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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.

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.

A logical formula separates $A$ from $B$ in $S$ if it is provable at the standard numeral tuple of every member of $A$ and refutable at that of every member of $B$:There is no obligation on tuples outside the two sets. In a first-order consistent first-order theory these two obligations make $A$ and $B$ disjoint.

Let $h$ be the Gödel number of $H(v_1)$ and take the literal diagonal first-order sentence $R=H(\bar{h})$. If $S⊢R$, then $h$ belongs to the second source set, so separation gives $S⊢\neg H(\bar{h})$, a contradiction. If $S⊢\neg R$, then $h$ belongs to the first source set, so separation gives $S⊢H(\bar{h})$, again a contradiction. HenceThe direction of separation matters: here $H$ is provable on refutable diagonal codes and refutable on provable diagonal codes.

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.

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.

Let $d(n)$ be the Gödel number of the diagonal first-order sentence obtained by substituting $\bar{n}$ into the one-free-variable logical formula with code $n$, with an arbitrary default value on noncodes. By the assumptions and (c), this total function has a strong defining logical formula $D(x,y)$.

For the specified $F$, form the one-variable logical formulaLet $h$ be the Gödel number of $H$ and put $C=H(\bar{h})$. Then $d(h)=┌C┐$. Strong definability of a numerical function at this particular standard input givesSubstitution into the existential definition yieldsThe permitted equivalence between the source's two substitution notations identifies this with its requested fixed point. Bound variables are renamed when necessary before forming $H$, so the substitution is capture-free. This constructs one particular first-order sentence from the code of the logical formula which applies $F$ to its own diagonal output.

Write $\square A$ for $P(┌A┐)$. A standard arithmetical provability predicate obeys the Hilbert-Bernays-Löb derivability conditions:The first condition is a rule about theorems; the latter two are statements proved within the first-order theory. When the terminology also includes external adequacy, require that $P(\bar{n})$ be true in the standard natural numbers exactly when the logical formula with code $n$ has an $S$-proof.

For Peano arithmetic, code finite strings, logical formulas and formal proof sequences by natural numbers. Formal proof checking is effective and has an arithmetized bounded checking relation with its computation data. Existentially assert the existence of a correct formal proof code and any required checking data. The resulting predicate symbol is Sigma-1, and its standard truth means exactly that an actual finite PA formal proof exists. Effective operations on formal proof codes combine an implication formal proof with an antecedent formal proof, and turn a formal proof into a formal proof that its code is a formal proof; PA verifies those constructions. They give distribution and positive introspection, while an actual theorem's finite formal proof gives its internal provability assertion. This is the requested rough construction, with the three internal formal proof properties distinguished from external agreement.

Use the supplied uniqueness theorem for the modalized logical formula $A(p)=\square p\to X$. Its occurrence of $p$ is boxed, and the source assumes $p$ does not occur in $X$. Choose another first-order sentence letter $q$ absent from $X$ and from $p$. Besides the uniqueness theorem and propositional reasoning, use that GL is closed under uniform substitution in modal logic.

Let $Y=\square X\to X$. Uniformly substitute $Y$ for the fresh letter $q$ in the uniqueness theorem. Since $q$ occurs nowhere in $X$, this givesPart (c) proves the second conjunct outright. Propositional reasoning removes that already-proved conjunct and yieldsThe freshness conditions make the substitution legal and ensure it changes only the placeholder intended for the second fixed point.

A prime model of a complete theory $T$ is a first-order model admitting an elementary embedding of first-order structures into every first-order model of $T$. Let $\mathcal{A}\models T$ be canonical and $\mathcal{B}\models T$. For each closed term $t$, defineEvery element of $A$ has a name, so this defines the desired map on all of $A$. If two first-order terms agree in $\mathcal{A}$, their logical equality first-order sentence holds in one first-order model of the complete theory and therefore in every first-order model of it; they agree in $\mathcal{B}$. This proves well-definedness. If they are distinct in $\mathcal{A}$, their inequality holds throughout the complete theory as well, proving injectivity.

For every logical formula $\varphi (\vec{x})$ and named tuple ${\vec{t}}^{\mathcal{A}}$, the substituted first-order sentence $\varphi (\vec{t})$ has the same truth value in the two first-order models by completeness of $T$. ConsequentlyThus $e$ is elementary. Since $\mathcal{B}$ was arbitrary, every such canonical first-order structure is prime. This formal proof also handles closed-term generation when function symbols are present; under the constant-name convention one simply uses constants as the first-order terms.

A sufficient condition is that $T$ be first-order consistent, complete, and full in its own first-order language. Work in the allowed function-free setting. Since $T$ proves the logically valid first-order sentence $\exists x(x=x)$, fullness guarantees at least one constant and a nonempty collection of names.

Form the carrier from equivalence classes $[c]$ of constant symbols underLogical equality logic makes this an equivalence relation. Interpret a constant by its class and interpret each relation symbol bySubstitution of equals makes this independent of representatives. This defines a canonical first-order structure because every carrier element is named by a constant.

The truth lemma, proved by induction on logical formulas with constants substituted for their free variables, saysAtomic formulas hold by construction; completeness and first-order consistency give the Boolean steps. For the existential step, a witness in the first-order structure is a named class, so its truth gives a proved instance and hence a proved existential sentence. Conversely a proved existential sentence has a constant witness by fullness, and the induction hypothesis realizes that instance in the first-order structure. The universal case follows by negation. Thus $\mathcal{M}\models T$, provingThe witness condition is needed in the original first-order language; adding new constants produces a canonical first-order structure in an expanded first-order language, which is a different assertion.

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.

Let $\vec{a}\in A^n$, with $A$ finite, and put $T=\mathrm{Th}(\mathcal{A})$. There are only finitely many competing tuples $\vec{b}\in A^n$. For each whose complete type differs from that of $\vec{a}$, choose a logical formula ${\varphi }_{\vec{b}}$ true at $\vec{a}$ and false at $\vec{b}$. Such a logical formula exists by the inequality of their complete types, negating a distinguishing logical formula if necessary.

Let $\theta$ be the conjunction of this finite list, or logical truth if the list is empty. It is true at $\vec{a}$, and every tuple satisfying it has the same partial type as $\vec{a}$. For each $\psi \in \mathrm{tp}(\vec{a})$, the first-order sentenceis therefore true in $\mathcal{A}$ and belongs to $T$. The logical formula $\theta$ is first-order consistent with $T$ because it is realized at $\vec{a}$. It is consequently a principal formula of a type of the partial type, proving every partial type realized in a finite first-order structure is principal. Finiteness of the first-order structure, rather than finiteness of the first-order language, is what makes the conjunction finite.

An $n$-type over a first-order theory $T$ is a set of logical formulas in the displayed $n$ free variables first-order consistent with $T$. A complete type chooses exactly one of $\varphi$ and $\neg \varphi$ for each such logical formula. Equivalently it is a maximal consistent such set. For parameter-free partial types the logical formulas use the original first-order language; partial types over parameters use additional names for those parameters.

A partial type $p(\vec{x})$ is principal if some logical formula $\theta (\vec{x})$ first-order consistent with $T$ implies every member of $p$ modulo $T$:Such a $\theta$ is a principal formula of a type of the partial type. For a complete type it isolates that partial type, so principal complete types are exactly isolated types.

For a tuple $\vec{a}$ in a first-order structure $\mathcal{A}$, its parameter-free partial type isIt is a complete type over $\mathrm{Th}(\mathcal{A})$ because the first-order structure decides the truth of every logical formula at that tuple.

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.

The definition of $a$ makes the diagonal instance $E_a[\bar{a}]$ exactly $G_S$. The representing logical formula therefore has the standard interpretationIf $S⊢G_S$, choose an actual witness $m$ to the left-hand existential statement. The bounded arithmetic sentence $A(\bar{a},\bar{m})$ is true and hence provable by ${\Sigma }_0$-completeness. But $G_S$ also gives $S⊢\neg A(\bar{a},\bar{m})$, a contradiction. ThusNow assume one-consistency. It implies first-order consistency, so no formal proof of $G_S$ exists and $A(a,m)$ is false for each standard $m$. Each bounded arithmetic sentence $\neg A(\bar{a},\bar{m})$ is consequently true and provable. A formal proof of $\neg G_S$ would give a formal proof of $\exists v_{2}A(\bar{a},v_2)$ while every standard witness instance is refuted, contrary to one-consistency. HenceThe two hypotheses play different roles: ordinary first-order consistency excludes the positive Gödel sentence, while the stronger hypothesis also excludes its negation.

By the standard interpretation of the representing logical formula,If $S$ is first-order consistent, part (b) supplies the right-hand side, so $G_S$ is true. If $S$ is inconsistent, explosion gives $S⊢G_S$, so the representing existential statement is true and $G_S$ is false. ThereforeThis is an external truth assertion about the standard natural numbers. Formalizing that assertion inside a first-order theory requires the additional facts explicitly supplied in part (d).

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.

An $L$-structure has a nonempty underlying set $A$. It assigns to each constant symbol $c_k$ an element $c_k^{\mathcal{A}}\in A$, to each $n$-ary function symbol $f_j$ a function $A^n\to A$, and to each $n$-ary predicate symbol $P_i$ a relation contained in $A^n$. Logical equality, when included as logical equality, is interpreted as actual identity.

The arities belong to the specification of the first-order language. An assignment of logical variables to elements of $A$ then evaluates first-order terms recursively and determines satisfaction of logical formulas through the logical connectives and quantifiers. A first-order sentence has no free variables, so its truth in the first-order structure is independent of the assignment.

The Downward Lowenheim-Skolem theorem says that for an $L$-structure $M$ and a subset $S\subseteq M$, there is an elementary substructure $N\preccurlyeq M$ containing $S$ withIf an infinite cardinal $\kappa$ satisfies $\max (|L|,{\aleph }_0)\le \kappa \le |M|$, one can in particular obtain size exactly $\kappa$ by starting with a subset of that size.

Choose a default element $a_0\in M$. For every logical formula $\exists x\varphi (x,\vec{y})$ choose a Skolem function $F_{\varphi }$ which returns a witness in $M$ when one exists, and returns $a_0$ otherwise. Begin with $S$, the interpretations of the first-order language's constants, and $a_0$. Successively adjoin the values of all the original functions and these Skolem functions on finite tuples already chosen. Let $N$ be the union of the resulting countable sequence of closure stages.

There are at most $\max (|L|,{\aleph }_0)$ logical formulas, and each function has finite arity. With $\kappa =\max (|S|,|L|,{\aleph }_0)$, each closure stage has size at most $\kappa$, and so does their countable union. Closure under the original functions makes $N$ an $L$-substructure. Whenever $M\models \exists x\varphi (x,\vec{a})$ for parameters in $N$, its chosen Skolem witness belongs to $N$, since the finitely many parameters lie in some common stage. The Tarski–Vaught test therefore gives $N\preccurlyeq M$.

For the exact-cardinality version, choose an initial set of $\kappa$ elements, enlarged to contain any prescribed subset of size at most $\kappa$. The construction gives an upper bound $\kappa$ while containment gives the matching lower bound.

If $\mathcal{A}\subseteq \mathcal{B}$, interpret each new constant $c_a$ in $\mathcal{B}$ as the same element $a$. Atomic logical formulas and their negations then have exactly their truth values in ${\mathcal{A}}_A$, so the expansion satisfies the diagram of a structure. If the inclusion is elementary, every logical formula with those named parameters has the same truth value, so the expansion satisfies the elementary diagram.

Conversely, from any expansion ${\mathcal{B}}^{+}$ satisfying the diagram of a structure defineThe diagram of a structure contains $c_a\ne c_{a^{\prime }}$ whenever $a\ne a^{\prime }$, so $e$ is injective. It contains every true function-value equation, so $e$ commutes with all operations and constants. Positive and negative relation instances ensure both preservation and reflection of the relations. Thus $e$ is an embedding of first-order structures. If the expansion satisfies the elementary diagram, for every logical formula $\varphi$ and tuple $\vec{a}$ it also givesso $e$ is an elementary embedding of first-order structures.

The precise conclusion for an arbitrary expansion is an embedding of first-order structures, rather than a literal inclusion of the original carriers. The source's inclusion notation is correct when one uses the customary identification of $\mathcal{A}$ with its embedded copy, or requires the expansion to interpret $c_a$ as the actual element $a$ of a carrier already containing $A$. With that naming convention, $e$ is the inclusion and the requested two equivalences follow exactly.

Without the convention the literal wording fails, even in the first-order language of logical equality: singleton first-order structures with carriers $\{0\}$ and $\{1\}$ have a diagram-satisfying expansion by sending the first first-order structure's named point to $1$, but the first carrier is not a subset of the second. The embedding of first-order structures formulation proved above states the invariant result and makes the intended inclusion formulation precise.