User blog comment:Plain'N'Simple/A question for proof-theory experts/@comment-35392788-20191029194318/@comment-35470197-20191030010257

@Syst3ms

> I conjecture that a function F outgrowing all functions provably terminating in some theory T outgrows ～

Literary it is not a precise statement, because f_α heavily depends on the choice of fundamental sequences. I note that we have no method to equip PTO(T) with a primitive recursive system of fundamental sequences canonically chosen using the data of T.

So in order to argue in a precise manner, we take a sufficiently large countable ordinal γ and equip it with a recursive syst3m of fundamental sequences. Your conjecture can be formalised in this way: For any recursive theory T including arithmetic, if PTO(T) is smaller than γ, does F_T (your F corresponding to T) surpasses f_α for any α < PTO(T) with respect to the fixed system of fundamental sequences below γ.

The answer is false. Since the system of fundamental sequences below γ has nothing to do with the provability of T, f_α can be very strong. A trivial counter example is given by setting ω[n] = F_PA(n). Then f_ω is strictly greater than F_PA(n), while ω < PTO(PA).

The point is that PA proves the well-foundedness of ω with respect to the canonical ordinal notation and the canonical fundamental sequences. Even if an ordinal is smaller than the PTO, the totality of the associated FGH is not necessarily provable with respect to a given ordinal notation and fundamental sequences. It is known that for any consistent recursive theory T including arithmetic, there is an ordinal notation whose ordinal type is ω such that its well-foundedness is not provable under T.

When we consider a specific situation, e.g. γ=ε_0 and the system of fundamental sequences is given by Wainer hierarchy, then I do not know the answer.