User blog comment:Mh314159/A hopefully powerful new system/@comment-35470197-20190628050713/@comment-39585023-20190628154214

I could not precisely understand cycle factorial rule, because it is formalised only for the specific cases [p]\[1][a] and [p]\[a][1]. Describing rule sets by the natural language is very ambiguous, and hence I could not understand why [1]\[3][2] = 1\[2]^{[3][1]}[2]. Could you explain the full rule sets of cycle factorial rule using mathematical formulae?

1 hour ago by P進大好きbot

Thanks a Moser! The cycle reduction applies to any [1][a] or [a][1] expressions to the right of the slash. Otherwise, the o through n rules apply. Only when [1][a] or [a][1] expression is reached is it replaced by an expression of the form [p-1][u] and everything else then continues to recurse using rules o through n. So 1\[3][2] recurses using rule nn to 1\[2]^{[3][1]}[2] just as if the 1\ wasn't there. Then I do think I made a mistake in the definitions of [p]\[1][a] and [p]\[a][1] by recursing p in the wrong spot, and I have corrected it. [p]\[1][a] = [p]\[u] where u = [p-1]\[1][a] etc. terminating at 0\[1][a] = [1][a]. So the [3][1] part of the expression is replaced by [1]\[u] with u = 0\[3][1] = [3][1] = 4 and since 1\[4] = [4,4,4,4,][4,4,4,4][4,4,4,4][4,4,4,4], this then becomes the functional exponent and to notate that it is a terminated number that will not be further recursed by the cycle rule, we retain the 0\, so we end up with 1\[2]^(0\[4,4,4,4,][4,4,4,4][4,4,4,4][4,4,4,4]0[3] which means a long string of 1\[2]...[2] functions.  To me this make sense in English, but I'm not sure how to formalize it mathematically, I guess I would need help with that.