Breaking of a commutator involving Dirac spinors and gamma matrices I'm trying to understand a particular step in the solution to problem 27 in THIS solution sheet. By the middle of the page, they start with the simplification of this expression
$$\left[s^{\mu}\left(x\right),s^{\nu}\left(y\right)\right]\overset{(1)}{=}e^{2}\left[\overline{\psi}\left(x\right)\gamma^{\mu}\psi\left(x\right),\overline{\psi}\left(y\right)\gamma^{\nu}\psi\left(y\right)\right]\overset{(2)}{=}\cdots$$
and I don't understand how they break this commutator at step (2). Is their algebra correct? Using $\left[AB,C\right]=A\left[B,C\right]+\left[A,C\right]B$,
and identifying $A=\overline{\psi}\left(x\right)\gamma^{\mu}$ and
$B=\psi\left(x\right)$, shouldn't it be
$$
\overset{2}{=}e^{2}\overline{\psi}\left(x\right)\gamma^{\mu}\left[\psi\left(x\right),\overline{\psi}\left(y\right)\gamma^{\nu}\psi\left(y\right)\right]+e^{2}\left[\overline{\psi}\left(x\right)\gamma^{\mu},\overline{\psi}\left(y\right)\gamma^{\nu}\psi\left(y\right)\right]\psi\left(x\right)
$$
The same kind of misplacement of $\gamma^{\nu}$ seems to happen again at step (3) when they break the first commutator. In here, if they were using 
$\left[A,BC\right]=\left\{ A,B\right\} C-B\left\{ A,C\right\} $ with
$B=\overline{\psi}\left(y\right)$, $C=\gamma^{\nu}\psi\left(y\right)$,
I guess one would have
$$
\left[\psi\left(x\right),\overline{\psi}\left(y\right)\gamma^{\nu}\psi\left(y\right)\right]=\left\{ \psi\left(x\right),\overline{\psi}\left(y\right)\right\} \gamma^{\nu}\psi\left(y\right)-\overline{\psi}\left(y\right)\left\{ \psi\left(x\right),\gamma^{\nu}\psi\left(y\right)\right\} 
$$
instead of what's there. I must be missing something, but I can't see what! Could you please help?
 A: Using the formula
\begin{equation}
    [AB,CD] = A\{B,C\}D - AC \{B,D\} + \{A,C\}DB - C \{A,D\}B \quad,
\end{equation}
and setting
\begin{equation}
    A = \bar{\psi}(x) \quad,\quad
    B = \gamma^\mu \psi(x) \quad,\quad
    C = \bar{\psi}(y) \quad,\quad
    D = \gamma^\nu \psi(y) \quad,
\end{equation}
one gets (spinor indices suppressed):
\begin{equation}\begin{alignedat}{9}
    &[\bar{\psi}(x) \gamma^\mu \psi(x),  \bar{\psi}(y) \gamma^\nu \psi(y)]
    \\&=
    \bar{\psi}(x)\{ \gamma^\mu \psi(x),\bar{\psi}(y)\}\gamma^\nu \psi(y) - \bar{\psi}(x)\bar{\psi}(y) \{ \gamma^\mu \psi(x),\gamma^\nu \psi(y)\}
    \\&+ \{\bar{\psi}(x),\bar{\psi}(y)\}\gamma^\nu \psi(y) \gamma^\mu \psi(x) - \bar{\psi}(y) \{\bar{\psi}(x),\gamma^\nu \psi(y)\} \gamma^\mu \psi(x)
    \\&=
    \gamma^\mu\bar{\psi}(x)\{ \psi(x),\bar{\psi}(y)\}\gamma^\nu \psi(y) - \gamma^\mu\bar{\psi}(x)\bar{\psi}(y) \{ \psi(x), \psi(y)\}\gamma^\nu
    \\&+ \{\bar{\psi}(x),\bar{\psi}(y)\}\gamma^\nu \psi(y) \gamma^\mu \psi(x) - \bar{\psi}(y) \{\bar{\psi}(x), \psi(y)\} \gamma^\nu\gamma^\mu \psi(x)=0\quad.
\end{alignedat}\end{equation}
Actually, a more general statement is true $-$ see the bottom line on page 5 here.
For the detailed solution see Problem 11 here.
