User blog comment:KthulhuHimself/Norminals Revisited./@comment-27173506-20160204132906/@comment-1605058-20160204175121

I personally don't see much similarity between definition of Orda and of N<0,a>. Additionally, although I reserve myself right to be wrong, I think FOOT with just single Ord is already stronger than all of N<0,a>. A justification for that is the following: while working in N<0>, we only consider sets, which are elements of $$V_{Ord}$$. When we go to N, we only get access to elements $$V_{Ord+a}$$, and N<0,1> lets us speak of elements of $$V_{Ord+Ord}=V_{Ord2}$$. Similarly with N<<0,1>,1> we can get to $$V_{Ord3}$$ and so on, with N<0,2> reaching $$V_{Ord\omega}$$. I don't know what stages of cumulative hierarchy we reach with N<0,a>, but I have no doubt that these are ordinals easily definable in FOOT with Ord only.