## FANDOM

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$10\underbrace{\uparrow\uparrow\cdots\uparrow\uparrow}_{10\underbrace{\uparrow\uparrow\cdots\uparrow\uparrow}_{10\uparrow^6 10\uparrow^6 6}10}10$

$8\uparrow^{10} 9 \rightarrow 8\uparrow^{10} 9 \rightarrow 3 \rightarrow 4$ .

$\alpha\uparrow\uparrow\omega=\epsilon_{\alpha+1}$と書かれるので、

この極限は$\vartheta(\epsilon_{\Omega+1})$となります。

$J(a_1,a_2,\cdots ,a_{n-1},a_n)=D(a_{n [J(a_1,a_2,\cdots ,a_{n-1})]})$

$D(a_{n [1]})$

$\underbrace{10\cdots10}_{\underbrace{10\cdots10}_{\underbrace{10\cdots10}_{\underbrace{10\cdots10}_{\underbrace{10\cdots10}_{\underbrace{10\cdots10}_{\underbrace{10\cdots10}_{\underbrace{10\cdots10}_{\underbrace{10\cdots10}_{10000000000}}}}}}}}}$

Well, I don't understand it wery well, but if if "x←[1@(y+1)/y]→x" goes $f_{\vartheta(\Omega_{x})}(y)$, the last suggestion of putting brackets to both sides of arrow, will go $f_{\vartheta(\Omega_{x})+1}(y)$.

$i\not\in\R$
$a\underbrace{\uparrow\uparrow\cdots\uparrow\uparrow}_{c}b$
$a\underbrace{\uparrow\uparrow\cdots\uparrow\uparrow}_{a\underbrace{\uparrow\uparrow\cdots\uparrow\uparrow}_{\underbrace{\vdots}_{a\underbrace{\uparrow\uparrow\cdots\uparrow\uparrow}_{c}b}}b}b$