Haskell

prove: concat $ map (map f) xss == map f $ concat xss
---------------------------------------------------
case I:xss = []:
       concat $ map (map f) []
    == concat []
    == []

       map f $ concat []
    == map f []
    == []
---------------------------------------------------
case II:xss = undefined
       concat $ map (map f) undefined
    == concat undefined
    == undefined

       map f $ concat undefined
    == map f undefined
    == undefined
---------------------------------------------------
case III:xss = ys : yss
       concat $ map (map f) (ys:yss)
    == concat $ map f ys : map (map f) yss
    == map f ys ++ (concat $ map (map f) yss)
    == map f ys ++ (map f $ concat yss)
    == map f $ ys ++ concat yss
    == map f $ concat (ys:yss)
    
QED.