Past exam of the model theory and Gödel's incompleteness theorems course of the University of Oxford / 2007 / A

Use the numeralwise enumeration in arithmetic convention in the first-order theory: membership supplies a standard stage with a provable enumerating instance, while nonmembership makes every standard-stage instance refutable. Put $D=\{n:P⊢E_n(\bar{n})\}$. By the definition of $a$, the diagonal instance $E_a(\bar{a})$ is exactly $G$.

If $P⊢G$, then $a\in D$. Numeralwise enumeration in arithmetic supplies a natural number $m$ such that $P⊢A(\bar{a},\bar{m})$. But universal instantiation of $G$ gives $P⊢\neg A(\bar{a},\bar{m})$. This contradicts first-order consistency. ThereforeThis uses the formal proof obligations in numeralwise enumeration in arithmetic in $P$, rather than merely the external assertion that a logical formula has a particular complexity.

One-consistency says that no bounded arithmetic formula $F(x)$ has a provable existential closure while each of its standard numeral instances is refutable. It implies first-order consistency, since an inconsistent classical first-order theory proves every logical formula.

If $P$ is one-consistent, (i) gives $P⊬G$, so $a\notin D$. Numeralwise enumeration in arithmetic therefore givesIf $P⊢\neg G$, classical logic gives $P⊢\exists v_{2}A(\bar{a},v_2)$. The matrix is bounded, and the preceding displayed derivations refute every standard instance. This violates one-consistency. ConsequentlyTogether with (i), this yields undecidability of this first-order sentence under the stronger hypothesis, without asserting that ordinary first-order consistency alone suffices for its negative half.

Bounded completeness of an arithmetic theory means that every bounded arithmetic first-order sentence true in the standard natural numbers is provable in $S$. If $S$ is first-order consistent and proves a false bounded arithmetic sentence $B$, bounded completeness of an arithmetic theory also proves its true bounded negation, a contradiction. Conversely, if every bounded theorem is true, $S$ cannot be inconsistent, because inconsistency would prove the false bounded arithmetic sentence $0=1$. ThusIf $S$ is one-consistent and proves $\exists xB(x)$ with bounded matrix, that existential sentence cannot be false: otherwise every $\neg B(\bar{n})$ would be a true bounded arithmetic sentence and hence provable, contrary to one-consistency. Thus $S$ has Sigma-1 soundness.

Conversely, Sigma-1 soundness excludes inconsistency. If $S⊢\exists xB(x)$, it supplies a standard witness $n$ with $B(n)$ true. Bounded completeness of an arithmetic theory proves $B(\bar{n})$, so first-order consistency forbids a formal proof of $\neg B(\bar{n})$. Hence there cannot simultaneously be refutations of all standard instances. ThereforeFinite blocks of existential logical variables are treated as finite tuples of witnesses, with the same argument.

Omega-consistency forbids a logical formula $B(x)$ of any complexity for which $S⊢\exists xB(x)$ and $S⊢\neg B(\bar{n})$ for every standard $n$. The difference from one-consistency is that the matrix is unrestricted. It implies first-order consistency.

For a complete first-order consistent classical first-order theory, theorem membership obeys the ordinary Boolean truth tables. For example, it proves a conjunction exactly when it proves both conjuncts; completeness and first-order consistency also make theoremhood of a negation the complement of theoremhood of the first-order sentence. Omega-consistency additionally gives the witness propertyThe reverse implication is existential introduction. For the forward implication, if no standard instance were provable, completeness would give a formal proof of each negated instance, violating omega-consistency. This works also after replacing additional free variables by numerals.

Now induct on logical formulas with their remaining free variables replaced by standard numerals. At the atomic level, the equivalence between theoremhood and standard truth follows from bounded completeness of an arithmetic theory and bounded soundness of an arithmetic theory from (b). The Boolean steps use the truth-table facts just proved. The existential step uses the witness property and the induction hypothesis on its matrix; universal quantification follows by negation. Hencefor every first-order sentence $X$. Completeness ensures it decides each first-order sentence, and this equivalence ensures every decision is correct.

Start with the supplied bounded-complete, one-consistent first-order theory $T$ proving a false first-order sentence $F$. Replace it by its deductive closure, which preserves these properties and contains $F$. Enumerate the arithmetic first-order sentences as $X_0,X_1,\dots$. Recursively extend the current first-order theory by either $X_n$ or $\neg X_n$, choosing an extension which remains one-consistent as permitted by the supplied lemma, and then take its deductive closure. Call the resulting increasing sequence $T_n$, with $T_0$ the initial first-order theory, and put $S={\bigcup }_n{T}_n$.

Every stage is still bounded-complete. By (b), every stage has Sigma-1 soundness. If $S$ proved a false Sigma-1 first-order sentence, the finite formal proof would use assumptions from one stage $T_n$, contradicting its Sigma-1 soundness. Thus $S$ has Sigma-1 soundness and is bounded-complete, so (b) makes it one-consistent.

This soundness of an arithmetic theory argument is essential: the definition of one-consistency involves infinitely many standard refutations, so it would be insufficient merely to say that a violation has a finite formal proof. What has a finite formal proof is the false existential sentence, and bounded completeness of an arithmetic theory is what turns one-consistency into the needed soundness of an arithmetic theory condition.

The union is deductively closed because each formal proof has finitely many assumptions. It contains either $X$ or $\neg X$ for every first-order sentence, by the numeralwise enumeration in arithmetic, and contains the false first-order sentence $F$. If it were omega-consistent, (c) would make it correct on all first-order sentences, contradicting $F$. We have therefore constructedOmega-consistency always implies one-consistency by restricting its matrices. This counterexample proves that the converse is strictly false.

A logical formula separates $A$ from $B$ in $S$ if it is provable at the standard numeral tuple of every member of $A$ and refutable at that of every member of $B$:There is no obligation on tuples outside the two sets. In a first-order consistent first-order theory these two obligations make $A$ and $B$ disjoint.

Let $h$ be the Gödel number of $H(v_1)$ and take the literal diagonal first-order sentence $R=H(\bar{h})$. If $S⊢R$, then $h$ belongs to the second source set, so separation gives $S⊢\neg H(\bar{h})$, a contradiction. If $S⊢\neg R$, then $h$ belongs to the first source set, so separation gives $S⊢H(\bar{h})$, again a contradiction. HenceThe direction of separation matters: here $H$ is provable on refutable diagonal codes and refutable on provable diagonal codes.

Write $H(\vec{v})$ for the proposed stage-comparison logical formula and fix a standard tuple $\vec{a}$. If $\vec{a}\in A$, numeralwise enumeration in arithmetic supplies a standard $m$ with $S⊢F(\overline{\stackrel{⃗}{a}},\bar{m})$. Disjointness gives $\vec{a}\notin B$, so $S⊢\neg G(\overline{\stackrel{⃗}{a}},\bar{j})$ for every standard $j$, in particular $j=0,\dots ,m$. The source's finite initial-segment axiom combines this finite list intoUse $\bar{m}$ as the existential witness to obtain $S⊢H(\overline{\stackrel{⃗}{a}})$.

If instead $\vec{a}\in B$, choose $n$ with $S⊢G(\overline{\stackrel{⃗}{a}},\bar{n})$. Disjointness gives refutations of $F(\overline{\stackrel{⃗}{a}},\bar{j})$ for all $j$, so the finitely many $j\le n$ giveFor an arbitrary $x$, the source's comparison axiom gives $x\le \bar{n}$ or $\bar{n}\le x$. In the first case the first conjunct of the proposed witness condition is false. In the second case $y=\bar{n}$ witnesses failure of its bounded universal conjunct, since $G(\overline{\stackrel{⃗}{a}},\bar{n})$ is provable. Generalization over $x$ therefore gives $S⊢\neg H(\overline{\stackrel{⃗}{a}})$.

Thus the logical formula separates $A$ from $B$. The formal proof uses the enumerators' numeralwise positive and negative formal proof properties and the two explicit order principles supplied in the source; it does not assume the first-order theory can decide arbitrary unbounded existential logical formulas.

A logical formula $D(\vec{x},y)$ weakly defines a total numerical function $f$ in $S$ if, for every standard input tuple $\vec{n}$ and $m=f(\vec{n})$,Strong definability of a numerical function additionally requires the internal uniqueness statementTogether with the positive instance, this is equivalent to proving $\forall y(D(\overline{\stackrel{⃗}{n}},y)\leftrightarrow y=\bar{m})$ for each standard input. The universal quantifier is the extra strength: it covers nonstandard potential outputs in first-order models of $S$. These are the weak and strong function conventions used here. ^[ref]

Suppose the graph of the total function is Sigma-1. Its complement is also Sigma-1, sinceReplace the graph assertion on the right by its Sigma-1 definition; existential quantifiers can be collected into one block. Bounded completeness of an arithmetic theory supplies numeralwise stage enumerators for each of these two disjoint complementary sets. Apply (b) to them. Its separating logical formula is provable exactly at the required graph tuples in the positive direction and refutable at all the other standard tuples, proving weak definability of a numerical function.

Using the graph logical formula alone would only supply positive instances from true existential witnesses; the separation construction is what supplies the negative-instance formal proofs without assuming Sigma-1 soundness or completeness for false Sigma-1 statements.

Let $d(n)$ be the Gödel number of the diagonal first-order sentence obtained by substituting $\bar{n}$ into the one-free-variable logical formula with code $n$, with an arbitrary default value on noncodes. By the assumptions and (c), this total function has a strong defining logical formula $D(x,y)$.

For the specified $F$, form the one-variable logical formulaLet $h$ be the Gödel number of $H$ and put $C=H(\bar{h})$. Then $d(h)=┌C┐$. Strong definability of a numerical function at this particular standard input givesSubstitution into the existential definition yieldsThe permitted equivalence between the source's two substitution notations identifies this with its requested fixed point. Bound variables are renamed when necessary before forming $H$, so the substitution is capture-free. This constructs one particular first-order sentence from the code of the logical formula which applies $F$ to its own diagonal output.

Write $\square A$ for $P(┌A┐)$. A standard arithmetical provability predicate obeys the Hilbert-Bernays-Löb derivability conditions:The first condition is a rule about theorems; the latter two are statements proved within the first-order theory. When the terminology also includes external adequacy, require that $P(\bar{n})$ be true in the standard natural numbers exactly when the logical formula with code $n$ has an $S$-proof.

For Peano arithmetic, code finite strings, logical formulas and formal proof sequences by natural numbers. Formal proof checking is effective and has an arithmetized bounded checking relation with its computation data. Existentially assert the existence of a correct formal proof code and any required checking data. The resulting predicate symbol is Sigma-1, and its standard truth means exactly that an actual finite PA formal proof exists. Effective operations on formal proof codes combine an implication formal proof with an antecedent formal proof, and turn a formal proof into a formal proof that its code is a formal proof; PA verifies those constructions. They give distribution and positive introspection, while an actual theorem's finite formal proof gives its internal provability assertion. This is the requested rough construction, with the three internal formal proof properties distinguished from external agreement.

Abbreviate $P(┌A┐)$ by $\square A$. Since $S⊢\neg X$, it proves $X\to G$. The theoremhood and distribution conditions giveThe fixed-point equation gives $S⊢G\to (\square G\to X)$, because $G$ implies $\neg \square G$ and a false antecedent implies any conclusion. Applying theoremhood and distribution yieldsPositive introspection gives $\square G\to \square \square G$; distribution then gives $\square G\to \square X$. Thus $S⊢\square G\leftrightarrow \square X$. Combining with $G\leftrightarrow \neg \square G$ provesIn particular, choosing a fixed logical contradiction for $X$ identifies this Gödel sentence with the formal consistency statement of $S$.

If $S⊢G$, theoremhood gives $S⊢\square G$, while the fixed-point equivalence also gives $S⊢\neg \square G$. Therefore first-order consistency implies $S⊬G$.

If $S\cup \{\neg G\}$ were inconsistent, the deduction theorem and classical propositional logic would give $S⊢\neg \neg G$, hence $S⊢G$. This is impossible by the preceding argument. ThereforeThis is a syntactic first-order consistency result. It does not claim that the extra axiom is true in the standard natural numbers or that the extended first-order theory is sound.

First establish the transfer of provability requested in the hint. The propositional theorem $X\to (\neg G\to X)$ and the derivability conditions giveThe supplied arithmetized deduction equivalence identifies the right-hand provability assertion, within $S^{\ast }$, with ${\square }^{\ast }X$. HenceChoose a fixed contradiction $\perp$ for $X$. Part (b) gives $S⊢G\leftrightarrow \neg \square \perp$, so adjoining $\neg G$ gives $S^{\ast }⊢\square \perp$. The transfer implication now gives $S^{\ast }⊢{\square }^{\ast }\perp$.

Apply (b) inside $S^{\ast }$ to its own Gödel sentence: it proves $G^{\ast }\leftrightarrow \neg {\square }^{\ast }\perp$. The last provability assertion therefore entailsThus the extension proves the negation of its own Gödel sentence as well. When $S$ is first-order consistent, (c) says this phenomenon occurs in a first-order consistent extension; the extension asserts an internal formal proof of contradiction without there being an actual standard formal proof of contradiction.

All implications in this argument are theorems of the extension $S$. From $Y\to (\square Y\to X)$, necessitation and normal modal distribution giveAxiom4 gives $\square Y\to \square \square Y$, soUnder the additional propositional antecedent $\square X\to X$, this theorem implies $\square Y\to X$. The reverse half of the fixed-point equivalence, $(\square Y\to X)\to Y$, consequently gives the theoremNecessitate that theorem and apply distribution to obtain $\square (\square X\to X)\to \square Y$. Combining with $\square Y\to \square X$ provesNecessitation was applied to a theorem after its propositional antecedent had been incorporated, never directly to a temporary assumption.

Take $\square A={\mathrm{Pr}}_{PA}(┌A┐)$ for the standard formal proof predicate symbol and putUse as given the arithmetization of syntax in PA, the derivability conditions, and the diagonal lemma. The latter supplies the fixed point needed in (a), and the former conditions interpret its normal modal distribution, necessitation on theorems and axiom4. Thus the argument in (a) proves the internal Löb implication in PA. With $X=\perp$ it readsThe inner implication is the first-order consistency first-order sentence $\neg \square \perp$, so contraposition givesThis is the arithmetic statement of the second incompleteness theorem proved within PA: if PA is first-order consistent, there is no PA formal proof of its first-order consistency first-order sentence. The displayed implication is a PA theorem without assuming that PA proves its antecedent.

For the corresponding external consequence, if a first-order consistent PA proved ${\mathrm{Con}}_{PA}$, theoremhood would give a formal proof of its boxed assertion. The displayed theorem together with the supposed formal proof of first-order consistency gives its negation, contradicting first-order consistency. Hence an externally first-order consistent PA does not prove its own first-order consistency.

Put $Y=\square X\to X$. The Löb axiom gives $\square Y\to \square X$. Consequently $Y$ implies $\square Y\to X$: if $\square Y$ holds, first obtain $\square X$ and then apply $Y$.

For the other direction, $X\to Y$ is a propositional theorem. Necessitation and distribution give $\square X\to \square Y$. Hence $\square Y\to X$ implies $\square X\to X$, which is $Y$. We have proved the unboxed theoremNecessitation yieldsThis is the explicit fixed point for the modalized operator $Z\mapsto (\square Z\to X)$.

Use the supplied uniqueness theorem for the modalized logical formula $A(p)=\square p\to X$. Its occurrence of $p$ is boxed, and the source assumes $p$ does not occur in $X$. Choose another first-order sentence letter $q$ absent from $X$ and from $p$. Besides the uniqueness theorem and propositional reasoning, use that GL is closed under uniform substitution in modal logic.

Let $Y=\square X\to X$. Uniformly substitute $Y$ for the fresh letter $q$ in the uniqueness theorem. Since $q$ occurs nowhere in $X$, this givesPart (c) proves the second conjunct outright. Propositional reasoning removes that already-proved conjunct and yieldsThe freshness conditions make the substitution legal and ensure it changes only the placeholder intended for the second fixed point.