+1. What does `head (tail (tail mylist))` evaluate to?
+
+2. Using the `length` function as a model, and using the predecessor
+function, write a function that computes factorials. (Recall that n!,
+the factorial of n, is n times the factorial of n-1.)
+
+Warning: my browser isn't able to compute factorials of numbers
+greater than 2 (it does't provide enough resources for the JavaScript
+interpreter; web pages are not supposed to be that computationally
+intensive).
+
+
+3. Write a function `listLenEq` that returns true just in case two lists have the
+same length. That is,
+
+ listLenEq mylist (makeList meh (makeList meh (makeList meh nil))) ~~> true
+
+ listLenEq mylist (makeList meh (makeList meh nil))) ~~> false
+
+
+4. Now write the same function, but don't use the length function (hint: use `leq` as a model).
+
+
+[The following should be correct, but won't run in my browser:
+
+let factorial = Y (\fac n. isZero n 1 (mult n (fac (predecessor n)))) in
+
+<pre>
+let reverse =
+ Y (\rev l. isNil l nil
+ (isNil (tail l) l
+ (makeList (head (rev (tail l)))
+ (rev (makeList (head l)
+ (rev (tail (rev (tail l))))))))) in
+
+reverse (makeList 1 (makeList 2 (makeList 3 nil)))
+</pre>
+
+It may require more resources than my browser is willing to devote to
+JavaScript.]
+