Coinciding path integrals need not imply a conservative field

Path independence requires the same line integral for every admissible path between fixed endpoints, not merely a few selected paths. For example, the smooth field $\mathbf{F}=(y-z,0,0)$ vanishes on the entire plane $y=z$. Every path in that plane from $(0,0,0)$ to $(1,1,1)$ therefore has zero integral. Yet the path through $(0,1,0)$ and $(1,1,0)$ has integral one: only the horizontal middle segment contributes. The curl is $(0,-1,-1)$, confirming that the field is not conservative. Special paths can hide a nonconservative component.