"(5) was intended to assume that $n \in \mathbb ..."

https://arbital.com/p/4y2

by Nate Soares Jun 29 2016


(5) was intended to assume that nR1, or possibly R0 if you want an easy way to prove (6). In that case, how does (8) not follow from (5)? (If f(xy)=yf(x) in general, then f(bn)=nf(b) and f(b)=1f(bn)=n, unless I'm missing something.)