User blog comment:Mh314159/A hopefully powerful new system/@comment-35470197-20190628050713/@comment-39585023-20190629023815

Hi P進大好きbot! I'm really enjoying learning from you!

[0]\[x] where [x] represents some expression of any form from [number][number] to [s][s][s] many strings, is evaluated as if the [0]\ was not there, using the set of rules from successor base function through o to generate what I call a terminated number. When using this number in an expression of the form [p]\[x], if it is going to remain in bracketed form, it should be written as [0]\[n] so it is understand that no further cycle reductions will be applied. Alternatively, it can be represented by an unbracketed symbol. Example: [1]\[[2][2] = [1]\[1](^[2][1])[2] and since [1]\[2][1] = [1]\[3] = [3,3,3][3,3,3][3,3,3] (because [0]\[2][1] = [2][1] = 3) we can now write [1]\[[2][2] = [1]\[1]^([0]\[3,3,3][3,3,3][3,3,3])[2] or we can set [3,3,3][3,3,3][3,3,3] = m and write [1]\[[2][2] = [1]\[1](^m)[2] and the cycle rule will no longer apply to m because it is a terminated number and not in brackets.