See here :


for the definition of the up-arrow function.

Can $10\uparrow^n m<2\uparrow^n (m+2)$ be formally proven for all $m\ge 1$ and $n\ge 3$ ?

With Saibians theorem we get $$10\uparrow^n m<(2\uparrow^n 3)\uparrow ^n m<2\uparrow^n (m+3)$$ Also the claim is trivially true for $m=1$, but I did not manage to complete the induction step.

  • $\begingroup$ Not to sound rude or anything, but does my answer suffice? $\endgroup$ – Simply Beautiful Art Aug 4 '18 at 21:44
  • $\begingroup$ Ah, sorry. I've edited to include that now :-) $\endgroup$ – Simply Beautiful Art Aug 6 '18 at 14:59
  • $\begingroup$ I will have to go through the details, but apparently, you have proven the claim :) $\endgroup$ – Peter Aug 6 '18 at 15:02

For $n=3$, we have:

For $m=1$, we have $10\uparrow^31=10$ and $2\uparrow^33=65536$.

Assume $10\uparrow^3m<2\uparrow^3(m+2)-3$ holds for some $m\ge1$. Then we have


Hence $10\uparrow^3m<2\uparrow^3(m+2)-3<2\uparrow^3(m+2)$.

$(0)$ is due to the fact that up-arrows are strictly increasing in all arguments.

Assume it holds for some $n\ge3$ and all $m\ge1$. Then we get


$(1)$ is the result of



Can $10\uparrow^n m<2\uparrow^n (m+2)$ be formally proven for all $m\ge 1$ and $n\ge 2$ ?

No, because $10\uparrow^2 2 > 2\uparrow^2 (2+2)$.

  • $\begingroup$ Sorry, I will edit my post. $\endgroup$ – Peter Jan 15 '18 at 18:17

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