If $$x-\frac{x}{2}=\frac{x}{2},$$ and $$\frac{x}{\sqrt{x}}=\sqrt{x},$$ and $$x-\uparrow(x-\uparrow^22)=x-\uparrow^22$$ when $(x\uparrow^n-[A])\uparrow^nA=x$, where $A$ is some constant, and one uses standard Knuth up-arrow notation[where ($\uparrow^{-1}x$)($-\uparrow^{-1}x$), ($\uparrow^0x$)($-\uparrow^{0}x$), ($\uparrow{x}$)($-\uparrow{x}$), ($\uparrow^2x$), and ($-\uparrow^2x$) all mean addition(and subtraction), multiplication(and division), exponentiation(and roots), and tetration(and super-roots) in that order],will $$x-\uparrow^n(x-\uparrow^{n+1}2)$$ always yield $$x-\uparrow^{n+1}2$$
for every degree $n$? If so, could someone please let me know where I can find an explanation of this identity?
Another, possibly more mathematically concise definition could be given by $$x-\uparrow^n\sum_{k=1}^{z-1}{x-\uparrow^{n+1}z}=x-\uparrow^{n+1}z$$
where the sigma could signify the sum, product, etc. depending on the degree of the up-arrow. $x\uparrow^az$ could be the $z$-th hyperpower($a$-power) of $x$ ($x$ $(n+2)$-ated to the $z$-th), or inversely, $x-\uparrow^az$ is the $z$-th hyperroot of x. Or in a different notation (MphLee's),
$$Hrt_n(x,\sum_{k=1}^{z-1}{Hrt_{n+1}(x,z))}=Hrt_{n+1}(x,z)$$
Where the $\sum$ still doesn't nessecarily represent the sum, but $z-1$ iterations of hyperoperations of degree $n$, and rank $z$.
Ex. $$Hrt_{1}(x,\sum_{k=1}^{2}{Hrt_2(x,3)})$$
Here, $Hrt_1$ represents a difference between $x$ and $\sum_{k=1}^{2}{Hrt_2(x,3)}$. Sigma ($\sum$) is operating in the first degree; addition, or $H_1$ (note that although we are using $Hrt$ and not $H$, only the subscript or 'degree' matters when defining Sigma ($\sum$)), and is the sum of 2 ($\sum_{}^{2}$) identical terms: $$Hrt_2(x,3)$$ that is $\frac{x}{3}$. Therefore,$$Hrt_{1}(x,\sum_{k=1}^{2}{Hrt_2(x,3)})=x-((\frac{x}{3})_1+(\frac{x}{3})_2)=x-2(\frac{x}{3})=\frac{x}{3}$$.
I am fairly certain that this equation is 'clumsy' to some extent, but if anyone could suggest a better form of notation, I would greatly appreciate it.
Here is a better rendering: $$Hrt_n(x,H_{n+1}(Hrt_{n+1}(x,z),[z-1]))=Hrt_{n+1}(x,z)$$ and if $Hrt_{n+1}(x,z)=T$, then$$Hrt_n(x,H_{n+1}(T,[z-1]))=T$$