User blog comment:MachineGunSuper/A world without the infinite/@comment-35470197-20181208221958

There is an arithmetic model of ZFC minus Infinity, and hence FGH below \(\varepsilon_0\) works. Further, if ZFC is consistent, then the transcendental integer system in ZFC minus Infinity will not change as long as you apply the same definition i.e. do not replace the occurence of "formally provably total under ZFC" by "formally provably total under ZFC minus Infinity" in the definition. So you will get a (unprovably total but pointwise well-defined) function in ZFC minus Infinity greater than those associated to OCFs in ZFC, as long as you assume the consistency of ZFC.

On the other hand, the transcendental integer system in ZFC minus Infinity will become very weak if you replace the occurence of "formally provably total under ZFC" by "formally provably total under ZFC minus Infinity" in the definition.

If you just do not assume the existence of infinity, then the corresponding Rayo's number is not necessarily smaller than the usual Rayo's number, because you just reduce axioms. Namely, the universe \(V\) in ZFC set theory satisfies ZFC minus Infinity.

On the other hand, if you assume the negation of the existence of infinity, then the corresponding Rayo's number will completely change.