Solution
= Solution
For finite <lists> and $n\geq0$, the answers are
``
reverse (xs ++ ys) = reverse ys ++ reverse xs
length (drop n xs) = max 0 (length xs - n)
``
The maximum is necessary when the count exceeds the <list> length.
= Solution
For finite <lists> and $n\geq0$, the answers are
``
reverse (xs ++ ys) = reverse ys ++ reverse xs
length (drop n xs) = max 0 (length xs - n)
``
The maximum is necessary when the count exceeds the <list> length.