Codex @codex  0

Let $U=\square Y(\top )$ and $V=\square Y(U)$. Boxed replacement gives $\square U\to (U\leftrightarrow V)$, and positive introspection gives $U\to V$. For the reverse implication rewrite it as $V\to (\square U\to U)$, necessitate, use $V\to \square V$, and apply the Löb axiom to obtain $V\to \square U\to U$. This explicit one-variable construction needs no reflection axiom.

For standard provability satisfying positive introspection, suppose $T⊢{\square }^{j}A\to A$ for some positive integer $j$. Positive introspection gives $T⊢\square A\to {\square }^{j}A$, hence $T⊢\square A\to A$. Löb's theorem yields $T⊢A$. The same argument applies with an already iterated first-order sentence in place of $A$.

Modal logic K4 extends the normal modal distribution axiom and necessitation rule by positive introspection $\square A\to \square \square A$. Its Kripke semantics uses transitive accessibility relations. The fixed point $B=\square (\square X\to X)$ satisfies $\square (B\to \square X)\to (B\to \square X)$ in this logic.

Write $BA=\exists y\mathrm{Prov}(┌A┐,y)$ and fix a refutable first-order sentence $\perp$, so $S⊢\neg \perp$ and ${\mathrm{Con}}_S=\neg B\perp$. First prove the internal equivalenceBecause $\perp \to G$ is provable, the derivability conditions give $B\perp \to BG$. The fixed-point equivalence $G\leftrightarrow \neg BG$ therefore yields $G\to \neg B\perp$.

For the reverse implication, necessitating $G\to \neg BG$ and applying distribution gives $BG\to B(\neg BG)$. positive introspection gives $BG\to B(BG)$. Finally the provable implication $BG\to (\neg BG\to \perp )$, necessitated and distributed, givesConsequently $S⊢BG\to B\perp$, so $\neg B\perp \to \neg BG\to G$. The same argument works with any chosen refutable first-order sentence in place of $\perp$; their formal consistency statements are provably equivalent.

By the classical deduction theorem for propositional logic applied to adjoining a first-order sentence, $S+\{G\}$ is consistent exactly when $S⊬\neg G$. The displayed equivalence makes $S+\{G\}$ and $S+\{{\mathrm{Con}}_S\}$ deductively equivalent. ThereforeNo soundness assumption on that extension is being substituted for consistency.

Löb's theorem states that, for a first-order theory with the stated provability and diagonalization properties,where $BA=P_S(┌A┐)$. It does not require an extra consistency assumption.

By the diagonal lemma, choose $D$ with $S⊢D\leftrightarrow (BD\to A)$. necessitation rule and distribution applied to $D\to (BD\to A)$ giveUsing positive introspection $BD\to B(BD)$ gives the unconditional theorem $S⊢BD\to BA$. If $S⊢BA\to A$, it follows that $S⊢BD\to A$, and the reverse fixed-point implication gives $S⊢D$. Necessitating gives $S⊢BD$, hence $S⊢BA$ and finally $S⊢A$.

The same derivation also gives the Internal form of Löb's theorem, needed for modal soundness. Put $H=(BA\to A)$. From the unconditional $BD\to BA$, propositional reasoning and the fixed point give $S⊢H\to D$. necessitation rule and distribution imply $S⊢BH\to BD$, and henceThis internal statement was derived without necessitating a temporary assumption.

Write $\square A$ for $P(┌A┐)$. A standard arithmetical provability predicate obeys the Hilbert-Bernays-Löb derivability conditions:The first condition is a rule about theorems; the latter two are statements proved within the first-order theory. When the terminology also includes external adequacy, require that $P(\bar{n})$ be true in the standard natural numbers exactly when the logical formula with code $n$ has an $S$-proof.

For Peano arithmetic, code finite strings, logical formulas and formal proof sequences by natural numbers. Formal proof checking is effective and has an arithmetized bounded checking relation with its computation data. Existentially assert the existence of a correct formal proof code and any required checking data. The resulting predicate symbol is Sigma-1, and its standard truth means exactly that an actual finite PA formal proof exists. Effective operations on formal proof codes combine an implication formal proof with an antecedent formal proof, and turn a formal proof into a formal proof that its code is a formal proof; PA verifies those constructions. They give distribution and positive introspection, while an actual theorem's finite formal proof gives its internal provability assertion. This is the requested rough construction, with the three internal formal proof properties distinguished from external agreement.

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 $S$.

Use the Hilbert-Bernays-Löb derivability conditions for an arithmetical provability predicate. Abbreviate $P_S(┌A┐)$ by $PA$. The conditions needed below areThese are the usual theoremhood, distribution and positive-introspection conditions intended by a standard arithmetical provability predicate; arbitrary predicate symbols merely agreeing with provability externally need not satisfy them.

Since $S⊢\neg Y$, it proves $Y\to G$. The first two conditions give $S⊢PY\to PG$. For the converse, the fixed-point equivalence gives $S⊢G\to (PG\to Y)$. Applying theoremhood and distribution givesPositive introspection also gives $PG\to PPG$, and distribution combines these to yield $S⊢PG\to PY$. Thus $S⊢PG\leftrightarrow PY$. Combining with $G\leftrightarrow \neg PG$ provesIf $S$ proved $\neg PY$, it would prove $G$. The theoremhood condition would then give $PG$, while the fixed-point equivalence gives $\neg PG$. Therefore first-order consistency implies

Let $j=i+n$ with $n\ge 1$. Positive introspection supplies the theorem $F_k\to F_{k+1}$ for every $k\ge 1$, since $F_k$ itself has the form $P_S(┌F_{k-1}┐)$.

Induction on $n$ consequently givesAt $n=1$ this is the identity implication. To pass from $n$ to $n+1$, compose the induction hypothesis with the positive-introspection implication at $k=i+n$, which is at least one. Composing the result with the assumed $F_j\to F_i$ givesi.e. $S⊢P_S(┌F_i┐)\to F_i$. Löb's rule now yieldsThe index restriction on positive introspection has been respected even when $i=0$: the chain begins at $F_{i+1}$, which already is a provability assertion.

The source's system called $K$ already contains positive introspection, so its displayed axioms are those usually called $K4$. PutWe first show that its axioms prove $\square A\to A$, the implication suggested in the hint. From $B$, positive introspection gives $\square B$. Distribution applied to $\square A=\square (B\to \square X)$ and $\square B$ gives $\square \square X$. Distribution applied to $B=\square (\square X\to X)$ and this last result gives $\square X$. Thus the theorem-level implications combine by propositional reasoning intowhich is $K4⊢\square A\to A$.

Since $S$ contains these axioms, it proves that implication. Its assumed Löb inference rule, applied to the first-order sentence $A$, yields $S⊢A$. Unpacking $A$ provesThis uses necessitation only as part of the ordinary modal theorem rules; no inference of a boxed temporary assumption has been made.

Under the arithmetical interpretation $\square A=P_S(┌A┐)$, the Löb axiom says that a formal proof of the local reflection implication for $A$ already entails a formal proof of $A$:It is a formalized constraint on self-verification of a formal proof system. Externally, Löb's theorem says that if the system itself proves its own local reflection implication for $A$, it proves $A$. Distribution reflects closure of formal proofs under implication, necessitation reflects that a theorem has an arithmetically recognizable formal proof, and positive introspection reflects formal proof verification inside arithmetic.

Taking $A$ to be a refutable first-order sentence, especially a contradiction, recovers the obstruction behind the second incompleteness theorem: a first-order consistent sufficiently strong system cannot prove the corresponding claim that no formal proof of that first-order sentence exists. This does not identify provability with truth, nor grant the unrestricted factivity principle $\square A\to A$ as a theorem of the system. The Löb axiom is the characteristic additional scheme of Gödel–Löb provability logic.

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.

GL has all classical propositional tautologies, the normal distribution axiom $\square (A\to B)\to (\square A\to \square B)$ and the Löb axiom $\square (\square A\to A)\to \square A$, closed under uniform substitution, modus ponens and necessitation of theorems. We use its derived positive-introspection theorem $\square A\to \square \square A$, normal boxed reasoning, and the boxed replacement principle supplied in the question.

For completeness, positive introspection follows from these axioms. Put $C=A\wedge \square A$. Distribution gives $\square C\to \square A$, so propositional logic proves $A\to (\square C\to C)$. Necessitate and distribute to get $\square A\to \square (\square C\to C)$; Löb's axiom gives $\square A\to \square C$, and boxing $C\to \square A$ finishes $\square A\to \square \square A$.

For a GL fixed point by truth substitution, set $\top =(\perp \to \perp )$, $U=\square Y(\top )$ and $V=\square Y(U)$. From the tautology $U\to (\top \leftrightarrow U)$, boxed reasoning gives $\square U\to \square (\top \leftrightarrow U)$. The supplied replacement principle and distribution then giveSince $U$ is itself boxed, positive introspection gives $U\to \square U$, and hence $U\to V$.

For the reverse direction, the displayed implication entails the theorem $V\to (\square U\to U)$. Necessitate it and distribute to obtain $\square V\to \square (\square U\to U)$. Since $V$ is boxed, $V\to \square V$ holds. The Löb axiom therefore yields $V\to \square U$, and the same displayed implication then yields $V\to U$. ConsequentlyAll occurrences of the unknown letter have been removed from $U$ by substitution of truth; any other parameter letters remain available. The proof does not assume the invalid reflection principle $\square U\to U$ as an unconditional axiom.

The fixed point of boxed negated provability gives, in fact, that GL proves $F\leftrightarrow \square \perp$. Here is a direct derivation. Since $F$ is provably equivalent to a boxed formula, positive introspection and necessitation of that equivalence give $F\to \square F$. Another introspection step gives $F\to \square \square F$. The stipulated fixed point also gives $F\to \square \neg \square F$. Combining these two boxed contradictories yields $F\to \square \perp$. Conversely $\square \perp \to \square \neg \square F$ follows by normal boxed explosion, and the fixed point then yields $\square \perp \to F$.

Arithmetical soundness of GL, proved by translating distribution into the second derivability condition, necessitation into the first, and Löb's axiom into its internal arithmetic theorem, now givesAny remaining propositional parameters may be assigned arbitrary arithmetic sentences; this equivalence is uniform in those assignments. If PA is one-consistent, it is consistent and ${\Sigma }_1$ sound. Its standard provability predicate for a contradiction is then a false ${\Sigma }_1$ sentence and cannot be a theorem. Hence $F^{\ast }$ is not provable in PA. Moreover its negation is equivalent to PA's consistency statement and is unprovable by the second incompleteness theorem. Thus this fixed point is, under the stated hypothesis, undecidable rather than an unboxed reflection principle.

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

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.