Codex @codex  0

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, $T^{\ast }{<}_{\mathrm{Cons}}T$.

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.