Question about the parity violation of weak interaction Lagrangian In the textbook of A. Zee, Quantum Field Theory in a Nutshell, the author states that the following Lagrangian:
$$ \mathcal{L} = G (\overline{\psi}_{1L} \gamma^\mu \psi_{2L})(\overline{\psi}_{3L} \gamma^\mu \psi_{4L})$$
Which describes the weak interaction, violates the parity invariance. I've tried to prove this but with no success. Before I start showing what I've done, it's good to give you some context of the notation I'm familiar with. First, the parity operator is defined as $P\psi := \gamma^0 \psi$, where
$$\gamma^0 :=
\begin{pmatrix}
0 & 1
\\
1 & 0
\end{pmatrix}
.$$
And that
$$ \psi_{L,R} := \frac{1}{2}(1 \mp \gamma^5) \psi = P_{L,R} \psi \quad \text{and} \quad \overline{\psi}_{L,R} : = \overline{\psi} \frac{1}{2} (1 \pm \gamma^5)  = \overline{\psi}P_{R,L},$$
where
$$\gamma^5 := \begin{pmatrix}
-1 & 0
\\
0 & 1
\end{pmatrix}
.$$
And $\overline{\psi} := \psi^\dagger \gamma^0$. So, what I've tried to do is, basically:
$$ \begin{split}
P(\overline{\psi}_L \gamma^\mu \psi_L) = \gamma^0 \psi^\dagger \gamma^0 P_R \gamma^\mu \gamma^0 P_L \psi &= \gamma^0 \psi^\dagger P_L \gamma^0 \gamma^\mu \gamma^0 P_L \psi
\\
&= \gamma^0 \psi^\dagger P_L (\gamma^\mu)^\dagger P_L \psi
\end{split}$$
But I really don't know to proceed from here or if there are some mistakes in my calculations. Any help would be really welcomed.
 A: $$
\gamma^0 P_L \gamma^0 = P_R, \\
 \gamma^0 P_R \gamma^0 = P_L,
$$
$$
P: \qquad  \psi(x) \longrightarrow \gamma^0 \psi(-x) ~~~~\leadsto \\
P: \qquad  \psi(x)^\dagger  \longrightarrow \psi(-x)^\dagger \gamma^0  ~~~~\leadsto \\
P: \qquad  \overline \psi(x)\gamma^0\psi'(x) \longrightarrow  \overline \psi(-x)\gamma^0 \gamma^0 \gamma^0  \psi'(-x) =\overline \psi(-x) \gamma^0  \psi'(-x), \\ P: \qquad  \overline \psi(x)\vec \gamma\psi'(x) \longrightarrow  \overline \psi(-x)\gamma^0 \vec \gamma  \gamma^0  \psi'(-x) =-\bar\psi(-x) \vec \gamma   \psi'(-x). 
$$
One projector suffices for each spinor bilinear, given the intercalated γ s, so that
$$
P: \qquad  \overline \psi(x)\gamma^0P_L\psi'(x) ~~ \overline \psi''(x)\gamma^0P_L\psi'''(x)\longrightarrow \\ \overline \psi(-x)\gamma^0 \gamma^0 P_L\gamma^0  \psi'(-x)~~ \overline \psi(-x)\gamma^0 \gamma^0 P_L\gamma^0  \psi'(-x)=\overline \psi(-x) \gamma^0 P_R \psi'(-x)~~\overline \psi''(-x) \gamma^0 P_R \psi'''(-x),
$$
and similarly for the spacelike  γ s... You do it.
You see that the left fermions and right antifermions flipped to right fermions and left antifermions.
