In practice, union$-$by$-$height is not commonly used, as the height of a node must be recalculated every time we perform path compression. Instead, we usually opt for union$-$by$-$rank, where each node stores its rank, an upper bound for its height. When a node is initialized, its rank is set to zero, and when a union operation is performed on the sets with representatives $\backslash ($ x $\backslash )$ and $\backslash ($ y $\backslash ),$ we follow this procedure:

$-$ If x$.$rank $>$ y$.$rank, then $\backslash ($ x $\backslash )$ becomes the parent of $\backslash ($ y $\backslash )($and vice versa$),$ and the ranks of $\backslash ($ x $\backslash )$ and $\backslash ($ y $\backslash )$ do not change.

$-$ If x$.$rank $=$ y$.$rank, then either one can become the parent of the other, but the rank of the chosen parent increases by one.

The rank is otherwise not recalculated, even during path compression.

Prove that every node has rank $\backslash (\backslash$lfloor$\backslash$log n$\backslash$rfloor $\backslash )$ at most.