# Bad Maths

$$
\begin{aligned}
\&e^{i \cdot Pi}=-1 \\
\&e^{i \cdot P \cdot i}=-1 \\
\&e^{P \cdot {i \cdot i}}=-1 \\
\&e^{P\left(i^{2}\right)}=-1 \\
\&e^{-P}=-1 \\\&e^{-2 P}=1 \\
\&P=0 \rightarrow \boxed{\therefore P=N P}
\end{aligned}
$$