A theorem about the anti-derivation of the algebra Sadri Hassani wrote down a theorem about the anti-derivation in his Mathematical Physics: A Modern Introduction to Its Foundations - Second Edition.
Theorem 3.4.10 Let $\Omega_1$ and $\Omega_2$ be anti-derivations with respect to two involutions $\omega_1$ and $\omega_2$. Suppose that $\omega_1\circ\omega_2=\omega_2\circ\omega_1$. Furthermore, assume that
$$
\omega_1\Omega_2=\pm\Omega_2\omega_1\quad \text{and} \quad \omega_2\Omega_1=\pm\Omega_1\omega_2
$$
Then $\Omega_1\Omega_2\mp\Omega_2\Omega_1$ is an anti-derivation with respect to the involution $\omega_1\circ\omega_2$.
I am trying to prove this theorem, but there are two strange terms that cannot vanish with each other. Follows are my attempts.
Firstly, the definition anti-derivation gives that
$$
\begin{aligned}
\Omega_1(\textbf a_1\textbf a_2)=\Omega_1(\textbf a_1)\cdot\textbf a_2+\omega_1(\textbf a_1)\cdot\Omega_1(\textbf a_2), \\
\Omega_2(\textbf a_1\textbf a_2)=\Omega_2(\textbf a_1)\cdot\textbf a_2+\omega_2(\textbf a_1)\cdot\Omega_1(\textbf a_2).
\end{aligned}
$$
Hence, we can calculate the $(\Omega_1\Omega_2\mp\Omega_2\Omega_1)(\textbf a_1\textbf a_2)$ as follows
$$
\begin{aligned}
&\ (\Omega_1\Omega_2\mp\Omega_2\Omega_1)(\textbf a_1\textbf a_2) \\
=&\ \Omega_1(\Omega_2(\textbf a_1)\cdot\textbf a_2+\omega_2(\textbf a_1)\cdot\Omega_2(\textbf a_2))\mp\Omega_2(\Omega_1(\textbf a_1)\cdot\textbf a_2+\omega_1(\textbf a_1)\cdot\Omega_1(\textbf a_2)) \\
=&\ (\Omega_1\Omega_2\mp\Omega_2\Omega_1)(\textbf a_1)\cdot\textbf a_2+\Omega_1\circ\omega_2(\textbf a_1)\cdot\Omega_2(\textbf a_2)+\omega_1\circ\omega_2(\textbf a_1)\cdot\Omega_1\circ\Omega_2(\textbf a_2) \\
&\quad\mp\Omega_2\circ\omega_1(\textbf a_1)\cdot\Omega_1(\textbf a_2)\mp\omega_2\circ\omega_1(\textbf a_1)\cdot\Omega_2\circ\Omega_1(\textbf a_2) \\
=&\ (\Omega_1\Omega_2\mp\Omega_2\Omega_1)(\textbf a_1)\cdot\textbf a_2+\omega_1\circ\omega_2(\textbf a_1)\cdot(\Omega_1\Omega_2\mp\Omega_2\Omega_1)(\textbf a_2) \\
&\quad+\Omega_1\circ\omega_2(\textbf a_1)\cdot\Omega_2(\textbf a_2)\mp\Omega_2\circ\omega_1(\textbf a_1)\cdot\Omega_1(\textbf a_2)
\end{aligned}
$$
If the theorem is true, the last two terms must vanish, which means
$$
\Omega_1\circ\omega_2(\textbf a_1)\cdot\Omega_2(\textbf a_2)\mp\Omega_2\circ\omega_1(\textbf a_1)\cdot\Omega_1(\textbf a_2)\equiv0
$$
But I don't know how to prove it.
 A: I think you've made some calculation error. Let me try this. 
\begin{align*}
& \quad \Omega_1 \Omega _2 (a_1 \cdot a_2) \\
&= \Omega_1 (\Omega_2 (a_1) \cdot a_2 + \omega _2 (a_1) \cdot \Omega_2(a_2)) \\
&= \Omega_1 \Omega_2 (a_1) \cdot a_2 + {\color {red} {\omega_ 1 \Omega_2 (a_1) \cdot \Omega_1 (a_2)}} + {\color {green} {\Omega_1\omega_2 (a_2) \cdot \Omega_2 (a_2)}} + {\color {blue}{ \omega_1 \omega_2 (a_1) \cdot \Omega_1 \Omega_2 (a_2)}}. 
\end{align*}
Note that exchanging $\Omega_j$'s only affects the subscript of $\omega_j$, thus we have 
$$
 \Omega_2 \Omega _1 (a_1 \cdot a_2) 
= \Omega_2 \Omega_1 (a_1) \cdot a_2 + {\color {green} {\omega_ 2 \Omega_1 (a_1) \cdot \Omega_2(a_2)}} + {\color {red} {\Omega_2\omega_1 (a_2) \cdot \Omega_1 (a_2)}} + {\color {blue}{ \omega_2 \omega_1 (a_1) \cdot \Omega_2 \Omega_1 (a_2)}}. 
$$
Then the condition given could help us cancel out the red and the green part, for example
\begin{align*}
&\quad (\Omega_1 \Omega_2 - \Omega_2 \Omega_1) (a_1\cdot a_2)\\
&= (\Omega_1 \Omega_2 -\Omega_2 \Omega_1) (a_1) \cdot a_2 + \color {green} {(\Omega_1 \omega_2 - \omega_2 \Omega_1) (a_1) \cdot \Omega_2 (a_2)} \\
&\phantom {== }+ \color{red} {(\omega_1 \Omega_2 - \Omega_2  \omega_1) (a_1) \cdot \Omega_1 (a_2)} + \color{blue} {\omega_1 \omega_2 (a_1) \cdot (\Omega_1 \Omega_2 - \Omega_2 \Omega_1) (a_2)} \\
&= (\Omega_1 \Omega_2 -\Omega_2 \Omega_1) (a_1) \cdot a_2 + \color{blue} {\omega_1 \omega_2 (a_1) \cdot (\Omega_1 \Omega_2 - \Omega_2 \Omega_1) (a_2)} 
\end{align*}
where we have make use of the commutativity of $\omega_1, \omega_2$. The theorem is proved. 
