Why does $\oint\mathbf{E}\cdot{d}\boldsymbol\ell=0$ imply $\nabla\times\mathbf{E}=\mathbf{0}$? I am looking at Griffith's Electrodynamics textbook and on page 76 he is discussing the curl of electric field in electrostatics. He claims that since
$$\oint_C\mathbf{E}\cdot{d}\boldsymbol\ell=0$$
then
$$\nabla\times\mathbf{E}=\mathbf{0}$$
I don't follow this logic. Although I know that curl of $\mathbf{E}$ in statics is $\mathbf{0}$, I can't see how you can simply apply Stokes' theorem to equate the two statements. 
If we take Stokes' original theorem, we have $\oint\mathbf{E}\cdot{d}\boldsymbol\ell=\int\nabla\times\mathbf{E}\cdot{d}\mathbf{a}=0$. How does this imply $\nabla\times\mathbf{E}=\mathbf{0}$?
Griffiths seem to imply that this step is pretty easy, but I can't see it!

The infinitesimal area $d\mathbf{a}$ is arbitrary which means that $\nabla\times\mathbf{E}=0$ must be true everywhere, and not just locally.
