Zero curl implies path around a closed loop is zero 
$\vec{\triangledown }\times \vec{F}=0 \Rightarrow \oint \vec{F} \cdot d\vec{l}=0$

Referring to a lemma:

$\vec{\triangledown }\times \vec{F}=0 \Leftrightarrow \vec{F}=-\bigtriangledown \vec{F}$

I get $\oint \left [ -\bigtriangledown V  \cdot d\vec{l} \right ]$ for some scalar function $V$
but how is this zero?
Doing some practice problems and was asked to show why this is true. Been thinking about this for a good while.
Some help please.
 A: Stokes' theorem implies that
$$\oint \vec F \cdot d\vec{\ell} = \iint_S \nabla \times \vec F \cdot d\vec S = 0$$
for an appropriately chosen surface $S$.
A: I thought it would be instructive to present an approach that does not rely on the use of Stokes' Theorem, but rather on the given Lemma in the OP.
Suppose that a vector field $\vec F$ is given by
$$\begin{align}
\vec F&=-\nabla V\\\\&=-\left(\hat x\frac{\partial V}{\partial x}+\hat y\frac{\partial V}{\partial y}+\hat z\frac{\partial V}{\partial z}\right)
\end{align}$$ 
for some scalar field $V$.  
Let $C$ be the closed contour with a parametric description given by $\vec r(t)=\hat xx(t)+\hat yy(t)+\hat zz(t)$, with $t\in[0,1]$ with $\vec r(t=0)=\vec r(t=1)$.
Then, using $d\vec \ell = \left(\hat x\frac{dx(t)}{dt}+\hat y\frac{dy(t)}{dt}+\hat z\frac{dz(t)}{dt}\right)\,dt$, we can write
$$\begin{align}
\oint_C \vec F\cdot d\vec \ell&=-\oint_C \nabla V\cdot d\vec \ell\\\\
&=-\int_0^1 \left(\frac{\partial V}{\partial x}\,\frac{dx}{dt}+\frac{\partial V}{\partial y}\,\frac{dy}{dt}+\frac{\partial V}{\partial z}\,\frac{dz}{dt}\right)\,dt\\\\
&=\int_0^1 \frac{dV(\vec r(t))}{dt}\,dt\\\\
&=V(\vec r(1))-V(\vec r(0))\\\\
&=0
\end{align}$$
as was to be shown!
