User blog:Hyp cos/DAN not into Taranovsky's notation

The (n+1)-separator in DAN cannot be embedded into the n-th system of Taranovsky's ordinal notation. In Taranovsky's ordinal notation, consider this kind of expressions: let \(f(x)=C(C(C(C(x,\beta_t),\beta_{t-1})\cdots,\beta_2),\beta_1)\), and \(\alpha>f(\alpha)\), then \(f(f(\cdots f(f(\alpha))))\) (with n+1 f's) is not standard in the n-th system.

Proof
Here ordinals are in postfix form. So \(f(x)=\beta_1\beta_2\cdots\beta_{t-1}\beta_txCC\cdots CC\).

Let \(\gamma_i=\beta_i\beta_{i+1}\cdots\beta_{t-1}\beta_tf^n(\alpha)CC\cdots CC\) for \(1\le i\le t\). So \(\gamma_1=f^{n+1}(\alpha)\). Let k be such that \(\gamma_k\le\gamma_i\) for all \(1\le i\le t\), then there is no \(\beta_i\beta_{i+1}\cdots\beta_{t-1}\beta_tf^j(\alpha)CC\cdots CC\) 0-built from below from \(\gamma_k\) (for all i and j).

Suppose \(\gamma_1\) is standard, then \(\gamma_k\) is standard, then \(\gamma_{k+1}\) is n-built from below from \(\gamma_k\). Let \(k_0=k+1\), \(k_{i+1}=\min\{j\le t|\gamma_j>\gamma_{k_i}\}\), until \(k_m\) (there is no such j that \(k_m\gamma_{k_m}\)). Let \(\delta_i=\beta_{k_m}\beta_{k_m+1}\cdots\beta_{t-1}\beta_tf^i(\alpha)CC\cdots CC\), then \(\gamma_{k_m}=\delta_n\), \(\delta_i\) is a subterm of \(\delta_j\) and \(\delta_i>\delta_j\) for i < j. So we have an antichain of \(\gamma_{k_0}<\gamma_{k_1}<\cdots<\gamma_{k_{m-1}}<\gamma_{k_m}=\delta_n<\delta_{n-1}<\cdots<\delta_1<\delta_0\) with m+n+1 different expressions; "\(\gamma_{k+1}\) is n-built from below from \(\gamma_k\)" needs \(\gamma_{k_n}\) or \(\delta_m\) to be "0-built from below from \(\gamma_k\)" but actually not, so \(\gamma_1\) is not standard.