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Assuming PA is consistent, the theory $S=\mathrm{PA}+\neg {\mathrm{Con}}_{\mathrm{PA}}$ is consistent by second incompleteness. Its false axiom $G=\neg {\mathrm{Con}}_{\mathrm{PA}}$ is provable, so the standard semantic biconditional $G\leftrightarrow \neg {\mathrm{Pr}}_S(┌G┐)$ is true with both sides false. Thus a true biconditional alone does not make an arbitrary sentence a true canonical Gödel sentence. A canonical Pi-1 construction adds the necessary syntactic or sound-base information. Truth, internal equivalence and canonical construction must be distinguished.

For a Gödel sentence $G\leftrightarrow \neg \square G$ with the Hilbert-Bernays-Löb derivability conditions, fix a refutable contradiction sentence $\perp$. Since $\perp \to G$ is provable, $\square \perp \to \square G$, giving $G\to \neg \square \perp$. For the reverse direction, $\square G$ implies both $\square \square G$ and $\square \neg \square G$, hence $\square \perp$. Contraposition gives $\neg \square \perp \to \neg \square G\to G$. This is an internal equivalence of sentences, not a rule asserting the theory's own consistency.

Assume the Hilbert-Bernays-Löb derivability conditions and the diagonal lemma. If $T+X$ proves ${\mathrm{Pr}}_T(┌Y┐)\to Y$ for every sentence $Y$, then $T⊢\neg X$. Take the single instance $Y=\neg X$ and discharge the assumption: $T⊢X\to (\square \neg X\to \neg X)$. This propositionally implies $T⊢\square \neg X\to \neg X$, and Löb's theorem gives $T⊢\neg X$. A diagonal proof is also possible using the Gödel sentence of the extension and formalized proof transfer, without appealing to Löb's theorem.

The preceding Gödel sentence argument gives $S⊢G\to {\mathrm{Con}}_S$. The permitted arithmetized positive half of the Rosser sentence argument isPure first-order logic gives $\neg BR\to R$: if there is no proof of $R$, each antecedent in its defining universal implication is false. ThereforeThis uses the supplied internal Rosser implication, rather than replacing an external nonprovability fact by an unproved internal first-order sentence.

Put $T=S+\{X\}$ and choose its Gödel sentence $Z$, so $T⊢Z\leftrightarrow \neg B_{T}Z$. The assumed reflection scheme gives the particular theoremThe given extension-provability implication is $S⊢B_{T}Z\to B_S(X\to Z)$. The fixed-point equivalence gives $T⊢\neg Z\to B_{T}Z$, so these implications compose toThe supplied propositional tautology $((\neg Z\to (X\to Z))\to (X\to Z))$ gives $T⊢X\to Z$, and $T⊢X$ then yields $T⊢Z$. The first of the derivability conditions for $T$ gives $T⊢B_{T}Z$, while its Gödel equivalence gives $T⊢\neg B_{T}Z$. Hence $T$ is inconsistent. By the deduction theorem for propositional logic,This proof uses diagonalization and proof transfer for the extension, and invokes neither Löb's theorem nor the second incompleteness theorem.

Use the preceding fixed point with $q=\perp$. If the chosen arithmetic interpretation sends $\perp$ to a PA-refutable sentence $Z$, it gives the arithmetic sentenceThis follows from Arithmetical soundness of GL. Since Peano arithmetic proves $\neg Z$, the two implications are equivalent to negated provability:Thus $C$ expresses the formal consistency statement for Peano arithmetic and is also a negative provability fixed point, like its Gödel sentence $G$.

Apply the supplied modal uniqueness formula to the context with a modalized occurrence $X(p)=(\square p\to \perp )$, with two fresh variables interpreted as $G$ and $C$. A modalized occurrence means that each indicated variable occurrence lies inside a box. Peano arithmetic proves both arithmetic fixed-point equations. The first derivability condition boxes those equations, so the translated uniqueness formula givesOne must not infer $G\leftrightarrow C$ by reflection. Instead, distribution and the provability of the two propositional projections give $\mathrm{PA}⊢BG\leftrightarrow BC$. Negating and using the two fixed-point equations now yieldsThis proves the Equivalence of a Gödel sentence and a consistency statement using the required modal soundness, fixed point and uniqueness results.

Here is the first incompleteness theorem in its modern consistency form: every consistent computably enumerable theory $T$ extending Robinson arithmetic is incomplete. The original Gödel sentence proof assumes omega-consistency to obtain both unprovabilities; we give that proof and the strengthening that removes this assumption. We also give the second incompleteness conclusion for theories extending Peano arithmetic.

First arithmetize syntax. Code strings and finite derivations by natural numbers. If the axioms of $T$ are merely computably enumerable, include the finite enumeration certificate for each nonlogical axiom used in a proof. The resulting proof-checking relation ${\mathrm{Prf}}_T(p,q)$ is primitive recursive. Finite strings and computation histories are coded by finite sequences, with bounded local verification of every step; these give arithmetic formulas representing the coding operations. In Robinson arithmetic every true or false instance with fixed numeral inputs is provable with the appropriate sign. Set

The diagonal lemma gives, for any formula $\theta (x)$, a sentence $D$ with $Q⊢D\leftrightarrow \theta (┌D┐)$. To see the construction, let $d(e)$ be the computable syntactic operation substituting the numeral for $e$ into the formula coded by $e$. Form the formula $\delta (x)=\theta (d(x))$, interpreting the displayed coding function through its arithmetic representation, and let $n=┌\delta ┐$. Then $D=\delta (\bar{n})$, and $d(n)=┌D┐$. The representation proves that fixed computation, giving the asserted equivalence.

Apply the diagonal lemma to $\neg {\mathrm{Prov}}_T(x)$, obtainingIf $T⊢G$, an actual finite proof has code $n$, so $Q$ proves its proof-checking instance and $T$ proves ${\mathrm{Prov}}_T(┌G┐)$. The fixed-point equivalence gives its negation, contradicting consistency. Hence $T⊬G$. For each standard numeral $n$, $Q$ proves $\neg {\mathrm{Prf}}_T(\bar{n},┌G┐)$ because there is no such proof. If $T⊢\neg G$, the equivalence gives $T⊢\exists p{\mathrm{Prf}}_T(p,┌G┐)$, contrary to omega-consistency. Thus the original proof gives two unprovabilities under that hypothesis. It also shows that $G$ is true in the standard natural numbers when $T$ is consistent.

For consistency alone use the Rosser sentenceIf $T⊢R$ with code $n$, consistency excludes every proof of $\neg R$. In particular the finitely many codes $q\le n$ all fail their primitive recursive proof tests. Robinson arithmetic verifies those finitely many failures and the actual proof at $n$, and hence proves $\neg R$, a contradiction. If $T⊢\neg R$ with code $m$, consistency excludes every proof of $R$. For $p<m$, Robinson arithmetic verifies all the finitely many failed proof tests. For $p\ge m$, it verifies that $q=m$ is a proof of $\neg R$ with $q\le p$. The elementary arithmetic dichotomy between $p<m$ and $m\le p$ therefore proves the entire displayed condition, hence $R$, again a contradiction. Thus

For the Gödel second incompleteness theorem, now assume $T\supseteq \mathsf{PA}$ and use its ordinary certified proof predicate. PA formalizes proof concatenation and the verification of a coded proof, giving the derivability conditions: if $T⊢A$ then $T⊢{\mathrm{Prov}}_T(┌A┐)$; and inside $T$,The first internal implication is certified by combining two proofs; the second is certified by producing the arithmetic proof that a given finite proof passes its check, with existential introduction. Apply them to $G\to \neg {\mathrm{Prov}}_T(┌G┐)$. From ${\mathrm{Prov}}_T(┌G┐)$, the two internal implications yield provability of both ${\mathrm{Prov}}_T(┌G┐)$ and its negation, and hence provability of $\perp$. ConsequentlyA proof of this consistency sentence in a consistent $T$ would prove $G$, which the first argument excludes. Therefore $\boxed{T⊬\mathrm{Con}(T)}$. This concerns the standard arithmetic consistency sentence and the stated effective, arithmetic hypotheses; it is not a prohibition on proving consistency in a stronger theory.

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 $S$.

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.

The definition of $a$ makes the diagonal instance $E_a[\bar{a}]$ exactly $G_S$. The representing logical formula therefore has the standard interpretationIf $S⊢G_S$, choose an actual witness $m$ to the left-hand existential statement. The bounded arithmetic sentence $A(\bar{a},\bar{m})$ is true and hence provable by ${\Sigma }_0$-completeness. But $G_S$ also gives $S⊢\neg A(\bar{a},\bar{m})$, a contradiction. ThusNow assume one-consistency. It implies first-order consistency, so no formal proof of $G_S$ exists and $A(a,m)$ is false for each standard $m$. Each bounded arithmetic sentence $\neg A(\bar{a},\bar{m})$ is consequently true and provable. A formal proof of $\neg G_S$ would give a formal proof of $\exists v_{2}A(\bar{a},v_2)$ while every standard witness instance is refuted, contrary to one-consistency. HenceThe two hypotheses play different roles: ordinary first-order consistency excludes the positive Gödel sentence, while the stronger hypothesis also excludes its negation.

Put $T=S\cup \{\neg G_S\}$. The first supplied equivalence, retained in $T$, combines with its new axiom to giveThe supplied implication $\mathrm{Con}(T)\to \mathrm{Con}(S)$ then yields, by contraposition,Finally the second supplied equivalence is $T⊢G_T\leftrightarrow \mathrm{Con}(T)$, so propositional reasoning givesNo extra first-order consistency assumption is needed for this formal derivation. If $S$ is first-order consistent, part (b) shows that adding $\neg G_S$ preserves first-order consistency, because inconsistency of the extension would imply $S⊢G_S$. Such a first-order consistent extension therefore proves the negation of its own true Gödel sentence, illustrating why first-order consistency alone does not imply one-consistency.

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

For the standard Gödel sentence and Rosser sentence, the formalized Rosser unprovability theorem and $G\leftrightarrow \mathrm{Con}(T)$ yield $T⊢G\to R$. If $T+\mathrm{Con}(T)$ is consistent, the converse is unprovable under the usual formalized incompleteness hypotheses. A proof of $R\to G$ would give $\square \neg G\to \square \neg R$. Formalized negative unprovability identifies $\neg \square \neg R$ with $\mathrm{Con}(T)$ and $\neg \square \neg G$ with $\mathrm{Con}(T+\mathrm{Con}(T))$. The strengthened theory would then prove its own consistency, contradicting the Gödel second incompleteness theorem. The consistency assumption belongs to this strictness conclusion.