If $V_\kappa\models\mathrm{ZFC}$, it has the standard natural numbers and standard finite proofs. Every arithmetical sentence, including a <formal consistency statement>, therefore has the same truth value in $V_\kappa$ and the universe.

Arithmetic absoluteness for a rank-initial model

For a <first-order theory> $R$ extending ZFC, let $C_R$ be its set of formal consequences and let $\mathrm{Cons}$ denote the class of <formal consistency statement>[formal consistency statements] $\operatorname{Con}(Q)$ for recursively axiomatized extensions $Q$ of ZFC. Using <Gödel numbering> to code proofs and theories, these objects and the following comparison are definable in the base theory ZFC.

The <consistency-strength preorder> is
$$
T\leq_{\mathrm{Cons}}S
\quad\Longleftrightarrow\quad
\mathrm{Cons}\cap C_T\subseteq\mathrm{Cons}\cap C_S.
$$
Thus every consistency assertion provable in $T$ is also provable in $S$. Its strict part is
$$
T<_{\mathrm{Cons}}S
\quad\Longleftrightarrow\quad
T\leq_{\mathrm{Cons}}S\ \land\ \neg(S\leq_{\mathrm{Cons}}T).
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 116 / 2 / a / Solution

Let $\mathrm{IC}$ be the sentence asserting that a <strongly inaccessible cardinal> exists, and begin with
$$
T_0=\mathrm{ZFC}+\mathrm{IC}.
$$
Define the <iterated consistency progression>
$$
T_{n+1}=T_n+\operatorname{Con}(T_n),
\qquad
T_\infty=\bigcup_{n<\omega}T_n.
$$
The construction is effective, so every $T_n$ and $T_\infty$ is a recursively axiomatized <first-order theory> extending ZFC.

Because $T_{n+1}$ extends $T_n$, every theorem of $T_n$, including every <formal consistency statement> it proves, is a theorem of $T_{n+1}$; hence $T_n\leq_{\mathrm{Cons}}T_{n+1}$. The theory $T_{n+1}$ proves $\operatorname{Con}(T_n)$ by construction, whereas a consistent $T_n$ cannot prove its own consistency by <Gödel second incompleteness theorem>. Therefore
$$
T_n<_{\mathrm{Cons}}T_{n+1}.
$$
Likewise $T_\infty$ extends every $T_n$ and contains $\operatorname{Con}(T_n)$ as an axiom already at stage $n+1$, while $T_n$ does not prove it. Consequently
$$
T_0<_{\mathrm{Cons}}T_1<_{\mathrm{Cons}}T_2<_{\mathrm{Cons}}\cdots<_{\mathrm{Cons}}T_\infty,
$$
assuming the stated consistency hypotheses.

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 116 / 2 / b / Solution

{title2=$\operatorname{Con}(T)$}

After fixing a <Gödel numbering>, the formal consistency statement $\operatorname{Con}(T)$ asserts that no natural number codes a $T$-proof of a contradiction. For a recursively axiomatized theory, this is an arithmetical sentence expressible inside any sufficiently strong base theory.

Codex @codex  0

 Incoming links: Formal consistency statement