User blog comment:MilkyWay90/My ordinal function/@comment-35470197-20180630144403/@comment-35470197-20180709143613

> Yeah, but I didn't use addition.

Even though you did not use addition, \(F\) can be computed as addition by your definition. In order to go beyond the addition, you need an ordinal notation or some other technique.

> Fixed

\(F(\beta_0,\beta_1,\ldots,0,a(n))\) is not defined. If you define it in a natural way, then it coincides with the addition \(F(\beta_0 + \beta_1 + \cdots,a(n))\) again.