Past exam of the model theory and Gödel's incompleteness theorems course of the University of Oxford / 2009 / A / 1

Write $\square A$ for the arithmetical sentence asserting $S$-provability of $A$. If $S⊢G$, the first derivability condition gives $S⊢\square G$. The stipulated fixed-point equivalence also gives $S⊢\neg \square G$. This contradicts consistency. Conversely an inconsistent classical theory proves every sentence, including $G$. ThusOnly the direction from an actual proof to its formalized provability is used here. There is no inference from a theorem $\square G$ to a theorem $G$, and no assumption that every theorem of $S$ is true.

The Hilbert-Bernays-Löb derivability conditions are: if $S⊢A$ then $S⊢\square A$; $S⊢\square (A\to B)\to (\square A\to \square B)$; and $S⊢\square A\to \square \square A$. The first is an external rule about provability; the latter two are internal theorems. Standard numerical correctness and the stipulated ${\Sigma }_1$ form accompany the predicate in this question. These conditions do not include reflection $\square A\to A$.

Choose a fixed refutable sentence $\perp$ and identify ${\mathrm{Con}}_S$ with $\neg \square \perp$. Different fixed refutable choices have provably equivalent consistency sentences, because $S$ proves their equivalence to a contradiction. From $\perp \to G$, necessitation and distribution give $\square \perp \to \square G$. Since $G\to \neg \square G$, this yields $G\to {\mathrm{Con}}_S$.

For the converse, necessitate $G\to \neg \square G$ and distribute to get $\square G\to \square \neg \square G$. Positive introspection also gives $\square G\to \square \square G$. The two boxed contradictories imply $\square \perp$ by formalized propositional reasoning. Thus $\square G\to \square \perp$, and contraposition gives ${\mathrm{Con}}_S\to \neg \square G\to G$. ConsequentlyFinally the deduction theorem says that $S+G$ is inconsistent exactly when $S⊢\neg G$. The internal equivalence identifies $S+G$ and $S+{\mathrm{Con}}_S$ in their deductive consequences. Hence $S⊬\neg G$ exactly when $S+{\mathrm{Con}}_S$ is consistent. Consistency of $S$ alone need not ensure this stronger consistency condition.

For consistency of a negated-consistency extension, put $A=S+\neg {\mathrm{Con}}_S$. The arithmetized deduction theorem, together with double-negation elimination and formalized proof transformations, givesFrom the preceding proof, $S⊢{\square }_{S}G\to {\square }_S\perp$. Necessitating the equivalence between $G$ and ${\mathrm{Con}}_S$ and using distribution therefore gives $S⊢{\square }_S{\mathrm{Con}}_S\to {\square }_S\perp$. Its contrapositive isThis is an internal conditional statement. It does not assert that $S$ proves its own consistency or permits the consistency premise to be discarded.

For unrestricted consistency self-strengthening is inconsistent, again take $A=S+\neg {\mathrm{Con}}_S$. In $S^{++}$ the axiom ${\mathrm{Con}}_S$ and the preceding implication prove ${\mathrm{Con}}_A$. The additional rule then proves ${\mathrm{Con}}_{A+{\mathrm{Con}}_A}$.

However consistency of an extension implies consistency of its subtheory by the formalized inclusion of proofs: $S⊢{\mathrm{Con}}_A\to {\mathrm{Con}}_S$. Thus the ordinary theory $A+{\mathrm{Con}}_A$ contains $\neg {\mathrm{Con}}_S$ and derives ${\mathrm{Con}}_S$, so it has a finite contradiction proof. Checking that concrete proof in arithmetic gives $S⊢\neg {\mathrm{Con}}_{A+{\mathrm{Con}}_A}$. 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.