rinse, repeat!)
+#Q. That reminds me, what about [[evaluation order]]?#
+
+A. For a recursive function that has a well-behaved base case, such as
+the factorial function, evaluation order is crucial. In the following
+computation, we will arrive at a normal form. Watch for the moment at
+which we have to make a choice about which beta reduction to perform
+next: one choice leads to a normal form, the other choice leads to
+endless reduction:
+
+ let prefac = \f n. iszero n 1 (mult n (f (pred n))) in
+ let fac = Y prefac in
+ fac 2
+ = [(\f.(\x.f(xx))(\x.f(xx))) prefac] 2
+ = [(\x.prefac(xx))(\x.prefac(xx))] 2
+ = [prefac((\x.prefac(xx))(\x.prefac(xx)))] 2
+ = [prefac(prefac((\x.prefac(xx))(\x.prefac(xx))))] 2
+ = [(\f n. iszero n 1 (mult n (f (pred n))))
+ (prefac((\x.prefac(xx))(\x.prefac(xx))))] 2
+ = [\n. iszero n 1 (mult n ([prefac((\x.prefac(xx))(\x.prefac(xx)))] (pred n)))] 2
+ = iszero 2 1 (mult 2 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] (pred 2)))
+ = mult 2 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] 1)
+ ...
+ = mult 2 (mult 1 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] 0))
+ = mult 2 (mult 1 (iszero 0 1 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] (pred 0))))
+ = mult 2 (mult 1 1)
+ = mult 2 1
+ = 2
+
+The crucial step is the third from the last. We have our choice of
+either evaluating the test `iszero 0 1 ...`, which evaluates to `1`,
+no matter what the ... contains;
+or we can evaluate the `Y` pump, `(\x.prefac(xx))(\x.prefac(xx))`, to
+produce another copy of `prefac`. If we postpone evaluting the
+`iszero` test, we'll pump out copy after copy of `prefac`, and never
+realize that we've bottomed out in the recursion. But if we adopt a
+leftmost/call-by-name/normal-order evaluation strategy, we'll always
+start with the `iszero` predicate, and only produce a fresh copy of
+`prefac` if we are forced to.
+