User blog comment:Ubersketch/Ordinals with transfinite FS expansions/@comment-35470197-20190701222433/@comment-35470197-20190702111303

> Since there is no minimal ordinal with infinite descending chain, then for any ordinal exists lesser ordinal with infinite descending chain. So, any ordinal has infinite descending chain, and there is no minimal ordinal.

Still wrong logic. Since there is no minimal ordinal with infinite descending chain, then just for any ordinal with infinite descending chain there exists lesser ordinal with infinite descending chain. It is impossible to remobe the bold condition. Then your proof does not work because 0 does no has an infinite descending chain.

The reason why I pointed out is because you has never used the axiom of regularity in your proof. Without using the regularity, it is impossible to verify the non-existence of the infinite descending chain unless you directly assume the corresponding axiom on ordinals. Moreover, if you use the regularity, then the non-existence immediately follows from it.