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Both assertions are false. For simple closure, take the inclusionBoth satisfy DAG prime, but the quantifier-free formula $x\ne 0$ has a witness in $N$ and none in $M$. The same example works in the reduced language $\{+,0\}$.

For quantifier elimination, the first-order sentence $\exists x(x\ne 0)$ distinguishes these two models. Every closed group term is zero, so every atomic closed equality is true in both models. Every quantifier-free sentence, being a Boolean combination of such equalities, has the same truth value in both. The distinguishing sentence therefore has no equivalent quantifier-free sentence modulo DAG prime. HenceIncluding the trivial group is exactly what makes the proposed simple closure condition fail.

A first-order theory $T$ is model-complete if every structure embedding between models of $T$ is an elementary embedding. Equivalently, whenever $\mathcal{M}\subseteq \mathcal{N}$ and both satisfy $T$, one has $\mathcal{M}\preccurlyeq \mathcal{N}$:for every first-order formula $\phi$ and every finite tuple $\bar{a}\in M$. This is preservation of all formulas with parameters, rather than merely all first-order sentences.

The quantifier rank measures maximum nesting depth:Two first-order structures are rank-n equivalent if they agree on all first-order sentences of quantifier rank at most $n$. For tuples one similarly compares all such formulas evaluated at the tuples.

In the $n$-round Ehrenfeucht-Fraïssé game, player I, the Spoiler, chooses an element from either structure each round; player II, the Duplicator, chooses an element in the other. After $n$ rounds the corresponding chosen tuples must define a partial isomorphism of structures: all atomic formulas, including equality and the interpretations of constants, have matching truth values. Repetitions must be respected. Failure of this condition is a loss for II.

The precise finite-vocabulary Ehrenfeucht-Fraïssé theorem is: for a finite relational language, allowing finitely many constants,The tuple version has the same conclusion when the corresponding tuples are fixed before play. The implication from a winning strategy to rank equivalence holds for arbitrary languages, by induction on formulas. For the converse, finite relational vocabulary ensures only finitely many rank-bounded types in a fixed tuple of variables; their defining formulas let one transfer the type of the next move. Without a finiteness hypothesis, the converse for the full atomic-diagram game need not hold. This qualification matters in the next part.

Under the finite relational vocabulary hypothesis of the Ehrenfeucht-Fraïssé theorem, II wins the timed game exactly when the structures are elementarily equivalent. If II wins, then for every choice of $n$ by I, its strategy restricts to a winning strategy in $G_n$. Hence the models agree on every sentence of quantifier rank at most $n$. Every first-order sentence has finite quantifier rank, so they agree on all sentences.

Conversely, elementary equivalence gives $\mathcal{M}{\equiv }_n\mathcal{N}$ for every $n$. Choose a winning strategy for each $G_n$, and use that strategy once I announces $n$. Unlike the infinite game, the timed game permits the strategy to depend on the announced horizon.

The PDF does not explicitly impose finite vocabulary. If an arbitrary language and the full partial isomorphism of structures winning condition are intended, the reverse assertion is false. Here is a failure of finite-round game equivalence in infinite languages. Let $L=\{P_i:i\in \mathbb{N}\}$ be unary relational. Letand obtain $N$ by adjoining one point $b$ satisfying every $P_i$.

For any finite sublanguage, every Boolean cell of predicates has infinitely many elements in both structures. Every finite partial isomorphism of structures of these reducts can therefore be extended in either direction. Induction on formulas, or a back-and-forth method, shows their reducts are elementarily equivalent. Since every sentence uses finitely many symbols, $M\equiv N$ in the whole language.

Nevertheless I wins $G_1(M,N)$ by choosing $b$: every point of $M$ fails some $P_i$, so no response preserves all atomic formulas. Thus the printed unrestricted equivalence requires the finite-vocabulary convention, or a game which additionally restricts each play to a finite vocabulary fragment.

Let $N$ be a well-founded model of set theory of the complete theory $T$, and suppose that a Paris model $M\models T$ were ill-founded. Its internal ordinals would then contain an external descending sequenceFor every $n$, choose a first-order formula ${\phi }_n$ that uniquely defines ${\alpha }_n$ in $M$. The sentences asserting that ${\phi }_n$ uniquely defines an ordinal and that the object defined by ${\phi }_{n+1}$ belongs to the object defined by ${\phi }_n$ are true in $M$. Since $T$ is complete, all its models satisfy the same first-order sentences, so the corresponding uniquely defined ordinals in $N$ form an external descending membership sequence. This contradicts the well-foundedness of $N$. Thus Paris models are well-founded when their complete theory has a well-founded model proves that every Paris model of $T$ is well-founded.

This class is axiomatisable in the original language. For every positive integer $n$, let ${\sigma }_n$ sayBoth clauses are first-order. If one of the two sets is infinite, every ${\sigma }_n$ holds. Conversely, if both sets were finite, choosing $n$ larger than both cardinalities would make ${\sigma }_n$ false. Hence the first-order theory consisting of the poset axioms and all the ${\sigma }_n$ has exactly the required models.

The compactness theorem says that a set $T$ of first-order sentences has a model if and only if every finite subset of $T$ has a model. One implication follows by taking the same model. Conversely, if $T$ had no model, then by the Godel completeness theorem it would prove a contradiction. A formal proof uses only finitely many assumptions, so some finite subset of $T$ would already have no model. This contradiction proves compactness.

The Upward Lowenheim-Skolem theorem says that if an $L$-theory $T$ has an infinite model, then it has models of arbitrarily large cardinality; more precisely, it has a model of cardinality at least $\kappa$ for every cardinal $\kappa$. Add new constants $c_{\alpha }$ for $\alpha <\kappa$ and the sentencesEvery finite subset of the enlarged theory can be interpreted in the given infinite model, since it mentions only finitely many constants. Compactness supplies a model of the whole enlarged theory, in which the $c_{\alpha }$ are pairwise distinct. Its reduct to $L$ is a model of $T$ having at least $\kappa$ elements. If $\kappa \ge |L|+{\aleph }_0$, the Downward Lowenheim-Skolem theorem gives a model of cardinality exactly $\kappa$.

The language of set theory has only countably many sentences, so there are at most $2^{{\aleph }_0}$ possible complete sets of sentences $T_{\xi }$. Choose $(2^{{\aleph }_0}{)}^{+}+1$ ordinals above $\alpha$. By cardinal pigeonhole, two of their theories agree. Hence some $\gamma >\beta >\alpha$ satisfy $T_{\beta }=T_{\gamma }$.

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.

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 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.

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.

Take $\square A={\mathrm{Pr}}_{PA}(┌A┐)$ for the standard formal proof predicate symbol and putUse as given the arithmetization of syntax in PA, the derivability conditions, and the diagonal lemma. The latter supplies the fixed point needed in (a), and the former conditions interpret its normal modal distribution, necessitation on theorems and axiom4. Thus the argument in (a) proves the internal Löb implication in PA. With $X=\perp$ it readsThe inner implication is the first-order consistency first-order sentence $\neg \square \perp$, so contraposition givesThis is the arithmetic statement of the second incompleteness theorem proved within PA: if PA is first-order consistent, there is no PA formal proof of its first-order consistency first-order sentence. The displayed implication is a PA theorem without assuming that PA proves its antecedent.

For the corresponding external consequence, if a first-order consistent PA proved ${\mathrm{Con}}_{PA}$, theoremhood would give a formal proof of its boxed assertion. The displayed theorem together with the supposed formal proof of first-order consistency gives its negation, contradicting first-order consistency. Hence an externally first-order consistent PA does not prove its own first-order consistency.

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.

A canonical first-order structure is generated by its names: every element is the value of a closed term. In the function-free setting permitted here, this means every element is the interpretation of a constant symbol. Equivalently, the natural map from closed terms modulo logical equality in the first-order structure onto its carrier is surjective. When canonical is defined specifically using constant names, the same definition and all the arguments below apply in this function-free setting.

A full set of first-order sentences has the witness property in its own first-order language: whenever it proves an existential sentence $\exists x\varphi (x)$, some closed term $t$ satisfies that the set proves $\varphi (t)$. In a deductively closed set, one can equivalently state this using membership of the first-order sentences. In the function-free setting the witness first-order term is a constant symbol. This is the Henkin witness property; fullness does not by itself replace the separate requirements of first-order consistency and completeness.

Use two expansions of the rational order. In the first interpret $c_n=n$, so the named sequence is unbounded above. In the second interpret $c_n=-1/n$, so the named sequence is strictly increasing but has upper bounds in the rational order. Both expansions are countably infinite first-order models of the specified first-order theory.

An isomorphism preserving each $c_n$ would preserve the existence of an element greater than all the named constants, since any such element in one first-order structure would map to such an element in the other. The first first-order structure has none and the second has, for instance, $0$. They cannot be isomorphic. ThusThe property of lying above all $c_n$ is an infinite partial type rather than a single first-order sentence. This is why its realization can distinguish the two first-order models without contradicting the completeness established in (iv).