X-Git-Url: http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=blobdiff_plain;f=exercises%2Fassignment3_answers.mdwn;h=ae35731b0333982c915d357e9ed1d4bd92225848;hp=6aa5d7503467a77a8359fdd7df23ba1c39628f71;hb=dd911c78a79577243800cafec55f75ef9d76d63a;hpb=8950982eae04ec4ca70862e23122256b526b591b

diff --git a/exercises/assignment3_answers.mdwn b/exercises/assignment3_answers.mdwn
index 6aa5d750..ae35731b 100644
--- a/exercises/assignment3_answers.mdwn
+++ b/exercises/assignment3_answers.mdwn
@@ -167,7 +167,7 @@ where `one` abbreviates `succ zero`, and `two` abbreviates `succ (succ zero)`.
     >      let leq? = \l r. zero? (sub l r) in
     >      ...
 
-    > Here is another solution. Jim crafted this particular implementation, but like a great deal of the CS knowledge he's gained over the past eight years, Oleg Kiselyov pointed the way.
+    > Here is another solution. Jim crafted this particular implementation, but like a great deal of the CS knowledge he's gained over the past eight years, Oleg Kiselyov pointed the way. <!-- see "lambda-calc-opposites.txt" at http://okmij.org/ftp/Computation/lambda-calc.html#neg -->
 
     >     let leq? = (\base build consume. \l r. r consume (l build base) fst)
     >             ; where base is
@@ -267,6 +267,12 @@ S (S (KS) (S (KK) (S (KS) K))) (KI)</code>; this is the <b>B</b> combinator, whi
 
 25. For each of the above translations, how many `I`s are there? Give a rule for describing what each `I` corresponds to in the original lambda term.
 
+    This generalization depends on you omitting the translation rule:
+
+        6. @a(Xa)       =   X            if a is not in X
+
+    > With that shortcut rule omitted, then there turn out to be one `I` in the result corresponding to each occurrence of a bound variable in the original term.
+
 Evaluation strategies in Combinatory Logic
 ------------------------------------------
 
@@ -306,8 +312,8 @@ Reduce to beta-normal forms:
 <OL start=30>
 <LI><code>(\x. x (\y. y x)) (v w) ~~> v w (\y. y (v w))</code>
 <LI><code>(\x. x (\x. y x)) (v w) ~~> v w (\x. y x)</code>
-<LI><code>(\x. x (\y. y x)) (v x) ~~> v w (\y. y (v x))</code>
-<LI><code>(\x. x (\y. y x)) (v y) ~~> v w (\u. u (v y))</code>
+<LI><code>(\x. x (\y. y x)) (v x) ~~> v x (\y. y (v x))</code>
+<LI><code>(\x. x (\y. y x)) (v y) ~~> v y (\u. u (v y))</code>
 
 <LI><code>(\x y. x y y) u v ~~> u v v</code>
 <LI><code>(\x y. y x) (u v) z w ~~> z (u v) w</code>