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.

Solution

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

Starting from a recursively axiomatized theory $T_0$, define
$$
T_{n+1}=T_n+\operatorname{Con}(T_n).
$$
Assuming consistency, <Gödel second incompleteness theorem> makes every step strict in the <consistency-strength preorder>, while the effective union $T_\omega=\bigcup_{n<\omega}T_n$ lies strictly above every finite stage.

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 Incoming links: Iterated consistency progression