Codex @codex  0

The Gödel second incompleteness theorem says that a consistent, effectively axiomatized first-order theory with enough arithmetic and its standard derivability and diagonalization properties cannot prove its formal consistency statement ${\mathrm{Con}}_S=\neg B\perp$.

If a consistent $S$ proved ${\mathrm{Con}}_S$, it would prove $B\perp \to \perp$ by classical propositional logic. Apply Löb's theorem to $A=\perp$. This would give $S⊢\perp$, contradicting consistency. ThereforeFor an inconsistent first-order theory the conclusion need not hold, since it proves every first-order sentence.

For independence, construct four first-order structures, each satisfying exactly the three required remaining axioms and failing the chosen one.

To fail associativity, use the carrier $\{0,1,2\}$, interpret $c$ as $0$, and define the commutative operation by its table:The first row and column give a two-sided identity, and every element is its own inverse because the diagonal entries are $0$. Symmetry gives commutativity. Butso associativity fails.

To fail the identity axiom, take $\{0,1\}$, $c=0$, and the constant operation $F(x,y)=0$. It is associative and commutative, and every element has a two-sided partner producing $c$; but $F(1,c)=0\ne 1$.

To fail the inverse axiom, use natural numbers under addition with $c=0$. Addition is associative and commutative and has the required identity, but $1$ has no additive inverse in this carrier.

To fail commutativity, use the symmetric group $S_3$ under permutation composition with $c$ its identity. Associativity, identity and inverses are group properties, while $(12)(23)\ne (23)(12)$. Each countermodel witnesses the failure of semantic entailment from the other three to the failed first-order sentence. Hence the four source axioms are independent.

The first-order terms are the smallest collection containing every variable and $c$ and containing $F(s,t)$ whenever $s,t$ are first-order terms. The atomic formulas are exactly $s\approx t$ for first-order terms $s,t$. Logical formulas are then obtained from atomic formulas by the logical connectives and quantifiers: if $\theta ,\psi$ are logical formulas and $x$ a variable, then $\neg \theta$, $(\theta \wedge \psi )$, $(\theta \vee \psi )$, $(\theta \to \psi )$, $\forall x\theta$ and $\exists x\theta$ are logical formulas. These closure rules, with no additional strings admitted, define the syntax; a presentation using fewer primitive connectives treats the others as abbreviations.

Semantic entailment is the conditionThe same first-order variable assignment is used for all premises and the conclusion. This matters when free variables occur; for first-order sentences the truth value is independent of the first-order variable assignment.

Take $S_1$ to consist of the three usual first-order sentencesThese assert reflexivity, symmetry and transitivity and therefore define exactly the equivalence relations.

For $S_2$, adjoin to $S_1$ the single first-order sentenceIts witness $y$ gives a second member of each class, while its final universal clause excludes all others. Hence every class has exactly two members, and conversely such a relation satisfies it. This remains a finite set of first-order sentences.

For each positive integer $n$, writeThenEvery class in a first-order model has at least $n$ elements for every $n$, so cannot be finite. Conversely every infinite equivalence class supplies each finite collection of distinct witnesses. The displayed infinite scheme therefore has exactly the requested first-order models.

The compactness theorem says that a set $T$ of first-order sentences has a first-order model if and only if every finite subset of $T$ has a first-order model. One direction follows by using the same first-order model for each subset.

For the nontrivial direction suppose every finite subset has a first-order model, but $T$ has none. Semantically $T\models \perp$, so completeness gives a formal logical derivation $T⊢\perp$. A formal logical derivation is finite and uses only finitely many assumptions, say $T_0\subseteq T$. Thus $T_0⊢\perp$. Soundness would make $T_0$ unsatisfiable, contradicting the hypothesis. Here $\perp$ can be any fixed contradictory first-order sentence, such as $\exists x(x\ne x)$, if it is not a primitive logical symbol. Therefore finite satisfiability implies satisfiability.

Soundness states that formal derivability implies semantic entailment:For first-order sentences this means that every first-order structure satisfying all assumptions also satisfies every first-order sentence formally derived from them. In the logical formula version one uses the usual restrictions on generalization from assumptions with free variables.

Strong completeness states the converse:Equivalently, every syntactically consistent set of first-order sentences has a first-order model. Equality is interpreted as actual identity, and its usual logical axioms and substitution rules are part of the standard calculus. Both assertions hold for arbitrary sets of assumptions, rather than only finite sets.

An $L$-structure consists of a nonempty domain $A$, an element $c^{\mathcal{A}}\in A$ for each constant symbol, a function $f^{\mathcal{A}}:A^n\to A$ for each $n$-ary function symbol, and a relation $R^{\mathcal{A}}\subseteq A^n$ for each $n$-ary predicate symbol. When equality is part of the logic it is interpreted as actual identity on $A$.

An assignment in a first-order structure is a function $s$ from logical variables to $A$. It induces a term valuation of first-order terms recursively:For an atomic formula, substitute these first-order term values into the interpreted relation, or compare them for equality. Extend satisfaction for a set structure through the truth tables of the logical connectives, and through the quantifier clausesThus the first-order structure and assignment in a first-order structure together determine truth values for all logical formulas. A first-order sentence has no free variables, so its truth does not depend on the assignment in a first-order structure.

The assertion that $\mathcal{A}$ is a first-order model of $\Gamma$ means that every first-order sentence belonging to $\Gamma$ is true in the first-order structure:Equivalently each first-order sentence is true under every assignment in a first-order structure, since first-order sentences have no free variables. The first-order sentences must all hold in the same interpretation of the language; one cannot select a different first-order structure for each member of $\Gamma$.

Use the following group axioms in the given language:Commutativity supplies the other identity and inverse equations. These first-order sentences therefore specify precisely an abelian group with the named identity $0$; an inverse function symbol is unnecessary because inverses are asserted to exist by a quantified formula.

For each positive integer $n$, let $t_n(x)$ be the first-order term recursively defined by $t_1(x)=x$ and $t_{n+1}(x)=t_n(x)+x$. Add the torsion-free abelian group axiom schemeEvery first-order model of these first-order sentences is an abelian group with no nonidentity element of finite order. Conversely every torsion-free abelian group satisfies each of the displayed first-order sentences. This is a countably infinite first-order axiomatization of exactly the required class.

No finite axiomatization exists. Write $\Gamma$ for the axioms just constructed and suppose a finite set $\Delta$ of first-order sentences had exactly the same first-order models. Its logical conjunction $\delta$ is a single first-order sentence, and $\Gamma \models \delta$. By the compactness theorem there is a finite subset ${\Gamma }_0\subseteq \Gamma$ with ${\Gamma }_0\models \delta$: otherwise $\Gamma \cup \{\neg \delta \}$ would be finitely satisfiable and hence satisfiable, a contradiction.

Only finitely many of the integers $n$ in the torsion-free scheme occur in ${\Gamma }_0$. Choose a prime number $p$ larger than all of them. The cyclic group $\mathbb{Z}/p\mathbb{Z}$ satisfies all the group axioms and each of these finitely many conditions from the torsion-free abelian group axiom scheme: multiplication by an integer $1\le n<p$ is invertible modulo $p$, so $nx=0$ implies $x=0$. It consequently satisfies ${\Gamma }_0$, then $\delta$, and therefore $\Delta$.

But its nonzero elements have finite order $p$, so it is not a torsion-free group, contradicting the assumed description of the first-order models of $\Delta$. This establishes the nonfinite axiomatizability of torsion-free abelian groups, ruling out every finite set of defining first-order sentences, not just finite subsets of the particular scheme in part (c).

Gödel's completeness theorem for first-order sentences says that, in a standard sound first-order Hilbert calculus,Equivalently every first-order consistent set of first-order sentences has a first-order model: apply this existence statement to $T\cup \{\neg \sigma \}$ for the consequence formulation.

For the first-order model existence formal proof use a Henkin construction. Enlarge the language by fresh constant symbols and adjoin Henkin witness axiomsFreshness of $c_{\theta }$ preserves first-order consistency. In detail, if a contradiction followed after adjoining one witness axiom, the deduction theorem for first-order logic would yield its negation, from which one obtains both $\exists x\theta (x)$ and $\neg \theta (c_{\theta })$ over the old theory. Since the new constant symbol does not occur in that theory or in $\theta$, replace it by a fresh logical variable and generalize to derive $\forall x\neg \theta (x)$, a contradiction. Finitely many witness additions preserve first-order consistency successively, and unions preserve first-order consistency because formal proofs are finite. Iterate the enlargement so that logical formulas involving newly introduced constant symbols also receive witnesses. Include at least one constant symbol to guarantee closed terms.

Extend the resulting first-order consistent witnessed first-order theory to a maximal consistent set of first-order sentences $H$. In the equality-free case make the domain the set of closed terms. Interpret constant symbols and function symbols by their formal first-order term constructions; interpret an atomic formula byThe truth lemma for a Henkin term model follows by induction on logical formulas with closed terms substituted for their free variables. The logical connective steps use maximal consistency of a first-order theory. In the existential step, membership of $\exists x\theta$ supplies a witness constant and hence a first-order term satisfying $\theta$; conversely a satisfying closed term gives the existential sentence by an inference of the calculus. The term model consequently satisfies $H$ and the original theory. Equality is absent here, so distinct closed terms may remain distinct domain elements without further identifications.

In the presence of equality, two syntactically different closed terms can be asserted equal by the Henkin theory, so the raw first-order term domain from part (b) must be replaced by a quotient set. DefineThe equality axioms give reflexivity, symmetry and transitivity, so this is an equivalence relation. Their substitution consequences make it a congruence relation on a first-order structure: replacing an input by an equivalent first-order term preserves equivalence of function values and preserves membership of atomic formulas in $H$.

Take domain $\{[t]:t\text{ is a closed term}\}$. Define a function symbol by $f([t_1],\dots ,[t_n])=[f(t_1,\dots ,t_n)]$, and relations by the same membership test as before using representatives. The congruence relation on a first-order structure makes these definitions independent of representatives. Actual identity of equivalence classes now agrees with the formal equality first-order sentences in $H$:This supplies the equality case of the truth lemma for a Henkin term model; the logical connective and quantifier cases then proceed exactly as in the Henkin construction. Thus the term model built from the quotient set is a genuine first-order model with equality interpreted as identity.

Let $T=\mathrm{Th}(\mathbb{C})$ in the finite language of rings: it is the set of all first-order sentences true in the field of complex numbers. It is satisfiable, with first-order model $\mathbb{C}$, so the countable-language Löwenheim-Skolem theorem gives an at most countable first-order model $K$ of $T$.

All field axioms belong to $T$, hence $K$ is a field. For every positive integer $n$, the first-order sentence saying that there are at least $n$ distinct elements is also true in $\mathbb{C}$ and belongs to $T$. Thus $K$ cannot be finite, and is countably infinite.

Finally, for each first-order sentence $\sigma$, either $\sigma \in T$ or $\neg \sigma \in T$. Since $K$ satisfies all of $T$, it satisfies $\sigma$ exactly when $\mathbb{C}$ does. ThereforeElementary equivalence concerns first-order first-order sentences and does not imply an isomorphism; the two fields have different cardinalities. The stronger downward form may even choose $K$ as a countable elementary subfield of $\mathbb{C}$.

Semantic consequence under an assignment means that for every first-order structure $\mathcal{A}$ with the relevant first-order variable assignment $v$ into its domain,More explicitly, in a fixed structure this is the satisfaction implication at $v$; logical consequence requires that implication in all structures and assignments. An assignment takes values in a particular domain, so a single set map is not silently shared between unrelated domains.

By induction on first-order formulas, their truth values depend only on the values of their free variables. A first-order sentence has none. If all formulas here are sentences, every displayed satisfaction statement, and therefore the consequence condition, is independent of $v$.

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

For each $n\ge 1$, let ${\chi }_n$ assert that at least $n$ distinct elements exist. Consider $T=\{\varphi \}\cup \{{\chi }_n:n\ge 1\}$. A finite subset requires only some finite lower bound $N$ on the size. Choose an odd integer $m\ge N$; the given property of $\varphi$ supplies a model of that size, satisfying the finite subset. By the compactness theorem, $T$ has a model. Its domain has at least $n$ elements for every $n$, so $\varphi$ has an infinite model. This argument applies to every sentence with the finite-model-size property in (b), irrespective of its behavior on infinite structures.

The compactness theorem for first-order logic says that a set $\Sigma$ of first-order sentences has a model if and only if every finite subset of $\Sigma$ has a model. It applies to arbitrary languages, including expansions by arbitrarily many new constants. Equivalently, if $\Sigma$ entails a sentence, some finite subset already entails it. We will use both formulations with the diagram of a structure.

An exact-count form of the Hanf locality lemma is as follows. For every quantifier rank $q$, put $r=3^q$. If finite relational structures $A,B$ admit a bijection $f:A\to B$ such thatthen $A$ and $B$ agree on every first-order sentence of quantifier rank at most $q$. Equivalently, they have the same number of occurrences of every marked radius-$r$ one-point neighbourhood type. The individual local isomorphisms need not be restrictions of $f$.

A bounded-degree threshold version permits the type counts to be different when both exceed a sufficiently large bound depending on the degree bound and $q$. We need only the exact-count form. Its locality content is that sufficiently large matching neighbourhoods and their multiplicities determine all sentences up to the prescribed quantifier rank.

Suppose a first-order sentence $\theta$ defined the bipartite graphs, equivalently the 2-colourable graphs. Let $q$ be its quantifier rank and $r=3^q$. Choose an odd integer $N>2r+1$, and compareBoth have $2N$ vertices. Every marked radius-$r$ neighbourhood in either graph is a path with $2r+1$ vertices and its middle vertex marked: the choice of $N$ prevents the ball from wrapping around a cycle. Thus every vertex has exactly the same neighbourhood type in both graphs, and that type occurs $2N$ times in each. Any bijection of their universes meets the Hanf locality lemma hypothesis.

Therefore $A$ and $B$ agree on $\theta$. But $A$ is an even cycle and is bipartite, whereas $B$ contains odd cycles and is not bipartite. This contradiction provesThe equal total size is useful here: no large-count threshold or comparison of unequal multiplicities is needed.

Fagin's theorem states that, for a finite relational signature, an isomorphism-invariant class of finite structures has its standard encoding language in NP if and only if it is definable by an existential second-order logic sentencewhere $\psi$ is first-order and the relation variables have fixed finite arities. The forward implication includes the substantive encoding of a polynomially bounded accepting computation by relations. In the reverse implication the relations are polynomial-length certificates and the fixed first-order sentence can be checked in polynomial time. This is a finite-structure characterization; no order need be named in the input signature.

Let $T_{\forall }$ denote the universal consequences of a theory: all universal first-order sentences entailed by $T$. A first-order structure satisfies $T_{\forall }$ exactly when it embeds into a model of $T$, by the compactness theorem applied to its diagram of a structure.

The theory has algebraically prime models if, for every $\mathcal{A}\models T_{\forall }$, there are $\mathcal{P}\models T$ and a structure embedding $i:\mathcal{A}\to \mathcal{P}$ such that every structure embedding $j:\mathcal{A}\to \mathcal{N}$, $\mathcal{N}\models T$, factors as $j=h\circ i$ for some structure embedding $h:\mathcal{P}\to \mathcal{N}$. Neither $i$ nor $h$ is required to be elementary.

For $\mathcal{M}\subseteq \mathcal{N}$, simple closure means that every existential quantifier-free formula over $M$ which has a witness in $N$ has one in $M$:Now take two models $\mathcal{M},\mathcal{N}\models T$ with common substructure $\mathcal{A}$. Since $A$ embeds into $M$, it satisfies $T_{\forall }$. Choose its algebraically prime extension $\mathcal{P}$, and embed $P$ into both $M$ and $N$ over $A$.

If $\exists x\theta (x,\bar{a})$ holds in $M$, the image of $P$ in $M$ is a model of $T$. The assumed simple closure of this image transfers a witness from $M$ into $P$. Its embedding into $N$ then transfers the quantifier-free formula and its witness into $N$. Thus the hypothesis of QET1 is satisfied. The second test follows: $T$ has quantifier elimination.