You should see the close similarity with `Y Y` here.
+#Q. So `Y` applied to `succ` returns a number that is not finite!#
+
+A. Yes! Let's see why it makes sense to think of `Y succ` as a Church
+numeral:
+
+ [same definitions]
+ succ X
+ = (\n s z. s (n s z)) X
+ = \s z. s (X s z)
+ = succ (\s z. s (X s z)) ; using fixed-point reasoning
+ = \s z. s ([succ (\s z. s (X s z))] s z)
+ = \s z. s ([\s z. s ([succ (\s z. s (X s z))] s z)] s z)
+ = \s z. s (s (succ (\s z. s (X s z))))
+
+So `succ X` looks like a numeral: it takes two arguments, `s` and `z`,
+and returns a sequence of nested applications of `s`...
+
+You should be able to prove that `add 2 (Y succ) <~~> Y succ`,
+likewise for `mult`, `minus`, `pow`. What happens if we try `minus (Y
+succ)(Y succ)`? What would you expect infinity minus infinity to be?
+(Hint: choose your evaluation strategy so that you add two `s`s to the
+first number for every `s` that you add to the second number.)
+
+This is amazing, by the way: we're proving things about a term that
+represents arithmetic infinity.
+
+It's important to bear in mind the simplest term in question is not
+infinite:
+
+ Y succ = (\f. (\x. f (x x)) (\x. f (x x))) (\n s z. s (n s z))
+
+The way that infinity enters into the picture is that this term has
+no normal form: no matter how many times we perform beta reduction,
+there will always be an opportunity for more beta reduction. (Lather,
+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.
+