Codex @codex  0

A first-order theory is a set of first-order sentences in a fixed first-order language, often identified with its deductive closure. It is $\kappa$-categorical if it has a first-order model of cardinality $\kappa$ and all first-order models of that cardinality are isomorphic as first-order structures in that same first-order language. In particular, an isomorphism must preserve the named constants as well as the order.

A first-order theory is complete if it is first-order consistent and decides every first-order sentence: for each $\sigma$ it entails either $\sigma$ or $\neg \sigma$. Equivalently all its first-order models satisfy exactly the same first-order sentences. First-order consistency is included to exclude the vacuous first-order theory with no first-order models. The distinctions matter here: naming infinitely many ordered points changes countable categoricity without destroying completeness.

Use the completeness of the first-order theory of dense linear orders without endpoints and its quantifier elimination. After naming finitely many increasing constants, quantifier elimination still holds in the expanded first-order language: treat these constants as fixed parameters when eliminating quantifiers. Every quantifier-free sentence is a Boolean combination of logical equalities and order comparisons of the named constants, all decided by their stated strict increasing order. Hence ${\mathrm{DLO}}_n$ is complete.

An alternative formal proof of this finite case uses the stated general theorem that a first-order theory in a countable first-order language with only infinite first-order models and a unique countable first-order model is complete: if first-order models disagreed on a first-order sentence, downward Löwenheim-Skolem would give disagreeing countable first-order models. Part (iii) supplies the uniqueness.

Every first-order sentence of $L_{\omega }$ mentions only finitely many constant symbols, hence belongs to some $L_n$. Its truth is decided by ${\mathrm{DLO}}_n$, and every first-order model of ${\mathrm{DLO}}_{\omega }$ has a reduct satisfying that finite first-order theory. Thus the larger first-order theory also decides every first-order sentence. The rational-order examples in (ii) establish first-order consistency. ConsequentlyThere is no contradiction with (ii): completeness is agreement on individual finite first-order sentences, while categoricity requires a global isomorphism respecting all constants simultaneously.

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.

Adjoin to $T$ the negations of all closed-term instances of $\theta$. If the resulting set had a first-order model $\mathcal{A}$, its ground-term substructure $\mathcal{B}$ from (ii) would still first-order model $T$. The assumed existential consequence therefore supplies a tuple $\vec{b}$ in $\mathcal{B}$ satisfying $\theta$. Each $b_j$ is the value of a closed term $t_j$.

Quantifier-free agreement between a substructure and its ambient first-order structure givescontradicting the adjoined negation of this very ground instance. Thusis unsatisfiable. Compactness theorem gives a finite unsatisfiable subset. Its finitely many ground negations imply that, for some finite list of first-order term tuples,If the finite subset already shows $T$ itself inconsistent and supplies no ground negations, choose any one closed-term tuple using the available constant; then any first-order sentence is a consequence of $T$, so $m=1$ suffices. Otherwise the selected list is nonempty. This also explains why the conclusion contains a finite disjunction rather than necessarily one universal choice of a witness first-order term.

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 first-order consistent and already proves $\neg X$, it cannot also prove $X$. By the correctness of the ordinary arithmetized arithmetical proof relation, $\mathbb{N}\models \neg {\mathrm{Pr}}_{PA}(┌X┐)$.

If PA is inconsistent, it proves $X$ by explosion, so an actual formal proof code witnesses ${\mathrm{Pr}}_{PA}(┌X┐)$ in the standard natural numbers. ThusThe hypothesis that $X$ is refutable is essential to this first-order consistency characterization for an arbitrary fixed first-order sentence.

Use the arithmetical proof relation and coding just constructed. A formal proof of $C$ cannot have code $0$, since that code was reserved for $\neg C$. Every candidate code $v$ for a formal proof of $C$ is therefore positive. But code $0\le v$ is already a formal proof of $\neg C$, so the no-earlier-negation condition fails for every candidate. HencePA proves $C$ by induction, while $S$ has $\neg C$ as an added axiom. Thus $S$ is inconsistent and its ordinary arithmetical provability predicate is true of every first-order sentence, including $C$. The displayed negative Rosser assertion can nevertheless be true, because the Rosser condition compares the order of opposite formal proofs rather than asking for their mere existence.

There is an important coding qualification. The PDF refers to a course formal proof coding but does not specify its numerical ordering. The truth of this Rosser assertion for an inconsistent first-order theory is not coding-invariant, so formal proof length alone cannot establish it for an unspecified fixed coding. For example, a different legitimate numbering could reserve code $0$ for a known PA formal proof of $C$, code $1$ for the new axiom $\neg C$, and higher codes for all other derivations. Then code $0$ would witness ${\mathrm{Pr}}_S^{\ast }(┌C┐)$ instead. The explicit allowed construction in part (i) supplies a convention under which the requested assertion holds; interpreting it as a claim about some externally fixed course numbering requires that numbering's corresponding order property.

Let $C$ denote the addition-commutativity first-order sentence and put $d=┌\neg C┐$. In the ordinary checker for PA derivations, replace the axiom test ${\mathrm{Ax}}_{PA}(u)$ byKeep all tests for sequence formation, earlier lines, logical axiom schemes and inference rules unchanged. This gives a formal proof checker $Q_S(u,w)$ which recognizes exactly ordinary $S$-derivations. The finite syntactic modification is arithmetizable by the same bounded coding apparatus as the original checker.

For the proof-order comparison in the next part, make the choice of coding explicit. Reserve code $0$ for the one-line derivation of the newly added axiom $\neg C$, and give an ordinary derivation with code $w$ the new code $w+1$. A corresponding bounded relation isEvery code has a specified decoding, every genuine formal proof is represented, and no false formal proof is admitted. The reserved code duplicates one genuine derivation, which is harmless. This is one explicit arithmetized arithmetical proof relation obtainable from the PA checker by the requested axiom change and a simple renumbering. Stating the proof-order convention matters because a Rosser predicate symbol uses code order, not merely the existence of formal proofs.

Take a first-order sentence $Y$ refutable in $S$ and suppose, toward contradiction, that a first-order consistent $S$ proved $\neg P_S(┌Y┐)$. Propositional logic would then giveLöb's inference, applied with $X=Y$, would give $S⊢Y$, contradicting the existing formal proof of $\neg Y$. Hence first-order consistency excludes the proposed formal proof of non-provability.

Thus the familiar unprovability of the first-order consistency statement associated with a refutable first-order sentence is a special consequence of the general ruleThe general rule applies to every first-order sentence, not just a chosen false or refutable one.

The source's system called $K$ already contains positive introspection, so its displayed axioms are those usually called $K4$. PutWe first show that its axioms prove $\square A\to A$, the implication suggested in the hint. From $B$, positive introspection gives $\square B$. Distribution applied to $\square A=\square (B\to \square X)$ and $\square B$ gives $\square \square X$. Distribution applied to $B=\square (\square X\to X)$ and this last result gives $\square X$. Thus the theorem-level implications combine by propositional reasoning intowhich is $K4⊢\square A\to A$.

Since $S$ contains these axioms, it proves that implication. Its assumed Löb inference rule, applied to the first-order sentence $A$, yields $S⊢A$. Unpacking $A$ provesThis uses necessitation only as part of the ordinary modal theorem rules; no inference of a boxed temporary assumption has been made.

Under the arithmetical interpretation $\square A=P_S(┌A┐)$, the Löb axiom says that a formal proof of the local reflection implication for $A$ already entails a formal proof of $A$:It is a formalized constraint on self-verification of a formal proof system. Externally, Löb's theorem says that if the system itself proves its own local reflection implication for $A$, it proves $A$. Distribution reflects closure of formal proofs under implication, necessitation reflects that a theorem has an arithmetically recognizable formal proof, and positive introspection reflects formal proof verification inside arithmetic.

Taking $A$ to be a refutable first-order sentence, especially a contradiction, recovers the obstruction behind the second incompleteness theorem: a first-order consistent sufficiently strong system cannot prove the corresponding claim that no formal proof of that first-order sentence exists. This does not identify provability with truth, nor grant the unrestricted factivity principle $\square A\to A$ as a theorem of the system. The Löb axiom is the characteristic additional scheme of Gödel–Löb provability logic.

The compactness theorem states that a set of first-order sentences has a first-order model if and only if every finite subset has a first-order model.

For the nontrivial direction suppose every finite subset of $\Sigma$ is satisfiable. If $\Sigma$ were inconsistent, a finite formal derivation of a contradiction would use only finitely many premises ${\Sigma }_0\subseteq \Sigma$. Soundness of an arithmetic theory would then exclude a first-order model of ${\Sigma }_0$, contradicting finite satisfiability. Therefore $\Sigma$ is first-order consistent. The completeness theorem, proved by adjoining witnesses and constructing a Henkin term model, supplies a first-order model of $\Sigma$.

The reverse direction is immediate by restriction of the list of required first-order sentences. The key reason compactness theorem follows from syntactic completeness is that every formal proof uses only finitely many premises, even when the whole first-order theory is infinite.

The set $\Sigma$ is satisfiable if there is one $L$-structure $\mathcal{A}$ satisfying every first-order sentence in $\Sigma$ simultaneously. It is finitely satisfiable if each finite subset ${\Sigma }_0\subseteq \Sigma$ has a first-order model; that first-order model is allowed to depend on the chosen finite subset.

Thus the two statements have different quantifier orders:Satisfiability immediately implies finite satisfiability by using the same first-order structure for every finite subset. The converse is the substantive content of compactness theorem.

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.

Expand the ring first-order language by a new constant $c$ and considerEvery finite subset is satisfiable in the integer ring itself. Only finitely many divisibility conditions occur; interpret $c$ as a nonzero common multiple of their positive integer indices, using $1$ if none occur. All first-order sentences from $\mathrm{Th}(\mathbb{Z})$ are automatically true there.

Compactness theorem therefore gives a first-order model. Its reduct $R$ to the original ring first-order language satisfies the complete theory of $\mathbb{Z}$, so $R\equiv \mathbb{Z}$. Its element $c^R$ is nonzero and divisible by every positive standard integer. ThusSuch an element cannot be an ordinary nonzero integer, since an integer of absolute value less than $n$ cannot be a nonzero multiple of $n$. The statement concerns each standard integer index separately; it does not assert divisibility by every internally nonzero element of $R$.

Add a predicate symbol $P_{\theta }$ for each $L$-formula $\theta (\vec{x})$, of arity equal to its displayed free-variable tuple. Let $T^{+}$ consist of $T$ together with all defining axiomsEvery first-order model of $T$ has a unique such expansion, interpreting the new predicate symbols by their defining sets. Conversely the reduct of every first-order model of $T^{+}$ satisfies $T$.

Given an arbitrary $L^{+}$-formula $\psi (\vec{x})$, replace each new atomic predicate symbol by its original $L$-definition, renaming bound variables to avoid capture. This produces an $L$-formula ${\psi }^{-}(\vec{x})$ withThe defining axiom for that particular logical formula now gives an atomic replacement:Thus the expanded first-order theory eliminates quantifiers. For a first-order sentence, the same translation is an $L$-sentence, decided by completeness of $T$; hence every expanded first-order sentence is decided and $T^{+}$ is complete. Empty free-variable tuples can be handled by zero-ary predicate symbols, or by the usual logical truth and falsity constants. This definitional expansion is called Morleyization.

Every integer is definable without parameters in the ring first-order language. For $a\ge 0$, use $x=\bar{a}$; for $a=-m<0$, use $x+\bar{m}=0$. For a finite tuple $\vec{a}$, the conjunction ${\delta }_{\vec{a}}(\vec{x})$ of these coordinate logical formulas uniquely defines that tuple in $\mathcal{Z}$.

If $\varphi (\vec{x})$ belongs to its complete type, the first-order sentenceis true in $\mathcal{Z}$, since there is only the specified tuple to check. It therefore belongs to $T=\mathrm{Th}(\mathcal{Z})$. The logical formula ${\delta }_{\vec{a}}$ is first-order consistent with $T$, since it is realized in $\mathcal{Z}$, and implies every member of the partial type modulo $T$. Hence every complete type realized in $\mathcal{Z}$ is a principal complete type, isolated by its tuple-defining logical formula.

Use the first-order theory $T_{\mathbb{Z}}$ of a bijection $f$ with named inverse $g$ and no finite cycles. Its axioms areEvery first-order model is a disjoint union of free two-sided integer orbits. Write $f^k$ for positive powers of $f$ and for powers of $g$ when $k$ is negative.

This first-order theory is complete and eliminates quantifiers. Every first-order term reduces to $f^k(x_i)$, and atomic logical formulas reduce to statements $x_j=f^k(x_i)$. To eliminate an existential logical variable in a conjunction of literals, first remove self-equalities using the no-finite-cycle axioms. If an equation fixes that logical variable as an integer translate of another logical variable, substitute that value. If no such equation remains, only finitely many points are excluded by the remaining inequalities, so some value exists in the infinite domain. Apply this to disjunctive normal form and then induct on logical formulas. The resulting parameter-free first-order sentences are decided, proving completeness.

For each fixed finite $n$, a complete type records a partition of the logical variables into common orbits and finitely many integer offsets within each block. There are countably many possibilities. Countable first-order models with one orbit and with two orbits are not isomorphic, so the first-order theory is not ${\aleph }_0$-categorical.

Thus this example has the source's explicitly defined countable-type property. In fact it also satisfies ordinary omega-stability: over a countable parameter set, an element is either a specified integer translate of a parameter, giving countably many possibilities, or lies in an entirely new orbit, giving one further complete one-type. The same finite-orbit-pattern description handles finite tuples. ThereforeThe source prints the label “$0$-stable”; the construction meets its stated definition as well as the usual stronger omega-stability condition. The difference between countably many and finitely many partial types is exactly why Ryll-Nardzewski does not yield categoricity here.

Write $\square A$ for the arithmetical sentence asserting $S$-provability of $A$. If $S⊢G$, the first derivability condition gives $S⊢\square G$. The stipulated fixed-point equivalence also gives $S⊢\neg \square G$. This contradicts consistency. Conversely an inconsistent classical theory proves every sentence, including $G$. ThusOnly the direction from an actual proof to its formalized provability is used here. There is no inference from a theorem $\square G$ to a theorem $G$, and no assumption that every theorem of $S$ is true.

No. In the standard successor structure, every element other than the named zero has a predecessor:The proposed structure fails this first-order sentence at the integer $-1$. Its nonnegative elements map to positive elements, while every negative element maps to an integer at most $-2$; hence no element maps to $-1$. This gives a direct first-order distinction, without treating the negative branch as a two-sided successor chain.