$\forall$$A$, $B$:Type, $f$:($A$$\rightarrow$$B$), ${\it as}$, ${\it as'}$:($A$ List).
\\[0ex]map($f$;${\it as}$ @ ${\it as'}$) = (map($f$;${\it as}$) @ map($f$;${\it as'}$)) $\in$ ($B$ List)