I have a proof that I need help with. Like the title says, the theorem is that every bit-reversal ring is $frac{1}{2}$-symmetric. The theorem is for Leader Election algorithm in synchronous ring. The things I know follow:

Bit reversal ring is defined as follows: We assign to each process $i$ the integer from ${0,ldots, n-1}$ whose $k$ bit representation is the reverse of the $k$ bit representation of $i$. $n$ is also a power of two, $n=2^{k}$.

Two segments $U$ and $V$ are order equivalent if they are the same length $k$, and for all $i$ and $j$ such that $1 leq i,j leq k$ we have that $u_{i} leq u_{j}$ if and only if $v_{i} leq v_{j}$.

Ring $R$ is $c$-symmetric if for every segment $S$ of $R$ there are at least $lfloor frac{cn}{l} rfloor$ segments that are order equivalent to $S$, including $S$ itself, where $l$ is length of the segment, and this holds for every $sqrt{n} leq l leq n$.

So after plugging all I know into formulas I get that $lfloor frac{2^{k-1}}{l} rfloor$ is formula for number of segments and $l$ is such that $2^{frac{k}{2}} leq l leq 2^{k}$.

Any hint or piece of information would be much appreciated! Thank you.