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A logical fixed point is a formula or sentence equivalent to the result of applying a specified formula operation to itself. In arithmetic, self-reference is mediated by a Gödel number and the diagonal lemma, yielding a first-order sentence $G$ with a provable equivalence $G\leftrightarrow H(┌G┐)$. In modal logic, a modal formula $A$ satisfies $A\leftrightarrow F(A)$ modulo the chosen logic. A merely true semantic equivalence is weaker than a specified provable or canonical construction, so its hypotheses must be kept distinct.

Mathematical logic studies formal languages, formal proofs, model theory, computability theory, and the foundations of mathematics. It makes the rules of reasoning explicit enough that reasoning itself becomes an object of mathematics. Proof theory studies deductions; model theory studies interpretations in which statements hold.

Three questions organize the subject: is an expression a well-formed first-order formula, is it true in a specified first-order structure, and does it have a formal proof from specified axioms? These questions are distinct. For example, $\exists n(n+n=1)$ is a first-order sentence in an appropriate first-order language; it is false in the natural numbers but true in the rational numbers, with $n=1/2$.

The beauty of mathematical logic is that the same logical connectives and rules of universal quantification and existential quantification organize arguments across mathematics. Classical propositional logic uses two truth values; Intuitionistic propositional logic interprets logical implication through constructions on proofs, without unrestricted law of excluded middle. This proof-based framework is developed in Frank Pfenning, Natural Deduction, sections 1–3.

A first-order consistent theory is maximally consistent when it has no proper consistent extension in its fixed language. It decides each first-order sentence or its negation.

Using the supplied counting function, an appropriate first-order sentence isHere the counting comparison expresses the threshold and the existential quantification expresses teaching at least one course. The word “if” supplies this direction of logical implication; it does not assert that these are the only reasons a faculty member can be busy. With the intended meanings of the nonlogical symbols, no further counting axioms are needed for this translation. If those meanings were unspecified, the displayed first-order sentence alone would not define arithmetic or connect the counting function to the supervision relation.

The supervisor can depend on the student, so the existential quantification belongs inside the universal quantification:The first-order sentence permits different students to share a supervisor. It requires neither that the supervisor be different from the student nor that the supervisor be a faculty member, since neither additional condition was given.

The condition about taking the course restricts which students the first-order sentence talks about. Within that condition we need both an inclusive logical disjunction and a prohibition on satisfying both alternatives:The logical conjunction on the right expresses exactly one of the two categories. It does not prohibit unrelated individuals from belonging to both categories: the restriction applies to the students in the antecedent.

For resolution, a first-order sentence is in clausal normal form when it is a universally quantified logical conjunction of first-order clauses:where every $L_{ij}$ is a first-order literal, meaning an atomic formula or the logical negation of an atomic formula. The quantifiers are often left implicit and the first-order clauses treated as a set. The empty first-order clause represents falsity; an empty set of first-order clauses represents truth.

A convention that allows an arbitrary quantifier prefix followed by a conjunctive matrix is weaker: that is a prenex normal form with a conjunctive matrix. The elimination of existential quantifiers by Skolemization is the extra step relevant to resolution and to part (ii).

A fragment has the finite model property if $\boxed{\text{every satisfiable sentence in the fragment has a finite model}}$. This does not mean that every model is finite, nor that there is one uniform bound on model size for the entire fragment.

ALC has the finite model property, including satisfiability with respect to finite TBoxes. A tableau algorithm for ALC uses finitely many possible labels and direct blocking in description logic to obtain a finite representation of a model. The resulting finite-model claim is a property of the logic; termination alone, for an arbitrary procedure, would not establish it.

Full first-order logic does not have the finite model property. For example, the first-order sentencerequires an irreflexive transitive relation with a successor for every element. It is satisfied by the usual strict order on the natural numbers. In a nonempty finite domain, repeatedly taking successors eventually repeats an element; transitivity would then force $R(x,x)$, contradicting irreflexivity. Hence this satisfiable first-order sentence has no finite model.

Using the absorbed TBox with its primitive inclusion, translate description logic concepts as unary predicates and description logic roles as binary predicates. Concept satisfiability means that some element belongs to $A\sqcap C$, while every element satisfies the TBox axioms. The required first-order sentence isOnly ${\mathcal{T}}_u$ is included here, as requested. There is no requirement to satisfy (1), (2) or (3). For instance, one element with $A=C=\{d\}$, $B=\varnothing$ and $P=\{(d,d)\}$ satisfies this first-order sentence.

Alternatively, unfolding the absorbed TBox first gives the existential first-order sentence $\exists x\exists y(A_0(x)\wedge \neg B(x)\wedge P(x,y))$. The interpretation of the eliminated names is then recovered by $A=A_0\setminus B$ and $C=\{x:\exists yP(x,y)\}$.

A first-order theory in a fixed first-order signature is a set of first-order sentences. Under the closed-theory convention used in the question, it contains all its semantic entailments: $T\models \theta$ implies $\theta \in T$. One can equivalently begin with a set of axioms and take its closure under consequence.

A complete theory is a consistent first-order theory that decides every first-order sentence $\theta$ in its first-order signature:Consistency ensures that exactly one of these alternatives holds.

A decidable theory is one for which an algorithm always terminates and determines whether a given first-order sentence is a consequence:If the first-order theory is already closed under consequence, this is its membership problem. It is a condition on the consequences, not merely on whether the chosen set of axioms has a decidable membership test.

A first-order theory $T$ admits quantifier elimination when every first-order formula $\phi (\bar{x})$ has a quantifier-free formula $\psi (\bar{x})$, in the same first-order signature and with no additional free variables, such thatThus all first-order models of $T$ agree that the two first-order formulas define the same tuples. This definition concerns first-order formulas with free variables, as well as first-order sentences; merely deciding closed assertions would be weaker. The existence of an effective procedure is an additional issue, supplied by the explicit construction in part (d).

It suffices to eliminate one existential quantifier from a quantifier-free formula $\theta (y,\bar{x})$. Indeed, logical connectives preserve equivalence, and a universal quantifier can be rewritten using $\forall y\alpha \equiv \neg \exists y\neg \alpha$. An induction on the construction of a first-order formula then eliminates all its quantifiers from the inside out.

Convert $\theta$ to disjunctive normal form. Since existential quantification distributes over a finite logical disjunction, consider a single logical conjunction of atomic first-order formulas and their negations. There are no function or constant symbols in this first-order signature, so its atoms are equalities between variables and instances of $E$ between variables.

For a tuple $\bar{x}=(x_1,\dots ,x_r)$, enumerate its possible complete atomic diagrams $\delta (\bar{x})$. Such a quantifier-free formula specifies which variables are equal and, between distinct equality classes, exactly which pairs are adjacent. Keep only the diagrams in which first-order equality is an equivalence relation and $E$ is symmetric and irreflexive. There are finitely many diagrams for each $r$, and every tuple in a first-order model of $T$ satisfies exactly one of them. Every retained diagram is realizable: insert one vertex for each equality class successively, using the extension axioms to prescribe its adjacency to earlier vertices.

Fix one diagram $\delta$. A potential existential witness for $y$ has one of two forms.

All first-order formulas involving only $\bar{x}$ are already decided by $\delta$. Therefore, for each diagram, the existence of a suitable $y$ has a definite answer, determined by a finite syntactic test. Let $\mathcal{D}$ be the set of diagrams for which at least one disjunct and at least one of the two witness cases succeeds. ThenThe right side is a quantifier-free formula. An empty logical disjunction means false, and for an empty tuple the empty atomic diagram means true. These logical truth constants also handle closed first-order sentences in this constant-free first-order signature.

For a concrete illustration of the fresh-witness case,If $x_1=x_2$, the edge requirements conflict. Otherwise the extension property supplies a vertex outside the finite set of parameters with the desired pattern; the extra avoidance of $x_3$ causes no obstruction. The general diagram construction handles all such equality coincidences systematically and proves quantifier elimination.

By part (c), $T$ has a first-order model, so it is consistent. By part (d), every first-order sentence is equivalent modulo $T$ to a quantifier-free sentence. In the first-order signature consisting only of $E$ and first-order equality, there are no constant symbols, no ground first-order terms, and no nullary predicates. Thus a closed quantifier-free formula reduces to true or false, using the logical truth constants. All first-order models of $T$ therefore agree on every first-order sentence:

The construction in part (d) is effective. Conversion to disjunctive normal form, enumeration of finite equality partitions and adjacency patterns, and testing the finitely many witness cases are all terminating operations. Apply them recursively to a given first-order sentence and evaluate the final truth constant. This decides whether the first-order sentence belongs to $T$, givingNo useful bound on the running time is needed for decidability; the potentially large enumeration still always terminates.

Use unary predicates $H$ and $M$ for happiness and merriness, a binary predicate $L$ with $L(x,y)$ meaning that $x$ loves $y$, and a constant symbol $\ell$ for Lucy. The domain consists of people. The following four first-order sentences use this same first-order signature.

The property of loving at least one person is the antecedent:An equivalent first-order sentence, useful for Skolemization, is $\forall x\forall y[L(x,y)\to L(x,\ell )]$. The person loved in the antecedent need not be Lucy; the conclusion concerns Lucy specifically.

First convert the four first-order sentences to clausal form. In the first, pull the existential quantifier across the logical disjunction, then replace its witness by a fresh unary Skolem function $f(x)$. The usual nonempty-domain convention for first-order logic makes this transformation valid even when $H(x)$ already holds. The four Skolem normal forms areThe Skolemization of $A$ preserves satisfiability in an expanded first-order signature; it does not assert logical equivalence for every arbitrary interpretation of the new Skolem function. The other three first-order sentences only require elimination of implications and movement of universal quantifiers.

For a ground resolution refutation, negate the desired conclusion and introduce a fresh constant $c$ for a person who is not merry. This adds the ground clause $\neg M(c)$. Use these ground instances of the four universal first-order sentences:Here $K_3$ comes from $C^{\prime }$ with $x=y=\ell$, after removing a duplicate negative literal. All first-order terms in these ground clauses are ground. Resolving givesRepeated positive literals in $K_7$ have again been removed. The empty ground clause proves unsatisfiability of the premises together with the negated conclusion. ThereforeThe key is to apply the first three premises to Lucy herself: either she is already happy, or she loves somebody, hence loves herself, which forces her happiness and activates the final premise.

For a first-order signature with at least one constant symbol, its Herbrand universe is the set of all ground terms built from its constant and function symbols. If necessary add a fresh constant to make this set nonempty. A Herbrand structure has exactly this domain; each constant denotes itself and each function symbol forms the corresponding syntactic first-order term. Its remaining choices are the truth values of the ground predicate atoms. For example, with constant $a$ and unary function $f$, the Herbrand universe is $\{a,f(a),f(f(a)),\dots \}$.

One useful clausal form of Herbrand's theorem is the following. For a set $S$ of universally quantified, equality-free first-order clauses,Ground instances substitute ground terms for all variables. On the right, treat the ground predicate atoms as independent Boolean variables. Equivalently, $S$ is satisfiable if and only if it has a Herbrand model. A general first-order sentence can first be put into Skolem normal form and clausal form, with the new symbols included in the Herbrand universe.

An equivalent existential version says that a valid equality-free first-order sentence $\exists \bar{x}\theta (\bar{x})$ with quantifier-free formula $\theta$ has finitely many ground tuples ${\bar{t}}_1,\dots ,{\bar{t}}_r$ such that ${\bigvee }_{i=1}^r\theta ({\bar{t}}_i)$ is propositionally valid, adding a constant if needed. This explains why a finite ground resolution refutation can establish a consequence involving infinitely many possible ground terms.

The equality-free qualification matters. If first-order equality is present, one must handle its axioms and use a term model modulo a suitable congruence, or use the corresponding theorem with equality. Literal syntactic term interpretation alone cannot model a consistent requirement such as $a=b$ for distinct constant symbols.

Write $BA=\exists y\mathrm{Prov}(┌A┐,y)$ and fix a refutable first-order sentence $\perp$, so $S⊢\neg \perp$ and ${\mathrm{Con}}_S=\neg B\perp$. First prove the internal equivalenceBecause $\perp \to G$ is provable, the derivability conditions give $B\perp \to BG$. The fixed-point equivalence $G\leftrightarrow \neg BG$ therefore yields $G\to \neg B\perp$.

For the reverse implication, necessitating $G\to \neg BG$ and applying distribution gives $BG\to B(\neg BG)$. positive introspection gives $BG\to B(BG)$. Finally the provable implication $BG\to (\neg BG\to \perp )$, necessitated and distributed, givesConsequently $S⊢BG\to B\perp$, so $\neg B\perp \to \neg BG\to G$. The same argument works with any chosen refutable first-order sentence in place of $\perp$; their formal consistency statements are provably equivalent.

By the classical deduction theorem for propositional logic applied to adjoining a first-order sentence, $S+\{G\}$ is consistent exactly when $S⊬\neg G$. The displayed equivalence makes $S+\{G\}$ and $S+\{{\mathrm{Con}}_S\}$ deductively equivalent. ThereforeNo soundness assumption on that extension is being substituted for consistency.

The preceding Gödel sentence argument gives $S⊢G\to {\mathrm{Con}}_S$. The permitted arithmetized positive half of the Rosser sentence argument isPure first-order logic gives $\neg BR\to R$: if there is no proof of $R$, each antecedent in its defining universal implication is false. ThereforeThis uses the supplied internal Rosser implication, rather than replacing an external nonprovability fact by an unproved internal first-order sentence.

Assume $S⊢\neg X$ and $S⊢BX\to X$. Combining these theorems gives $S⊢\neg BX$. Because $X$ is refutable, $\neg BX$ is a legitimate formal consistency statement for $S$: the standard contradiction first-order sentence and $X$ are provably interderivable, and the derivability conditions transfer those implications to their provability statements.

Thus if $S$ were consistent, the Gödel second incompleteness theorem would contradict $S⊢\neg BX$. Consequently the assumed reflection theorem forces $S$ to be inconsistent; explosion then gives $S⊢X$. This establishes the instance of Löb's theoremfor every $X$ refutable in $S$, using only the second incompleteness theorem.