User blog:Alemagno12/Edwin Shade's Pair Sequence System Analysis in a more readable format

original blog post

Up to ε_0
First, there is the single element pair sequence (0,0)[n]. When (0,0) is at the end of the pair sequence, we delete it and increment n by one, so (0,0)[n] = [n+1] = n+1. In HH, H1(n) = H0(n+1) = n+1, so (0,0) = 1.

Concatenating (0,0) multiple times allows us to represent all natural numbers: adding a (0,0) at the end is equal to adding 1 to the ordinal.

Next, there is (0,0)(1,0)[n], which evaluates to (0,0)(0,0)(0,0)...[n] with n concatenations, and eventually, to 2n. In HH, Hω(n) = Hn(n) = 2n, so (0,0)(1,0) = ω.

[WIP]