It is well known that the Laplacian is invariant under traslations and rotations. I'm trying to prove that the Laplacian is also invariant under reflections across the $y$-axis: \begin{align*} \Delta\left(u\left(-x, y\right)\right) = \left(\Delta u\right)\left(-x, y\right) \end{align*} whenever $u$ is a smooth function of two variables. Could anyone help to prove this? Moreover, I would also ask: the Laplacian is also invariant under "all" reflections, i mean, not only under reflections across the $y$-axis?
EDIT (after the answer of @Paul) Thank you! Could you tell if this is true? \begin{align*} \frac{\partial^2 u}{\partial x^{\prime 2}} = \frac{\partial}{\partial x^{\prime}}\left(\frac{\partial u}{\partial x^{\prime}}\right) = \frac{\partial}{\partial x^{\prime}}\left( -\frac{\partial u}{\partial x}\right) = - \frac{\partial^2 u}{\partial x^2}\frac{\partial x}{\partial x^{\prime}} = \frac{\partial ^2 u}{\partial x^2}. \end{align*} Then \begin{align*} \left(\Delta u\right)(-x, y) = \frac{\partial^2 u}{\partial (-x)^2} + \frac{\partial^2 u}{\partial y^2} = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}, \end{align*} \begin{align*} \Delta\left(u(-x, y)\right) = \frac{\partial^2 u}{\partial x^{\prime^2}} + \frac{\partial^2 u}{\partial y^2} = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}. \end{align*}
Let $x'=-x$. By the chain rule
$$\frac{\partial u}{\partial x'}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial x'}=-\frac{\partial u}{\partial x}$$
And just do that again for the second derivative. Reflections about the $y-$axis work exactly the same.