User blog comment:Wythagoras/Dollar function formal definition/@comment-5529393-20130624121851

Unfortunately, it looks like a$[0,1]_2 is smaller than a$[0,1].

For we have a$[0,1]_2 = a$[...[[0,0]_2]...] = a$[...[[0_2]...] = a$[1]_2.

SImilarly, a$[0,1]_X = a$[1]_X. So a$[0,1]_X is less than a$[0,1] for X less than [0,1], and a$[0,1]_[0,1] will have ordinal Gamma_0, same as a$[0,1].

Similarly, a$[0,1]_[0,1]_X is less than a$[0,1]_[0,1] for X less than [0,1], so a$[0,1]_[0,1]_[0,1] will also have ordinal Gamma_0. So a$[0,1]_[0,1]...[0,1] will have ordinal Gamma_0 for any number of [0,1]'s.

So a$[0,2] = a$0,1]_[0,1]...[0,1] has ordinal Gamma_0 as well. I'm pretty sure this is not what you want, so your notation needs some fixing.