Proposition 1

Let $d_1, d_2$ be two $\varepsilon$-derivations of $(A, A', A'')$ of respective degrees $\delta_1, \delta_2$. Then the $\varepsilon$-bracket

is an $\varepsilon$-derivation of degree $\delta_1 + \delta_2$. Moreover, if $d$ is an $\varepsilon$-derivation of $(A, A', A'')$ of degree $\delta$ and if $\varepsilon_{\delta, \delta} = -1$, then $d^2 = d \circ d$ is a derivation.

Suppose $x \in A$ is homogeneous of degree $\xi$; for all $y \in A'$,

taking account of formulae (1) of no. 1. Exchanging the roles of $d_1$ and $d_2$, we obtain, after simplifications again using (1) (no. 1), that is, writing $d = [d_1, d_2]_\varepsilon$ and $\delta = \delta_1 + \delta_2$, which proves that $d$ is an $\varepsilon$-derivation.

On the other hand, if, in (11), we let $d_1 = d_2 = d$, $\delta_1 = \delta_2 = \delta$ and $\varepsilon_{\delta, \delta} = -1$, we obtain, since then $\varepsilon_{\delta, \delta + \xi} = -\varepsilon_{\delta, \xi}$ by (1),

and as $\varepsilon_{2\delta, \xi} = 1$ it is seen that $d^2$ is a derivation.