# Is predicting (in the limit) computable sequences as hard as a dominating function?

Define a "predicting oracle" to be an oracle that does as described in this question.

default (weak) version:

Is it the case that, for every predicting oracle $O$, there exists an
oracle machine $M$ such that when $M$ operates on $O$, the result
computes a function that dominates all (ordinarily) computable functions?

uniform (strong) version:

Does there exist an oracle machine $M$ such that for every predicting oracle $O$, when $M$ operates on $O$, the result computes a function that dominates all (ordinarily) computable functions?

The idea behind the proof is that, for every total computable function $h$ there is a total computable function $g$ whose running time dominates $h$. So, you use the oracle to, on input x, find the greatest of the running time of all functions inferable from a prefix of length $x$. In order not to run forever, you stop counting the steps of the function predicted by the oracle if you find that it will change its prediction when a finite number of 0s are appended to the prefix (this eventually happens because the oracle has to predict every finite variation of the constantly 0 function). The trick is that for every $h$ and every sufficiently large $z$, the output is at least as big as the number of steps to compute $g(z)$, for the respective $g$.