If we have simple closed curves $\alpha$ and $\beta$ on a surface $\Sigma_g$, the intersection number $i(\alpha ,\beta)$ is defined to be the minimal cardinality of $\alpha_1\cap\beta_1$ as $\alpha_1$ and $\beta_1$ range over all simple closed curves isotopic to $\alpha$ and $\beta$, respectively. We say $\alpha$ and $\beta$ intersect minimally if $i(\alpha ,\beta) = |\alpha\cap\beta|\,$.
How to see that $\alpha$ and $\beta$ intersect minimally if there are no pairs of $p,q\in\alpha\cap\beta$ such that the arc joining $p$ to $q$ along $\alpha$ followed by the arc from $q$ back to $p$ along $\beta$ bounds a disk in $\Sigma_g$?
Maybe a sketch of the proof idea?
I think the converse is also true : "that $\alpha$ and $\beta$ intersect minimally only if there are no pairs of $p,q\in\alpha\cap\beta$ such that the arc joining $p$ to $q$ along $\alpha$ followed by the arc from $q$ back to $p$ along $\beta$ bounds a disk in $\Sigma_g$."