User blog:Wythagoras/BEAF:structures and ordinals

First we define \((X\uparrow^n\#)\uparrow^nX= X\uparrow^n(\#+X)\)

Proving \(X\uparrow^{n+1}X=\varphi(n,0)\)
'''Lemma 1. \(\#\uparrow\uparrow X = \varepsilon_{\alpha+1}\), where \(\alpha\) is the ordinal for \(\#\)''' First, we shall prove \(\#\uparrow\uparrow X = \varepsilon_{\alpha+1}). We know that \(\#\uparrow\uparrow X = lim(\#,(\#)^{\#},(\#)^{(\#)^{\#}}...)\) and \(\varepsilon_{\alpha+1} = lim(\varepsilon_{\alpha},\varepsilon_{\alpha}^{\varepsilon_{\alpha}},\varepsilon_{\alpha}^{\varepsilon_{\alpha}^{\varepsilon_{\alpha}}}...)\).

Since \(\varepsilon_0 = X \uparrow\uparrow X\), \(\varepsilon_k = X \uparrow\uparrow X(k+1)\), by induction.

Then \(X\uparrow\uparrow X^2 = \varepsilon_\omega\), \(X\uparrow\uparrow X^3 = \varepsilon_{\omega^2}\), \(X\uparrow\uparrow X^X = \varepsilon_{\omega^\omega}\), \(X\uparrow\uparrow X\uparrow\uparrow X = \varepsilon_{\varepsilon_0}\) and \(X\uparrow\uparrow\uparrowX= \zeta_0\)

\(X\uparrow\uparrow\uparrowX\uparrow\uparrowX= \varepsilon_{\zeta_0+1}\) (by lemma 1), \(X\uparrow\uparrow\uparrowX\uparrow\uparrow(X\uparrow\uparrow\uparrowX)= \varepsilon_{\zeta_02}\), \(X\uparrow\uparrow\uparrowX\uparrow\uparrowX= \varepsilon_{\varepsilon_{\zeta_0+1}}\) and \(X\uparrow\uparrow\uparrowX2= \zeta_1\). We can make a proof similiar to lemma 2 for zeta, eta, and beyond. Therefore \(X\uparrow^{n+1}X=\varphi(n,0)\). (A more formal proof will come tomorrow.