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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

Σ

is satisfiable. If

Σ

were inconsistent, a finite formal derivation of a contradiction would use only finitely many premises

Σ_{0} \subseteq Σ

. Soundness of an arithmetic theory would then exclude a first-order model of

Σ_{0}

, contradicting finite satisfiability. Therefore

Σ

is first-order consistent. The completeness theorem, proved by adjoining witnesses and constructing a Henkin term model, supplies a first-order model of

Σ

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.

Past exam of the model theory and Gödel's incompleteness theorems course of the University of Oxford / 2008 / B / 7 / iv / Solution 2026-10-31

Take the parameter-free complete type

p (x) = tp^{Z^{*}} (c / \emptyset)

of the nonstandard prime constructed in part (i). It is first-order consistent and complete because it consists of exactly the logical formulas true of an actual element in a first-order model of

T

. It contains

x \neq = \overset{a}{ˉ}

for every integer

a

, and consequently no element of

Z

realizes it. Thus

p is a complete type over T omitted in Z .

(2)

It is necessarily a nonprincipal complete type. If a first-order consistent logical formula isolated it, completeness of

T

would force the existential closure of that logical formula to hold in every first-order model of

T

, including

Z

; a witness there would realize the partial type, a contradiction. This connects the explicit omission with the hypothesis of the omitting types theorem.

Past exam of the model theory and Gödel's incompleteness theorems course of the University of Oxford / 2009 / A / 1 / b / Solution 2026-10-31

The Hilbert-Bernays-Löb derivability conditions are: if

S ⊢ A

then

S ⊢ □ A

;

S ⊢ □ (A \to B) \to (□ A \to □ B)

; and

S ⊢ □ A \to □□ A

. The first is an external rule about provability; the latter two are internal theorems. Standard numerical correctness and the stipulated

Σ_{1}

form accompany the predicate in this question. These conditions do not include reflection

□ A \to A

Choose a fixed refutable sentence

⊥

and identify

Con_{S}

with

\neg □ ⊥

. Different fixed refutable choices have provably equivalent consistency sentences, because

S

proves their equivalence to a contradiction. From

⊥ \to G

, necessitation and distribution give

□ ⊥ \to □ G

. Since

G \to \neg □ G

, this yields

G \to Con_{S}

For the converse, necessitate

G \to \neg □ G

and distribute to get

□ G \to □ \neg □ G

. Positive introspection also gives

□ G \to □□ G

. The two boxed contradictories imply

□ ⊥

by formalized propositional reasoning. Thus

□ G \to □ ⊥

, and contraposition gives

Con_{S} \to \neg □ G \to G

. Consequently

S ⊢ G \leftrightarrow Con_{S} .

Finally the deduction theorem says that

S + G

is inconsistent exactly when

S ⊢ \neg G

. The internal equivalence identifies

S + G

and

S + Con_{S}

in their deductive consequences. Hence $S ⊬ \neg G$ exactly when $S + Con_{S}$ is consistent. Consistency of

S

alone need not ensure this stronger consistency condition.

Past exam of the model theory and Gödel's incompleteness theorems course of the University of Oxford / 2009 / A / 1 / d / Solution 2026-10-31

For unrestricted consistency self-strengthening is inconsistent, again take

A = S + \neg Con_{S}

. In

S^{++}

the axiom

Con_{S}

and the preceding implication prove

Con_{A}

. The additional rule then proves

Con_{A + Con_{A}}

However consistency of an extension implies consistency of its subtheory by the formalized inclusion of proofs:

S ⊢ Con_{A} \to Con_{S}

. Thus the ordinary theory

A + Con_{A}

contains

\neg Con_{S}

and derives

Con_{S}

, so it has a finite contradiction proof. Checking that concrete proof in arithmetic gives

S ⊢ \neg Con_{A + Con_{A}}

. This theorem is also present in

S^{++}

, contradicting the result of the new rule. Therefore $S^{++}$ is inconsistent.

The error in the added rule is its unrestricted internal use on arbitrary arithmetically presented systems, including systems containing false consistency assertions. A true external consistency progression for a specified sound system is not this unrestricted inference rule.

Past exam of the model theory and Gödel's incompleteness theorems course of the University of Oxford / 2009 / A / 2 / a / Solution 2026-10-31

For separation of number sets in an arithmetic theory, a formula separates

A

from

B

S

n \in A

implies

S ⊢ F (\overline{n})

, and

n \in B

implies

S ⊢ \neg F (\overline{n})

. No demand is made for numbers outside their union. Consistency forces disjointness when such a separator exists.

Let

r

be the code of the fixed-point sentence. If

S ⊢ R

, then

r

is in the provable set, so separation gives

S ⊢ H (\overline{r})

. The fixed-point equivalence gives its negation, a contradiction. If

S ⊢ \neg R

, then

r

is in the refutable set, so separation gives

S ⊢ \neg H (\overline{r})

, and the fixed-point equivalence gives

S ⊢ R

, again a contradiction. Thus

S ⊬ R and S ⊬ \neg R .

The argument is syntactic: it needs only the separating implications and consistency, not omega-consistency or general soundness.

Past exam of the model theory and Gödel's incompleteness theorems course of the University of Oxford / 2009 / A / 3 / d / Solution 2026-10-31

The fixed point of boxed negated provability gives, in fact, that GL proves

F \leftrightarrow □ ⊥

. Here is a direct derivation. Since

F

is provably equivalent to a boxed formula, positive introspection and necessitation of that equivalence give

F \to □ F

. Another introspection step gives

F \to □□ F

. The stipulated fixed point also gives

F \to □ \neg □ F

. Combining these two boxed contradictories yields

F \to □ ⊥

. Conversely

□ ⊥ \to □ \neg □ F

follows by normal boxed explosion, and the fixed point then yields

□ ⊥ \to F

Arithmetical soundness of GL, proved by translating distribution into the second derivability condition, necessitation into the first, and Löb's axiom into its internal arithmetic theorem, now gives

PA ⊢ F^{*} \leftrightarrow Pr_{PA} (┌ ⊥ ┐) .

Any remaining propositional parameters may be assigned arbitrary arithmetic sentences; this equivalence is uniform in those assignments. If PA is one-consistent, it is consistent and

Σ_{1}

sound. Its standard provability predicate for a contradiction is then a false

Σ_{1}

sentence and cannot be a theorem. Hence $F^{*}$ is not provable in PA. Moreover its negation is equivalent to PA's consistency statement and is unprovable by the second incompleteness theorem. Thus this fixed point is, under the stated hypothesis, undecidable rather than an unboxed reflection principle.

Past exam of the model theory and Gödel's incompleteness theorems course of the University of Oxford / 2009 / B / 4 / c / ii / β / Solution 2026-10-31

B ⊨ T

and

B \subseteq A^{*} \equiv A

, every universal sentence of

A

holds in

A^{*}

and is preserved by passage to the substructure

B

. Thus

B

witnesses satisfiability of the joint theory.

For universal theory and substructures of elementarily equivalent models, conversely let

B

satisfy

T

together with the universal theory of

A

. Every existential sentence true in

B

must be true in

A

: if an existential

σ

were false in

A

, its universal negation would belong to

Th_{\forall} (A)

and hold in

B

, a contradiction. Hence

Th_{\exists} (B) \subseteq Th (A)

. In particular

Th (A) \cup Th_{\exists} (B)

is satisfiable, with model

A

. Apply the preceding embedding result to the theory

Th (A)

and structure

B

. It supplies a structure

A^{*} ⊨ Th (A)

containing

B

. Therefore $A^{*} \equiv A$ and $B \subseteq A^{*}$ , as required. Elementary equivalence here does not assert that

A

itself contains a copy of

B

Past exam of the model theory and Gödel's incompleteness theorems course of the University of Oxford / 2009 / B / 4 / d / Solution 2026-10-31

Existential sentences are preserved under extensions: witnesses remain present and quantifier-free truth is unchanged by an embedding. Therefore any existential axiomatization implies the specified extension-closure property.

For existential axiomatization from extension closure, let

E

be all existential consequences of

T

, and suppose

A ⊨ E

. We claim that

T \cup Th_{\forall} (A)

is satisfiable. Otherwise compactness supplies finitely many universal sentences

θ_{1}, \dots, θ_{m}

true in

A

such that

T ⊨ \neg (θ_{1} \land \dots \land θ_{m})

. The finite conjunction is universal after renaming bound variables, so its negation is existential. It belongs to

E

but is false in

A

, a contradiction.

The preceding universal-theory embedding result now yields

B ⊨ T

and

A^{*} \equiv A

with

B \subseteq A^{*}

. By the assumed extension closure,

A^{*} ⊨ T

; elementary equivalence implies

A ⊨ T

. Thus

E

axiomatizes exactly the models of

T

, proving existential axiomatizability is equivalent to closure under extensions. This includes the inconsistent case, whose empty model class is axiomatized by an existential contradiction.

Past exam of the model theory and Gödel's incompleteness theorems course of the University of Oxford / 2009 / B / 5 / a / Solution 2026-10-31

A theory is

κ

-categorical if it has a model of cardinality

κ

and any two models of that cardinality are isomorphic. The existence clause prevents vacuous categoricity at cardinalities where there is no model.

To show that infinite-cardinal categoricity forces completeness, suppose

T

were incomplete. Then some sentence

σ

has both

T + σ

and

T + \neg σ

satisfiable, by completeness of first-order logic. Their models must be infinite because

T

has no finite models. For an infinite

κ \geq ∣ L ∣

, the upward and downward Löwenheim-Skolem theorems produce a model of each extension of cardinality exactly

κ

: pass to an elementary submodel of size at most

max (∣ L ∣, ℵ_{0})

and then to the desired size, or reduce a larger model to that size. Categoricity makes these two models isomorphic, but one satisfies

σ

and the other its negation. Contradiction. Therefore $T$ is complete. If language size is defined without automatically including

ℵ_{0}

, the infinite-cardinal qualification is explicit; the assumed existence of a

κ

-sized model and absence of finite models already force it here.

Past exam of the model theory and Gödel's incompleteness theorems course of the University of Oxford / 2009 / B / 5 / b / Solution 2026-10-31

Let

F_{n}

be the formulas whose free variables are among

x_{1}, \dots, x_{n}

. Define

ϕ E_{n} (T) ψ

when

T ⊨ \forall \overset{x}{ˉ} (ϕ \leftrightarrow ψ)

. The n-variable Lindenbaum algebra quotient is a Boolean algebra, with meet, join and complement induced by conjunction, disjunction and negation; its zero and one are contradiction and truth classes. Its order is

[ϕ] \leq [ψ]

exactly when

T ⊨ \forall \overset{x}{ˉ} (ϕ \to ψ)

The Ryll-Nardzewski theorem states that for a complete theory with infinite models in a countable language, the following are equivalent: the theory is countably categorical; for every finite

n

there are only finitely many complete

n

-types over the empty set; and every Boolean algebra

F_{n} / E_{n} (T)

is finite. Equivalently each such type is isolated. The countable-language and completeness hypotheses are essential parts of this formulation.

Past exam of the model theory and Gödel's incompleteness theorems course of the University of Oxford / 2009 / B / 5 / c / iv / Solution 2026-10-31

No. If a finite conjunction

ϕ

axiomatized the infinite spaces, it would follow from the vector-space axioms together with all the

D_{n}

. Compactness would make it follow from finitely many of those axioms and dimension requirements. Let

N

be at least every dimension mentioned there. The finite-dimensional space

K^{N}

satisfies all vector-space axioms and those finitely many

D_{n}

, hence would satisfy

ϕ

. But it is a finite set and so cannot be a model of the purported axiomatization. This contradiction proves failure of finite axiomatizability, including finite axiomatizability relative to vector spaces.

Past exam of the model theory and Gödel's incompleteness theorems course of the University of Oxford / 2009 / B / 6 / b / Solution 2026-10-31

A model is prime if it has an elementary embedding into every model of

T

. A model is atomic if every finite tuple realizes an isolated complete type over the empty set.

A

is prime, the downward Löwenheim-Skolem theorem gives a countable model

M

T

, and an elementary embedding

A \to M

makes

A

countable. If a tuple of

A

realized a nonisolated type, the omitting types theorem would give a model of

T

omitting that type. The prime embedding into it would preserve the tuple's full type and realize the omitted type, a contradiction. Thus a prime model is countable and atomic.

Conversely enumerate a countable atomic model as

a_{0}, a_{1}, \dots

, and take any

B ⊨ T

. Build finite partial elementary maps into

B

. Suppose the current tuple

\overset{a}{ˉ}

maps to

\overset{ˉ}{b}

. Choose a formula

θ (\overset{x}{ˉ}, y)

isolating the type of

(\overset{a}{ˉ}, a_{n})

. The assertion

\exists y θ (\overset{a}{ˉ}, y)

holds in

A

, so its counterpart holds at

\overset{ˉ}{b}

B

by partial elementarity. Choose a witness

b_{n}

. Since

θ

isolates the entire extended tuple type, the extended map is elementary on its finite domain; its equality formulas ensure distinct Elements remain distinct. The union is a total elementary embedding

A \to B

. The empty initial tuple works because

T

is complete. Hence

A is prime ⟺ A is countable and atomic .

Past exam of the year 1 of the computer science course of the University of Oxford / 2014 / Digital systems, linear algebra, introduction to formal proof / 7 / c / iii / Solution 2026-10-31

The premise turns an assumed

\neg P

into a contradiction. Use the negation introduction rule to obtain

\neg\neg P

, then the supplied classical double-negation elimination rule:

Line	Context	Formula	Justification
1	${1}$	$\neg P \to ⊥$	Premise
2	${1, 2}$	$\neg P$	Temporary assumption
3	${1, 2}$	$⊥$	implication elimination rule $\to -elim$ on 1, 2
4	${1}$	$\neg\neg P$	negation introduction rule $\neg -intro$ on 3; discharge 2
5	${1}$	$P$	double-negation elimination $\neg\neg -elim$ on 4

Hence

(\neg P \to ⊥) ⊢ P

. The final classical rule is essential: deriving

\neg\neg P

alone is not yet a proof of

P

in Intuitionistic propositional logic.

Past exam of the year 1 of the computer science course of the University of Oxford / 2014 / Digital systems, linear algebra, introduction to formal proof / 7 / c / iv / Solution 2026-10-31

Assuming

\neg P

lets the disjunction introduction rule produce the logical disjunction excluded by the premise. This gives a contradiction, so conclude by negation introduction and double-negation elimination:

Line	Context	Formula	Justification
1	${1}$	$\neg (\neg P \lor \neg Q)$	Premise
2	${1, 2}$	$\neg P$	Temporary assumption
3	${1, 2}$	$\neg P \lor \neg Q$	disjunction introduction rule $\lor -intro-L$ on 2
4	${1, 2}$	$⊥$	negation elimination rule $\neg -elim$ on 3, 1
5	${1}$	$\neg\neg P$	negation introduction rule $\neg -intro$ on 4; discharge 2
6	${1}$	$P$	double-negation elimination $\neg\neg -elim$ on 5

Thus

\neg (\neg P \lor \neg Q) ⊢ P .

No separate result about

Q

or use of a derived De Morgan rule is needed.

Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Digital systems, linear algebra, introduction to formal proof / 8 / c / Solution 2026-10-25

Assume the existential statement and eliminate it using a fresh witness

u

. Let

A

abbreviate

R (u, u)

. Instantiating its universal quantification at the witness itself gives two incompatible conditional requirements.

1  | ∃x ∀y ((R(x,y) → ¬R(y,y)) ∧ (¬R(y,y) → R(y,y)))     assume
2  | | ∀y ((R(u,y) → ¬R(y,y)) ∧ (¬R(y,y) → R(y,y)))        fresh witness
3  | | (A → ¬A) ∧ (¬A → A)                                 ∀-elim, 2, y := u
4  | | A → ¬A                                               ∧-elim-L, 3
5  | | ¬A → A                                               ∧-elim-R, 3
6  | | | A                                                  assume
7  | | | ¬A                                                 →-elim, 6 and 4
8  | | | ⊥                                                  ¬-elim, 6 and 7
9  | | ¬A                                                   ¬-intro, discharge 6–8
10 | | A                                                    →-elim, 9 and 5
11 | | ⊥                                                    ¬-elim, 10 and 9
12 | ⊥                                                      ∃-elim, 1, subproof 2–11
13 ¬∃x ∀y ((R(x,y) → ¬R(y,y)) ∧ (¬R(y,y) → R(y,y)))        ¬-intro, discharge 1–12

The contradiction at line 12 contains no free witness variable, so the existential quantifier elimination rule meets its eigenvariable condition. This argument is already valid without classical first-order logic double-negation elimination.

Past exam of the year 1 of the computer science course of the University of Oxford / 2021 / Digital systems, linear algebra, introduction to formal proof / 7 / b / i / Solution 2026-10-25

Let

N (ϕ, ψ)

abbreviate

\neg ϕ \land \neg ψ

. Suitable rules are

\frac{\neg ϕ \neg ψ}{N ( ϕ , ψ )} N I, \frac{N ( ϕ , ψ )}{\neg ϕ} N E_{L}, \frac{N ( ϕ , ψ )}{\neg ψ} N E_{R} .

Their derivations require no new axiom. For introduction, apply the conjunction introduction rule to the two premises, obtaining

\neg ϕ \land \neg ψ

, which is exactly the abbreviation. For left and right elimination, unfold the abbreviation and apply the corresponding conjunction elimination rule. These are therefore derived rules for the NOR connective.

An equivalent subproof presentation of introduction permits one subproof from assumption

ϕ

⊥

and another from assumption

ψ

⊥

: use the negation introduction rule to discharge each assumption, then apply the conjunction introduction rule. If elimination is desired in contradiction form, combine the left projection with

ϕ

, or the right projection with

ψ

, using the negation elimination rule. All these presentations reduce directly to the permitted rules.

Proof by minimal counterexample 2026-10-31

Assume that a proposed statement fails for some natural number, and choose the least failing index using the well-ordering principle. Prove the necessary base cases and reduce the least failure to a smaller index that must satisfy the statement. If that smaller case implies the alleged failure also satisfies it, there is a contradiction. Reductions by a step larger than one require enough consecutive base cases to cover every possible residue class.

Propositional consistency 2026-10-31

A collection of propositional formulas is consistent if it does not derive a contradiction. Equivalently, in a standard classical calculus, it does not derive both a logical formula and its negation.

Propositional logic Created 2026-09-24 Updated 2026-10-14

Propositional logic studies reasoning with whole propositions represented by letters such as

p

and

q

, and propositional formulas built from them using logical connectives. The classical two-valued, truth-functional presentation below assigns each primitive proposition either true or false: each compound value depends only on its component values. This is classical propositional logic; other systems, including Intuitionistic propositional logic, have different semantics and rules.

Starting with

p

and

q

, one can build

\neg p

by logical negation,

p \land q

by logical conjunction,

p \lor q

by logical disjunction, and

p \to q

by logical implication, then combine these into larger propositional formulas. Parentheses specify the construction:

(p \to q) \land p

has logical conjunction as its outermost logical connective. Primitive propositions have no internal structure in this language; a first-order formula can additionally describe objects and quantify over them. The formation rules are given in forall x: Calgary, chapter 6.

The following truth table shows the four basic logical connectives. Logical disjunction is inclusive:

p \lor q

is true when either or both inputs are true. Classical logical implication has exactly the same values as

\neg p \lor q

; its only false row has

p

true and

q

false. These truth conditions are specified in forall x: Calgary, chapter 9.

Figure 1.
Classical logical connectives on all four Boolean valuations, with matching columns for $p \to q$ and $\neg p \lor q$
. **Details**.

Evaluate

(p \to q) \land p

from the inside out: first the logical implication, then the outer logical conjunction. The final column also tests the logical implication expressing the modus ponens rule:

p T T F F q T F T F p \to q T F T T (p \to q) \land p T F F F ((p \to q) \land p) \to q T T T T

p = T, q = F

, the logical implication is false, so the logical conjunction is false. The abstract syntax tree below separates the construction of the propositional formula from the values propagated through it. Both occurrences of

p

receive the same value from the Boolean valuation. With the opposite assignment

p = F, q = T

, the logical implication is true but the logical conjunction is still false because its second input

p

is false. Recursive evaluation and exhaustive truth tables are explained in forall x: Calgary, chapter 11.

Figure 2.
An abstract syntax tree evaluating $(p \to q) \land p$ to false at the Boolean valuation $p = T, q = F$
. **Details**.

The last column is always true:

((p \to q) \land p) \to q

is a tautology. This expresses the logical validity of the modus ponens rule: whenever both premises

p \to q

and

p

are true,

q

is true. It does not say the premises are true under every Boolean valuation; their logical conjunction is true only on the first row. For a short natural deduction proof outline, assume

(p \to q) \land p

, extract both conjuncts, apply the implication elimination rule to obtain

q

, then discharge the assumption by the implication introduction rule. See forall x: Calgary, sections 17.3–17.5.

Three tests answer different questions: Boolean satisfiability requires at least one true row, a tautology requires every row true, and a contradiction has every row false. Thus

p \lor \neg p

is a tautology,

p \land \neg p

is a contradiction, and

p \land q

is satisfiable but not a tautology: it is true only at

p = q = T

. See forall x: Calgary, sections 12.1–12.4.

Propositional logic is useful because an inference pattern can be checked once and reused for many choices of propositions. Its beauty is that a finite truth table completely settles classical truth-functional questions about a fixed propositional formula. But logical validity does not verify real atomic claims. If

p

means “the switch is closed” and

q

means “the lamp is lit”, a valid use of the modus ponens rule still requires establishing its premises about the apparatus; the truth table supplies neither an observation nor a causal explanation. See forall x: Calgary, chapter 2 and chapter 10.

Reductio ad absurdum Created 2026-10-14 Updated 2026-10-31

A reductio rejects an assumption by deriving an absurd or contradictory consequence from it. In a classical mathematical proof of

P

, assume its negation and derive a contradiction. Constructively the same argument yields the double negation of

P

; removing it requires an additional logical principle. This differs from proving a negation directly by deriving a contradiction from the positive assumption.