User blog comment:BlauesWasser/A function I have created (still unnamed)/@comment-35392788-20180823063358/@comment-35470197-20180823120310

@Syst3ms

Let me assist your analysis, because I am the one who recommended the OP to construct functions in order to understand fgh.

@BlauesWasser

After creating new functions, it is good to analyse them. So let us prove Syst3ms' analysis. Although I write a "full" proof here, I hope that you will try a similar (not necessarilly full, possibly more rough, but sufficiently effective) analysis by yourself again. Then you will understand more about \(f_{\omega + 1}\), which is the first brick wall.

We have \begin{eqnarray*} f_0(n) & = & n+1 \\ f_1(n) & = & f_0^n(n) = n+n =2n \\ f_2(n) & = & f_1^n(n) = 2^n n. \end{eqnarray*} This is the start point. By \(n \leq 2^n\), we have \begin{eqnarray*} n^n \leq (2^n)^{2^n} = 2^{2^n n} = 2^{f_2(n)} \leq 2^{f_2(n+1)}(n+1) = f_2^2(n+1) = f_2^2(f_0(n)) \leq f_2^3(n). \end{eqnarray*} Therefore we have \begin{eqnarray*} a^b \leq (\max \{a,b\})^{\max(a,b)} \leq f_2^3(\max(a,b)). \end{eqnarray*} and \begin{eqnarray*} a \uparrow^1 b & = & a^b \leq f_2^3(\max(a,b)) \\ a \uparrow^1 a & \leq & f_2^3(a) \\ a \uparrow^2 b & \leq & (f_2^3)^b(a) = f_2^{3b}(a) \\ a \uparrow^2 a & \leq & f_2^{3a}(a) = (f_2^a)^3(a) \leq f_3^3(a) \\ a \uparrow^3 b & \leq & (f_3^3)^b(a) = f_3^{3b}(a) \\ a \uparrow^3 a & \leq & f_3^{3b}(a) = (f_3^a)^3(a) \leq f_4^3(a) \\ & \vdots & (\textrm{induction on } c) \\ a \uparrow^c b & \leq & f_c^{3b}(a). \end{eqnarray*} We obtain \begin{eqnarray*} \{a,b\} & = & a^b \leq f_2^3(\max(a,b)) \\ \{a,b,c\} & = & a \uparrow^c b \leq f_c^{3b}(a) \leq (f_c^{\max \{a,b\}})^3(\max (a,b)) \leq f_{c+1}^3(\max(a,b)) \\ & \leq & f_{\max(a,b,c)+1}^3(\max(a,b,c)+1) \leq f_{\omega}^3(\max(a,b,c)+1) f_{\omega}^3(f_2(\max(a,b,c))) \\ \{a,b,c,d\} & = & \{a,b,\{c,d\}\} \leq \{a,b,f_2^3(\max(c,d))\} \leq f_{\omega}^3(f_2(\max(a,b,f_2^3(\max(c,d))))) \\ & \leq & f_{\omega}^3(f_2^4(\max(a,b,c,d))) \\ \{a,b,c,d,e\} & = & \{a,b,\{c,d,e\}\} \leq \{a,b,f_{\omega}^3(f_2(\max(c,d,e)))\} \leq f_{\omega}^3(f_2(\max(a,b,f_{\omega}^3(f_2(\max(c,d,e)))))) \\ & leq & f_{\omega}^3(f_2(f_{\omega}^3(f_2^3(\max(a,b,c,d,e))))) \\ \{a_1,\ldots,a_6\} & = & \{a_1,a_2,\{a_3,\ldots,a_6\}\} \leq \{a_1,a_2,f_{\omega}^3(f_2^4(\max(a_3,\ldots,a_6)))\} \\ & \leq & f_{\omega}^3(f_2(\max(a_1,a_2,f_{\omega}^3(f_2^4(\max(a_3,\ldots,a_6)))))) \leq (f_{\omega}^3 f_2^4)^2(\max(a_1,\ldots,a_6)) \\ \{a_1,\ldots,a_7\} & = & \{a_1,a_2,\{a_3,\ldots,a_7\}\} \leq \{a_1,a_2,f_{\omega}^3(f_2(f_{\omega}^3(f_2^3(\max(a_3,\ldots,a_7)))))\} \\ & \leq & f_{\omega}^3(f_2(a_1,a_2,f_{\omega}^3(f_2(f_{\omega}^3(f_2^3(\max(a_3,\ldots,a_7))))))) \leq (f_{\omega}^3f_2^3)^3(\max(a_1,\ldots,a_7)) \\ & \vdots & (\textrm{induction on } n) \\ \{a_1,\ldots,a_{2n+2}\} & \leq & (f_{\omega}^3 f_2^4)^n(\max(a_1,\ldots,a_{2n+2})) \\ \{a_1,\ldots,a_{2n+3}\} & = & (f_{\omega}^3f_2^3)^{n+1}(\max(a_1,\ldots,a_{2n+3})). \end{eqnarray*} Therefore the limit of your notation is \(f_{\omega + 1}\). I hope that this consequence helps you how strong \(\omega + 1\) is and what you should do in order to go beyond the first brick wall.

I emphasise that you do not have to write such a full proof many times. You can think "Arrows are bounded by \(f_{\omega}\), and hence this repetition is bounded by f_{\omega + 1}", once you understand how to analyse.