Past exam of the model theory and Gödel's incompleteness theorems course of the University of Oxford / 2009 / A / 2 / b

For Rosser separation by a bounded stage comparison, use the numeralwise enumeration in arithmetic properties meant by enumeration in $S$: membership supplies some standard stage with a provable positive instance, while nonmembership makes every standard-stage instance refutable. These are stronger than merely attaching a ${\Sigma }_1$ syntactic label to an arbitrary formula. The finite-initial-segment axiom lets us combine finitely many numeralwise refutations into a bounded universal assertion.

If $n$ is provable, choose a standard stage $p$ with $S⊢P(\overline{n},\overline{p})$. Consistency means $n$ is not refutable. Therefore $S⊢\neg R(\overline{n},\overline{j})$ for each $j\le p$. Using the displayed finite-segment axiom and equality substitution, combine these proofs to get $S⊢\forall z\le \overline{p}\neg R(\overline{n},z)$. Existential introduction with stage $p$ proves the separator at $n$.

If $n$ is refutable, choose a standard stage $q$ with $S⊢R(\overline{n},\overline{q})$. Consistency implies that $n$ is not provable, so all standard instances of $P(\overline{n},\cdot )$ are refutable. In particular their finitely many instances through $q$ give $S⊢\forall z\le \overline{q}\neg P(\overline{n},z)$. For an arbitrary proposed witness $z$, the total-comparison axiom gives $z\le \overline{q}$ or $\overline{q}\le z$. In the first case its $P$ conjunct fails; in the second case the known $R$ instance at $q$ contradicts its bounded no-refutation conjunct. Generalize on $z$ to refute the existential separator. Hence the proposed first-proof-before-refutation formula separates the two sets.