have:

-   The algebras $A_i$ and $A_j$ commute, if ${\varepsilon}_{ij}=1$.

-   Moreover, for any $k\in{ {\mathbb N} }$ and any choice $a_1,\ldots,a_k\in A$ with $a_j\in A_{i(j)}$ and the properties

    -   ${\varphi}(a_j)=0$ for all $j=1,\ldots,k$,

    -   and for any $1\leq p<r\leq k$ with $i(p)=i(r)$ there is a $q$ with $p<q<r$ such that ${\varepsilon}_{i(p)i(q)}=0$ and $i(p)\neq i(q)$,

    we have ${\varphi}(a_1\cdots a_k)=0$.

(Selfadjoint) variables $x_1,\ldots,x_n\in A$ are ${\varepsilon}$-independent, if the algebras $\textnormal{alg}(x_j,1)\subset A$ are ${\varepsilon}$-independent.

\[ExEps\] Here are a few examples of ${\varepsilon}$-independent variables.

-   Let ${\varepsilon}_{\operatorname{comm}}\in M_n(\{0,1\})$ be the matrix defined by ${\varepsilon}_{ij}=1$ for all $i\neq j$ and ${\varepsilon}_{ii}=0$ for all $i$. Variables $x_1,\ldots, x_n$ are ${\varepsilon}$-independent with respect to ${\varepsilon}_{\operatorname{comm}}$ if and only if they all commute and are classically independent. Indeed, the constraint on the indices in Definition \[DefEpsInd\](ii) yields that all indices must be mutually different, in case ${\varepsilon}={\varepsilon}_{\operatorname{comm}}$. Now, by the usual centering trick on $a_j:=x_{i(j)}^{m_j}-{\varphi}(x_{i(j)}^{m_j})$ with mutually different indices $i(j)$, we infer $${\varphi}(x_{i(1)}^{m_1}\cdots x_{i(k)}^{m_k})=\prod_{j=1}^k {\varphi}(x_{i(j)}^{m_j})$$ and hence classical independence. See also [@SpW Prop. 3.2].

-   Let ${\varepsilon}_{\operatorname{free}}\in M_n(\{0,1\})$ be defined by ${\varepsilon}_{ij}=0$ for all $i,j$. Variables $x_1,\ldots, x_n$ are ${\varepsilon}$-independent with respect to ${\varepsilon}_{\operatorname{free}}$ if and only if they are freely independent. Note that if ${\varepsilon}={\varepsilon}_{\operatorname{free}}$, the 