Several secondary methods for constructing bent–negabent functions

Abstract

In this paper, we present three secondary methods for constructing bent–negabent functions under the frameworks of the indirect sum construction (proposed by Carlet in 2004), the modified indirect sum construction (proposed by Hod$\check{\text {z}}$i$\acute{\text {c}}$ et al in Des Codes Cryptogr 88(10):2007–2035, 2020) and Rothaus’ construction. We first give a construction of bent–negabent functions by using the indirect sum construction, and specify some sufficient conditions on initial functions so that these constructed functions are provably outside the completed Maiorana–McFarland class. Here, we correct some results on the construction of bent–negabent functions proposed by Zhang et al. (IEEE Trans Inf Theory 61(3):1496–1506, 2015). Then, we investigate the nega–Hadamard transform of the composition of a multi-output function and a Boolean function. Using this tool, with initial primary bent–negabent functions, we analyze and find out the necessary and sufficient conditions for the modified indirect sum construction and Rothaus’ construction to generate bent–negabent functions. Furthermore, by developing methods to obtain initial functions satisfying the required necessary and sufficient conditions, we propose several constructions of bent–negabent functions under these two frameworks of the secondary constructions.

Appendices

Appendix A: Proof of Theorem 8

To prove Theorem 8, we need the following two lemmas.

Lemma 7

Let $M_1, M_2, M_3, M_4 \in \{\pm 1, \pm \imath \}$. The following four equations have the same solutions of $(M_1, M_2, M_3, M_4)$:

$$\begin{aligned} |\imath \cdot M_1 + M_2 + M_3 + \imath \cdot M_4|= 2;&\ |M_1 + \imath \cdot M_2 -\imath \cdot M_3 - M_4|= 2; \\ |M_1 -\imath \cdot M_2 + \imath \cdot M_3 - M_4|= 2;&\ |-\imath \cdot M_1 + M_2 + M_3 - \imath \cdot M_4|= 2. \end{aligned}$$

Proof

The proof can be completed easily by exhaustive search and we omit it. $\square$

Lemma 8

Let n be a positive even integer, and $a, b \in \mathcal {B}_n$ be negabent functions. Then, ${{\,\textrm{N}\,}}_b(\varvec{\textrm{u}}) = \pm \imath \cdot {{\,\textrm{N}\,}}_a(\varvec{\textrm{u}})$ if and only if $\widetilde{a_\sigma }(\varvec{\textrm{u}}) \oplus \widetilde{a_\sigma }(\bar{\varvec{\textrm{u}}}) \oplus \widetilde{b_\sigma }(\varvec{\textrm{u}}) \oplus \widetilde{b_\sigma }(\bar{\varvec{\textrm{u}}}) = 1$ for $\varvec{\textrm{u}} \in \mathbb {F}_2^n$.

Proof

Since a and b are negabent functions, then $a_\sigma$ and $b_\sigma$ are bent functions by Lemma 3. By Lemma 4, we have

$$\begin{aligned} \left\{ \begin{aligned}&\imath \cdot {{\,\textrm{N}\,}}_a(\varvec{\textrm{u}}) = \frac{\imath }{2} \left[ {{\,\textrm{W}\,}}_{a_\sigma }(\varvec{\textrm{u}}) + {{\,\textrm{W}\,}}_{a_\sigma }(\bar{\varvec{\textrm{u}}})\right] - \frac{1}{2} \left[ {{\,\textrm{W}\,}}_{a_\sigma }(\varvec{\textrm{u}}) - {{\,\textrm{W}\,}}_{a_\sigma }(\bar{\varvec{\textrm{u}}})\right] , \\&{{\,\textrm{N}\,}}_b(\varvec{\textrm{u}}) = \frac{1}{2} \left[ {{\,\textrm{W}\,}}_{b_\sigma }(\varvec{\textrm{u}}) + {{\,\textrm{W}\,}}_{b_\sigma }(\bar{\varvec{\textrm{u}}})\right] + \frac{\imath }{2} \left[ {{\,\textrm{W}\,}}_{b_\sigma }(\varvec{\textrm{u}}) - {{\,\textrm{W}\,}}_{b_\sigma }(\bar{\varvec{\textrm{u}}})\right] . \end{aligned} \right. \end{aligned}$$

The condition ${{\,\textrm{N}\,}}_b(\varvec{\textrm{u}}) = \imath \cdot {{\,\textrm{N}\,}}_a(\varvec{\textrm{u}}) \Leftrightarrow$

$$\begin{aligned} \left\{ \begin{aligned}&{{\,\textrm{W}\,}}_{a_\sigma }(\varvec{\textrm{u}}) + {{\,\textrm{W}\,}}_{a_\sigma }(\bar{\varvec{\textrm{u}}}) = {{\,\textrm{W}\,}}_{b_\sigma }(\varvec{\textrm{u}}) - {{\,\textrm{W}\,}}_{b_\sigma }(\bar{\varvec{\textrm{u}}}), \\&-{{\,\textrm{W}\,}}_{a_\sigma }(\varvec{\textrm{u}}) + {{\,\textrm{W}\,}}_{a_\sigma }(\bar{\varvec{\textrm{u}}}) = {{\,\textrm{W}\,}}_{b_\sigma }(\varvec{\textrm{u}}) + {{\,\textrm{W}\,}}_{b_\sigma }(\bar{\varvec{\textrm{u}}}), \end{aligned} \right. \end{aligned}$$

$\Leftrightarrow$ ${{\,\textrm{W}\,}}_{a_\sigma }(\varvec{\textrm{u}}) = -{{\,\textrm{W}\,}}_{b_\sigma }(\bar{\varvec{\textrm{u}}})$ and ${{\,\textrm{W}\,}}_{a_\sigma }(\bar{\varvec{\textrm{u}}}) = {{\,\textrm{W}\,}}_{b_\sigma }(\varvec{\textrm{u}})$, which are further equivalent to $(-1)^{\widetilde{a_\sigma }(\varvec{\textrm{u}})} = -(-1)^{\widetilde{b_\sigma }(\bar{\varvec{\textrm{u}}})}$ and $(-1)^{\widetilde{a_\sigma }(\bar{\varvec{\textrm{u}}})} = (-1)^{\widetilde{b_\sigma }(\varvec{\textrm{u}})}$, $\Leftrightarrow$ $\widetilde{a_\sigma }(\varvec{\textrm{u}}) = \widetilde{b_\sigma }(\bar{\varvec{\textrm{u}}}) \oplus 1$ and $\widetilde{a_\sigma }(\bar{\varvec{\textrm{u}}}) = \widetilde{b_\sigma }(\varvec{\textrm{u}})$.

Similarly, the condition ${{\,\textrm{N}\,}}_b(\varvec{\textrm{u}}) = -\imath \cdot {{\,\textrm{N}\,}}_a(\varvec{\textrm{u}})$ $\Leftrightarrow$ $\widetilde{a_\sigma }(\varvec{\textrm{u}}) = \widetilde{b_\sigma }(\bar{\varvec{\textrm{u}}})$ and $\widetilde{a_\sigma }(\bar{\varvec{\textrm{u}}}) = \widetilde{b_\sigma }(\varvec{\textrm{u}}) \oplus 1$.

Together with the above two conditions, the result follows immediately. $\square$

Proof of Theorem 8

Rewrite f as $f(\varvec{\textrm{x}}, \varvec{\textrm{y}}, z_1, z_2) = h (a(\varvec{\textrm{x}}), b(\varvec{\textrm{x}}), c(\varvec{\textrm{y}}), d(\varvec{\textrm{y}}), z_1, z_2)$, where h is a 6-variable Boolean function with the ANF

$$\begin{aligned} h(x_1, \cdots , x_6) = x_2\oplus x_4\oplus (x_1\oplus x_2\oplus x_5)(x_3\oplus x_4\oplus x_6). \end{aligned}$$

By a direct calculation, we have

$$\begin{aligned}&{{\,\textrm{W}\,}}_h(0,1,0,1,0,0)=2^2,\ {{\,\textrm{W}\,}}_h(0,1,1,0,0,1)=2^2,\\&{{\,\textrm{W}\,}}_h(1,0,0,1,1,0)=2^2,\ {{\,\textrm{W}\,}}_h(1,0,1,0,1,1)=-2^2, \end{aligned}$$

and the Walsh–Hadamard transform of h vanishes at other positions. By Theorem 7, the nega–Hadamard transform of f at $(\varvec{\textrm{u}}, \varvec{\textrm{v}}, w_1, w_2)\in \mathbb {F}_2^n \times \mathbb {F}_2^m \times \mathbb {F}_2 \times \mathbb {F}_2$ is given by

$$\begin{aligned}&{{\,\textrm{N}\,}}_f(\varvec{\textrm{u}}, \varvec{\textrm{v}}, w_1, w_2) = 2^{-3} \sum _{\varvec{{\omega }} \in {\mathbb {F}_2^{6}}}{{\,\textrm{W}\,}}_h(\varvec{{\omega }}) {{\,\textrm{N}\,}}_{\varvec{{\omega }}\cdot (a(\varvec{\textrm{x}}), b(\varvec{\textrm{x}}), c(\varvec{\textrm{y}}), d(\varvec{\textrm{y}}), z_1, z_2)}(\varvec{\textrm{u}}, \varvec{\textrm{v}}, w_1, w_2) \\&= 2^{-1} \left[ {{\,\textrm{N}\,}}_b(\varvec{\textrm{u}}){{\,\textrm{N}\,}}_d(\varvec{\textrm{v}}){{\,\textrm{N}\,}}_0(w_1, w_2) + {{\,\textrm{N}\,}}_b(\varvec{\textrm{u}}){{\,\textrm{N}\,}}_c(\varvec{\textrm{v}}){{\,\textrm{N}\,}}_{z_2}(w_1, w_2) \right. \\&\quad \left. + {{\,\textrm{N}\,}}_a(\varvec{\textrm{u}}){{\,\textrm{N}\,}}_d(\varvec{\textrm{v}}){{\,\textrm{N}\,}}_{z_1}(w_1, w_2) - {{\,\textrm{N}\,}}_a(\varvec{\textrm{u}}){{\,\textrm{N}\,}}_c(\varvec{\textrm{v}}){{\,\textrm{N}\,}}_{z_1\oplus z_2}(w_1, w_2)\right] \\&= {\left\{ \begin{array}{ll} 2^{-1} \left[ \imath \cdot {{\,\textrm{N}\,}}_b(\varvec{\textrm{u}}){{\,\textrm{N}\,}}_d(\varvec{\textrm{v}}) + {{\,\textrm{N}\,}}_b(\varvec{\textrm{u}}){{\,\textrm{N}\,}}_c(\varvec{\textrm{v}}) + {{\,\textrm{N}\,}}_a(\varvec{\textrm{u}}){{\,\textrm{N}\,}}_d(\varvec{\textrm{v}}) +\imath \cdot {{\,\textrm{N}\,}}_a(\varvec{\textrm{u}}){{\,\textrm{N}\,}}_c(\varvec{\textrm{v}}) \right] ,\\ (w_1, w_2) = (0, 0); \\ 2^{-1} \left[ {{\,\textrm{N}\,}}_b(\varvec{\textrm{u}}){{\,\textrm{N}\,}}_d(\varvec{\textrm{v}}) + \imath \cdot {{\,\textrm{N}\,}}_b(\varvec{\textrm{u}}){{\,\textrm{N}\,}}_c(\varvec{\textrm{v}}) - \imath \cdot {{\,\textrm{N}\,}}_a(\varvec{\textrm{u}}){{\,\textrm{N}\,}}_d(\varvec{\textrm{v}}) - {{\,\textrm{N}\,}}_a(\varvec{\textrm{u}}){{\,\textrm{N}\,}}_c(\varvec{\textrm{v}}) \right] , \\ (w_1, w_2) = (0, 1); \\ 2^{-1} \left[ {{\,\textrm{N}\,}}_b(\varvec{\textrm{u}}){{\,\textrm{N}\,}}_d(\varvec{\textrm{v}}) - \imath \cdot {{\,\textrm{N}\,}}_b(\varvec{\textrm{u}}){{\,\textrm{N}\,}}_c(\varvec{\textrm{v}}) + \imath \cdot {{\,\textrm{N}\,}}_a(\varvec{\textrm{u}}){{\,\textrm{N}\,}}_d(\varvec{\textrm{v}}) - {{\,\textrm{N}\,}}_a(\varvec{\textrm{u}}){{\,\textrm{N}\,}}_c(\varvec{\textrm{v}}) \right] , \\ (w_1, w_2) = (1, 0); \\ 2^{-1} \left[ -\imath \cdot {{\,\textrm{N}\,}}_b(\varvec{\textrm{u}}){{\,\textrm{N}\,}}_d(\varvec{\textrm{v}}) + {{\,\textrm{N}\,}}_b(\varvec{\textrm{u}}){{\,\textrm{N}\,}}_c(\varvec{\textrm{v}}) + {{\,\textrm{N}\,}}_a(\varvec{\textrm{u}}){{\,\textrm{N}\,}}_d(\varvec{\textrm{v}}) -\imath \cdot {{\,\textrm{N}\,}}_a(\varvec{\textrm{u}}){{\,\textrm{N}\,}}_c(\varvec{\textrm{v}}) \right] , \\ (w_1, w_2) = (1, 1). \end{array}\right. } \end{aligned}$$

Since a, b, c, d are all negabent functions, then all the values ${{\,\textrm{N}\,}}_a(\varvec{\textrm{u}}), {{\,\textrm{N}\,}}_b(\varvec{\textrm{u}}), {{\,\textrm{N}\,}}_c(\varvec{\textrm{v}}), {{\,\textrm{N}\,}}_d(\varvec{\textrm{v}})$ lie in $\{\pm 1, \pm \imath \}$. Furthermore, the values ${{\,\textrm{N}\,}}_b(\varvec{\textrm{u}}){{\,\textrm{N}\,}}_d(\varvec{\textrm{v}}), {{\,\textrm{N}\,}}_b(\varvec{\textrm{u}}){{\,\textrm{N}\,}}_c(\varvec{\textrm{v}}), {{\,\textrm{N}\,}}_a(\varvec{\textrm{u}}){{\,\textrm{N}\,}}_d(\varvec{\textrm{v}}), {{\,\textrm{N}\,}}_a(\varvec{\textrm{u}}){{\,\textrm{N}\,}}_c(\varvec{\textrm{v}})$ still lie in $\{\pm 1, \pm \imath \}$. f is negabent if and only if $|{{\,\textrm{N}\,}}_f(\varvec{\textrm{u}}, \varvec{\textrm{v}}, w_1, w_2)|= 1$ for all $(\varvec{\textrm{u}}, \varvec{\textrm{v}}, w_1, w_2)\in \mathbb {F}_2^n \times \mathbb {F}_2^m \times \mathbb {F}_2 \times \mathbb {F}_2$. By Lemma 7, we know that the four equations $|{{\,\textrm{N}\,}}_f(\varvec{\textrm{u}}, \varvec{\textrm{v}}, w_1, w_2)|= 1$ have the same solutions of $({{\,\textrm{N}\,}}_a(\varvec{\textrm{u}}), {{\,\textrm{N}\,}}_b(\varvec{\textrm{u}}), {{\,\textrm{N}\,}}_c(\varvec{\textrm{v}}), {{\,\textrm{N}\,}}_d(\varvec{\textrm{v}}))$ for the four cases of $(w_1, w_2) \in \mathbb {F}_2^2$. Hence, we only study the first equation:

$$\begin{aligned} |\imath \cdot {{\,\textrm{N}\,}}_b(\varvec{\textrm{u}}){{\,\textrm{N}\,}}_d(\varvec{\textrm{v}}) + {{\,\textrm{N}\,}}_b(\varvec{\textrm{u}}){{\,\textrm{N}\,}}_c(\varvec{\textrm{v}}) + {{\,\textrm{N}\,}}_a(\varvec{\textrm{u}}){{\,\textrm{N}\,}}_d(\varvec{\textrm{v}}) +\imath \cdot {{\,\textrm{N}\,}}_a(\varvec{\textrm{u}}){{\,\textrm{N}\,}}_c(\varvec{\textrm{v}})|= 2, \end{aligned}$$

which is equivalent to

$$\begin{aligned} |{{\,\textrm{N}\,}}_c(\varvec{\textrm{v}})\left( \imath \cdot {{\,\textrm{N}\,}}_a(\varvec{\textrm{u}}) + {{\,\textrm{N}\,}}_b(\varvec{\textrm{u}}) \right) + {{\,\textrm{N}\,}}_d(\varvec{\textrm{v}})\left( {{\,\textrm{N}\,}}_a(\varvec{\textrm{u}}) + \imath \cdot {{\,\textrm{N}\,}}_b(\varvec{\textrm{u}}) \right) |= 2. \end{aligned}$$

Then we have the following discussions.

• (1) If $|\imath \cdot {{\,\textrm{N}\,}}_a(\varvec{\textrm{u}}) + {{\,\textrm{N}\,}}_b(\varvec{\textrm{u}})|= 0$ or 2 for all $\varvec{\textrm{u}}\in \mathbb {F}_2^n$, we obtain $|{{\,\textrm{N}\,}}_a(\varvec{\textrm{u}}) + \imath \cdot {{\,\textrm{N}\,}}_b(\varvec{\textrm{u}})|= 2$ or 0. Then we have ${{\,\textrm{N}\,}}_b(\varvec{\textrm{u}}) = \pm \imath \cdot {{\,\textrm{N}\,}}_a(\varvec{\textrm{u}})$ for all $\varvec{\textrm{u}}\in \mathbb {F}_2^n$.

• (2) If $|\imath \cdot {{\,\textrm{N}\,}}_a(\varvec{\textrm{u}}) + {{\,\textrm{N}\,}}_b(\varvec{\textrm{u}})|=\sqrt{2}$ for some $\varvec{\textrm{u}}\in \mathbb {F}_2^n$, we obtain ${{\,\textrm{N}\,}}_d(\varvec{\textrm{v}}) = \pm \imath \cdot {{\,\textrm{N}\,}}_c(\varvec{\textrm{v}})$ for all $\varvec{\textrm{v}}\in \mathbb {F}_2^n$.

According to Lemma 8, f is negabent if and only if $\widetilde{a_\sigma }(\varvec{\textrm{x}}) \oplus \widetilde{a_\sigma }(\bar{\varvec{\textrm{x}}}) \oplus \widetilde{b_\sigma }(\varvec{\textrm{x}}) \oplus \widetilde{b_\sigma }(\bar{\varvec{\textrm{x}}}) = 1$ or $\widetilde{c_\sigma }(\varvec{\textrm{y}}) \oplus \widetilde{c_\sigma }(\bar{\varvec{\textrm{y}}}) \oplus \widetilde{d_\sigma }(\varvec{\textrm{y}}) \oplus \widetilde{d_\sigma }(\bar{\varvec{\textrm{y}}}) = 1$. $\square$

Appendix B: Proof of Theorem 10

To prove Theorem 10, we need the following two lemmas.

Lemma 9

Let n be a positive even integer, and $a, d \in \mathcal {B}_n$ be arbitrary negabent functions. Then, for $\varvec{\textrm{u}} \in \mathbb {F}_2^n$, we have

• (i) ${{\,\textrm{N}\,}}_a(\varvec{\textrm{u}}) = {{\,\textrm{N}\,}}_d(\varvec{\textrm{u}})$ if and only if $(\widetilde{a_\sigma }(\varvec{\textrm{u}}), \widetilde{a_\sigma }(\bar{\varvec{\textrm{u}}})) \oplus (\widetilde{d_\sigma }(\varvec{\textrm{u}}), \widetilde{d_\sigma }(\bar{\varvec{\textrm{u}}})) = (0, 0)$;

• (ii) ${{\,\textrm{N}\,}}_a(\varvec{\textrm{u}}) = -{{\,\textrm{N}\,}}_d(\varvec{\textrm{u}})$ if and only if $(\widetilde{a_\sigma }(\varvec{\textrm{u}}), \widetilde{a_\sigma }(\bar{\varvec{\textrm{u}}})) \oplus (\widetilde{d_\sigma }(\varvec{\textrm{u}}), \widetilde{d_\sigma }(\bar{\varvec{\textrm{u}}})) = (1, 1)$.

Proof

• (i) From Lemma 3, we know that $a_\sigma , d_\sigma$ are bent functions since a, d are negabent functions. By Lemma 4 we know ${{\,\textrm{N}\,}}_a(\varvec{\textrm{u}}) = {{\,\textrm{N}\,}}_d(\varvec{\textrm{u}})$ if and only if ${\left\{ \begin{array}{ll} {{\,\textrm{W}\,}}_{a_\sigma }(\varvec{\textrm{u}}) + {{\,\textrm{W}\,}}_{a_\sigma }(\bar{\varvec{\textrm{u}}}) = {{\,\textrm{W}\,}}_{d_\sigma }(\varvec{\textrm{u}}) + {{\,\textrm{W}\,}}_{d_\sigma }(\bar{\varvec{\textrm{u}}}), \\ {{\,\textrm{W}\,}}_{a_\sigma }(\varvec{\textrm{u}}) - {{\,\textrm{W}\,}}_{a_\sigma }(\bar{\varvec{\textrm{u}}}) = {{\,\textrm{W}\,}}_{d_\sigma }(\varvec{\textrm{u}}) - {{\,\textrm{W}\,}}_{d_\sigma }(\bar{\varvec{\textrm{u}}}). \end{array}\right. }$ That is to say, ${{\,\textrm{W}\,}}_{a_\sigma }(\varvec{\textrm{u}}) = {{\,\textrm{W}\,}}_{d_\sigma }(\varvec{\textrm{u}})$ and ${{\,\textrm{W}\,}}_{a_\sigma }(\bar{\varvec{\textrm{u}}}) = {{\,\textrm{W}\,}}_{d_\sigma }(\bar{\varvec{\textrm{u}}})$. Furthermore, it is equivalent to $(-1)^{\widetilde{a_\sigma }(\varvec{\textrm{u}})} = (-1)^{\widetilde{d_\sigma }(\varvec{\textrm{u}})}$ and $(-1)^{\widetilde{a_\sigma }(\bar{\varvec{\textrm{u}}})} = (-1)^{\widetilde{d_\sigma }(\bar{\varvec{\textrm{u}}})}$, i.e., $(\widetilde{a_\sigma }(\varvec{\textrm{u}}), \widetilde{a_\sigma }(\bar{\varvec{\textrm{u}}})) \oplus (\widetilde{d_\sigma }(\varvec{\textrm{u}}), \widetilde{d_\sigma }(\bar{\varvec{\textrm{u}}})) = (0, 0)$.

• (ii) Item (ii) can be proved similarly to item (i) and we omit it.

$\square$

Lemma 10

Let n be a positive even integer, and $a, b, c, d \in \mathcal {B}_n$ be negabent functions. Then, ${{\,\textrm{N}\,}}_b(\varvec{\textrm{u}}) + {{\,\textrm{N}\,}}_c(\varvec{\textrm{u}})=\pm ({{\,\textrm{N}\,}}_a(\varvec{\textrm{u}}) + {{\,\textrm{N}\,}}_d(\varvec{\textrm{u}}))$ if and only if $(\widetilde{b_\sigma }(\varvec{\textrm{u}}), \widetilde{b_\sigma }(\bar{\varvec{\textrm{u}}})) \oplus (\widetilde{c_\sigma }(\varvec{\textrm{u}}), \widetilde{c_\sigma }(\bar{\varvec{\textrm{u}}})) = (\widetilde{a_\sigma }(\varvec{\textrm{u}}), \widetilde{a_\sigma }(\bar{\varvec{\textrm{u}}})) \oplus (\widetilde{d_\sigma }(\varvec{\textrm{u}}), \widetilde{d_\sigma }(\bar{\varvec{\textrm{u}}}))$ for $\varvec{\textrm{u}} \in \mathbb {F}_2^n$.

Proof

Since a, b, c, d are negabent functions, then $a_\sigma , b_\sigma , c_\sigma , d_\sigma$ are bent functions by Lemma 3.

For the case ${{\,\textrm{N}\,}}_b(\varvec{\textrm{u}}) + {{\,\textrm{N}\,}}_c(\varvec{\textrm{u}}) = {{\,\textrm{N}\,}}_a(\varvec{\textrm{u}}) + {{\,\textrm{N}\,}}_d(\varvec{\textrm{u}})$, by Lemma 4 we have

$$\begin{aligned} {\left\{ \begin{array}{ll} &{}\left( {{\,\textrm{W}\,}}_{b_\sigma }(\varvec{\textrm{u}}) + {{\,\textrm{W}\,}}_{c_\sigma }(\varvec{\textrm{u}}) \right) + \left( {{\,\textrm{W}\,}}_{b_\sigma }(\bar{\varvec{\textrm{u}}}) + {{\,\textrm{W}\,}}_{c_\sigma }(\bar{\varvec{\textrm{u}}})\right) \\ &{}\quad = \left( {{\,\textrm{W}\,}}_{a_\sigma }(\varvec{\textrm{u}}) + {{\,\textrm{W}\,}}_{d_\sigma }(\varvec{\textrm{u}}) \right) + \left( {{\,\textrm{W}\,}}_{a_\sigma }(\bar{\varvec{\textrm{u}}}) + {{\,\textrm{W}\,}}_{d_\sigma }(\bar{\varvec{\textrm{u}}})\right) , \\ &{}\left( {{\,\textrm{W}\,}}_{b_\sigma }(\varvec{\textrm{u}}) + {{\,\textrm{W}\,}}_{c_\sigma }(\varvec{\textrm{u}}) \right) - \left( {{\,\textrm{W}\,}}_{b_\sigma }(\bar{\varvec{\textrm{u}}}) + {{\,\textrm{W}\,}}_{c_\sigma }(\bar{\varvec{\textrm{u}}})\right) \\ &{}\quad = \left( {{\,\textrm{W}\,}}_{a_\sigma }(\varvec{\textrm{u}}) + {{\,\textrm{W}\,}}_{d_\sigma }(\varvec{\textrm{u}}) \right) - \left( {{\,\textrm{W}\,}}_{a_\sigma }(\bar{\varvec{\textrm{u}}}) + {{\,\textrm{W}\,}}_{d_\sigma }(\bar{\varvec{\textrm{u}}})\right) , \end{array}\right. } \end{aligned}$$

$\Leftrightarrow$

$$\begin{aligned} {\left\{ \begin{array}{ll} {{\,\textrm{W}\,}}_{b_\sigma }(\varvec{\textrm{u}}) + {{\,\textrm{W}\,}}_{c_\sigma }(\varvec{\textrm{u}}) = {{\,\textrm{W}\,}}_{a_\sigma }(\varvec{\textrm{u}}) + {{\,\textrm{W}\,}}_{d_\sigma }(\varvec{\textrm{u}}), \\ {{\,\textrm{W}\,}}_{b_\sigma }(\bar{\varvec{\textrm{u}}}) + {{\,\textrm{W}\,}}_{c_\sigma }(\bar{\varvec{\textrm{u}}}) = {{\,\textrm{W}\,}}_{a_\sigma }(\bar{\varvec{\textrm{u}}}) + {{\,\textrm{W}\,}}_{d_\sigma }(\bar{\varvec{\textrm{u}}}), \end{array}\right. } \end{aligned}$$

$\Leftrightarrow$

$$\begin{aligned} {\left\{ \begin{array}{ll} (-1)^{\widetilde{b_\sigma }(\varvec{\textrm{u}})} + (-1)^{\widetilde{c_\sigma }(\varvec{\textrm{u}})} = (-1)^{\widetilde{a_\sigma }(\varvec{\textrm{u}})} + (-1)^{\widetilde{d_\sigma }(\varvec{\textrm{u}})}, \\ (-1)^{\widetilde{b_\sigma }(\bar{\varvec{\textrm{u}}})} +(-1)^{\widetilde{c_\sigma }(\bar{\varvec{\textrm{u}}})} = (-1)^{\widetilde{a_\sigma }(\bar{\varvec{\textrm{u}}})} +(-1)^{\widetilde{d_\sigma }(\bar{\varvec{\textrm{u}}})}. \end{array}\right. } \end{aligned}$$

(B1)

Similarly, the case ${{\,\textrm{N}\,}}_b(\varvec{\textrm{u}}) + {{\,\textrm{N}\,}}_c(\varvec{\textrm{u}}) = -({{\,\textrm{N}\,}}_a(\varvec{\textrm{u}}) + {{\,\textrm{N}\,}}_d(\varvec{\textrm{u}}))$ $\Leftrightarrow$

$$\begin{aligned} {\left\{ \begin{array}{ll} (-1)^{\widetilde{b_\sigma }(\varvec{\textrm{u}})} + (-1)^{\widetilde{c_\sigma }(\varvec{\textrm{u}})} = (-1)^{\widetilde{a_\sigma }(\varvec{\textrm{u}})\oplus 1} + (-1)^{\widetilde{d_\sigma }(\varvec{\textrm{u}})\oplus 1}, \\ (-1)^{\widetilde{b_\sigma }(\bar{\varvec{\textrm{u}}})} +(-1)^{\widetilde{c_\sigma }(\bar{\varvec{\textrm{u}}})} = (-1)^{\widetilde{a_\sigma }(\bar{\varvec{\textrm{u}}})\oplus 1} +(-1)^{\widetilde{d_\sigma }(\bar{\varvec{\textrm{u}}})\oplus 1}. \end{array}\right. } \end{aligned}$$

(B2)

Finally, the condition (B1) or (B2) is equivalent to $(\widetilde{b_\sigma }(\varvec{\textrm{u}}), \widetilde{b_\sigma }(\bar{\varvec{\textrm{u}}})) \oplus (\widetilde{c_\sigma }(\varvec{\textrm{u}}), \widetilde{c_\sigma }(\bar{\varvec{\textrm{u}}})) = (\widetilde{a_\sigma }(\varvec{\textrm{u}}), \widetilde{a_\sigma }(\bar{\varvec{\textrm{u}}})) \oplus (\widetilde{d_\sigma }(\varvec{\textrm{u}}), \widetilde{d_\sigma }(\bar{\varvec{\textrm{u}}}))$. $\square$

Proof of Theorem 10

Rewrite f as $f(\varvec{\textrm{x}}, y_1, y_2)=h(a(\varvec{\textrm{x}}), b(\varvec{\textrm{x}}), c(\varvec{\textrm{x}}), y_1, y_2),$ where h is a 5-variable Boolean function with ANF

$$\begin{aligned} h(x_1, \ldots , x_5) = x_1x_2\oplus x_1x_3\oplus x_2x_3\oplus x_4x_5\oplus (x_1\oplus x_2)x_4\oplus (x_1\oplus x_3)x_5. \end{aligned}$$

By a direct calculation, we have

$$\begin{aligned}&{{\,\textrm{W}\,}}_h(1,0,0,0,0)=2^\frac{3}{2},\ {{\,\textrm{W}\,}}_h(0,1,0,0,1)=2^\frac{3}{2},\\&{{\,\textrm{W}\,}}_h(0,0,1,1,0)=2^\frac{3}{2},\ {{\,\textrm{W}\,}}_h(1,1,1,1,1)=-2^\frac{3}{2}, \end{aligned}$$

and the Walsh–Hadamard transform of h vanishes at other positions. By Theorem 7, the nega–Hadamard transform of f at $(\varvec{\textrm{u}}, v_1, v_2) \in \mathbb {F}_2^n \times \mathbb {F}_2 \times \mathbb {F}_2$ is given by

$$\begin{aligned} {{\,\textrm{N}\,}}_f(\varvec{\textrm{u}}, v_1, v_2)&= 2^{-\frac{5}{2}}\sum _{\varvec{{\omega }} \in {\mathbb {F}_2^{5}}}{{\,\textrm{W}\,}}_h(\varvec{{\omega }}) {{\,\textrm{N}\,}}_{\varvec{{\omega }}\cdot (a(\varvec{\textrm{x}}), b(\varvec{\textrm{x}}), c(\varvec{\textrm{x}}), y_1, y_2)}(\varvec{\textrm{u}}, v_1, v_2) \\&= 2^{-1} \left[ {{\,\textrm{N}\,}}_a(\varvec{\textrm{u}}){{\,\textrm{N}\,}}_0(v_1, v_2) + {{\,\textrm{N}\,}}_b(\varvec{\textrm{u}}){{\,\textrm{N}\,}}_{y_2}(v_1, v_2) \right. \\&\quad \left. + {{\,\textrm{N}\,}}_c(\varvec{\textrm{u}}){{\,\textrm{N}\,}}_{y_1}(v_1, v_2) - {{\,\textrm{N}\,}}_d(\varvec{\textrm{u}}){{\,\textrm{N}\,}}_{y_1\oplus y_2}(v_1, v_2)\right] \\&= {\left\{ \begin{array}{ll} 2^{-1} \left[ \imath \cdot {{\,\textrm{N}\,}}_a(\varvec{\textrm{u}})+ {{\,\textrm{N}\,}}_b(\varvec{\textrm{u}}) + {{\,\textrm{N}\,}}_c(\varvec{\textrm{u}}) +\imath \cdot {{\,\textrm{N}\,}}_d(\varvec{\textrm{u}}) \right] , \quad (v_1, v_2) = (0, 0); \\ 2^{-1} \left[ {{\,\textrm{N}\,}}_a(\varvec{\textrm{u}})+ \imath \cdot {{\,\textrm{N}\,}}_b(\varvec{\textrm{u}}) -\imath \cdot {{\,\textrm{N}\,}}_c(\varvec{\textrm{u}}) - {{\,\textrm{N}\,}}_d(\varvec{\textrm{u}}) \right] , \quad (v_1, v_2) = (0, 1); \\ 2^{-1} \left[ {{\,\textrm{N}\,}}_a(\varvec{\textrm{u}})- \imath \cdot {{\,\textrm{N}\,}}_b(\varvec{\textrm{u}}) +\imath \cdot {{\,\textrm{N}\,}}_c(\varvec{\textrm{u}}) - {{\,\textrm{N}\,}}_d(\varvec{\textrm{u}}) \right] , \quad (v_1, v_2) = (1, 0); \\ 2^{-1} \left[ -\imath \cdot {{\,\textrm{N}\,}}_a(\varvec{\textrm{u}})+ {{\,\textrm{N}\,}}_b(\varvec{\textrm{u}}) + {{\,\textrm{N}\,}}_c(\varvec{\textrm{u}}) -\imath \cdot {{\,\textrm{N}\,}}_d(\varvec{\textrm{u}}) \right] , \ (v_1, v_2) = (1, 1). \end{array}\right. } \end{aligned}$$

Since a, b, c, d are all negabent, then all the values of ${{\,\textrm{N}\,}}_a(\varvec{\textrm{u}})$, ${{\,\textrm{N}\,}}_b(\varvec{\textrm{u}})$, ${{\,\textrm{N}\,}}_c(\varvec{\textrm{u}})$, ${{\,\textrm{N}\,}}_d(\varvec{\textrm{u}})$ lie in $\{\pm 1, \pm \imath \}$.

f is negabent if and only if $|{{\,\textrm{N}\,}}_f(\varvec{\textrm{u}}, v_1, v_2)|=1$ for all $(\varvec{\textrm{u}}, v_1, v_2)\in \mathbb {F}_2^n \times \mathbb {F}_2 \times \mathbb {F}_2$. From Lemma 7, for the four cases of $(v_1, v_2)\in \mathbb {F}_2^2$, the above four equations $|{{\,\textrm{N}\,}}_f(\varvec{\textrm{u}}, v_1, v_2)|=1$ have the same solutions of $({{\,\textrm{N}\,}}_a(\varvec{\textrm{u}}), {{\,\textrm{N}\,}}_b(\varvec{\textrm{u}}), {{\,\textrm{N}\,}}_c(\varvec{\textrm{u}}), {{\,\textrm{N}\,}}_d(\varvec{\textrm{u}}))$. Hence, we only study the first equation:

$$\begin{aligned} |\imath \cdot \left[ {{\,\textrm{N}\,}}_a(\varvec{\textrm{u}}) + {{\,\textrm{N}\,}}_d(\varvec{\textrm{u}})\right] + \left[ {{\,\textrm{N}\,}}_b(\varvec{\textrm{u}}) + {{\,\textrm{N}\,}}_c(\varvec{\textrm{u}})\right] |= 2. \end{aligned}$$

(B3)

The solutions of (B3) lie in the following three cases:

• (1) ${{\,\textrm{N}\,}}_b(\varvec{\textrm{u}}) = - {{\,\textrm{N}\,}}_c(\varvec{\textrm{u}})$, ${{\,\textrm{N}\,}}_a(\varvec{\textrm{u}}) = {{\,\textrm{N}\,}}_d(\varvec{\textrm{u}})$;

• (2) ${{\,\textrm{N}\,}}_b(\varvec{\textrm{u}}) = {{\,\textrm{N}\,}}_c(\varvec{\textrm{u}})$, ${{\,\textrm{N}\,}}_a(\varvec{\textrm{u}}) = - {{\,\textrm{N}\,}}_d(\varvec{\textrm{u}})$;

• (3) ${{\,\textrm{N}\,}}_c(\varvec{\textrm{u}}) = \pm \imath \cdot {{\,\textrm{N}\,}}_b(\varvec{\textrm{u}})$, ${{\,\textrm{N}\,}}_b(\varvec{\textrm{u}}) + {{\,\textrm{N}\,}}_c(\varvec{\textrm{u}})=\pm ({{\,\textrm{N}\,}}_a(\varvec{\textrm{u}}) + {{\,\textrm{N}\,}}_d(\varvec{\textrm{u}}))$.

From Lemma 9, the cases (1) and (2) are respectively equivalent to

$$\begin{aligned} {\left\{ \begin{array}{ll} (\widetilde{b_\sigma }(\varvec{\textrm{u}}), \widetilde{b_\sigma }(\bar{\varvec{\textrm{u}}})) \oplus (\widetilde{c_\sigma }(\varvec{\textrm{u}}), \widetilde{c_\sigma }(\bar{\varvec{\textrm{u}}})) = (1,1), \\ (\widetilde{a_\sigma }(\varvec{\textrm{u}}), \widetilde{a_\sigma }(\bar{\varvec{\textrm{u}}})) \oplus (\widetilde{d_\sigma }(\varvec{\textrm{u}}), \widetilde{d_\sigma }(\bar{\varvec{\textrm{u}}})) = (0, 0), \end{array}\right. } \end{aligned}$$

(B4)

and

$$\begin{aligned} {\left\{ \begin{array}{ll} (\widetilde{b_\sigma }(\varvec{\textrm{u}}), \widetilde{b_\sigma }(\bar{\varvec{\textrm{u}}})) \oplus (\widetilde{c_\sigma }(\varvec{\textrm{u}}), \widetilde{c_\sigma }(\bar{\varvec{\textrm{u}}})) = (0,0), \\ (\widetilde{a_\sigma }(\varvec{\textrm{u}}), \widetilde{a_\sigma }(\bar{\varvec{\textrm{u}}})) \oplus (\widetilde{d_\sigma }(\varvec{\textrm{u}}), \widetilde{d_\sigma }(\bar{\varvec{\textrm{u}}})) = (1, 1). \end{array}\right. } \end{aligned}$$

(B5)

For the case (3), by Lemma 8, ${{\,\textrm{N}\,}}_c(\varvec{\textrm{u}}) = \pm \imath \cdot {{\,\textrm{N}\,}}_b(\varvec{\textrm{u}})$ is equivalent to $\widetilde{b_\sigma }(\varvec{\textrm{u}}) \oplus \widetilde{b_\sigma }(\bar{\varvec{\textrm{u}}}) \oplus \widetilde{c_\sigma }(\varvec{\textrm{u}}) \oplus \widetilde{c_\sigma }(\bar{\varvec{\textrm{u}}}) = 1$, i.e., $(\widetilde{b_\sigma }(\varvec{\textrm{u}}), \widetilde{b_\sigma }(\bar{\varvec{\textrm{u}}})) \oplus (\widetilde{c_\sigma }(\varvec{\textrm{u}}), \widetilde{c_\sigma }(\bar{\varvec{\textrm{u}}}))\in \{(0, 1), (1, 0) \}$. Furthermore, by Lemma 10, we know that ${{\,\textrm{N}\,}}_b(\varvec{\textrm{u}}) + {{\,\textrm{N}\,}}_c(\varvec{\textrm{u}})=\pm ({{\,\textrm{N}\,}}_a(\varvec{\textrm{u}}) + {{\,\textrm{N}\,}}_d(\varvec{\textrm{u}}))$ is equivalent to

$$\begin{aligned} (\widetilde{b_\sigma }(\varvec{\textrm{u}}), \widetilde{b_\sigma }(\bar{\varvec{\textrm{u}}})) \oplus (\widetilde{c_\sigma }(\varvec{\textrm{u}}), \widetilde{c_\sigma }(\bar{\varvec{\textrm{u}}}))&= (\widetilde{a_\sigma }(\varvec{\textrm{u}}), \widetilde{a_\sigma }(\bar{\varvec{\textrm{u}}})) \oplus (\widetilde{d_\sigma }(\varvec{\textrm{u}}), \widetilde{d_\sigma }(\bar{\varvec{\textrm{u}}})) \nonumber \\&\quad \in \{(0,1), (1,0) \}. \end{aligned}$$

(B6)

It follows immediately that (B4) together with (B5) are equivalent to the C-1, and (B6) is equivalent to C-2 in Theorem 10. The proof is completed. $\square$