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Let $T=\mathrm{ZFC}+\text{“there is a worldly cardinal”}$ and $T^{\ast }=\mathrm{ZFC}+\mathrm{Con}(\mathrm{ZFC})$. A worldly $\kappa$ makes $V_{\kappa }$ a model of ZFC, and arithmetic absoluteness for a rank-initial model makes it a model of $T^{\ast }$, so $T$ proves $\mathrm{Con}(T^{\ast })$. If $T^{\ast }$ proved $\mathrm{Con}(T)$, then the stronger theory $T$ would prove its own consistency, contradicting the Gödel second incompleteness theorem. Thus, assuming consistency, $T^{\ast }{<}_{\mathrm{Cons}}T$.

Starting from a recursively axiomatized arithmetic first-order theory $T_0$ with the usual arithmetical provability predicates, defineIf every stage is first-order consistent, the Gödel second incompleteness theorem makes every step strict in the consistency-strength preorder: the next stage proves the preceding stage's formal consistency statement, which that preceding stage cannot prove. Every formal proof from the effective union $T_{\omega }={\bigcup }_{n<\omega }{T}_n$ uses only finitely many axioms and lies in one stage. Thus the union is first-order consistent under this hypothesis and lies strictly above every finite stage. Arithmetic soundness of $T_0$ is a sufficient starting assumption, since each added formal consistency statement is then true in the standard natural numbers.

Consistency of $T_0$ alone does not ensure consistency of the next stage. Assuming Peano arithmetic is first-order consistent, $T_0=\mathrm{PA}+\neg \mathrm{Con}(\mathrm{PA})$ is first-order consistent by the Gödel second incompleteness theorem. Proof inclusion makes it prove $\neg \mathrm{Con}(T_0)$, so adding $\mathrm{Con}(T_0)$ produces an inconsistent next stage. The hypotheses for a progression must therefore distinguish a consistent starting theory from consistency of all its iterates.

Let $T=S+\{{\mathrm{Con}}_S\}$ and assume $T$ is consistent. The allowed arithmetizations of the previous equivalences giveSuppose $S⊢R\to G$. Its contrapositive is $\neg G\to \neg R$, and the Hilbert-Bernays-Löb derivability conditions then giveCombining with the two internal equivalences yields $S⊢{\mathrm{Con}}_S\to \mathrm{Con}(T)$. Since $T$ contains ${\mathrm{Con}}_S$, it would prove $\mathrm{Con}(T)$, contrary to the given Gödel second incompleteness theorem for this consistent extension. HenceTogether with the preceding implication, this is the Strict Gödel-to-Rosser implication under the stated consistency hypothesis.

Assume $S⊢B_{S}X\to X$ and put $T=S+\{\neg X\}$. Then $T⊢\neg B_{S}X$. The given extension-provability implication, with $Y=\perp$, givesThe provable classical implication $(\neg X\to \perp )\to X$, together with distribution, gives $S⊢B_S(\neg X\to \perp )\to B_{S}X$. HenceIf $T$ were consistent, this would violate its Gödel second incompleteness theorem. Thus $T$ is inconsistent. The deduction theorem for propositional logic and double-negation elimination give $S⊢\neg \neg X$, hence $S⊢X$. Therefore the second incompleteness theorem for $S+\{\neg X\}$ yields Löb's theorem for arbitrary $X$.

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.

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.

The consistency implication needs more than the fact that the finite tree statement is unprovable in Peano arithmetic. Its mechanism is a concrete encoding of ordinal descent by bad finite-tree sequences, combined with the controlled-termination form of Gentzen ordinal reduction for arithmetic.

First define the natural-sum ordinal rank of a finite rooted tree. A leaf has rank zero, and a tree with immediate child subtrees $T_1,\dots ,T_m$ haswhere $\#$ is Hessenberg natural sum. By induction every such rank is below epsilon zero. Conversely, if $\alpha ={\sum }_{j=1}^r{\omega }^{{\alpha }_j}{c}_j$ is in Cantor normal form, represent it by a root with $c_j$ copies of the recursively constructed tree for ${\alpha }_j$. The leaf represents zero. Denote the resulting canonical tree by $\tau (\alpha )$ and its vertex count by $\Vert \alpha \Vert$.

We claim that a rooted-tree homeomorphic embedding $T⪯U$ implies $o(T)\le o(U)$. If the roots match, the source child subtrees embed into distinct target branches. By induction their ranks do not decrease; exponentiation and Hessenberg natural sum preserve these comparisons, and additional target branches only increase the sum. If the source root maps below the host root, its entire image lies in an immediate subtree $V$ of $U$. Induction gives $o(T)\le o(V)$, whileThis also holds when $o(V)=0$. Thus the claim follows by induction on the host tree. In particular,

Consequently Friedman's finite form of Kruskal's theorem supplies slow well-ordering below epsilon zero: for each $k$, descending ordinal sequences with $\Vert {\alpha }_i\Vert \le k+i$ have uniformly bounded length. Otherwise their canonical trees would be arbitrarily long admissible bad sequences. The notation system and its comparisons are primitive recursive, and there are only finitely many notations of bounded norm.

The proof-theoretic input is the following precise miniaturization lemma: for the standard effective Cantor normal form notation system below ${\epsilon }_0$, slow well-ordering with linear norm control implies the formal consistency statement for Peano arithmetic. The Friedman–Smith controlled-termination theorem, stated in its general ordinal-analysis form in Section 2.2 of formal.hknu.ac.kr/Publi/kruskalJKMS.pdf , supplies this implication over arithmetical comprehension theory ${\mathrm{ACA}}_0$, whose first-order part is conservative over Peano arithmetic. This is an established proof-theoretic lemma, separate from the elementary tree-rank calculation above.

Here is why that lemma is a consistency result. Gentzen ordinal reduction for arithmetic assigns an alleged arithmetic contradiction an effective ordinal below ${\epsilon }_0$. Reducing cuts and induction inferences preserves the contradictory end-sequent while strictly decreasing the assigned ordinal. A terminal normal derivation of a contradiction is excluded by inspection of the rules. The miniaturization argument converts this reduction/induction analysis into finite controlled descents: a contradictory proof would yield arbitrarily long descending notation sequences with a common initial allowance and linear norm control. Slow well-ordering excludes these. The control conversion is substantive; ordinary proof reductions must not simply be asserted to have linear-size growth. Gentzen's ordinal-reduction framework is discussed in math.stanford.edu/~feferman/papers/reductive.pdf .

Combining the coding with that precise termination lemma givesThus the finite combinatorial principle proves that no PA proof of a contradiction exists in the specified arithmetic metatheory. If consistent Peano arithmetic proved FFF, it would also prove its own formal consistency statement: the displayed implication is arithmetical and the first-order conservativity of ${\mathrm{ACA}}_0$ transfers it to PA. This contradicts the Gödel second incompleteness theorem. It is the consistency implication, rather than unprovability by itself, that explains the strength of the finite tree principle.

A large cardinal axiom says that there is an uncountable cardinal with a strong closure, reflection, combinatorial or embedding property. Its interest is not merely the magnitude of the cardinal: it packages a universe-like portion of set theory, or a coherent way of comparing the universe with a proper inner model. Such axioms form a hierarchy extending ordinary ZFC, compared through relative consistency implications. The consistency qualifications are essential; an existence axiom is not a theorem of ZFC simply because its consequences are attractive.

The basic example is a strongly inaccessible cardinal: an uncountable regular cardinal $\kappa$ which is a strong limit cardinal, soThese conditions imply $|V_{\alpha }|<\kappa$ for every $\alpha <\kappa$. Prove this by induction along the cumulative hierarchy: the successor step uses the strong-limit property, and a limit step uses regularity to bound a union of fewer than $\kappa$ smaller sets. Thus every member of $V_{\kappa }$ has cardinality below $\kappa$.

Consequently $V_{\kappa }$ is a transitive model of ZFC. Pairing, union, power set, separation, axiom of infinity and foundation stay within the hierarchy. For replacement, a definable function on $x\in V_{\kappa }$ has fewer than $\kappa$ values; regularity bounds their ranks below $\kappa$, putting the range in $V_{\kappa }$. Ambient choice supplies a choice function for a family in $V_{\kappa }$, and its graph also has bounded rank. This verifies the substantial closure axiom rather than only saying that the rank is large. It also shows why inaccessible existence cannot be proved in a consistent ZFC: it would yield a set first-order model of ZFC and hence its consistency, contrary to the Gödel second incompleteness theorem.

An uncountable regular limit cardinal is a weakly inaccessible cardinal; requiring the strong-limit inequality is stronger without additional hypotheses. Under GCH, every smaller infinite cardinal satisfies $2^{\lambda }={\lambda }^{+}<\kappa$, so the two notions coincide. This distinction separates cardinal-arithmetic assumptions from the regularity requirement.

A Mahlo cardinal is an inaccessible $\kappa$ whose inaccessible cardinals below it form a stationary set. Thus every club set meets those inaccessibles, which strengthens mere unboundedness. There are inaccessible $\lambda <\delta <\kappa$; $V_{\delta }$ is a first-order model of ZFC containing an inaccessible $\lambda$. Mahloness therefore supplies a consistency statement stronger than the bare existence of one inaccessible. Iterating the demand that smaller cardinals have the same reflection property leads to further levels of the hierarchy.

A weakly compact cardinal is a logical and combinatorial strengthening of inaccessibility. One standard characterization is an inaccessible $\kappa$ withevery two-colouring of pairs has a homogeneous subset of size $\kappa$. Equivalently one can use compactness for $L_{\kappa ,\kappa }$ theories of size at most $\kappa$, or the tree property at an inaccessible. These viewpoints explain the terminology: a local collection of compatible requirements has a global realization, and tall narrow trees cannot avoid cofinal branches. They are characterization statements; the elementary closure and measure arguments below are separate proofs of the consequences used here.

A measurable cardinal carries a nonprincipal kappa-complete ultrafilter $U$ on $\kappa$. Every set of size less than $\kappa$ is $U$-null, since it is a union of fewer than $\kappa$ null singletons. Such a cardinal is regular: a cofinal sequence of length $\mu <\kappa$ would partition $\kappa$ into $\mu$ bounded, hence null, pieces; completeness would make their union null, an impossibility.

It is also strong limit. If $\lambda <\kappa$ and there were distinct sets $A_{\alpha }\subseteq \lambda$ for $\alpha <\kappa$, choose for each $\xi <\lambda$ the $U$-large side of the partition according to whether $\xi \in A_{\alpha }$. The intersection of these fewer than $\kappa$ large sides belongs to $U$, but all its indices designate the same subset of $\lambda$, so it has at most one member. This contradicts nonprincipality. Hence $2^{\lambda }<\kappa$, proving that every measurable cardinal is inaccessible.

The measure gives an ultrapower embedding. Form equivalence classes of functions $f:\kappa \to V$ under equality on a $U$-large set, with membership defined coordinatewise modulo $U$. Induction on first-order formulas proves elementarity: Boolean operations use the ultrafilter laws, and the existential step chooses coordinate witnesses on a large set. Countable completeness makes the ultrapower well-founded. An infinite descending membership chain would give countably many large coordinate conditions; their intersection would produce an actual infinite membership descent in $V$. Collapse the ultrapower to a transitive class $M$, obtaining $j:V\to M$.

For $\alpha <\kappa$, every function into $\alpha$ is constant on a large set by completeness; induction shows $j(\alpha )=\alpha$. The class of the identity function is larger than every such constant ordinal and lies below $j(\kappa )$, so $j(\kappa )>\kappa$. Thus the critical point of an elementary embedding is $\kappa$. The measure viewpoint has become a precise self-similarity principle for the universe.

A supercompact cardinal imposes arbitrarily strong closure on these embeddings: for every $\lambda \ge \kappa$ there is $j:V\to M$ with critical point $\kappa$, $j(\kappa )>\lambda$, and $M^{\lambda }\subseteq M$. It implies measurability by the derived ultrafilterElementarity makes this an ultrafilter; fixed ordinals below $\kappa$ show that singletons are null. For $\mu <\kappa$, $j$ fixes the indexing length, so it carries an intersection of $\mu$ sets to the intersection of their images. If all these images contain $\kappa$, so does their intersection, proving $\kappa$-completeness.

These examples illustrate three connected uses of large cardinals: obtaining set first-order models and stronger consistency statements, extending finite or countable compactness and partition phenomena to uncountable domains, and constructing elementary embeddings with controlled closure. They also interact with inner models: an inaccessible remains inaccessible in the constructible universe, since it cannot acquire a new short cofinal sequence there and there are fewer subsets of smaller ordinals. A measurable cardinal's measure need not belong to that inner model, so preservation of the stronger property is a different issue. Large cardinal axioms are thus structured additional hypotheses, with carefully distinguishable consequences and relative-consistency claims, rather than a single claim that sufficiently big cardinals exist.

Here is the first incompleteness theorem in its modern consistency form: every consistent computably enumerable theory $T$ extending Robinson arithmetic is incomplete. The original Gödel sentence proof assumes omega-consistency to obtain both unprovabilities; we give that proof and the strengthening that removes this assumption. We also give the second incompleteness conclusion for theories extending Peano arithmetic.

First arithmetize syntax. Code strings and finite derivations by natural numbers. If the axioms of $T$ are merely computably enumerable, include the finite enumeration certificate for each nonlogical axiom used in a proof. The resulting proof-checking relation ${\mathrm{Prf}}_T(p,q)$ is primitive recursive. Finite strings and computation histories are coded by finite sequences, with bounded local verification of every step; these give arithmetic formulas representing the coding operations. In Robinson arithmetic every true or false instance with fixed numeral inputs is provable with the appropriate sign. Set

The diagonal lemma gives, for any formula $\theta (x)$, a sentence $D$ with $Q⊢D\leftrightarrow \theta (┌D┐)$. To see the construction, let $d(e)$ be the computable syntactic operation substituting the numeral for $e$ into the formula coded by $e$. Form the formula $\delta (x)=\theta (d(x))$, interpreting the displayed coding function through its arithmetic representation, and let $n=┌\delta ┐$. Then $D=\delta (\bar{n})$, and $d(n)=┌D┐$. The representation proves that fixed computation, giving the asserted equivalence.

Apply the diagonal lemma to $\neg {\mathrm{Prov}}_T(x)$, obtainingIf $T⊢G$, an actual finite proof has code $n$, so $Q$ proves its proof-checking instance and $T$ proves ${\mathrm{Prov}}_T(┌G┐)$. The fixed-point equivalence gives its negation, contradicting consistency. Hence $T⊬G$. For each standard numeral $n$, $Q$ proves $\neg {\mathrm{Prf}}_T(\bar{n},┌G┐)$ because there is no such proof. If $T⊢\neg G$, the equivalence gives $T⊢\exists p{\mathrm{Prf}}_T(p,┌G┐)$, contrary to omega-consistency. Thus the original proof gives two unprovabilities under that hypothesis. It also shows that $G$ is true in the standard natural numbers when $T$ is consistent.

For consistency alone use the Rosser sentenceIf $T⊢R$ with code $n$, consistency excludes every proof of $\neg R$. In particular the finitely many codes $q\le n$ all fail their primitive recursive proof tests. Robinson arithmetic verifies those finitely many failures and the actual proof at $n$, and hence proves $\neg R$, a contradiction. If $T⊢\neg R$ with code $m$, consistency excludes every proof of $R$. For $p<m$, Robinson arithmetic verifies all the finitely many failed proof tests. For $p\ge m$, it verifies that $q=m$ is a proof of $\neg R$ with $q\le p$. The elementary arithmetic dichotomy between $p<m$ and $m\le p$ therefore proves the entire displayed condition, hence $R$, again a contradiction. Thus

For the Gödel second incompleteness theorem, now assume $T\supseteq \mathsf{PA}$ and use its ordinary certified proof predicate. PA formalizes proof concatenation and the verification of a coded proof, giving the derivability conditions: if $T⊢A$ then $T⊢{\mathrm{Prov}}_T(┌A┐)$; and inside $T$,The first internal implication is certified by combining two proofs; the second is certified by producing the arithmetic proof that a given finite proof passes its check, with existential introduction. Apply them to $G\to \neg {\mathrm{Prov}}_T(┌G┐)$. From ${\mathrm{Prov}}_T(┌G┐)$, the two internal implications yield provability of both ${\mathrm{Prov}}_T(┌G┐)$ and its negation, and hence provability of $\perp$. ConsequentlyA proof of this consistency sentence in a consistent $T$ would prove $G$, which the first argument excludes. Therefore $\boxed{T⊬\mathrm{Con}(T)}$. This concerns the standard arithmetic consistency sentence and the stated effective, arithmetic hypotheses; it is not a prohibition on proving consistency in a stronger theory.

Use the following form of the omitting types theorem. Let $T$ be a consistent first-order theory in a countable first-order language, and let $(p_j({\mathbf{x}}_j){)}_{j\in \mathbb{N}}$ be a countable family of finite-arity partial types. Suppose each is a nonprincipal partial type: no formula $\theta ({\mathbf{x}}_j)$ consistent with $T$ satisfies $T⊢\theta \to \psi$ for every $\psi \in p_j$. Then $T$ has a countable model omitting all the $p_j$.

For the proof, add countably many fresh constants and perform a Henkin construction. Enumerate three kinds of tasks: deciding each sentence, providing a fresh constant witness for each existential sentence, and making every constant tuple omit each $p_j$ of the corresponding arity. Starting with the empty set, keep a finite set ${\Gamma }_s$ such that $T\cup {\Gamma }_s$ is consistent. A sentence can be decided in one of its two ways while preserving consistency. For an existential sentence, add $\exists x\alpha (x)\to \alpha (c)$ with $c$ fresh; inconsistency would contradict freshness and the existential sentence itself.

For an omission task at a tuple $\mathbf{c}$, let $\gamma$ be the conjunction of ${\Gamma }_s$. Replace its constants by variables, identify the occurrences of the tuple's constants with the corresponding variables $\mathbf{x}$, and existentially quantify the other variables. If the tuple repeats a constant, include the corresponding equalities between its variables. This gives $\theta (\mathbf{x})$ with $T\cup \{\exists \mathbf{x}\theta \}$ consistent. Nonprincipality supplies $\psi \in p_j$ such thatis consistent. Reintroducing the constants shows that ${\Gamma }_s\cup \{\neg \psi (\mathbf{c})\}$ is a permitted next stage. Thus each constant tuple will fail some formula of each prescribed type.

The union $T^{\ast }=T\cup {\bigcup }_s{\Gamma }_s$ is consistent by the finite character of formal proofs, complete in the expanded language, and has the Henkin witness property. Its term model has closed terms modulo provable equality as elements, and satisfies the truth lemma for a Henkin term model, proved by induction on formulas. It is countable. Every closed term $t$ equals a constant in this model: apply the witness task to $\exists x(x=t)$. Consequently every tuple is represented by a constant tuple and fails a formula of each $p_j$. This proves simultaneous omission.

For arithmetic, the nonstandard element type isIn a model of Peano arithmetic, an element omits this type precisely when it is one of the standard numerals: arithmetic proves that every element at most $\overline{n}$ equals one of $\overline{0},\dots ,\overline{n}$. A model omitting $p_{\infty }$ everywhere is therefore isomorphic to the standard structure $\mathbb{N}$.

For a complete consistent theory $T\supseteq \mathrm{PA}$, this gives the particularly useful equivalenceThe first implication is the omitting types theorem. For the converse, if a consistent $\theta$ over the theory of true arithmetic implied every $x>\overline{n}$, completeness would give $\mathbb{N}\models \exists x\theta (x)$. A standard witness $m$ contradicts the implication $\theta \to x>\overline{m}$.

Completeness matters in this formulation. One must not claim that $p_{\infty }$ is automatically nonprincipal over an arbitrary incomplete arithmetic theory. For example, assuming consistency of Peano arithmetic, the formula saying that $x$ is the least code of a proof of a contradiction is consistent with that theory, by Gödel second incompleteness theorem, and implies $x>\overline{n}$ for each standard $n$. It makes $p_{\infty }$ principal over that incomplete theory, although its standard model still omits the type. The theorem is a sufficient omission criterion, not a necessary one for arbitrary partial types over incomplete theories.

WriteIf $\kappa$ is a worldly cardinal, then $V_{\kappa }\models \mathrm{ZFC}$. The existence of this set model proves $\mathrm{Con}(\mathrm{ZFC})$ in the universe. By arithmetic absoluteness for a rank-initial model, the same formal consistency statement holds in $V_{\kappa }$. Hence $V_{\kappa }\models T^{\ast }$, so $T$ proves $\mathrm{Con}(T^{\ast })$.

Conversely, suppose $T^{\ast }$ proved $\mathrm{Con}(T)$. The theory $T$ proves every axiom of $T^{\ast }$, since a worldly cardinal proves $\mathrm{Con}(\mathrm{ZFC})$. It would therefore also prove $\mathrm{Con}(T)$, contrary to the Gödel second incompleteness theorem when $T$ is consistent. Thus $T^{\ast }$ cannot prove $\mathrm{Con}(T)$, andThis is the consistency strength of a worldly cardinal comparison.

Let $\mathrm{IC}$ be the sentence asserting that a strongly inaccessible cardinal exists, and begin withDefine the iterated consistency progressionThe construction is effective, so every $T_n$ and $T_{\infty }$ is a recursively axiomatized first-order theory extending ZFC.

Because $T_{n+1}$ extends $T_n$, every theorem of $T_n$, including every formal consistency statement it proves, is a theorem of $T_{n+1}$; hence $T_n{\le }_{\mathrm{Cons}}{T}_{n+1}$. The theory $T_{n+1}$ proves $\mathrm{Con}(T_n)$ by construction, whereas a consistent $T_n$ cannot prove its own consistency by Gödel second incompleteness theorem. ThereforeLikewise $T_{\infty }$ extends every $T_n$ and contains $\mathrm{Con}(T_n)$ as an axiom already at stage $n+1$, while $T_n$ does not prove it. Consequentlyassuming the stated consistency hypotheses.

This University of Oxford Part C paper is cross-listed for Mathematics and Mathematics and Philosophy. Section A concerns Gödel second incompleteness theorem and provability logic; section B concerns model theory. ^[ref] The archive contains the complete editable four-page transcription. Bounded official-source searches found no advertised answers to this exact paper. All 31 source leaves are answered below. The enumeration and function-definability conventions are stated explicitly, and the canonical-model construction uses the function-free setting permitted by the source.

This University of Oxford Part C paper is cross-listed for Mathematics and Mathematics and Philosophy. Section A concerns Gödel second incompleteness theorem and provability logic; section B concerns model theory. ^[ref] The archive contains a complete editable six-page transcription. Bounded official-source searches found no advertised solutions to this exact paper. All 37 source leaves are answered below. The solutions state the proof-numbering convention in A2(d), the naming convention for diagram inclusions in B6(ii), and the literal printed stability definition in B8(ii)(e).

The literal semantic hypothesis is insufficient: a semantic Gödel fixed point need not be canonical. A sentence can be false and provable in a consistent, unsound arithmetic theory, in which case both sides of the stipulated semantic biconditional can be false.

For a concrete counterexample, assume PA is consistent and set $S=\mathrm{PA}+\neg {\mathrm{Con}}_{\mathrm{PA}}$. Gödel's second incompleteness theorem makes this extension consistent: otherwise PA would prove its own consistency. It is still bounded-complete. Use ordinary numeralwise proof and refutation enumerators for this recursively axiomatized theory, and take $G=\neg {\mathrm{Con}}_{\mathrm{PA}}$. This $G$ is false in the standard natural numbers but is an axiom of $S$. Since it is provable, $\exists yP(┌G┐,y)$ is true; its negation is false. Thus the semantic fixed-point biconditional required in this part is true, with both sides false.

Choose $R$ by the usual canonical Rosser construction for these proof matrices. Its undecidability follows from the preceding parts, and it is a true ${\Pi }_1$ sentence: if it were false, its true ${\Sigma }_1$ negation would be provable in this bounded-complete theory, contradicting undecidability. Hence $G$ is false and $R$ true, so their proposed semantic equivalence is false. This counterexample meets the literal assumptions and does not exploit a missing source page.

Under the intended canonical ${\Pi }_1$ Gödel/Rosser convention, the expected answer does hold. Consistency plus bounded completeness implies ${\Pi }_1$ soundness: a false ${\Pi }_1$ theorem has a standard bounded counterexample whose negation the theory proves. If canonical $G$ were false, its truth fixed point would make it provable, contradicting this soundness; thus $G$ is true. Canonical $R$ is true by the preceding ${\Sigma }_1$-negation argument, so both canonical sentences are true and their equivalence is true. This last conclusion needs that canonical construction or an equivalent additional restriction; truth of a biconditional alone does not supply it.

Use the two formal consistency equivalences expressly supplied for this part. If $S⊢G\leftrightarrow R$, the derivability conditions give $S⊢\square \neg G\leftrightarrow \square \neg R$. Taking negations and applying those supplied equivalences yieldsOne-consistency implies ${\Sigma }_1$ soundness under bounded completeness. Hence $S+{\mathrm{Con}}_S$ is consistent: if not, $S$ would prove $\neg {\mathrm{Con}}_S$, a false ${\Sigma }_1$ sentence. But in that consistent extension the last equivalence and its consistency axiom prove its own consistency ${\mathrm{Con}}_{S+{\mathrm{Con}}_S}$. Gödel's second incompleteness theorem rules this out. ThereforeThis establishes the requested unprovability using the explicit formal hypotheses, without inferring internal provability from the semantic equivalence discussed in the preceding part. The hypotheses for the second theorem are the standard effective arithmetic presentation and adequate provability predicates stipulated here.

Gödel's second incompleteness theorem says that a consistent effectively presented arithmetic theory with an adequate proof predicate cannot prove its formal consistency statement. The background strength must support the diagonal lemma and the derivability conditions, as in the present arithmetic setting.

Löb's theorem saysIt is a theorem about derivability for every sentence $A$, not merely an external soundness assertion about the interpretation of $\square$. Taking $A=\perp$, a proof of consistency gives $S⊢\square \perp \to \perp$; Löb then gives $S⊢\perp$. Thus a consistent $S$ cannot prove consistency. This is how the second incompleteness theorem is the contradiction-sentence instance of Löb's more general result.

The fixed point of boxed negated provability gives, in fact, that GL proves $F\leftrightarrow \square \perp$. Here is a direct derivation. Since $F$ is provably equivalent to a boxed formula, positive introspection and necessitation of that equivalence give $F\to \square F$. Another introspection step gives $F\to \square \square F$. The stipulated fixed point also gives $F\to \square \neg \square F$. Combining these two boxed contradictories yields $F\to \square \perp$. Conversely $\square \perp \to \square \neg \square F$ follows by normal boxed explosion, and the fixed point then yields $\square \perp \to F$.

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 givesAny 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 ${\Sigma }_1$ sound. Its standard provability predicate for a contradiction is then a false ${\Sigma }_1$ sentence and cannot be a theorem. Hence $F^{\ast }$ 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.

Source release: Oxford examination archive release thread, Part C bundle pages 300–304. This is one combined paper covering both courses: section A concerns Gödel second incompleteness theorem and provability logic, and section B concerns model theory. The cover cross-lists Mathematics and Mathematics and Philosophy, both Part C, Thursday 10 June 2010, 2.30pm–5.30pm.

Answer availability checked on 28 October 2026: bounded official Oxford searches by exact year, paper label and both subjects found no advertised exact-paper answers. The official Part C syllabus for the 2010 examination confirms the paired course identity. No later-year answer-release policy was assumed applicable.

Assume additionally that the internal modus-ponens condition requested in part (c) holds for this $R$. If positive introspection also held for every sentence, then $R$ would satisfy all three Hilbert-Bernays-Löb derivability conditions: external necessitation rule was proved in part (c), modus ponens rule is the added condition, and introspection is the supposition.

The diagonal lemma applies to $R$ just as to any arithmetic formula. The usual Gödel second incompleteness theorem for a predicate with these conditions would then implywhen $PA$ is consistent. Part (b) proves the opposite. Therefore the three conditions cannot all hold, and under the added modus-ponens hypothesis there must be some sentence $X$ such thatThis is the intended contradiction argument and requires no identification of a particular $X$.

Without that extra hypothesis, the valid conclusion is only that at least one of internal modus ponens rule and positive introspection fails. Rosser predicates can be constructed that satisfy positive introspection, so a failure of that condition cannot be inferred from the printed general assumptions alone. ^[ref]. The distinction concerns proof coding, not a missing page of the exam.