How to prove that $x\uparrow \uparrow 1/2 = \sqrt x_s$ [closed]

This may be a stupid question but when we work with exponentiation we can see that $x^{\frac 12}=\sqrt x$ because:

$x^{\frac 12}\times x^{\frac 12}=x^{\frac 12+\frac 12}=x^1=x$

and

$\sqrt x \times \sqrt x={\sqrt x}^2=x$

Now it seems obvious working with tetration that $x\uparrow \uparrow \frac 12 = \sqrt x_s$ (where $\sqrt x_s$ is the super square root so that $\sqrt x_s^{\sqrt x_s}=x$) but I'm not sure so how do I actually prove/disprove this ? Can it be generalized for higher degree operations like $x\uparrow \uparrow \uparrow \frac 12=\sqrt x_{ss}$ (where $\sqrt x_{ss}\uparrow \uparrow \sqrt x_{ss}=x$) and so on ?

closed as off-topic by Did, Jonas Meyer, C. Falcon, Leucippus, ShaileshJul 3 '16 at 1:01

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This would only be obvious if there's a rule that $(a\uparrow\uparrow b)\uparrow\uparrow c = a\uparrow\uparrow bc$, because in that case $a\uparrow\uparrow (1/b)$ ought to be some number $x$ such that $x\uparrow\uparrow b=a\uparrow\uparrow 1$. However such a rule does not hold in general. For example with $a=b=c=2$, $$(2\uparrow\uparrow 2)\uparrow\uparrow 2 = 4^4 = 256$$ but $$2\uparrow\uparrow (2\cdot 2) = 2^{2^{2^2}} = 65536$$