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For standard effective proof presentations, the syntactic deduction theorem and its reverse give effective transformations between proofs in a one-sentence extension and proofs of an implication in the base theory. Arithmetic can verify these transformations, yielding the displayed internal equivalence under the usual representation and proof-predicate hypotheses. It distinguishes internal proof transformations from an unlicensed reflection step taking provability to truth.

Gödel's completeness theorem for first-order sentences says that, in a standard sound first-order Hilbert calculus,Equivalently every first-order consistent set of first-order sentences has a first-order model: apply this existence statement to $T\cup \{\neg \sigma \}$ for the consequence formulation.

For the first-order model existence formal proof use a Henkin construction. Enlarge the language by fresh constant symbols and adjoin Henkin witness axiomsFreshness of $c_{\theta }$ preserves first-order consistency. In detail, if a contradiction followed after adjoining one witness axiom, the deduction theorem for first-order logic would yield its negation, from which one obtains both $\exists x\theta (x)$ and $\neg \theta (c_{\theta })$ over the old theory. Since the new constant symbol does not occur in that theory or in $\theta$, replace it by a fresh logical variable and generalize to derive $\forall x\neg \theta (x)$, a contradiction. Finitely many witness additions preserve first-order consistency successively, and unions preserve first-order consistency because formal proofs are finite. Iterate the enlargement so that logical formulas involving newly introduced constant symbols also receive witnesses. Include at least one constant symbol to guarantee closed terms.

Extend the resulting first-order consistent witnessed first-order theory to a maximal consistent set of first-order sentences $H$. In the equality-free case make the domain the set of closed terms. Interpret constant symbols and function symbols by their formal first-order term constructions; interpret an atomic formula byThe truth lemma for a Henkin term model follows by induction on logical formulas with closed terms substituted for their free variables. The logical connective steps use maximal consistency of a first-order theory. In the existential step, membership of $\exists x\theta$ supplies a witness constant and hence a first-order term satisfying $\theta$; conversely a satisfying closed term gives the existential sentence by an inference of the calculus. The term model consequently satisfies $H$ and the original theory. Equality is absent here, so distinct closed terms may remain distinct domain elements without further identifications.

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.

Suppose $S+A$ proves every specified reflection instance. Use only the instance with $X=\neg A$. The deduction theorem givesClassical propositional reasoning turns this into $S⊢\square \neg A\to \neg A$: if $\neg A$ already holds the implication is true, while otherwise $A$ holds and the displayed theorem supplies it. Löb's theorem now yields $S⊢\neg A$, making $S+A$ inconsistent. Therefore no sentence consistent with $S$ can supply all the local reflection instances in a one-sentence extension. The proof uses just one well-chosen instance, and never applies necessitation to a derivation with the extra assumption still open.