Yes! That's exactly right. And which formula this is will depend on the particular way you've implemented the successor function.
-Let's pick a way of defining the successor function and reason about it.
-Here is one way that is compatible with the constraints given in
-homework 2: `succ := \nfz.f(nfz)`. This takes a Church
-number, and returns the next Church number. For instance,
-
- succ 2 == succ (\fz.f(fz))
- == (\nfz.f(nfz)) (\fz.f(fz))
- ~~> \fz.f((\fz.f(fz))fz)
- ~~> \fz.f(f(fz))
- == 3
-
-Using logic similar to the discussion above of the fixed point for K,
-we can say that for any Church number argument to the successor
-function, the result will be the next Church number. Assume that
-there is some Church number `n` that is a fixed point. Then
-`succ n <~~> n` (because `n` is a fixed point) and `succ n <~~> n + 1`
-(since that's what the successor function does). By the Church Rosser
-theorem, `n <~~> n + 1`. What kind of `n` could satisfy that
-requirement?
-
-Let's run the recipe:
-
- H := \f . succ (ff)
- == \f . (\nfz.f(nfz)) (ff)
- ~~> \h . (\nfz.f(nfz)) (hh)
- ~~> \hfz.f(hhfz)
-
- H H == (\hfz.f(hhfz)) (\hfz.f(hhfz))
- ~~> \fz.f((\hfz.f(hhfz))(\hfz.f(hhfz))fz)
- ~~> \fz.f(f((\hfz.f(hhfz))(\hfz.f(hhfz))fz))
- ~~> \fz.f(f(f((\hfz.f(hhfz))(\hfz.f(hhfz))fz))
-
-We can see that the fixed point generates an endless series of `f`'s.
-In terms of Church numbers, this is a way of representing infinity:
-if the size of a Church number is the number `f`'s it contains, and
-this Church number contains an unbounded number of `f`'s, then its
-size is unbounded.
-
-We can also see how this candidate for infinity behaves with respect
-to our other arithmetic operators.
-
- add 2 (HH) == (\mnfz.mf(nfz)) (\fz.f(fz)) (H H)
- ~~> \fz.(\fz.f(fz)) f ((HH)fz)
- ~~> \fz.\z.f(fz) ((HH)fz)
- ~~> \fz.f(f((HH)fz))
- == \fz.f(f(((\hfz.f(hhfz)) (\hfz.f(hhfz)))fz))
- ~~> \fz.f(f((\fz.f((\hfz.f(hhfz)) (\hfz.f(hhfz))))fz))
- ~~> \fz.f(f(f((\hfz.f(hhfz)) (\hfz.f(hhfz)))))
-
-So `2 + (HH) <~~> (HH)`. This is what we expect from arithmetic infinity.
-You can check to see if `2 * (HH) <~~> (HH)`.
-
-So our fixed point recipe has delivere a reasonable candidate for
-arithmetic infinity.
-
One (by now obvious) upshot is that the recipes that enable us to name
fixed points for any given formula aren't *guaranteed* to give us
*terminating* fixed points. They might give us formulas X such that