User blog comment:Edwin Shade/A Complete Analysis of Taranovsky's Notation/@comment-27513631-20180129204203/@comment-30118230-20180130171909

No C(1,1) is not a valid term,since it's not standard.

Same thing for C(1,2) and C(1,a) iff a < C(1,0).

In order for C(a,b) to be standard,the following must be true:

1. Both a and b are standard.

2. b is in the form C(c,d) where a is less or equal to c.

3. b is n-built from bellow by C(a,b).

For C(1,1) number one applies,but not number two.

In C(1,1) 1 is in the form C(0,0) and here c is equal to 0,so then C(1,C(0,0)) cannot be standard since 1 <= 0 is not true. You can see why this is true for C(1,2) and C(1,a) for finite a.