How to prove or disprove $\forall x\in\Bbb{R}, \forall n\in\Bbb{N},n\gt 0\implies \lfloor\frac{\lfloor x\rfloor}{n}\rfloor=\lfloor\frac{x}{n}\rfloor$.

Let $[\frac{x}{n}]=q$ then $q\leqslant \frac{x}{n}< q+1$. Then $nq\leqslant x < n(q+1)$. Hence $nq\leqslant [x] < n(q+1)$ and $q\leqslant [\frac{[x]}{n}]< q+1$. Thus $[\frac{[x]}{n}]=q$

Since $n$ is a natural number we can divide $x$ (with remainder) by $n$ in order to express $x = nb +r_1$ for some $b \in \mathbb{N}$ and $r_1 < n$. Now $\lfloor \frac{x}{n} \rfloor = \lfloor \frac{nb+r}{n} \rfloor = \lfloor b + \frac{r}{n} \rfloor = b$ since $r<n$. On the other hand we have that $\lfloor x \rfloor = bn + r_2$ for the $\textbf{same}$ b and with $r_2<n$. This is because n cannot divide the fractional part of x since it is a natural number. Then $\lfloor \frac{\lfloor x \rfloor}{n} \rfloor = \lfloor \frac{ bn + r_2}{n} \rfloor = b$. Therefore, $\lfloor \frac{\lfloor x \rfloor}{n} \rfloor = \lfloor \frac{x}{n} \rfloor$.