any positive integer $m$.

\[lem-z-13\] Let $v$ be an element of $Z_2^{4m+2}$. If $v\in S_0\cup S_3$, then $$|C(v,S_0\cup S_3)|=2^{4m}-(-1)^m2^{2m}-2.$$

If $v\in S_0$, one can verify that there are $\sum_{t=0}^m\sum_{j=0}^{2k}{4k\choose 2j}{4m-4k+2\choose 4t-2j}-2$ pairs of $(x,y)$ such that $v=x+y$ and $x,y\in S_0$, and there are $\sum_{t=0}^m\sum_{j=0}^{2k}{4k\choose 2j}{4m-4k+2\choose 4t+3-2j}$ pairs of $(x,y)$ such that $v=x+y$ and $x,y\in S_3$. Therefore, by Theorem \[thm-z-3\] (viii), we have $$\begin{array}{lll}
|C(v,S_0\cup S_3)|&=&\displaystyle\sum_{t=0}^m\sum_{j=0}^{2k}{4k\choose 2j}{4m-4k+2\choose 4t-2j}+\sum_{t=0}^m\sum_{j=0}^{2k}{4k\choose 2j}{4m-4k+2\choose 4t-2j+3}-2\\
&=&\displaystyle\sum_{t=0}^m\sum_{j=0}^{2k}{4k\choose 2j}{4m-4k+3\choose 4t-2j}-2\\
&=&2^{4m}-(-1)^m2^{2m}-2.
\end{array}$$

If $v\in S_3$, one can verify that there are $\sum_{t=0}^m\sum_{j=0}^{2k+1}{4k+3\choose 2j}{4m-4k-1\choose 4t-2j}-1$ pairs of $(x,y)$ such that $v=x+y$ and $x\in S_0$, $y\in S_3$, and there are also $\sum_{t=0}^m\sum_{j=0}^{2k+1}{4k+3\choose 2j}{4m-4k-1\choose 4t-2j}-1$ pairs of $(x,y)$ such that $v=x+y$ and $x\in S_3$, $y\in S_0$. Therefore, by Theorem \[thm-z-3\] (ix), we have $$|C(v,S_0\cup S_3)|=2\sum_{t=0}^m\sum_{j=0}^{2k+1}{4k+3\choose 2j}{4m-4k-1\choose 4t-2j}-2=2^{4m}-(-1)^m2^{2m}-2.$$

It completes the proof.

\[lem-z-14\] Let $v$ be an element of $Z_2^{4m+2}$. If $v\in \overline{S_0\cup S_3}=S_1\cup S_2$, then $$|C(v,S_0\cup S_3)|=2^{4m}-(-1)^m2^{2m}.$$

If $v\in S_1$, there are $\sum_{t=0}^m\sum_{j=0}^{2k}{4k+1\choose 2j+1}{4m-4k+1\choose 4t-2j-1}$ pairs of $(x,y)$ such that $v=x+y$ and $x\in S_0$, $y\in S_3$, and there are also $\sum_{t=0}^m\sum_{j=0}^{2k}{4k+1\choose 2j+1}{4m-4k+1\choose 4t-2j-1}$ pairs of $(x,y)$ such that $v=x+y$ and $x\in S_3$, $y\in S_0$. Therefore, by Theorem \[thm-z-3\] (x), we have $$|C(v,S_0\cup S_3)|=2\sum_{t=0}^m\sum_{j=0}^{2k}{4k+1\choose 2j+1}{4m-4k+1\choose 4t-2j-1}=2^{4m}-(-1)^m2^{2m}.$$

If $v\in S_2$, there are $\sum_{t=0}^m\sum_{j=0}^{2k}{4k+2\choose 