ধরি, ভেক্টর ক্ষেত্র $\vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k}$। প্রথমে কার্ল $\vec{\nabla} \times \vec{A}$ নির্ণয় করি: $$ \vec{\nabla} \times \vec{A} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ A_x & A_y & A_z \end{vmatrix} $$ $$ = \left( \frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z} \right)\hat{i} + \left( \frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x} \right)\hat{j} + \left( \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y} \right)\hat{k} $$ এখন, এর সাথে ডাইভারজেন্স $\vec{\nabla} \cdot (\vec{\nabla} \times \vec{A})$ প্রয়োগ করি: $$ \vec{\nabla} \cdot (\vec{\nabla} \times \vec{A}) = \frac{\partial}{\partial x} \left( \frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z} \right) + \frac{\partial}{\partial y} \left( \frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x} \right) + \frac{\partial}{\partial z} \left( \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y} \right) $$ অন্তরীকরণের নিয়ম অনুযায়ী সাজিয়ে পাই: $$ \begin{aligned} &= \frac{\partial^2 A_z}{\partial x \partial y} - \frac{\partial^2 A_y}{\partial x \partial z} + \frac{\partial^2 A_x}{\partial y \partial z} - \frac{\partial^2 A_z}{\partial y \partial x} + \frac{\partial^2 A_y}{\partial z \partial x} - \frac{\partial^2 A_x}{\partial z \partial y} \\ &= \left( \frac{\partial^2 A_z}{\partial x \partial y} - \frac{\partial^2 A_z}{\partial y \partial x} \right) + \left( \frac{\partial^2 A_y}{\partial z \partial x} - \frac{\partial^2 A_y}{\partial x \partial z} \right) + \left( \frac{\partial^2 A_x}{\partial y \partial z} - \frac{\partial^2 A_x}{\partial z \partial y} \right) \end{aligned} $$ যেহেতু আংশিক অন্তরীকরণের ক্রম পরিবর্তনশীল (Clairaut's theorem), তাই প্রতিটি বন্ধনীর মান শূন্য: $$ = 0 + 0 + 0 = 0 $$ অতএব, $$ \vec{\nabla} \cdot (\vec{\nabla} \times \vec{A}) = 0 $$ প্রমাণিত।