User blog comment:Hyp cos/Fundamental Sequences in Taranovsky's Notation/@comment-32697988-20190204142752/@comment-11227630-20190205001230

The \(\beta\)'s are "terms" but may not be standard. Having the expression with "full amount of C's" is not the end of the process, because step 2 can still apply.

In "Problem 1", let \(\alpha=C(C(C(0,\Omega_2),0),0)\), to get \(\alpha[1]\), we have these changes: Finally C(C(C(0,Ω2),0),0)[1] = C(C(Ω2,C(Ω2,C(Ω2,0))),0).
 * Step 2. C(C(C(0,Ω2),0),0) -> C(C(Ω2,C(Ω2,0)),0)
 * Step 3. C(C(Ω2,C(Ω2,0)),0) -> C(C(C(Ω2,Ω2),C(Ω2,0)),0)
 * Step 2. C(C(C(Ω2,Ω2),C(Ω2,0)),0) -> C(C(C(0,Ω2),C(Ω2,0)),0)
 * Step 2. C(C(C(0,Ω2),C(Ω2,0)),0) -> C(C(Ω2,C(Ω2,C(Ω2,0))),0)

In "Problem 2", C(C(C(C(0,Ω2),0),C(Ω2,0)),0) can further (by step 2) change into C(C(C(Ω2,C(Ω2,0)),C(Ω2,0)),0).