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[lambda.git] / exercises / assignment3_answers.mdwn

diff --git a/exercises/assignment3_answers.mdwn b/exercises/assignment3_answers.mdwn
index ea0256d..ae35731 100644 (file)
--- 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
@@ -312,8 +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>