In the study of experimental design, we find that $T_1$ and $T_2$ are closely related to the subarrays $W$ or $W^{\text{*}}$ of the regular $2^{m-p}$ design. According to the definition of $T_1$ in the formula, when the subarray $W$ composed of any $s$ columns in design $D$ does not contain defining words, when calculating $T_1$, the contribution of $W$ to $T_1$ is $N/2^s$. When the subarray $W$ composed of any $s$ columns in design $D$ contains defining words, that is, several columns in $W$ form defining words, the situation becomes complicated. The existence of defining words means that some column combinations in $W$ have special structures, which will affect the contribution of $W$ to $T_1$. In addition, similar situations need to be analyzed when calculating $T_2$. Therefore, investigating whether the subarray $W$ contains defining words is crucial for simplifying the calculation of $K_s$. For the convenience of the subsequent analysis of the $K_s$ formula, the following lemmas are given to provide theoretical explanations for all possible cases of defining words.

Lemma 1. Suppose $D$ is a regular $2^{m-p}$ design with resolution $t$, let $W\in {\Omega }_{t+3}\left(D\right)$, then

(i) When $t=3$, $W$ contains at most three independent defining words;

(ii) When $4\le t\le 6$, $W$ contains at most two independent defining words;

(iii) When $t\ge 7$, $W$ contains at most one defining word,

where $m\ge t+3$.

Proof:

(i) When $t=3$, $W=\left\{m_1,m_2,m_3,m_4,m_5,m_6\right\}$, $W\in {\Omega }_6\left(D\right)$, then the number of independent defining words contained in $W$ is as follows:

(a1) $W$ does not contain independent defining words;

(a2) $W$ contains one independent defining word, the length of which may be 3 or 4 or 5 or 6;

(a3) $W$ contains two independent defining words, their lengths may be 3 and 3 or 3 and 4;

(a4) $W$ contains three independent defining words, all of whose lengths are 3.

Situations (a1) and (a2) are obvious. In situation (a3), when the lengths of the two independent defining words are 3 and 3, there are two cases for their composition, that is, they contain a common column (for example, $m_1{m}_2{m}_3=m_3{m}_4{m}_5$) and do not contain a common column (for example, $m_1{m}_2{m}_3=m_4{m}_5{m}_6$); when the lengths of the two independent defining words are 3 and 4, their composition is that they contain a common column (for example, $m_1{m}_2{m}_3=m_3{m}_4{m}_5{m}_6$). In situation (a4), if the length of one of the independent defining words exceeds 3, then it is impossible to have two other independent defining words with lengths ≤3. Therefore, when $W$ contains three independent defining words, the length of the independent defining words can only be 3. Since 6 factors form 3 independent defining words, there must be a common column between every two independent defining words. Let $m_1,m_2,m_3$ be the common column factors between the three independent defining words, $\ast$ represents an unknown factor, then the relationship between the three independent defining words can be obtained as: $m_1\ast m_3=m_1\ast m_2=m_2\ast m_3$, such arrangement can supplement the $\ast$ place with the other three factors. Since these three independent defining words will generate a non-independent defining word with a length of 3, so when $t=3$, $W$ cannot contain four independent defining words.

(ii) When $t=4$, $W=\left\{m_1,m_2,m_3,m_4,m_5,m_6,m_7\right\}$, $W\in {\text{Ω}}_7\left(D\right)$, then the number of independent defining words contained in $W$ is as follows:

(a1) $W$ does not contain independent defining words;

(a2) $W$ contains one independent defining word, the length of which may be 4 or 5 or 6 or 7;

(a3) $W$ contains two independent defining words, their lengths may be 4 and 4 or 4 and 5.

Situations (a1) and (a2) are obvious. In situation (a3), the independent defining words with lengths 4 and 4 may have one common column or two common columns. The independent defining words with lengths 4 and 5 have two common columns. Consider whether $W$ may contain three independent defining words. Suppose $W$ contains three independent defining words, all of which have a length of 4, and two of the independent defining words have two common columns, denoted as $m_3{m}_4$, and there is the following defining relationship, that is, $m_1{m}_2{m}_3{m}_4=m_3{m}_4{m}_5{m}_6$, which can generate a non-independent defining word $m_1{m}_2{m}_5{m}_6$, so that the other independent defining word has at most two common columns with these three defining words, thus two additional factors need to be added, exceeding the range of seven factors. Therefore, when $t=4$, $W$ contains at most two independent defining words. The cases of $t=5$ and $t=6$ can be proved in the same way.

(iii) When $t=7$, $W$ cannot have two independent defining words. Selecting two independent defining words with a length of 7 from 10 factors will inevitably obtain a non-independent defining word with a length $\le 6$, which does not conform to the condition of resolution 7, so when the resolution is 7, $W$ contains at most one independent defining word. The cases of $t>7$ can be proved in the same way.

Lemma 2. Consider a regular matrix $D$, which is an orthogonal matrix of strength 2. If this design is used to create a baseline design, then the values of $T_1$ and $T_2$ in the expression satisfy the following conditions:

(i) When $v\le t-2$, $T_1=C_m^v{\left(\frac{N}{2^v}\right)}^2$, $T_2=0$

(ii) When $v=t-1$, $T_1=C_m^{t-1}{\left(\frac{N}{2^{t-1}}\right)}^2$, $T_2=t\left(\frac{N^2}{2^{2t-2}}\right)B_t$

(iii) When $v=t$ and $t$ is odd

$\begin{array}{l}{T}_1=\frac{N^2}{2^{2t}}\left(C_m^t-A_t^0+3A_t^1\right)\\ T_2=\frac{N^2}{2^{2t}}\left(\left(t+1\right)A_{t+1}+t\left(m-t\right)A_t^0+t\left(m-t\right)A_t^1\right)\end{array}$

(iv) When $v=t$ and $t$ is even

$\begin{array}{l}{T}_1=\frac{N^2}{2^{2t}}+\left(C_m^t+3A_t^0-A_t^1\right)\\ T_2=\frac{N^2}{2^{2t}}\left(\left(t+1\right)A_{t+1}+t\left(m-t\right)A_t^0+t\left(m-t\right)A_t^1\right)\end{array}$

Proof:

(i) For any regular two-level design, let $W$ be the submatrix composed of any $v$ columns of the design matrix. Let $t$ be the resolution of the design, then when $v\le t-1$, $\alpha \left(w\right)=N/2^v$. Substituting it into the expressions of $T_1$ and $T_2$, we can get $T_1=C_m^v{\left(\frac{N}{2^v}\right)}^2$, $T_2={\displaystyle {\sum }_{w^{*}\in {\Omega }^{v+1}}{\displaystyle {\sum }_{w^0\in {\Omega }^v\left(w^{*}\right)}{\left(\frac{2N}{2^{v+1}}-\frac{N}{2^v}\right)}^2}}=0$, proved. (ii) When $v=t-1$, $\alpha \left(w\right)$ still satisfies the above condition, that is, $\alpha \left(w\right)=N/2^{t-1}$, so $T_1=C_m^{t-1}{\left(\frac{N}{2^{t-1}}\right)}^2$.

But for $T_2$, $w^{*}\in {\Omega }^{v+1}$ that is, $w^{*}\in {\Omega }^t$, then it is necessary to consider whether this $t$ columns contain defining words. Let $J_k\left(w\right)=\left|2\Phi \left(w\right)-N\right|$, where $\Phi \left(w\right)$ represents the value of the sum of the rows of the $W$ matrix modulo 2 and then added together. When $W$ is a non-defining word matrix, $\Phi \left(w\right)=N/2$, $J_k\left(w\right)=0$. When $W$ is a defining word matrix, $\Phi \left(w\right)=0$ or $N$, $J_k\left(w\right)=N$. Therefore, when this $t$ columns do not contain defining words, $J_k\left(w\right)=0$, $\alpha \left(w^{\text{*}}\right)=N/2^t$, $\alpha \left(w^0\right)=N/2^{t-1}$, it can be concluded that $T_2=0$ when this $t$ columns contain defining words, $J_k\left(w\right)=N$. Among them, this submatrix contains $\left(N-J_t\left(w\right)\right)/2^t$ $2^t$ designs and $\left(J_t\left(w\right)\right)/2^t$ half of the $2^t$ designs, that is, 0 $2^t$ designs and $N/2^{t-1}$ half of the $2^t$ designs. Whether the $N/2^{t-1}$ half of the $2^t$ designs contain all 1 rows needs to consider the parity of $t$ and the $\varphi$ value of the defining words. The following four situations are divided:

(a1) When $t$ is odd, $\varphi \left(w\right)=0_N$, that is, there is no all 1 row, $\alpha \left(w^{\text{*}}\right)=0$, $\alpha \left(w^0\right)=N/2^{t-1}$, $2\alpha \left(w^{\text{*}}\right)-\alpha \left(w^0\right)=-N/2^{t-1}$;

(a2) When $t$ is odd, $\varphi \left(w\right)=1_N$, that is, there is an all 1 row, $\alpha \left(w^{\text{*}}\right)=N/2^{t-1}$, $\alpha \left(w^0\right)=N/2^{t-1}$, $2\alpha \left(w^{\text{*}}\right)-\alpha \left(w^0\right)=N/2^{t-1}$;

(a3) When $t$ is even, $\varphi \left(w\right)=0_N$, that is, there is an all 1 row, $\alpha \left(w^{\text{*}}\right)=N/2^{t-1}$, $\alpha \left(w^0\right)=N/2^{t-1}$, $2\alpha \left(w^{\text{*}}\right)-\alpha \left(w^0\right)=N/2^{t-1}$;

(a4) When $t$ is even, $\varphi \left(w\right)=1_N$, that is, there is no all 1 row, $\alpha \left(w^{\text{*}}\right)=0$, $\alpha \left(w^0\right)=N/2^{t-1}$, $2\alpha \left(w^{\text{*}}\right)-\alpha \left(w^0\right)=-N/2^{t-1}$.

The $\left|2\alpha \left(w^{\text{*}}\right)-\alpha \left(w^0\right)\right|$ obtained in the four situations are all $N/2^{t-1}$, so we can get:

$\begin{array}{c}{T}_2=\underset{w^{*}\in {\Omega }^t}{{\displaystyle \sum }}\underset{w^0\in {\Omega }^{t-1}\left(w^{*}\right)}{{\displaystyle \sum }}\frac{N^2}{2^{2t-2}}\\ =\underset{w^{*}\in {\Omega }^t}{{\displaystyle \sum }}t\frac{N^2}{2^{2t-2}}\\ =t\frac{N^2}{2^{2t-2}}\underset{w^{*}\in {\Omega }^t}{{\displaystyle \sum }}{\left(\frac{J_t\left(w^{*}\right)}{N}\right)}^2\\ =t\frac{N^2}{2^{2t-2}}{B}_t\end{array}$

(iii) The calculation of $T_1$ needs to consider three situations:

$n_W$: The number of W in each situation.

According to Table 1 , the calculation result of $T_1$ can be obtained.

$\begin{array}{c}{T}_1=\left[\left(\begin{array}{c}m\\ t\end{array}\right)-A_t^0-A_t^1\right]\cdot {\left(\frac{N}{2^t}\right)}^2+0\cdot A_t^0+{\left(\frac{N}{2^{t-1}}\right)}^2\cdot A_t^1\\ =\frac{N^2}{2^{2t}}\left(C_m^t-A_t^0+3A_t^1\right)\end{array}$

The calculation of $T_2$ needs to consider four situations:

(a1) $W^{\text{*}}$ is a defining word with a length of $t+1$ and $\varphi =0_N$;

(a2) $W^{\text{*}}$ is a defining word with a length of $t+1$ and $\varphi =1_N$;

(a3) $W^{\text{*}}$ contains a defining word with a length of $t$ and $\varphi =0_N$;

(a4) $W^{\text{*}}$ contains a defining word with a length of $t$ and $\varphi =1_N$.

$n_{W^{\text{*}}}$: The number of each $\alpha \left(W^0\right)$ value corresponding to $W^{\text{*}}$. $n_W$: The number of $W^{\text{*}}$ in (a1)-(a4).

According to Table 2 , the calculation result of $T_2$ can be obtained.

$T_2=\frac{N^2}{2^{2t}}\left(\left(t+1\right)A_{t+1}+t\left(m-t\right)A_t^0+t\left(m-t\right)A_t^1\right)$

(iii) is proved, and the result of (iv) can be obtained in the same way.

Lemma 3. The $\varphi$ value of the defining word with a length of 5 obtained by the combination of independent defining words with lengths 3 and 4 containing a common column is determined. Let the $\varphi$ values of the above defining words with lengths 3 and 4 be ${\varphi }_1$ and ${\varphi }_2$ respectively, and the $\varphi$ value of the defining word with a length of 5 obtained by the combination be ${\varphi }_3$, the following results can be obtained:

(i) If ${\varphi }_1$ and ${\varphi }_2$ are both $0_N$, then ${\varphi }_3$ is $0_N$;

(ii) If ${\varphi }_1$ and ${\varphi }_2$ are both $1_N$, then ${\varphi }_3$ is $0_N$;

(iii) If one of ${\varphi }_1$ and ${\varphi }_2$ is $0_N$ and the other is $1_N$, then ${\varphi }_3$ is $1_N$.

Proof:

Let $W=\left\{m_1,m_2,m_3,m_4,m_5,m_6\right\}$, the defining relationship is $I=m_1{m}_2{m}_3=m_3{m}_4{m}_5{m}_6=m_1{m}_2{m}_4{m}_5{m}_6$, let $W_1=\left\{m_1,m_2,m_3\right\}$, $W_2=\left\{m_3,m_4,m_5,m_6\right\}$, where the three factors in $W_1$ form a defining word with a length of 3, $m_1{m}_2{m}_3$, the four factors in $W_2$ form a defining word with a length of 4, $m_3{m}_4{m}_5{m}_6$, their common column is $m_3$.

${\varphi }_1=0_N$ indicates that there are an even number of 1 s in each row of the three columns $m_1,m_2,m_3$, ${\varphi }_2=0_N$ indicates that there are an even number of 1 s in each row of the three columns $m_3,m_4,m_5,m_6$. When the column $m_3$ is 1, the corresponding two columns $m_1,m_2$ have an odd number of 1 s, and $m_4,m_5,m_6$ have an even number of 1 s. When the column $m_3$ is 0, the corresponding two columns $m_1,m_2$ have an even number of 1 s, and $m_4,m_5,m_6$ have an even number of 1 s, that is, $m_1,m_2,m_4,m_5,m_6$ have an even number of 1 s, so when ${\varphi }_1$ and ${\varphi }_2$ are both $0_N$, then ${\varphi }_3$ is $0_N$. Thus, (i) is proved, and (ii) and (iii) can be proved in the same way.

This section mainly studies the relationship between $K_5$ and WLP in regular $2^{m-p}$ designs with resolution $t=3$. By discussing the defining words contained in $W\in {\Omega }_5\left(D\right)$ and $W\in {\Omega }_6\left(D\right)$, the analysis process of $T_1$ and $T_2$ is given, and finally the relationship between $K_5$ and WLP in regular $2^{m-p}$ designs with resolution $t=3$ is obtained.

Theorem 1. Suppose $D$ is a regular $2^{m-p}$ design with resolution $t=3$, then

$K_5=\frac{4}{N^2}\left(5T_1+T_2\right)$

where

$\begin{array}{c}{T}_1=\underset{w\in {\Omega }^5}{{\displaystyle \sum }}{\left(\alpha \left(w\right)\right)}^2\\ ={\left(\frac{N}{8}\right)}^2\cdot A_3^{1,1}+{\left(\frac{N}{16}\right)}^2\cdot \left[\left(\frac{m-3}{2}\right)A_3^1+\left(m-4\right)A_4^0-A_3^{0,0}-3A_3^{1,1}-A_3^{0,1}+A_5^1\right]\\ +{\left(\frac{N}{32}\right)}^2\cdot I_1^5\end{array}$

$\begin{array}{c}{T}_2={\left(\frac{N}{2^5}\right)}^2\left[6\left(A_6^0+A_6^1\right)+5\left(m-5\right)\left(A_5^0+A_5^1\right)\right]\\ +4\left(\begin{array}{c}m-4\\ 2\end{array}\right)\left(A_4^0+A_4^1\right)+3\left(\begin{array}{c}m-3\\ 3\end{array}\right)\left(A_3^0+A_3^1\right)\\ -10A_3^{0,0}-2A_3^{0,1}+10A_3^{1,1}+6\left(\text{Δ}{A}_3^{\left(0,0\right)}-\text{Δ}{A}_3^{\left(0,1\right)}+\text{Δ}{A}_3^{\left(1,1\right)}\right)\\ -21A_3^{\left(0,0,0\right)}-21A_3^{\left(1,0,0\right)}-5A_3^{\left(1,1,0\right)}+75A_3^{\left(1,1,1\right)}-8A_{\left(3,4\right)}^{\left(0,1\right)}+12A_{\left(3,4\right)}^{\left(1,0\right)}\end{array}$

Proof: First, calculate the expression of $T_1$, let $W=\left\{m_1,m_2,m_3,m_4,m_5\right\}\in {\text{Ω}}_5\left(D\right)$, then there are ten possible situations for the columns in $W$:

(a1) $W$ contains two independent defining words with a length of 3 and $\varphi =0_N$;

(a2) $W$ contains two independent defining words with a length of 3 and $\varphi =1_N$;

(a3) $W$ contains two independent defining words with a length of 3, one of which has $\varphi =0_N$ and the other has $\varphi =1_N$;

(a4) $W$ only contains one defining word with a length of 3 and $\varphi =0_N$;

(a5) $W$ only contains one defining word with a length of 3 and $\varphi =1_N$;

(a6) $W$ only contains one defining word with a length of 4 and $\varphi =0_N$;

(a7) $W$ only contains one defining word with a length of 4 and $\varphi =1_N$;

(a8) $W$ only contains one defining word with a length of 5 and $\varphi =0_N$;

(a9) $W$ only contains one defining word with a length of 5 and $\varphi =1_N$;

(a10) $W$ contains 5 independent columns.

The reason why (a1), (a2), (a3) appear independently from other situations is that two defining words can combine to form other situations, and to avoid repeated calculations, they are listed separately. The defining words in (a1) and (a2) can combine to form an independent defining word with a length of 4 and $\varphi =0_N$. The defining words in (a3) can combine to form an independent defining word with a length of 4 and $\varphi =1_N$.

Below are the proofs of the values of $\alpha \left(w\right)$ and the number of each $W$ in situations (a1)-(a10).

For (a1), since there exists a defining word with a length of 3 and $\varphi =0_N$, there cannot be an all-one row in $W$, $\alpha \left(w\right)$ is 0, the number is $A_3^{0,0}$.

For (a2), two independent defining words with a length of 3 and $\varphi =1_N$ can combine to form an independent defining word with a length of 4 and $\varphi =0_N$. The cross column of the two independent defining words is affected by the other two columns of the defining words. When the other two columns of the defining words are all 1, the cross column must be 1, so there is an all-one row in this $W$, $\alpha \left(w\right)$ is $N/8$, the number is $A_3^{1,1}$.

For (a3) and (a4), the situation is the same as (a1), $\alpha \left(w\right)$, the numbers are $A_3^{0,1}$ and $\left(\begin{array}{c}m-3\\ 2\end{array}\right)A_3^0-2A_3^{0,0}-A_3^{0,1}$ respectively. In situation (a4), each $W$ has and only has one defining word with a length of 3 and $\varphi =0_N$. The number of such defining words is $A_3^0$, $W\in {\Omega }_5\left(D\right)$, that is, selecting 5 factors from $m$ factors that meet the conditions of (a4), which is $\left(\begin{array}{c}m-3\\ 2\end{array}\right)A_3^0$. To avoid the appearance of repeated situations, it is necessary to remove the situations where $\varphi =0_N$ in (a1) and (a2). Since (a1) contains two $\varphi =0_N$ defining words in $W$, the final number in situation (a4) is $\left(\begin{array}{c}m-3\\ 2\end{array}\right)A_3^0-2A_3^{0,0}-A_3^{0,1}$.

For (a5), each $W^{\text{*}}$ contains a defining word with a length of 3 and $\varphi =1_N$, so the number of all-one rows in the defining word, combined with the other two independent columns, is $\frac{N}{16}$. In this situation, each $W^{\text{*}}$ contains 6 $W^0$, two of which are combinations of the defining word and one independent column, their $\alpha \left(W^0\right)$ is $\frac{N}{16}$, and the other four $W^0$ are independent columns, their $\alpha \left(W^0\right)$ is $\frac{N}{32}$. The number is based on $\left(\begin{array}{c}m-3\\ 2\end{array}\right)A_3^1$ and needs to remove the non-independent defining words with a length of 5 and $\varphi =0_N$ generated in (a9), (a10), (a15), (a16), (a18). Thus, the number is $\left(\begin{array}{c}m-3\\ 2\end{array}\right)A_3^1-2A_3^{1,1}-A_3^{0,1}$.

For (a6), each $W$ contains a defining word with a length of 4 and $\varphi =0_N$, so the number of all-one rows in the defining word is $N/8$, and the number of all-one rows combined with another independent column is $N/16$, that is, $\alpha \left(w\right)$ is $N/16$, the number is $\left(m-4\right)A_4^0-2A_3^{0,0}-A_3^{1,1}$.

For (a7), since there exists a defining word with a length of 4 and $\varphi =1_N$, there cannot be an all-one row in $W$, $\alpha \left(w\right)$ is 0, the number is $\left(m-4\right)A_4^1-A_3^{0,1}$.

For (a8), since there exists a defining word with a length of 5 and $\varphi =0_N$, there cannot be an all-one row in $W$, $\alpha \left(w\right)$ is 0, the number is $A_5^0$.

For (a9), $W$ contains an all-one row, five factors form a defining word, so $\alpha \left(w\right)$ is $N/16$, the number is $A_5^1$.

For (a10), since $W$ contains 5 independent columns, $\alpha \left(w\right)$ is $N/32$. Selecting 5 factors from $m$ factors has $C_m^5$ possibilities, removing all previous situations, the number is

$\left(\begin{array}{c}m\\ 5\end{array}\right)+2A_3^{0,0}+2A_3^{1,1}+2A_3^{0,1}-\left(\begin{array}{c}m-3\\ 2\end{array}\right)\left(A_3^0+A_3^1\right)-\left(m-4\right)A_4^1-A_5^0-A_5^1$, denoted as $I_1^5$.

Summarizing the above situations, we obtain Table 3 :

$n_W$: The number of $W$ in each situation.

Thus, the expression of $T_1$ is obtained.

Next, calculate the expression of $T_2$. When $t=3$, $W\in {\Omega }_6\left(D\right)$ has the following twenty situations:

(a1) $W$ contains only one defining word with a length of 6 and $\varphi =0_N$;

(a2) $W$ contains only one defining word with a length of 6 and $\varphi =1_N$;

(a3) $W$ contains only one defining word with a length of 5 and $\varphi =0_N$;

(a4) $W$ contains only one defining word with a length of 5 and $\varphi =1_N$;

(a5) $W$ contains only one defining word with a length of 4 and $\varphi =0_N$;

(a6) $W$ contains only one defining word with a length of 4 and $\varphi =1_N$;

(a7) $W$ contains only one defining word with a length of 3 and $\varphi =0_N$;

(a8) $W$ contains only one defining word with a length of 3 and $\varphi =1_N$;

(a9) $W$ contains two independent defining words with a length of 3 and $\varphi =0_N$, sharing one common column;

(a10) $W$ contains two independent defining words with a length of 3 and $\varphi =1_N$, sharing one common column;

(a11) $W$ contains two independent defining words with a length of 3, one with $\varphi =0_N$ and the other with $\varphi =1_N$, sharing one common column;

(a12) $W$ contains two independent defining words with a length of 3 and $\varphi =0_N$, without sharing any common column;

(a13) $W$ contains two independent defining words with a length of 3 and $\varphi =1_N$, without sharing any common column;

(a14) $W$ contains two independent defining words with a length of 3, one with $\varphi =0_N$ and the other with $\varphi =1_N$, without sharing any common column;

(a15) $W$ contains three independent defining words with a length of 3, each pair sharing one common column, divided into the following four cases:

(1) $W$ does not contain any defining word with $\varphi =1_N$;

(2) $W$ contains only one defining word with $\varphi =1_N$;

(3) $W$ contains only two defining words with $\varphi =1_N$;

(4) $W$ contains only three defining words with $\varphi =1_N$.

(a16) $W$ contains one independent defining word with a length of 3 and $\varphi =0_N$ and one independent defining word with a length of 4 and $\varphi =0_N$, sharing one common column;

(a17) $W$ contains one independent defining word with a length of 3 and $\varphi =0_N$ and one independent defining word with a length of 4 and $\varphi =1_N$, sharing one common column;

(a18) $W$ contains one independent defining word with a length of 3 and $\varphi =1_N$ and one independent defining word with a length of 4 and $\varphi =0_N$, sharing one common column;

(a19) $W$ contains one independent defining word with a length of 3 and $\varphi =1_N$ and one independent defining word with a length of 4 and $\varphi =1_N$, sharing one common column;

(a20) $W$ contains 6 independent columns.

Next, using Lemma 2’s calculation method, the values of $\alpha \left(w_{\text{*}}\right)$ and $\alpha \left(w_0\right)$ and the number of each $W$ in situations (a1)-(a20) are proved.

For (a1), each $W^{\text{*}}$ contains a defining word with a length of 6 and $\varphi =0_N$, so the number of all-one rows in the defining word is $\frac{N}{2^5}$. In this situation, each $W^{\text{*}}$ contains 6 $W^0$ which are all independent columns, that is, $\alpha \left(W^0\right)$ is $\frac{N}{2^5}$. The number is based on $A_6^0$ and needs to remove the non-independent defining words with a length of 6 and $\varphi =0_N$ generated by two independent defining words with a length of 3 in (a12) and (a14). Thus, the number is $A_6^0-\text{Δ}{A}_3^{\left(0,0\right)}-\text{Δ}{A}_3^{\left(1,1\right)}$;

For (a2), since there exists a defining word with a length of 6 and $\varphi =0_N$, there cannot be an all-one row in $W^{\text{*}}$, $\alpha \left(W_{\text{*}}\right)$ is 0. In this situation, each $W^{\text{*}}$ contains 6 $W^0$ which are all independent columns, that is, $\alpha \left(W^0\right)$ is $\frac{N}{2^5}$. The number is based on $A_6^1$ and needs to remove the non-independent defining words with a length of 6 and $\varphi =1_N$ generated by two independent defining words with a length of 3 in (a13). Thus, the number is $A_6^0-\text{Δ}{A}_3^{\left(0,1\right)}$;

For (a3), since there exists a defining word with a length of 5 and $\varphi =0_N$, there cannot be an all-one row in $W^{\text{*}}$, $\alpha \left(W^{\text{*}}\right)$ is 0. In this situation, each $W_{\text{*}}$ contains 6 $w_0$, one of which is a defining word, its $\alpha \left(W^0\right)$ is 0, and the other five $w^0$ are independent columns, their $\alpha \left(W^0\right)$ is $\frac{N}{2^5}$. The number is based on $\left(m-5\right)A_5^0$ and needs to remove the non-independent defining words with a length of 5 and $\varphi =0_N$ generated by two independent defining words with lengths of 3 and 4 in (a16) and (a19). Thus, the number is $\left(m-5\right)A_5^0-A_{\left(3,4\right)}^{\left(0,0\right)}-A_{\left(3,4\right)}^{\left(1,1\right)}$;

For (a4), each $W^{\text{*}}$ contains a defining word with a length of 5 and $\varphi =1_N$, so the number of all-one rows in the defining word, combined with another independent column, is $\frac{N}{2^5}$. In this situation, each $W^{\text{*}}$ contains 6 $W^0$, one of which is a defining word, its $\alpha \left(W^0\right)$ is $\frac{N}{2^4}$, and the other five $W^0$ are independent columns, their $\alpha \left(W^0\right)$ is $\frac{N}{2^5}$. The number is based on $\left(m-5\right)A_5^1$ and needs to remove the non-independent defining words with a length of 5 and $\varphi =1_N$ generated by two independent defining words with lengths of 3 and 4 in (a16) and (a19). Thus, the number is $\left(m-5\right)A_5^1-A_{\left(3,4\right)}^{\left(0,1\right)}-A_{\left(3,4\right)}^{\left(1,0\right)}$;

For (a5), each $W^{\text{*}}$ contains a defining word with a length of 4 and $\varphi =0_N$, so the number of all-one rows in the defining word, combined with the other two independent columns, is $\frac{N}{2^5}$. In this situation, each $W^{\text{*}}$ contains 6 $W^0$, two of which are combinations of defining words with a length of 4 and one independent column, their $\alpha \left(W^0\right)$ is $\frac{N}{2^4}$, and the other four $W^0$ are independent columns, their $\alpha \left(W^0\right)$ is $\frac{N}{2^5}$. The number is based on $\left(m-5\right)A_5^1$ and needs to remove the non-independent defining words with a length of 4 and $\varphi =0_N$ generated in (a9), (a10), (a15), (a16), (a18). Thus, the number is $\left(\begin{array}{c}m-4\\ 2\end{array}\right)A_4^0-A_{\left(3,4\right)}^{\left(0,0\right)}-A_{\left(3,4\right)}^{\left(1,0\right)}-3A_3^{\left(0,0,0\right)}-A_3^{\left(1,0,0\right)}-A_3^{\left(1,1,0\right)}-3A_3^{\left(1,1,1\right)}-A_3^{0,0}-A_3^{1,1}$, denoted as $N_{a5}$;

For (a6), since there exists a defining word with a length of 4 and $\varphi =1_N$, there cannot be an all-one row in $W^{\text{*}}$, $\alpha \left(W^{\text{*}}\right)$ is 0. In this situation, each $W^{\text{*}}$ contains 6 $W^0$, two of which are combinations of defining words with a length of 4 and one independent column, their $\alpha \left(W^0\right)$ is 0, and the other four $W^0$ are independent columns, their $\alpha \left(W^0\right)$ is $\frac{N}{2^5}$. The number is similar to the analysis in (a5), which is $\left(\begin{array}{l}m-4\hfill \\ 2\hfill \end{array}\right)A_4^1-A_{\left(3,4\right)}^{\left(0,1\right)}-A_{\left(3,4\right)}^{\left(1,1\right)}-2A_3^{\left(1,0,0\right)}-2A_3^{\left(1,1,0\right)}-A_3^{0,1}$, denoted as $N_{a6}$;

For (a7), since there exists a defining word with a length of 3 and $\varphi =0_N$, there cannot be an all-one row in $W^{\text{*}}$, $\alpha \left(W^{\text{*}}\right)$ is 0. In this situation, each $W^{\text{*}}$ contains 6 $W^0$, three of which are combinations of defining words with a length of 3 and two independent columns, their $\alpha \left(W^0\right)$ is 0, and the other three $W^0$ are independent columns, their $\alpha \left(W^0\right)$ is $\frac{N}{2^5}$. The number needs to remove (a9), (a11), (a12), (a14), (a15), (a16), (a17), which is $\left(\begin{array}{c}m-3\\ 3\end{array}\right)A_3^0-2A_3^{0,0}-A_3^{0,1}-2\text{Δ}{A}_3^{\left(0,0\right)}-\text{Δ}{A}_3^{\left(0,1\right)}-3A_3^{\left(0,0,0\right)}-2A_3^{\left(1,0,0\right)}-A_3^{\left(1,1,0\right)}-A_{\left(3,4\right)}^{\left(0,0\right)}-A_{\left(3,4\right)}^{\left(0,1\right)}$, denoted as $N_{a7}$;

For (a8), each $W^{\text{*}}$ contains a defining word with a length of 3 and $\varphi =1_N$, so the number of all-one rows in the defining word, combined with the other three independent columns, is $\frac{N}{2^5}$. In this situation, each $W^{\text{*}}$ contains 6 $W^0$, three of which are combinations of defining words with a length of 3 and two independent columns, their $\alpha \left(W^0\right)$ is $\frac{N}{2^4}$, and the other three $W^0$ are independent columns, their $\alpha \left(W^0\right)$ is $\frac{N}{2^5}$. The number is similar to the situation of (a8), which is $\left(\begin{array}{c}m-3\\ 3\end{array}\right)A_3^1-A_3^{0,1}-2A_3^{1,1}-2\text{Δ}{A}_3^{\left(1,1\right)}-\text{Δ}{A}_3^{\left(0,1\right)}-A_3^{\left(1,0,0\right)}-2A_3^{\left(1,1,0\right)}-3A_3^{\left(0,0,0\right)}-A_{\left(3,4\right)}^{\left(1,0\right)}-A_{\left(3,4\right)}^{1,1)}$, denoted as $N_{a8}$;

For (a9), since there exists a defining word with a length of 3 and $\varphi =0_N$, there cannot be an all-one row in $W^{\text{*}}$, $\alpha \left(W^{\text{*}}\right)$ is 0. In this situation, each $W^{\text{*}}$ contains 6 $W^0$, their $\alpha \left(W^0\right)$ are all 0, and there is no all-one column in the case without common columns. The number is $A_3^{0,0}$;

For (a10), $W^{\text{*}}$ contains an all-one row, $\alpha \left(W^{\text{*}}\right)$ is $\frac{N}{2^4}$. In this situation, each $W^{\text{*}}$ contains 6 $W^0$, the $\alpha \left(W^0\right)$ in the case without common columns is $\frac{N}{8}$, and the $\alpha \left(W^0\right)$ in the other five cases is $\frac{N}{2^4}$. The number is $A_3^{1,1}$;

For (a11), since there exists a defining word with a length of 3 and $\varphi =0_N$, there cannot be an all-one row in $W^{\text{*}}$, $\alpha \left(W^{\text{*}}\right)$ is 0. In this situation, each $W^{\text{*}}$ contains 6 $W^0$, two of which have $\alpha \left(W^0\right)$ as $\frac{N}{2^4}$, and four have $\alpha \left(W^0\right)$ as 0. The number is $A_3^{0,1}$;

For (a12), since there exists a defining word with a length of 3 and $\varphi =0_N$, there cannot be an all-one row in $W^{\text{*}}$, $\alpha \left(W^{\text{*}}\right)$ is 0. In this situation, each $W^{\text{*}}$ contains 6 $W^0$ which are all 0. The number is $\text{Δ}{A}_3^{\left(0,0\right)}$;

For (a13), $W^{\text{*}}$ contains an all-one row, $\alpha \left(W^{\text{*}}\right)$ is $\frac{N}{2^5}$. In this situation, each $W^{\text{*}}$ contains 6 $W^0$ which are all $\frac{N}{2^4}$. The number is $\text{Δ}{A}_3^{\left(1,1\right)}$;

For (a14), since there exists a defining word with a length of 3 and $\varphi =0_N$, there cannot be an all-one row in $W^{\text{*}}$, $\alpha \left(W^{\text{*}}\right)$ is 0. In this situation, each $W^{\text{*}}$ contains 6 $W^0$, three of which are 0, and the other three are $\frac{N}{2^4}$. The number is $\text{Δ}{A}_3^{\left(0,1\right)}$;

For (a15), three independent defining words with a length of 3 will generate a non-independent defining word with a length of 3, the $\varphi$ value of which is determined by the $\varphi$ values of the three independent defining words. As long as there exists a defining word with a length of 3 and $\varphi =0_N$, there cannot be an all-one row in $W^{\text{*}}$, so the $\alpha \left(W^{\text{*}}\right)$ values of situations (1), (2), and (3) are all 0, and the $\alpha \left(W^{\text{*}}\right)$ value of situation (4) is $\frac{N}{8}$. When there are two or more defining words with $\varphi =0_N$, each $W^{\text{*}}$ contains 6 $W^0$ which are all 0, so the $\alpha \left(W^0\right)$ of situations (1) and (2) are all 0. In situation (3), each $W^{\text{*}}$ contains 6 $W^0$, one of which is $\frac{N}{8}$, and the other five are 0. In situation (4), each $W^{\text{*}}$ contains 6 $W^0$ whose $\alpha \left(W^0\right)$ are all $\frac{N}{8}$. Their numbers are denoted as $A_3^{\left(0,0,0\right)}$, $A_3^{\left(1,0,0\right)}$, $A_3^{\left(1,1,0\right)}$, $A_3^{\left(1,1,1\right)}$ respectively;

For (a16), since there exists a defining word with a length of 3 and $\varphi =0_N$, there cannot be an all-one row in $W^{\text{*}}$, $\alpha \left(W^{\text{*}}\right)$ is 0. In this situation, each $W^{\text{*}}$ contains 6 $W^0$, three of which are 0, and the other three are $\frac{N}{2^4}$. The number is $A_{\left(3,4\right)}^{\left(0,0\right)}$;

For (a17), since there exists a defining word with a length of 3 and $\varphi =0_N$, there cannot be an all-one row in $W^{\text{*}}$, $\alpha \left(W^{\text{*}}\right)$ is 0. In this situation, each $W^{\text{*}}$ contains 6 $W^0$ whose $\alpha \left(W^0\right)$ are all $\frac{N}{2^4}$. The number is $A_{\left(3,4\right)}^{\left(0,1\right)}$;

For (a18), according to Lemma 3, a defining word with a length of 3 and $\varphi =1_N$ and a defining word with a length of 4 and $\varphi =0_N$ will generate a defining word with a length of 5 and $\varphi =1_N$, $\alpha \left(W^{\text{*}}\right)$ is $\frac{N}{2^4}$. In this situation, each $W^{\text{*}}$ contains 6 $W^0$, five of which are 0, and the other one is $\frac{N}{2^4}$. The number is $A_{\left(3,4\right)}^{\left(1,0\right)}$;

For (a19), since there exists a defining word with a length of 4 and $\varphi =1_N$, there cannot be an all-one row in $W^{\text{*}}$, $\alpha \left(W^{\text{*}}\right)$ is 0. In this situation, each $W^{\text{*}}$ contains 6 $W^0$, three of which are 0, and the other three are $\frac{N}{2^4}$. The number is $A_{\left(3,4\right)}^{\left(1,1\right)}$;