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diff --git a/hints/assignment_4_hint_2.mdwn b/hints/assignment_4_hint_2.mdwn
index 6bcd7701..f9f4cf25 100644
--- a/hints/assignment_4_hint_2.mdwn
+++ b/hints/assignment_4_hint_2.mdwn
@@ -1,3 +1,18 @@
+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 =
@@ -22,3 +37,24 @@ let list_equal =
                 ; when fold is finished, check sofar-pair
                 (\might_be_equal right_tail. and might_be_equal (isempty right_tail))
 -->
+
+<!--
+list_equal [] right      = isempty right
+list_equal (hd:tl) right = and (not (isempty right))
+                            (hd == head right)
+                            (list_equal tl (tail right))
+
+Rearangement:
+
+list_equal []      = \right -> isempty right
+list_equal (hd:tl) = let prev = list_equal tl
+                     in \right -> and (not (isempty right))
+                                   (hd == head right)
+                                   (prev (tail right))
+
+Now it fits the pattern of fold_right
+
+let list_equal = \left. left (\hd sofar. \right. and (and (not (isempty right)) (eq hd (head right))) (sofar (tail right))) isempty
+
+-->
+