Let $O(x)$ denote a bilinear field operator, such that

\begin{equation}\label{eq:eq1}

O(x) = \bar{\psi}(x)\psi(x)

\end{equation}

where $\bar{\psi} \equiv \psi^\dagger \gamma^0$. I'm supposed to prove that

\begin{equation}\label{eq:eq2}

(\partial^2-m^2) \bar{\psi}(x)\psi(x) = 0

\end{equation}

**My attempt**

I did the following:

\begin{equation}

(\partial^2-m^2) \bar{\psi}(x)\psi(x) = 0 \\

\Leftrightarrow (\partial^2 \bar{\psi})\psi+\bar{\psi}\partial^2\psi-m^2\bar{\psi}\psi = 0

\end{equation}

Using the slash notation,

\begin{equation}

i \displaystyle{\not} \partial(-i \displaystyle{\not}\partial\bar{\psi})\psi+\bar{\psi}(-i\displaystyle{\not}\partial)i\displaystyle{\not}\partial\psi-m^2\bar{\psi}\psi = 0 \\

\Leftrightarrow i \displaystyle{\not} \partial(m\bar{\psi})\psi+\bar{\psi}(-i\displaystyle{\not}\partial)m\psi-m^2\bar{\psi}\psi = 0 \\

\Leftrightarrow -m^2\bar{\psi}\psi-m^2\bar{\psi}\psi-m^2\bar{\psi}\psi = 0 \\

\Leftrightarrow -3m^2\bar{\psi}\psi = 0

\end{equation}

which would imply that $m = 0$. What am I doing wrong?