Path-independence implies no circulation
If the vector field F is a path-independent vector field, then the path integral
| ∫ CF ⋅ ds |
Suppose, for example, we have two paths C1 and C2 connecting point p with point q.

Then we know that
| ∫ C1F ⋅ ds = ∫ C2F ⋅ ds | (1) |
Now, what would happen is we turned the path C2 around, so that it starts at point q and ends at point p? (We will denote this path as C2-.) At any point, the tangent vector of C2- will be the opposite of the tangent vector of C 2 because the path is going in the opposite direction. Consequently
| ∫ C2-F ⋅ ds = -∫ C2F ⋅ ds. | (2) |
If we define the path C to be the path C1 followed by the path C2-, the path C will start at the point p, go to the point q (via C1) and then return to the point p (via C2). In other words, the path C will be a closed path.

Then path integral over C is simply the path integral over C1 plus the path integral over C2-. By combining equation (1) with equation (2), we see these last two integrals are opposites
| ∫ C2-F ⋅ ds = -∫ C1F ⋅ ds. |
| ∫ CF ⋅ ds = ∫ C1F ⋅ ds + ∫ C2-F ⋅ ds = 0. |
If C is a closed path, we call the integral
| ∫ CF ⋅ ds |
We can use this result as a test for path-dependence. If we can find a single path C where
| ∫ CF ⋅ ds≠0, |
The key point to remember is that path-independence means there is no circulation around any curve.