and $y\in S_0$.

By arguments above, we have $$|C(v,S_0\cup S_1)|=2\sum_{t=0}^m\sum_{j=0}^{2k}{4k+1\choose 2j}{4m-4k-1\choose 4t-2j}-2.$$ By Theorem \[thm-z-3\] (i), we have $$|C(v,S_0\cup S_1)|=2^{4m-2}+(-1)^m2^{2m-1}-2.$$

It completes the proof.

\[lem-z-10\] Let $v$ be an element of $Z_2^{4m}$. If $v\in \overline{S_0\cup S_1}=S_2\cup S_3$, then $$|C(v,S_0\cup S_1)|=2^{4m-2}+(-1)^m2^{2m-1}.$$

We divide two cases to discuss. [**Case 1.** ]{} $v\in S_2$.\
In this case, assume that $|v|=4k+2$ where $0\le k\le m-1$. For $(x,y)\in C(v,S_0\cup S_1)$, we have $v=x+y$ and thus $x,y\in S_0$ or $x,y\in S_1$. Suppose that $x$ has $i$’s coordinates of $1$ coinciding with that of $v$, that is, $i=v^Tx$.

If $x,y\in S_0$ then $|x|=4t$ and $|y|=4s$ for some $0\le t,s\le m$. Note that $y=v+x$ due to $v=x+y$. By our assumption, $4s=|y|=|v+x|=(4k+2-i)+(4t-i)$. It follows that $i$ is odd, say $i=2j+1$ where $0\le j\le r_{k,t}=\min\{2k,2t\}$. Therefore, $x$ has exactly ${4k+2\choose 2j+1}{4m-4k-2\choose 4t-2j-1}$ possible choices and $y$ is uniquely determined whenever $x$ is chosen. Thus, there are exactly $\sum_{t=0}^m\sum_{j=0}^{r_{k,t}}{4k+2\choose 2j+1}{4m-4k-2\choose 4t-2j-1}=\sum_{t=0}^m\sum_{j=0}^{2k}{4k+2\choose 2j+1}{4m-4k-2\choose 4t-2j-1}$ pairs of such $(x,y)$.

If $x,y\in S_1$ then $|x|=4t+1$ and $|y|=4s+1$ for some $0\le t,s\le m$. Note that $y=v+x$ due to $v=x+y$. By our assumption, $4s+1=|y|=|v+x|=(4k+2-i)+(4t+1-i)$. It follows that $i$ is also odd, say $i=2j+1$ where $0\le j\le r_{k,t}=\min\{2k,2t\}$. Therefore, $x$ has exactly ${4k+2\choose 2j+1}{4m-4k-2\choose 4t-2j}$ possible choices 