Path independence is equivalent to zero integral around every closed piecewise-C1 path

Assume condition 1, and let $\gamma$ be closed at $x$. The path $\gamma$ and the constant path at $x$ have the same endpoints, so [L1] and [L3] give ${\int }_{\gamma }F\cdot d\mathbf{r}=0$. Thus condition 2 holds.

Conversely, assume condition 2. Let $\alpha$ and $\beta$ be paths from $x$ to $y$. The increasing affine bijection of $[0,1]$ onto a path's domain is an orientation-preserving oriented reparametrization, so by [L4] we may replace each path by its reparametrization on $[0,1]$ without changing either integral. With both domains $[0,1]$, the concatenation in [L2] is defined and $\alpha \ast {\beta }^{-}$ is closed.

By condition 2 and [L2], $0={\int }_{\alpha \ast {\beta }^{-}}F\cdot d\mathbf{r}={\int }_{\alpha }F\cdot d\mathbf{r}-{\int }_{\beta }F\cdot d\mathbf{r}\mathrm{.}$

Hence the two integrals agree, and [L1] gives path independence.

Step 1.1 proves the forward direction, and steps 1.2, 2.1, and 3.1 prove the reverse direction. Piecewise-$C^1$ path-connectedness guarantees that the comparison paths relevant to condition 1 exist between any two points of $U$.