First we define \((\alpha\uparrow^n\beta)\uparrow^nX= \alpha\uparrow^n(\beta+X)\) for \(n > 1\)


Lemma 1.  \(\#\uparrow\uparrow X = \varepsilon_{\alpha+1}\), where \(\varepsilon_{\alpha}\) is the ordinal for \(\#\)

We know that

\(\#\uparrow\uparrow X = lim(\#,(\#)^{\#},(\#)^{(\#)^{\#}}...)\) and

\(\varepsilon_{\alpha+1} = lim(\varepsilon_{\alpha},\varepsilon_{\alpha}^{\varepsilon_{\alpha}},\varepsilon_{\alpha}^{\varepsilon_{\alpha}^{\varepsilon_{\alpha}}}...)\).

Lemma 2. \(\{X,X,\#+1\}=\varphi(\beta+1,0)\) where \(\beta\) is the ordinal for \(\#\)

Proof by induction:

Base case: \(\varepsilon_0 = X \uparrow\uparrow X\), this has been proven by FB100Z.

If we have \(\{X,X,\#\}\) with level \(\varphi(\beta,0)\), we have \(\{X,X2,\#\}\) with level \(\varphi(\beta,1)\). This is actually easy to prove, since the function is applied to \(\varphi(\beta,0)\) instead of \(\varphi(\beta-1,0)\), and that is exactly what the definiton is of \(\varphi(\beta,1)\).

\(\{X,X^2,\#\} = \varphi(\beta,\omega)\),

\(\{X,X^X,\#\} = \varphi(\beta,\omega^\omega)\)

\(\{X,\{X,X,\#\},\#\} = \varphi(\beta,\varphi(\beta,0))\)


This lemma can be generalized with the theta function and can be extended to:

Up to \(\{X,X,X\}\)

\(\varepsilon_0 = X \uparrow\uparrow X\),

\(\varepsilon_k = X \uparrow\uparrow X(k+1)\), using lemma 1 multiple times.


\(X\uparrow\uparrow X^2 = \varepsilon_\omega\),

\(X\uparrow\uparrow X^3 = \varepsilon_{\omega^2}\),

\(X\uparrow\uparrow X^X = \varepsilon_{\omega^\omega}\),

\(X\uparrow\uparrow X\uparrow\uparrow X = \varepsilon_{\varepsilon_0}\) and

\(X\uparrow\uparrow\uparrow X= \zeta_0\)

\(X\uparrow\uparrow\uparrow X\uparrow\uparrow X= \varepsilon_{\zeta_0+1}\) (by lemma 1),

\(X\uparrow\uparrow\uparrow X\uparrow\uparrow(X\uparrow\uparrow\uparrow X)= \varepsilon_{\zeta_02}\), \(X\uparrow\uparrow\uparrow X\uparrow\uparrow(X\uparrow\uparrow\uparrow X\uparrow\uparrow X)= \varepsilon_{\varepsilon_{\zeta_0+1}}\) and \(X\uparrow\uparrow\uparrow X2= \zeta_1\).

See also lemma 2.

Up to \(\{X,X(1)2\}\)

\(\{X,X,X\} = \varphi(\omega,0)\)

\(\{X,X,X+1\} = \varphi(\omega+1,0)\) by lemma 2.

\(\{X,X,X2\} = \varphi(\omega2,0)\)

\(\{X,X,\{X,X,X\}\} = \varphi(\varphi(\omega,0),0)\)

\(\{X,X,1,2\} = \varphi(1,0,0)\) can be proved using limit sequence.

\(\{X,X,2,2\} = \varphi(1,1,0)\) by lemma 2.

\(\{X,X,1,3\} = \varphi(2,0,0)\) can be proved using limit sequence.

\(\{X,X,1,1,2\} = \varphi(1,0,0,0)\) can be proved using limit sequence.

\(\{X,X(1)2\} = \vartheta(\Omega^\omega)\)

With lemma 2 we can easily prove this.

Up to \(\{X,X,2\}\text{&}X\)

\(\{X,X(1)2\} = \vartheta(\Omega^\omega)\)

\(\{X,X2(1)2\} = \vartheta(\Omega^{\omega2})\), we can prove it with an limit point. In general, when there are only Xs, we have to change the omegas.

\(\{X,X^2(1)2\} = \vartheta(\Omega^{\omega^2})\), see above

\(\{X,X,2(1)2\} = \vartheta(\Omega^\Omega)\)

\(\{X,X,2(1)2\}\uparrow\uparrow X = \varepsilon_{\vartheta(\Omega^\Omega)+1}\)

\(\{X,X2,2(1)2\} = \theta(\Omega^\Omega,1)\)

\(\{X,X,3(1)2\} = \vartheta(\Omega^\Omega+1)\)

\(\{X,X,4(1)2\} = \vartheta(\Omega^\Omega+2)\)

\(\{X,X,X(1)2\} = \vartheta(\Omega^\Omega+\omega)\)

\(\{X,X,1,2(1)2\} = \vartheta(\Omega^\Omega+\Omega)\)

\(\{X,X(1)3\} = \vartheta(\Omega^\Omega+\Omega^\omega)\)

\(\{X,X(1)X\} = \vartheta(\Omega^\Omega\omega)\)

\(\{X,X(1)1,2\} = \vartheta(\Omega^{\Omega+1})\)

\(\{X,X(1)1,3\} = \vartheta(\Omega^{\Omega+1}2)\)

\(\{X,X(1)1,1,2\} = \vartheta(\Omega^{\Omega+2})\)

\(\{X,X(1)(1)2\} = \vartheta(\Omega^{\Omega2})\)

\(\{X,X(1)(1)(1)2\} = \vartheta(\Omega^{\Omega3})\)

\(\{X,X(2)2\} = \vartheta(\Omega^{\Omega^2})\)

\(\{X,X(3)2\} = \vartheta(\Omega^{\Omega^3})\)

\(\{X,X(0,1)2\} = \vartheta(\Omega^{\Omega^\omega})\)

\(\{X,X((1)1)2\} = \vartheta(\Omega^{\Omega^{\Omega^\omega}})\)

\(\{X,X,2\}\text{&}X = \vartheta(\varepsilon_{\Omega+1})\)

Ad blocker interference detected!

Wikia is a free-to-use site that makes money from advertising. We have a modified experience for viewers using ad blockers

Wikia is not accessible if you’ve made further modifications. Remove the custom ad blocker rule(s) and the page will load as expected.