M(m,n) map

m(m,n) map is a function which maps maps to maps. It was defined by Japanese googologist Fish in 2007 and used to define Fish number 6. It is a 2 variable extention of m(n) map and has a growth rate of \(f_{\zeta_0}(n)\).

Definition
\(M[m,n]\) map \((m=0,1,...; n=1,2,...)\) is defined as follows.


 * \(M[0,1]\) is a set of function of natural numbers to natural numbers.
 * \(M[m+1,1] (n=1,2,...)\) is a set that has one element from each \(M[m,n]\); \(M[m+1,1]\) is a of \(M[m,n]\).
 * Element of \(M[m,1]\) also works as a function in \(M[0,1]\) that is element of element of ... element of \(M[m,1]\).
 * \(M[m,n+1] (n=1,2,...)\) is a set of maps from \(M[m,n]\) to \(M[m,n]\).

Element of \(M[m,n]\), \(m(m,n)\) is defined as follows. Here, \(a_i, b_i,f_i\) are elements of \(m(m,i)\) and strict structure of definition is same as m(n) map.

\begin{eqnarray*} m(0,1) (x) & := & x+1 \\ m(m,n+1) f_n f_{n-1} ...f_1 (x) & := & {f_n}^x f_{n-1}... f_1 (x) \\ & & (m=0; n=1,2,... \text{ or } m=1,2,...; n=2,3,…) \\ m(m+1,1) & := & [m(m,1),m(m,2),m(m,3),…] \\ m(m+1,2)[a_1,a_2,...] & := & [b_1,b_2,…] \text{ where } b_n \text{ is defined as:} \\ b_n f_{n-1}...f_1(x) & := & a_y a_{y-1}...a_n f_{n-1}…f_1(x) (y=max(x,n)) \end{eqnarray*}

Calculation
\begin{eqnarray*} m(1,1)(x) & = & [m(0,1),m(0,2),m(0,3),…](x) \\ & = & m(0,1)(x) = x+1 \\ \end{eqnarray*} Let \(m(1,2) m(1,1) = [a_1,a_2,a_3,…]\) and \begin{eqnarray*} a_1(x) & = & m(0,x) m(0,x-1) … m(0,1) (x) \\ & \approx & f_{\varepsilon_0}(x) \\ \therefore m(1,2) m(1,1)(x) & \approx & f_{\varepsilon_0}(x) \\ m(0,2) a_1(x) & \approx & f_{\varepsilon_0 + 1}(x) \\ m(0,3) m(0,2) a_1 (x) & \approx & f_{\varepsilon_0 + \omega}(x) \\ m(0,4) m(0,3) m(0,2) a_1 (x) & \approx & f_{\varepsilon_0 + \omega^{\omega}}(x) \\ m(0,5) m(0,4) m(0,3) m(0,2) a_1 (x) & \approx & f_{\varepsilon_0 + \omega^{\omega^{\omega}}}(x) \\ a_2 a_1(x) & = & m(0,y) m(0,y-1) …m(0,2) a_1(x) (y=max(x,2)) \\ & \approx & f_{\varepsilon_0 \times 2}(x) \\ \end{eqnarray*}

Then,

\begin{eqnarray*} m(0,3) a_2 a_1 (x) & \approx & f_{\varepsilon_0 \times \omega}(x) \\ m(0,4) m(0,3) a_2 a_1 (x) & \approx & f_{\varepsilon_0 \times \omega^{\omega}}(x) \\ m(0,5) m(0,4) m(0,3) a_2 a_1 (x) &\approx & f_{\varepsilon_0 \times \omega^{\omega^{\omega}}}(x) \\ a_3 a_2 a_1(x) & = & m(0,y) m(0,y-1) ... m(0,3) a_2 a_1 (x) (y=max(x,3)) \\ & \approx & f_{\varepsilon_0 ^2}(x) \end{eqnarray*} As for \(a_4\), \begin{eqnarray*} m(0,4) a_3 a_2 a_1 (x) & \approx & f_{\varepsilon_0 ^\omega}(x) \\ m(0,5) m(0,4) a_3 a_2 a_1 (x) & \approx & f_{\varepsilon_0 ^{\omega \times 2}}(x) \\ m(0,6) m(0,5) m(0,4) a_3 a_2 a_1 (x) & \approx & f_{\varepsilon_0 ^{\omega ^ 2}}(x) \\ m(0,7) m(0,6) m(0,5) m(0,4) a_3 a_2 a_1 (x) & \approx & f_{\varepsilon_0 ^{\omega^{\omega}}}(x) \\ a_4 a_3 a_2 a_1(x) & = & m(0,y) m(0,y-1)...m(0,4) a_3 a_2 a_1(x) (y=max(x,4)) \\ & \approx & f_{\varepsilon_0 ^{\varepsilon_0}}(x) \end{eqnarray*}

And similarly, \(a_5\), \(a_6\) can be calculated as \begin{eqnarray*} a_5 a_4 a_3 a_2 a_1(x) & \approx & f_{\varepsilon_0 ^{\varepsilon_0^{\varepsilon_0}}}(x) \\ a_6 a_5 a_4 a_3 a_2 a_1(x) & \approx & f_{\varepsilon_0 ^{\varepsilon_0^{\varepsilon_0^{\varepsilon_0}}}}(x) \\ \end{eqnarray*}

Therefore \begin{eqnarray*} m(1,2)^2 m(1,1) (x) &=& m(1,2)[a_1,a_2,...](x) \\ & \approx & f_{\varepsilon_0^\omega}(x) \\ &=& f_{\varepsilon_1}(x) \end{eqnarray*}

Now let \begin{eqnarray*} m(1,2)^3 m(1,1)(x) = [b_1,b_2,b_3,...](x) \end{eqnarray*} and for calculating \(b_i\), \(\varepsilon_0\) in the calculation of \(a_i\) is changed to \(\varepsilon_1\). Therefore,

\begin{eqnarray*} m(1,2)^3 m(1,1)(x) & \approx & f_{\varepsilon_2}(x) \\ m(1,2)^4 m(1,1)(x) & \approx & f_{\varepsilon_3}(x) \\ m(1,2)^n m(1,1)(x) & \approx & f_{\varepsilon_{n-1}}(x) \\ \end{eqnarray*}

Then calculation goes on similar to \(m(n)\) map. \begin{eqnarray*} m(1,3) m(1,2) m(1,1) (x) & \approx &  f_{\varepsilon_\omega} \\ m(1,4) m(1,3) m(1,2) m(1,1) (x) & \approx &  f_{\varepsilon_{\omega^\omega}} \\ m(1,5) m(1,4) m(1,3) m(1,2) m(1,1) (x) & \approx &  f_{\varepsilon_{\omega^{\omega^\omega}}} \\ m(2,2) m(2,1) (x) &=& m(1,x) m(1,x-1) ... m(1,2) m(1,1) (x) \\ & \approx & f_{\varepsilon_{\varepsilon_0}} \\ \end{eqnarray*}

And then \begin{eqnarray*} m(2,2)^2 m(2,1) (x) & \approx & f_{\varepsilon_{\varepsilon_1}}(x) \\ m(2,2)^3 m(2,1) (x) & \approx & f_{\varepsilon_{\varepsilon_2}}(x) \\ m(2,2)^4 m(2,1) (x) & \approx & f_{\varepsilon_{\varepsilon_3}}(x) \\ m(2,3) m(2,2) m(2,1) (x) & \approx &  f_{\varepsilon_{\varepsilon_\omega}}(x) \\ m(3,2) m(3,1) (x) &=& m(2,x) m(2,x-1) ... m(2,2) m(2,1) (x) \\ & \approx & f_{\varepsilon_{\varepsilon_{\varepsilon_0}}}(x) \end{eqnarray*}

As we have calculated to \(m(3,2) m(3,1) (x)\), it is easy to see that \begin{eqnarray*} m(1,2) m(1,1) (x) & \approx & f_{\varepsilon_0} (x) \\ m(2,2) m(2,1) (x) & \approx & f_{\varepsilon_{\varepsilon_0}} (x) \\ m(3,2) m(3,1) (x) & \approx & f_{\varepsilon_{\varepsilon_{\varepsilon_0}}} (x) \\ m(4,2) m(4,1) (x) & \approx & f_{\varepsilon_{\varepsilon_{\varepsilon_{\varepsilon_0}}}} (x) \\ \end{eqnarray*} and therefore

\[m(x,2)m(x,1)(x) \approx f_{\zeta_0}(x)\]