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diff --git a/assignment3.mdwn b/assignment3.mdwn
index 9f64d808..f93afd61 100644
--- a/assignment3.mdwn
+++ b/assignment3.mdwn
@@ -4,7 +4,7 @@ Assignment 3
 Once again, the lambda evaluator will make working through this
 assignment much faster and more secure.
 
-*Writing recursive functions on version 1 style lists*
+##Writing recursive functions on version 1 style lists##
 
 Recall that version 1 style lists are constructed like this:
 
@@ -46,7 +46,7 @@ eq 3 3
 
 Then `length mylist` evaluates to 3.
 
-1. What does `head (tail (tail mylist))` evaluate to?
+1. Warm-up: 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!,
@@ -57,83 +57,31 @@ greater than 2 (it does't provide enough resources for the JavaScript
 interpreter; web pages are not supposed to be that computationally
 intensive).
 
-3. (Easy) Write a function `listLenEq` that returns true just in case two lists have the
+
+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).
 
-4. (Still easy) Now write the same function, but don't use the length function (hint: use `leq` as a model).
-
-5. In assignment 2, we discovered that version 3-type lists (the ones that
-work like Church numerals) made it much easier to define operations
-like map and filter.  But now that we have recursion in our toolbox,
-reasonable map and filter functions for version 3 lists are within our
-reach.  Give definitions for such a map and a filter.
-
-6. Linguists analyze natural language expressions into trees.  
-We'll need trees in future weeks, and tree structures provide good
-opportunities for learning how to write recursive functions.
-Making use of the resources we have at the moment, we can approximate
-trees as follows: instead of words, we'll use Church numerals.
-Then a tree will be a (version 1 type) list in which each element is
-itself a tree.  For simplicity, we'll adopt the convention that 
-a tree of length 1 must contain a number as its only element.  
-Then we have the following representations:
-
-<pre>
-   (a)           (b)             (c)  
-    .
-   /|\            /\              /\
-  / | \          /\ 3             1/\
-  1 2  3        1  2               2 3
-
-[[1];[2];[3]]  [[[1];[2]];[3]]   [[1];[[2];[3]]]
-</pre>
-
-Limitations of this scheme include the following: there is no easy way
-to label a constituent (typically a syntactic category, S or NP or VP,
-etc.), and there is no way to represent a tree in which a mother has a
-single daughter.
-
-When processing a tree, you can test for whether the tree contains
-only a numeral (in which case the tree is leaf node) by testing for
-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 trees.
-
-Write a function that sums the number of leaves in a tree.
-Expected behavior:
-
-let t1 = (make-list 1 nil) 
-let t2 = (make-list 2 nil)
-let t3 = (make-list 3 nil)
-let t12 = (make-list t1 (make-list t2 nil))
-let t23 = (make-list t2 (make-list t3 nil))
-let ta = (make-list t1 t23)
-let tb = (make-list t12 t3)
-let tc = (make-list t1 (make-list t23 nil))
+##Trees##
 
-count-leaves t1 ~~> 1
-count-leaves t2 ~~> 2
-count-leaves t3 ~~> 3
-count-leaves t12 ~~> 3
-count-leaves t23 ~~> 5
-count-leaves ta ~~> 6
-count-leaves tb ~~> 6
-count-leaves tc ~~> 6
-
-Write a function that counts the number of leaves.
+Since we'll be working with linguistic objects, let's approximate
+trees as follows: a tree is a version 1 list
+a Church number is a tree, and 
+if A and B are trees, then (make-pair A B) is a tree.
 
 
 
 
 [The following should be correct, but won't run in my browser:
 
-<pre>
 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