Past exam of the model theory and Gödel's incompleteness theorems course of the University of Oxford / 2010 / A

Write $\square A$ for the arithmetical provability predicate applied to the Gödel number of $A$. Use the Hilbert-Bernays-Löb derivability conditions that belong to the course meaning of a provability predicate: external necessitation rule, internal modus ponens rule, and positive introspection. No incompleteness theorem is needed here.

If $S⊢G$, necessitation rule gives $S⊢\square G$. The displayed Gödel sentence equivalence also gives $S⊢\neg \square G$. These two theorems contradict the assumption that $S$ is a consistent first-order theory. ThereforeFor this part alone, the external necessitation rule condition suffices; the stronger internal conditions are used in the next part.

Again abbreviate provability by $\square$. The Gödel sentence equivalence gives $S⊢G\to \neg \square G$. Internalizing this implication using the first two Hilbert-Bernays-Löb derivability conditions givesPositive introspection also gives $S⊢\square G\to \square \square G$. Internal modus ponens rule combines proofs of a sentence and its negation into a proof of a contradiction; a contradiction implies any sentence. ConsequentlyThese are syntactic manipulations of provability, not an appeal to soundness. If $S⊢\neg \square \neg G$, taking the contrapositive of the last implication gives $S⊢\neg \square G$, and the diagonal equivalence gives $S⊢G$. Part (a) rules that out for a consistent theory. Hence

An arithmetic theory is one-consistent if there is no bounded arithmetic formula $\delta (x)$ for which it proves $\exists x\delta (x)$ and also proves $\neg \delta (\overline{n})$ for every standard natural number $n$. A finite tuple of existential witnesses can be coded by one natural number. The associated Sigma-1 soundness property says that every ${\Sigma }_1$ sentence it proves is true in the standard natural numbers.

Suppose first that the theory is one-consistent, but proves a false ${\Sigma }_1$ sentence $\exists x\delta (x)$. For every standard $n$, $\neg \delta (\overline{n})$ is then a true bounded sentence. Bounded completeness makes each such sentence a theorem, violating one-consistency.

Conversely suppose the theory is ${\Sigma }_1$-sound. If it proved both the existential sentence and every negated standard instance, soundness would supply an actual witness $n$. Its positive instance $\delta (\overline{n})$ is a true bounded sentence and hence provable. Together with the assumed negative instance this would make the theory inconsistent; an inconsistent theory proves a false ${\Sigma }_1$ sentence, contrary to soundness. Thusunder the stated bounded-completeness assumption. In particular either condition implies that the theory is a consistent first-order theory.

By part (c), one-consistency gives Sigma-1 soundness and hence ordinary consistency. First, $S$ cannot prove $\neg G$: if it did, the diagonal equivalence would give $S⊢\square G$. This is a ${\Sigma }_1$ sentence, so soundness would make it true. The stipulated standard interpretation of the arithmetical provability predicate would then supply an actual proof of $G$, contradicting consistency.

Now suppose $S⊢\square \neg G$. This too is a ${\Sigma }_1$ theorem. Soundness and the standard correctness of the predicate would imply that there is an actual $S$-proof of $\neg G$, which was just excluded. ThereforeThe two applications of soundness concern the provability assertions, not the truth of arbitrary arithmetic sentences.

Write $\mathrm{Prf}(n,m)$ for the specified proof relation and $R(n)$ for its Rosser provability predicate. If $R(\overline{n})$ is true in the standard natural numbers, its existential witness $m$ is the code of an actual proof of the expression with code $n$. Thus that expression is provable.

Conversely, suppose an actual proof has code $m$. If Peano arithmetic is consistent, there is no proof of the negation at all, in particular none with code at most $m$. The same $m$ therefore witnesses the bounded proof-order condition as well as the positive proof condition. Henceunder consistency. For numbers that do not code well-formed expressions, neither side holds under the usual proof coding. This is an external equivalence of theoremhood; it does not establish every internal derivability conditions for $R$.

Let $\perp$ abbreviate the fixed false arithmetic equality, and fix an actual proof of $\neg \perp$ with code $m$. For a proposed proof code $y\ge m$, that known proof of the negation already violates the bounded negative-proof test in $R(┌\perp ┐)$.

For the finitely many codes $y<m$, either one is an actual proof of $\perp$, in which case Peano arithmetic is inconsistent and proves the requested conclusion by explosion, or none is. In the latter case the finitely many true bounded facts $\neg \mathrm{Prf}(┌\perp ┐,\overline{y})$ are all provable by bounded completeness. Their conjunction excludes every code below $m$. The positive bounded fact that $m$ proves $\neg \perp$ is likewise provable. Splitting $y<m$ and $y\ge m$ inside arithmetic now givesThus the consistency statement for this Rosser provability predicate is provable. The argument uses only a fixed finite initial segment of proof codes; it does not prove the usual unbounded consistency statement.

The external necessitation rule assertion is valid. Suppose $PA⊢X$ and fix a proof code $m$. If $PA$ is inconsistent the assertion follows by explosion. Otherwise there is no actual proof of $\neg X$. Each of the finitely many bounded checks up to $m$ is therefore a true bounded arithmetic fact and is provable. Together with the positive proof check at $m$ this gives

The requested internal modus-ponens assertion needs an additional coding property. Combining a proof of $X$ and a proof of $X\to Y$ produces a proof of $Y$, but its code can exceed both original codes. The two Rosser hypotheses exclude proofs of $\neg X$ and $\neg (X\to Y)$ only up to their respective witness bounds. They provide no bound excluding proofs of $\neg Y$ below the new proof code. Ordinary internal modus ponens rule for the unbounded provability predicate does not fill this gap.

Consequently the premises printed here do not, just by themselves, justifyThis property varies with the Rosser proof-order dependence: some specially chosen proof predicates satisfy it, while general Rosser proof predicates need not. ^[ref].

The reference to a proof predicate “as given in the course” would need an explicit additional proof-order lemma to establish this part for that particular coding. Standard correctness on numeral instances and the ordinary Hilbert-Bernays-Löb derivability conditions alone are insufficient. The next solution gives the intended argument with the missing condition stated explicitly.

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.

Abbreviate the arithmetical provability predicate by $\square$, with the Hilbert-Bernays-Löb derivability conditions understood. Löb's theorem saysChoose a diagonal sentence $D$ with $S⊢D\leftrightarrow (\square D\to A)$. Internalizing its forward implication gives $S⊢\square D\to \square (\square D\to A)$. Internal modus ponens rule gives $\square (\square D\to A)\to (\square \square D\to \square A)$, while positive introspection gives $\square D\to \square \square D$. Combining these yields $S⊢\square D\to \square A$.

Now use the assumed theorem $\square A\to A$. It follows that $S⊢\square D\to A$, and the reverse direction of the diagonal equivalence gives $S⊢D$. External necessitation rule gives $S⊢\square D$, and the previously derived implication gives $S⊢A$. This proves the theorem from the supplied fixed-point property and the derivability conditions.

The Gödel second incompleteness theorem states that a consistent arithmetic theory in this setting cannot prove its own consistency sentence ${\mathrm{Con}}_S=\neg \square \perp$. If it could, it would prove $\square \perp \to \perp$; applying Löb's theorem with $A=\perp$ would make it inconsistent. Therefore $\boxed{S⊬{\mathrm{Con}}_S}$ when $S$ is consistent.

The language of Gödel-Löb logic has propositional variables, $\perp$, implication, and the unary operator $\square$; the other logical connectives are abbreviations. Its axioms are all classical tautologies, the distribution scheme $\square (A\to B)\to (\square A\to \square B)$, and the Löb scheme $\square (\square A\to A)\to \square A$. Its rules are modus ponens rule, necessitation rule of theorems, and uniform substitution.

For a single boxed equation $p\leftrightarrow \square C(p)$, set $F=\square C(\top )$. The identity supplied in the source gives directly $GL⊢F\leftrightarrow \square C(F)$. If $p$ does not occur, take $F=\square C$ instead.

Proceed by induction for a system of $m$ equations, allowing other propositional variables as parameters. Treat $p_1,\dots ,p_{m-1}$ as parameters in the last equation and putThe one-variable result proves $D\leftrightarrow \square C_m(p_1,\dots ,p_{m-1},D)$. Substitute this $D$ for $p_m$ in the first $m-1$ equations. They are still boxed equations, so induction supplies solutions $F_1,\dots ,F_{m-1}$. Define $F_m=D(F_1,\dots ,F_{m-1})$.

By construction, the first $m-1$ equations hold after this substitution; substituting into the theorem for $D$ proves the last. HenceThe elimination introduces no new parameter letters and removes every unknown letter. This gives a constructive simultaneous boxed fixed-point elimination in GL.

Take the outermost boxed subformulas of $X(p)$ containing the distinguished letter. Because $X$ is modalized in $p$, every occurrence of $p$ lies inside one of these boxes. The supplied decomposition therefore has the formwhere $H$ has no occurrence of $p$. Other letters and subformulas not involving $p$ may be retained as parameters in $H$ and the $C_i$.

Introduce temporary letters $q_1,\dots ,q_m$ and solve the boxed systemby part (b), obtaining $Q_i$ without the temporary letters or $p$. Put $F=H(Q_1,\dots ,Q_m)$. Each solved equation gives $GL⊢Q_i\leftrightarrow \square C_i(F)$. The supplied replacement-of-equivalents principle then givesOnly letters already present in $X$, apart from the removed $p$, occur in the constructed $F$. If $p$ is absent from $X$ to begin with, take $F=X$. This is the modalized fixed-point theorem for GL, proved by finite elimination rather than a semantic completeness argument.

One can takeFor the first formula, the Löb axiom with conclusion $\perp$ gives $\square (\square \perp \to \perp )\to \square \perp$, that is, $\square F_1\to \square \perp$. Conversely, the tautology $\perp \to F_1$, necessitation rule, and distribution give $\square \perp \to \square F_1$. Thus $GL⊢\square F_1\leftrightarrow \square \perp$. Negating both sides gives $GL⊢\neg \square F_1\leftrightarrow F_1$.

For the second formula, $\neg F_2=\top$ is a theorem, so $GL⊢\square \neg F_2$. Hence $\neg \square F_2\wedge \neg \square \neg F_2$ is false in the logic and is equivalent to $\perp =F_2$. A fixed point is a formula satisfying the equivalence; it need not itself express a nontrivial undecidable assertion.