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\section{Vectorial Boolean functions} \begin{Definition}\label{VectBool}  A vectorial Boolean function is a function  $$F:(\Fb)^n\rightarrow (\Fb)^m,$$  with $m,n \in \NN$.  \end{Definition}  We can represent a vectorial Boolean function as a vector of Boolean functions:  $$F:(\Fb)^n\Rightarrow (\Fb)^m,$$  $$(x_1,...,x_n)\mapsto \begin{matrix} F1(x_1,...,x_n)\\ \vdots \\ Fm(x_1,...,x_n) \end{matrix},$$   \begin{Example}\label{ExVectBool}  The function $F:(\Fb)^3\rightarrow (\Fb)^2,$  given by $(x,y,z)\mapsto (xy+y, y+z)$  is a vectorial Boolean function.  \end{Example}  We notice that every component of a vectorial Boolean function is a Boolean function, so each component can be represented in its absolute normal form.  \begin{Corollary}\label{VectANF}  Let $$F:(\Fb)^n\rightarrow (\Fb)^m, n,m \in |NN$$ be a vectorial Boolean function. Then  $F=\begin{matrix} F1(x_1,...,x_n)\\ \vdots \\ Fm(x_1,...,x_n) \end{matrix},$ with   $$F_i =\sum_{S \subset \{1,...,n\}} a_S^{(i)}x_S^{(i)},$$  where $a_S^{(i)} \in \Fb$ and $x_S^{(i)}=\prod_{j \in S} x_j$ for every $j \in \{1,...,m\}$.  \end{Corollary}  Since the size of the set of all functions from a set $A$ to a set $B$ i.e. $\mathfrak{F}=\{f:B\rightarrow B\}$ is $\vert \mathfrak{F} \vert = \vert B\vert ^{\vert AQ\vert }$, then the number of all vectorial Boolean functions from $(\Fb)^n$ to $(\Fb)^m$ is   $(2^m)^{2^n}=2^{m2^n}$. We observe that the same number comes from counting all vectors of length $m$ with components which are Boolean functions in $x_1,...,x_n$.  A special case of Boolean functions arise when $m=n$, i.e. the domain and the codomain of the function coincide:  $$F:(\Fb)^n\rightarrow (\Fb)^n, n\in \NN$$   Some of these could be permutations (i.e. bijections). Since a vectorial Boolean function is a  function between two finite sets, then it is a invertible if and only if it is  injective or surjective.  \begin{Theorem}\label{DegVectBoolFunct}  Let $n\in \NN$. If $F:(\Fb)^n\rightarrow (\Fb)^n$ is a permutation, then   $$\deg(F_i)\leq n-1,\, \forall i \in \{1,...,n\}.$$  \end{Theorem}