User blog comment:Mh314159/A hopefully powerful new system/@comment-35470197-20190628050713/@comment-39585023-20190629185413

Yes, in my f-less notation, I intend for [2]\[5][8] = [2]\[4]^([2]\[5][7])[8] if the parentheses indicate a number that is the result of completely recursing [2]\[5][7].

I made a conscious decision to avoid using traditional function notation with subscripts and to use braces instead, so that I could nest expressions inside braces to build recursions and avoid having to write out long nested subscripts (which I don't even know how to format in the comment section here -- where are the tools?). In effect, [a] replaces fsub-a. I hope there is a way to keep this. Would something as simple as always distributing the cycle number, as above, do the job?

So using a less intimidating number, [1]\[2][2] = [1]\[1]^([1]\[2][1])[2] where [1]\[2][1] = [1][u] where u = [0]\[2][1] = 3 and therefore [1]\[2][1] = [1]\[3] = [0]\[3,3,3][3,3,3][3,3,3] which recurses to a number, making the functional exponent well defined.

So then we would have [1]\[1][1]..[2] but the cycle rule applies at each step, so how to express this? [1]\[1][([1]\[1][([1]\[1][2])])] with [3,3,3][3,3,3][3,3,3] nestings? It makes expressions much more difficult to write and interpret, I think. I prefer the rule where expressions that appear to be functional recursion numbers are really just superscripts and not function recursions until the cycle rule is applied. I thought that something like the distributive property for cycle numbers would be acceptable as long as it's clear what the superscripts really mean.

I really should send you a case of beer for all the work you've done on this, thanks.