+
+## Q: I still don't fully understand the Y combinator. Can you
+ explain it in a different way?
+
+Sure! Here is another way to derive Y. We'll start by choosing a
+specific goal, and at each decision point, we'll make a reasonable
+guess. The guesses will all turn out to be lucky, and we'll arrive at
+a fixed point combinator.
+
+Given an arbitrary term f, we want to find a fixed point X such that
+
+ X <~~> f X
+
+Our strategy will be to seek an X such that X ~~> f X. Because X and
+f X are not the same, the only way that X can reduce to f X is if X
+contains at least one redex. The simplest way to satisfy this
+constraint would be for the fixed point to itself be a redex:
+
+ X == ((\u.M) N) ~~> f X
+
+The result of beta reduction on this redex will be M with some
+substitutions. We know that after these substitutions, M will have
+the form `f X`, since that is what the reduction arrow tells us. So we
+can refine the picture as follows:
+
+ X == ((\u.f(___)) N) ~~> f X
+
+Here, the ___ has to be something that reduces to the fixed point X.
+It's natural to assume that there will be at least one occurrence of
+"u" in the body of the head abstract:
+
+ X == ((\u.f(__u__)) N) ~~> f X
+
+After reduction of the redex, we're going to have
+
+ X == f(__N__) ~~> f X
+
+Apparently, `__N__` will have to reduce to X. Therefore we should
+choose a skeleton for N that is consistent with what we have decided
+so far about the internal structure of X. We might like for N to
+match X in its entirety, but this would require N to contain itself as
+a subpart. So we'll settle for the more modest assumption that N
+matches the head of X:
+
+ X == ((\u.f(__u__)) (\u.f(__u__))) ~~> f X
+
+At this point, we've derived a skeleton for X on which it contains two
+so-far identical halves. We'll guess that the halves will be exactly
+identical. Note that at the point at which we perform the first
+reduction, `u` will get bound N, which now corresponds to a term
+representing one of the halves of X. So in order to produce a full X,
+we simply make a second copy of `u`:
+
+ X == ((\u.f(uu))(\u.f(uu))) ~~> f((\u.f(uu))(\u.f(uu))) == f X
+
+Success.
+
+So the function `\f.(\u.f(uu))(\u.f(uu))` maps an arbtirary function
+`f` to a fixed point for `f`.