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One-consistency says that no bounded arithmetic formula $F(x)$ has a provable existential closure while each of its standard numeral instances is refutable. It implies first-order consistency, since an inconsistent classical first-order theory proves every logical formula.

If $P$ is one-consistent, (i) gives $P⊬G$, so $a\notin D$. Numeralwise enumeration in arithmetic therefore givesIf $P⊢\neg G$, classical logic gives $P⊢\exists v_{2}A(\bar{a},v_2)$. The matrix is bounded, and the preceding displayed derivations refute every standard instance. This violates one-consistency. ConsequentlyTogether with (i), this yields undecidability of this first-order sentence under the stronger hypothesis, without asserting that ordinary first-order consistency alone suffices for its negative half.

Use the numeralwise enumeration in arithmetic convention in the first-order theory: membership supplies a standard stage with a provable enumerating instance, while nonmembership makes every standard-stage instance refutable. Put $D=\{n:P⊢E_n(\bar{n})\}$. By the definition of $a$, the diagonal instance $E_a(\bar{a})$ is exactly $G$.

If $P⊢G$, then $a\in D$. Numeralwise enumeration in arithmetic supplies a natural number $m$ such that $P⊢A(\bar{a},\bar{m})$. But universal instantiation of $G$ gives $P⊢\neg A(\bar{a},\bar{m})$. This contradicts first-order consistency. ThereforeThis uses the formal proof obligations in numeralwise enumeration in arithmetic in $P$, rather than merely the external assertion that a logical formula has a particular complexity.

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.

Write $H(\vec{v})$ for the proposed stage-comparison logical formula and fix a standard tuple $\vec{a}$. If $\vec{a}\in A$, numeralwise enumeration in arithmetic supplies a standard $m$ with $S⊢F(\overline{\stackrel{⃗}{a}},\bar{m})$. Disjointness gives $\vec{a}\notin B$, so $S⊢\neg G(\overline{\stackrel{⃗}{a}},\bar{j})$ for every standard $j$, in particular $j=0,\dots ,m$. The source's finite initial-segment axiom combines this finite list intoUse $\bar{m}$ as the existential witness to obtain $S⊢H(\overline{\stackrel{⃗}{a}})$.

If instead $\vec{a}\in B$, choose $n$ with $S⊢G(\overline{\stackrel{⃗}{a}},\bar{n})$. Disjointness gives refutations of $F(\overline{\stackrel{⃗}{a}},\bar{j})$ for all $j$, so the finitely many $j\le n$ giveFor an arbitrary $x$, the source's comparison axiom gives $x\le \bar{n}$ or $\bar{n}\le x$. In the first case the first conjunct of the proposed witness condition is false. In the second case $y=\bar{n}$ witnesses failure of its bounded universal conjunct, since $G(\overline{\stackrel{⃗}{a}},\bar{n})$ is provable. Generalization over $x$ therefore gives $S⊢\neg H(\overline{\stackrel{⃗}{a}})$.

Thus the logical formula separates $A$ from $B$. The formal proof uses the enumerators' numeralwise positive and negative formal proof properties and the two explicit order principles supplied in the source; it does not assume the first-order theory can decide arbitrary unbounded existential logical formulas.

Every automorphism preserves all logical formulas by induction on their construction, so an automorphism taking $\vec{a}$ to $\vec{b}$ necessarily gives equal partial types.

For the converse, first note that a finite first-order structure is atomic by (ii). Thus in both cases every realized finite-tuple partial type has an isolating logical formula. Suppose two finite tuples $\vec{u},\vec{v}$ have the same complete type, and choose a new element $x$ to extend the first tuple. Let $\theta (\vec{z},w)$ isolate the realized partial type of $(\vec{u},x)$. Sincelogical equality of partial types gives $\mathcal{A}\models \exists w\theta (\vec{v},w)$. Choose a witness $y$. Isolation ensures $(\vec{u},x)$ and $(\vec{v},y)$ have the same complete type. In particular, if $x$ is distinct from the entries already used, their inequality logical formulas ensure $y$ is distinct from its corresponding entries. The resulting finite map is injective and partial elementary.

Start from $\vec{a}\mapsto \vec{b}$, which is well-defined even for repeated coordinates because their logical equality patterns agree. Enumerate the carrier and extend alternately to include the next missing element in the domain and the next missing element in the range. For a range step apply the same extension argument to the inverse finite map. The union is an elementary bijection from $A$ onto itself: every finite tuple and every logical formula involving it is handled at some stage. It is therefore an automorphism taking $\vec{a}$ to $\vec{b}$. In the finite case the construction terminates once all elements have been included.

HenceCountability makes the back-and-forth numeralwise enumeration in arithmetic exhaust the entire first-order structure; it cannot simply be dropped.

For Rosser separation by a bounded stage comparison, use the numeralwise enumeration in arithmetic properties meant by enumeration in $S$: membership supplies some standard stage with a provable positive instance, while nonmembership makes every standard-stage instance refutable. These are stronger than merely attaching a ${\Sigma }_1$ syntactic label to an arbitrary formula. The finite-initial-segment axiom lets us combine finitely many numeralwise refutations into a bounded universal assertion.

If $n$ is provable, choose a standard stage $p$ with $S⊢P(\overline{n},\overline{p})$. Consistency means $n$ is not refutable. Therefore $S⊢\neg R(\overline{n},\overline{j})$ for each $j\le p$. Using the displayed finite-segment axiom and equality substitution, combine these proofs to get $S⊢\forall z\le \overline{p}\neg R(\overline{n},z)$. Existential introduction with stage $p$ proves the separator at $n$.

If $n$ is refutable, choose a standard stage $q$ with $S⊢R(\overline{n},\overline{q})$. Consistency implies that $n$ is not provable, so all standard instances of $P(\overline{n},\cdot )$ are refutable. In particular their finitely many instances through $q$ give $S⊢\forall z\le \overline{q}\neg P(\overline{n},z)$. For an arbitrary proposed witness $z$, the total-comparison axiom gives $z\le \overline{q}$ or $\overline{q}\le z$. In the first case its $P$ conjunct fails; in the second case the known $R$ instance at $q$ contradicts its bounded no-refutation conjunct. Generalize on $z$ to refute the existential separator. Hence the proposed first-proof-before-refutation formula separates the two sets.