bounded completeness implies it, since a false universal bounded statement has a standard counterexample whose arithmetic verification is provable. It is weaker than full soundness and does not prevent a consistent arithmetic theory from proving a false Sigma-1 sentence.