User blog comment:Alemagno12/How strong is this function? (2)/@comment-5529393-20171111024810

Maybe I'm missing something, but it seems like f(n) =  0 for all n. The complexity of an element of S will simply be its nesting depth, i.e. the how far deep the parentheses stack in the ordinal pair. I see no reason why, for any ordinal pair s, we can't simply start our well-ordering with s, and then make sure S does not contain either subterm of s or 0. If you insist that S must contain 0, then we have f(n) = 1 for all n > 0.