by Theorem \[keythm-1\].

\[delta-1\] Let $G(x,y)=(x^{2^i}\pi(y)+g(y),h(y))$, where $\pi(y)$ is a permutation over $\mathbb{F}_{2^{k}}$, and $g(y)$ and $h(y)$ are arbitrary $(k,k)$-function and $(k, m-k)$-function respectively. Then the differential spectrum of the function $G$ is given by $$\begin{aligned}
\delta_{G}((u,v),(a,b))\!= \! \begin{cases} 0, &\text{if ~} v\!=\!0 \text{~and~ } b\!\neq\! 0,\\
2^{k}, &\text{if ~} v\!=\!0 \text{~and~ } b\!=\!0, \\  \delta_{h}(v,b),  &\text{otherwise. }
\end{cases}\end{aligned}$$ In particular, $G$ is differentially $2^k$-uniform.

We have $$\begin{aligned}
&&G(x,y)\!+\!G(x\!+\!u,y\!+\!v)\\
&\!=\!&(x^{2^{i}}\pi(y)+g(y),h(y))\!+\!(x\!+\!u)^{2^{i}}\pi(y\!+\!v)+g(y\!+\!v),h(y\!+\!v)) \\
             &\!=\!&(x^{2^{i}}\pi(y)\!+\!(x\!+\!u)^{2^{i}}\pi(y\!+\!v)\!+\!g(y)+\!g(y\!+\!v),h(y)\!+\!h(y\!+\!v))\\
             &\!=\!&((\pi(y)\!+\!\pi(y\!+\!v))x^{2^{i}}\!+\!u^{2^{i}}\pi(y\!+\!v)+\!g(y)+\!g(y\!+\!v),h(y)\!+\!h(y\!+\!v)).
             \end{aligned}$$

Then $G(x,y)+G(x+u,y+v)=(a,b)$ is equivalent to $$\begin{aligned}
&((\pi(y)\!+\!\pi(y\!+\!v))x^{2^{i}}\!+\!u^{2^{i}}\pi(y\!+\!v)+\!g(y)+\!g(y\!+\!v)\!=\!a,\label{eq1}\\
&h(y)\!+\!h(y\!+\!v)\!=\!b. \label{eq2}\end{aligned}$$ If $v=0$ (and then $u\neq 0$) and $b=0$, then (\[eq2\]) holds for any $y\in \mathbb{F}_{2^{k}}$. Note that (\[eq1\]) is reduced to $u^{2^{i}}\pi(y)=a$, which has exactly one solution $y=\pi^{-1}(au^{-2^{i}})$. Therefore, the solutions of $G(x,y)+G(x+u,y+v)=(a,b)$ are exactly those $(x,\pi^{-1}(au^{-2^{i}}))$ with $x\in \mathbb{F}_{2^{k}}$, and thus $\delta_{G}((u,0),(a,0))=2^{k}$.

If $v=0$ and $b\neq 0$, (\[eq2\]) has no solutions, then $\delta_{G}((u,0),(a,b))=0$.

When $v\neq 0$, one has $\pi(y)+\pi(y+v)\neq 0$, since $\pi$ is a permutation function. Then Eq. (\[eq1\]) can be rewritten as $x^{2^{i}}=(a+u^{2^{i}}\pi(y+v)+g(y)+g(y+v))[\pi(y)+\pi(y+v)]^{-1}$, so that all the solutions to $G(x,y)+G(x+u,y+v)=(a,b)$ are $(((a+u^{2^{i}}\pi(y+v)+g(y)+g(y+v))[\pi(y)+\pi(y+v)]^{-1})^{2^{-i}},y)$, where $y$ satisfies (\[eq2\]) . Therefore in this case $\delta_{G}((u,v),(a,b))=\delta_{h}(v,b)$.

One can see that at least most of the functions given by Theorem \[keythm-1\] are [CCZ]{}-inequivalent to those trivial ones, since they have 