typo

practical.tex

@@ -183,7 +183,7 @@ application. Here by iterating a second time, the algorithm deduces
 that {\tt x} has type $\Empty$ (i.e., $\textsf{Empty}$), that is that the first branch can never
 be selected (and our implementation warns the user accordingly). In hindsight, the only way for a well-typed overloaded function to have
 type $(\Int{\to}\Int)\land(\Any{\to}\Bool)$ is to diverge when the
-argument is of type \Int: since this intersection types states that
+argument is of type \Int: since this intersection type states that
 whenever the input is \Int, {\em both\/} branches can be selected,
 yielding a result that is at the same time an integer and a Boolean.
 This is precisely reflected by the case $\Int\to\Empty$ in the result.