+Hints for `list_equal`.
+
+* If `left` is `[]`, what does `right` have to be for `left` and `right` to be equal? (Come on, it's not too hard, you can figure it out.)
+
+* Suppose on the other hand that `left` has head `left_hd` and tail `left_tl`.
+
+ <OL type=a>
+ <LI>If `right` is then `[]`, are `left` and `right` equal?
+ <LI>If `right` isn't `[]`, and its head isn't equal to `left_hd`, are `left` and `right` equal?
+ <LI>If `right` isn't `[]` and its head *is* equal to `left_hd`, what else has to be the case for `left` and `right` to be equal?
+ </OL>
+
+* Can you now write a recursive definition of the `list_equal` function?
+What's your base case?
+
<!--
let list_equal =
; when fold is finished, check sofar-pair
(\might_be_equal right_tail. and might_be_equal (isempty right_tail))
-->
+
+<!--
+eql [] l2 = empty l2
+eql (l1h:l1t) l2 = and (not (empty l2))
+ (l1h == head l2)
+ (eql l1t (tail l2))
+
+Rearangement:
+
+eql [] = \l2 -> empty l2
+eql (l1h:l1t) = let prev = eql l1t
+ in \l2 -> and (not (empty l2))
+ (l1h == head l2)
+ (prev (tail l2))
+
+Now it fits the pattern of foldr
+
+let list_equal = \lst. lst (\hd sofar. \lst. and (and (not (isempty lst)) (eq hd (head lst))) (sofar (tail lst))) isempty
+
+-->
+