Let's do one more example to illustrate. We'll do `K`, since we
wondered above whether it had a fixed point.
+Before we begin, we can reason a bit about what the fixed point must
+be like. We're looking for a fixed point for `K`, i.e., `\xy.x`. `K`
+ignores its second argument. That means that no matter what we give
+`K` as its first argument, the result will ignore the next argument
+(that is, `KX` ignores its first argument, no matter what `X` is). So
+if `KX <~~> X`, `X` had also better ignore its first argument. But we
+also have `KX == (\xy.x)X ~~> \y.X`. This means that if `X` ignores
+its first argument, then `\y.X` will ignore its first two arguments.
+So once again, if `KX <~~> X`, `X` also had better ignore at least its
+first two arguments. Repeating this reasoning, we realize that `X`
+must be a function that ignores an infinite series of arguments.
+Our expectation, then, is that our recipe for finding fixed points
+will build us a function that somehow manages to ignore an infinite
+series of arguments.
+
h := \xy.x
H := \f.h(ff) == \f.(\xy.x)(ff) ~~> \fy.ff
H H := (\fy.ff)(\fy.ff) ~~> \y.(\fy.ff)(\fy.ff)
-Ok, it doesn't have a normal form. But let's check that it is in fact
-a fixed point:
+Let's check that it is in fact a fixed point:
K(H H) == (\xy.x)((\fy.ff)(\fy.ff)
~~> \y.(\fy.ff)(\fy.ff)
Yep, `H H` and `K(H H)` both reduce to the same term.
-This fixed point is bit wierd. Let's reduce it a bit more:
+To see what this fixed point does, let's reduce it a bit more:
H H == (\fy.ff)(\fy.ff)
~~> \y.(\fy.ff)(\fy.ff)
~~> \yy.(\fy.ff)(\fy.ff)
~~> \yyy.(\fy.ff)(\fy.ff)
-It appears that where `K` is a function that ignores (only) the second
-argument you feed to it, the fixed point of `K` ignores an endless,
-infinite series of arguments. It's a write-only memory, a black hole.
+Sure enough, this fixed point ignores an endless, infinite series of
+arguments. It's a write-only memory, a black hole.
+
+Now that we have one fixed point, we can find others, for instance,
+
+ (\fy.fff)(\fy.fff)
+ ~~> \y.(\fy.fff)(\fy.fff)(\fy.fff)
+ ~~> \yy.(\fy.fff)(\fy.fff)(\fy.fff)(\fy.fff)
+ ~~> \yyy.(\fy.fff)(\fy.fff)(\fy.fff)(\fy.fff)(\fy.fff)
+Continuing in this way, you can now find an infinite number of fixed
+points, all of which have the crucial property of ignoring an infinite
+series of arguments.
##What is a fixed point for the successor function?##