# How to prove something that works with each operator?

For example: $2+2=4$, $2*2=4$, $2^2=4$, even $2$ in tetration of $2$ equals $4$. (Tetration example: $3$ of $4$ equals to $(((3)^3)^3)^3$)

Mathematic have a way to prove those things like: $2$ [any operator here] $2 = 4$? But any form of this case. (Like replace those $2$ with $x$ and $y$ and $4$ with $z$)

I need an example of how to prove the example given here to understand the basic I think.

In general, these operations are known as hyperoperations, we'll here denote them $x H_n y$. It's fairly tautological that $2H_n2=4$, but it can be seen as inductively following from the definitions:
Base case: $2 + 2 = 4$.
Inductive step: Suppose $2H_n2=4$. Then $2H_{n+1}2=2H_n2$ by definition, and $2H_{n+1}2=4$. $\square$
• How by definition $2H_{n+1}2=2H_n2$ is true? Sep 29, 2014 at 18:53
• @KugBuBu In the most typical case, $aH_nb = aH_{n-1}(aH_n(b-1))$. Plugging in, you get $$2H_{n+1}2=2H_n(2H_{n+1}(1)) = 2H_n2$$ with the last equality following again from the definition. Sep 29, 2014 at 22:16
• Just to note that your example is wrong - 3 of 4 is $3^\left(3^\left(3^3\right)\right)$ , not $((3^3)^3)^3)$. Jan 13, 2019 at 10:59