User blog comment:Wythagoras/Dollar function formal definition/@comment-5529393-20130616091711/@comment-5529393-20130616110202

You're sure it does because you are plugging the ordinal for the array into phi(_,0), but you have to make sure that your rules actually work that way. As they currently stand, they don't.

You said yourself that

a$[0]_([0][0]) = a$[0]_(a[0]) = a$[0]_(a-1[0])  [0]_(a-1[0])  ... [0]_(a-1[0])  [0]_(a-1[0])

so, following your lead, you get

a$[0]_(1[0]) = a$[0]_[0] [0]_[0] ,,, [0]_[0] = f_{phi(omega, 0) + 1} (a).

and therefore

a$[0]_(2[0]) = a$[0]_1[0] [0]_1[0] ,,, [0]_1[0] = f_{phi(omega, 0) + 2} (a).

and in general

a$[0]_(b[0]) = f_{phi(omega, 0) + b} (a).

so

a$[0]_([0][0]) = f_{phi(omega, 0) + omega} (a), not f_{phi(omega*2, 0)} (a).

If there is something wrong with the above, please point it out. But your comparisons must stem from the rules - you can't just extrapolate without checking that they follow the rules.