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diff --git a/assignment3.mdwn b/assignment3.mdwn
index 72c6f28d..e240b732 100644
--- a/assignment3.mdwn
+++ b/assignment3.mdwn
@@ -1,6 +1,19 @@
 Assignment 3
 ------------
 
+Erratum corrected 11PM Sun 3 Oct: the following line
+
+	let tb = (make_list t12 (make_list t3 empty)) in
+
+originally read 
+
+	let tb = (make_list t12 t3) in
+
+This has been corrected below, and in the preloaded evaluator for 
+working on assignment 3, available here: [[assignment 3 evaluator]].
+
+<hr>
+
 Once again, the lambda evaluator will make working through this
 assignment much faster and more secure.
 
@@ -14,32 +27,32 @@ Recall that version 1 style lists are constructed like this (see
 	let false = \x y. y in
 	let and = \l r. l (r true false) false in
 
-	; version 1 lists
 	let make_pair = \f s g. g f s in
-	let fst = true in
-	let snd = false in
+	let get_fst = true in
+	let get_snd = false in
 	let empty = make_pair true junk in
-	let isempty = \x. x fst in
+	let isempty = \x. x get_fst in
 	let make_list = \h t. make_pair false (make_pair h t) in
-	let head = \l. isempty l err (l snd fst) in
-	let tail = \l. isempty l err (l snd snd) in
-
+	let head = \l. isempty l err (l get_snd get_fst) in
+	let tail = \l. isempty l err (l get_snd get_snd) in
+	
 	; a list of numbers to experiment on
 	let mylist = make_list 1 (make_list 2 (make_list 3 empty)) in
-
-	; a fixed-point combinator for defining recursive functions
-	let Y = \f. (\h. f (h h)) (\h. f (h h)) in
-
+	
 	; church numerals
 	let iszero = \n. n (\x. false) true in
 	let succ = \n s z. s (n s z) in
-	let mult = \m n s. m (n s) in
-	let length = Y (\length l. isempty l 0 (succ (length (tail l)))) in
-	let pred = \n. iszero n 0 (length (tail (n (\p. make_list junk p) empty)))
-	in
+	let add = \l r. l succ r in
+	let mul = \m n s. m (n s) in
+	let pred = (\shift n. n shift (make\_pair 0 0) get\_snd) (\p. p (\x y. make\_pair (succ x) x)) in
 	let leq = \m n. iszero(n pred m) in
 	let eq = \m n. and (leq m n)(leq n m) in
-
+	
+	; a fixed-point combinator for defining recursive functions
+	let Y = \f. (\h. f (h h)) (\h. f (h h)) in
+	let length = Y (\length l. isempty l 0 (succ (length (tail l)))) in
+	let fold = Y (\f l g z. isempty l z (g (head l)(f (tail l) g z))) in
+	
 	eq 2 2 yes no
 
 
@@ -51,14 +64,14 @@ Then `length mylist` evaluates to 3.
 function, write a function that computes factorials.  (Recall that n!,
 the factorial of n, is n times the factorial of n-1.)
 
-Warning: it takes a long time for my browser to compute factorials larger than 4!
+	Warning: it takes a long time for my browser to compute factorials larger than 4!
 
 3. (Easy) Write a function `equal_length` that returns true just in case
 two lists have the same length.  That is,
 
-	equal_length mylist (make_list junk (make_list junk (make_list junk empty))) ~~> true
+		equal_length mylist (make_list junk (make_list junk (make_list junk empty))) ~~> true
 
-	equal_length mylist (make_list junk (make_list junk empty))) ~~> false
+		equal_length mylist (make_list junk (make_list junk empty))) ~~> false
 
 
 4. (Still easy) Now write the same function, but don't use the length
@@ -108,7 +121,8 @@ whether the length of the list is less than or equal to 1.  This will
 be your base case for your recursive functions that operate on these
 trees.
 
-1.    Write a function that sums the number of leaves in a tree.
+<OL start=6>
+<LI>Write a function that sums the values at the leaves in a tree.
 
 Expected behavior:
 
@@ -118,7 +132,7 @@ Expected behavior:
 	let t12 = (make_list t1 (make_list t2 empty)) in
 	let t23 = (make_list t2 (make_list t3 empty)) in
 	let ta = (make_list t1 t23) in
-	let tb = (make_list t12 t3) in
+	let tb = (make_list t12 (make_list t3 empty)) in
 	let tc = (make_list t1 (make_list t23 empty)) in
 
 	sum-leaves t1 ~~> 1
@@ -131,5 +145,7 @@ Expected behavior:
 	sum-leaves tc ~~> 6
 
 
-2.   Write a function that counts the number of leaves.
+<LI>Write a function that counts the number of leaves.
+
+</OL>