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Let $T=\mathrm{ZFC}+\text{“there is a worldly cardinal”}$ and $T^{\ast }=\mathrm{ZFC}+\mathrm{Con}(\mathrm{ZFC})$. A worldly $\kappa$ makes $V_{\kappa }$ a model of ZFC, and arithmetic absoluteness for a rank-initial model makes it a model of $T^{\ast }$, so $T$ proves $\mathrm{Con}(T^{\ast })$. If $T^{\ast }$ proved $\mathrm{Con}(T)$, then the stronger theory $T$ would prove its own consistency, contradicting the Gödel second incompleteness theorem. Thus, assuming consistency, $T^{\ast }{<}_{\mathrm{Cons}}T$.

Starting from a recursively axiomatized theory $T_0$, defineAssuming 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.

WriteIf $\kappa$ is a worldly cardinal, then $V_{\kappa }\models \mathrm{ZFC}$. The existence of this set model proves $\mathrm{Con}(\mathrm{ZFC})$ in the universe. By arithmetic absoluteness for a rank-initial model, the same formal consistency statement holds in $V_{\kappa }$. Hence $V_{\kappa }\models T^{\ast }$, so $T$ proves $\mathrm{Con}(T^{\ast })$.

Conversely, suppose $T^{\ast }$ proved $\mathrm{Con}(T)$. The theory $T$ proves every axiom of $T^{\ast }$, since a worldly cardinal proves $\mathrm{Con}(\mathrm{ZFC})$. It would therefore also prove $\mathrm{Con}(T)$, contrary to the Gödel second incompleteness theorem when $T$ is consistent. Thus $T^{\ast }$ cannot prove $\mathrm{Con}(T)$, andThis is the consistency strength of a worldly cardinal comparison.

Let $\mathrm{IC}$ be the sentence asserting that a strongly inaccessible cardinal exists, and begin withDefine the iterated consistency progressionThe 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 $T_n{\le }_{\mathrm{Cons}}{T}_{n+1}$. The theory $T_{n+1}$ proves $\mathrm{Con}(T_n)$ by construction, whereas a consistent $T_n$ cannot prove its own consistency by Gödel second incompleteness theorem. ThereforeLikewise $T_{\infty }$ extends every $T_n$ and contains $\mathrm{Con}(T_n)$ as an axiom already at stage $n+1$, while $T_n$ does not prove it. Consequentlyassuming the stated consistency hypotheses.