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

I tried the method explained in the "Reality use" section and it does not seem to work. I'm not sure if I'm misinterpreting your explanation or the method is wrong.

Problem 1.

Let \(\alpha=C(C(C(0,\Omega_2),0),0)\).

According to step 2, we have to change \(C(C(0,\Omega_2),0)\) to \(C(\Omega_2,C(\Omega_2,0))\), which gives \(\beta=C(C(\Omega_2,C(\Omega_2,0)),0)\). Then, because of step 5, \(\beta\) will become \(C(C(C(\Omega_2,\Omega_2),C(\Omega_2,0)),0)\). This is nonstandard so we have to go back to step 1, which gives \(\beta=C(C(C(0,\Omega_2),C(\Omega_2,0)),0)\). This is nonstandard, and since the expression already has 4 \(C\)s, we can never find \(\alpha[1]\).

Problem 2.

Let \(\alpha=C(C(\Omega_2,C(\Omega_2,0)),0)\).

Step 2 changes \(\beta\) to \(C(C(0,C(\Omega_2,0)),0)\). Then step 5 changes 0 to \(C(\Omega_2,0)\), which leads to \(\beta=C(C(C(\Omega_2,0),C(\Omega_2,0)),0)\). Since this is standard this will be \(\alpha[1]\). However, to find \(\alpha[2]\), we have to repeat step 4 and step 5 once more. \(\Omega_2\) will become \(C(\Omega_2,\Omega_2)\) so now \(\beta=C(C(C(C(\Omega_2,\Omega_2),0),C(\Omega_2,0)),0)\). This is nonstandard so we have to go back to step 1. Step 1 gives \(\beta=C(C(C(C(0,\Omega_2),0),C(\Omega_2,0)),0)\). This is nonstandard, and since the expression already has 5 \(C\)s, we can never find \(\alpha[2]\).