last commit before class, there are still some comments/TODOs for later

diff --git a/topics/_week4_fixed_point_combinator.mdwn b/topics/_week4_fixed_point_combinator.mdwn
index fdf19a3..fecc653 100644 (file)
--- a/topics/_week4_fixed_point_combinator.mdwn
+++ b/topics/_week4_fixed_point_combinator.mdwn
@@ -1,13 +1,4 @@
-[[!toc levels=2]]
-
-~~~~
-
-**Chris:** I'll be working on this page heavily until 11--11:30 or so. Sorry not to do it last night, I crashed.
-
-ξ
-Θ′
-≡
-~~~~
+[[!toc levels=3]]

 
 #Recursion: fixed points in the Lambda Calculus#
 
 
 #Recursion: fixed points in the Lambda Calculus#
 


@@ -29,7 +20,8 @@ How could we compute the length of a list? Without worrying yet about what Lambd
 In OCaml, you'd define that like this:
 
     let rec length = fun xs ->
 In OCaml, you'd define that like this:
 
     let rec length = fun xs ->

-                       if xs = [] then 0 else 1 + length (List.tl xs)
+                       if xs = [] then 0
+                                  else 1 + length (List.tl xs)

     in ... (* here you go on to use the function "length" *)
 
 In Scheme you'd define it like this:
     in ... (* here you go on to use the function "length" *)
 
 In Scheme you'd define it like this:


@@ -44,7 +36,7 @@ Some comments on this:
 
 1. `null?` is Scheme's way of saying `empty?`. That is, `(null? xs)` returns true (which Scheme writes as `#t`) iff `xs` is the empty list (which Scheme writes as `'()` or `(list)`).
 
 
 1. `null?` is Scheme's way of saying `empty?`. That is, `(null? xs)` returns true (which Scheme writes as `#t`) iff `xs` is the empty list (which Scheme writes as `'()` or `(list)`).
 

-2. `cdr` is function that gets the tail of a Scheme list. (By definition, it's the function for getting the second member of a [[dotted pair|week3_unit#imp]]. As we discussed in notes for last week, it just turns out to return the tail of a list because of the particular way Scheme implements lists.)
+2. `cdr` is function that gets the tail of a Scheme list. (By definition, it's the function for getting the second member of a [[dotted pair|week3_unit#imp]]. As we discussed in notes for last week, it just turns out to return the tail of a list because of the particular way Scheme implements lists.) `List.tl` is the function that gets the tail of an OCaml list.

 
 3.  We alternate between `[ ]`s and `( )`s in the Scheme code just to make it more readable. These have no syntactic difference.
 
 
 3.  We alternate between `[ ]`s and `( )`s in the Scheme code just to make it more readable. These have no syntactic difference.
 


@@ -54,7 +46,8 @@ The main question for us to dwell on here is: What are the `let rec` in the OCam
 Answer: These work a lot like `let` expressions, except that they let you use the variable `length` *inside* the body of the function being bound to it --- with the understanding that it will there be bound to *the same function* that you're *then* in the process of binding `length` to. So our recursively-defined function works the way we'd expect it to. Here is OCaml:
 
     let rec length = fun xs ->
 Answer: These work a lot like `let` expressions, except that they let you use the variable `length` *inside* the body of the function being bound to it --- with the understanding that it will there be bound to *the same function* that you're *then* in the process of binding `length` to. So our recursively-defined function works the way we'd expect it to. Here is OCaml:
 
     let rec length = fun xs ->

-                       if xs = [] then 0 else 1 + length (List.tl xs)
+                       if xs = [] then 0
+                                  else 1 + length (List.tl xs)

     in length [20; 30]
     (* this evaluates to 2 *)
 
     in length [20; 30]
     (* this evaluates to 2 *)
 


@@ -69,7 +62,8 @@ Here is Scheme:
 If you instead use an ordinary `let` (or `let*`), here's what would happen, in OCaml:
 
     let length = fun xs ->
 If you instead use an ordinary `let` (or `let*`), here's what would happen, in OCaml:
 
     let length = fun xs ->

-                   if xs = [] then 0 else 1 + length (List.tl xs)
+                   if xs = [] then 0
+                              else 1 + length (List.tl xs)

     in length [20; 30]
     (* fails with error "Unbound value length" *)
 
     in length [20; 30]
     (* fails with error "Unbound value length" *)
 


@@ -97,7 +91,8 @@ We can verify this by wrapping the whole expression in a more outer binding of `
 
     let length = fun xs -> 99
     in let length = fun xs ->
 
     let length = fun xs -> 99
     in let length = fun xs ->

-                      if xs = [] then 0 else 1 + length (List.tl xs)
+                      if xs = [] then 0
+                                 else 1 + length (List.tl xs)

     in length [20; 30]
     (* evaluates to 1 + 99 *)
 
     in length [20; 30]
     (* evaluates to 1 + 99 *)
 


@@ -111,12 +106,13 @@ We've defined all of `empty?`, `0`, `succ`, and `tail` in earlier discussion. Bu
 
 What we really want to do is something like this:
 
 
 What we really want to do is something like this:
 

-    \xs. (empty? xs) 0 (succ (... (tail xs)))
+    \xs. (empty? xs) 0 (succ ((...) (tail xs)))

 
 where this very same formula occupies the `...` position:
 
 
 where this very same formula occupies the `...` position:
 

-    \xs. (empty? xs) 0 (succ (\xs. (empty? xs) 0 (succ (... (tail xs)))
-                                   (tail xs)))
+    \xs. (empty? xs) 0 (succ (
+    \xs. (empty? xs) 0 (succ ((...) (tail xs)))
+                                  ) (tail xs)))

 
 but as you can see, we'd still have to plug the formula back into itself again, and again, and again... No dice.
 
 
 but as you can see, we'd still have to plug the formula back into itself again, and again, and again... No dice.
 


@@ -243,11 +239,13 @@ Think about how it might be true.  We'll return to this point below.)
 
 ###How fixed points help define recursive functions###
 
 
 ###How fixed points help define recursive functions###
 

-Recall our initial, abortive attempt above to define the `length` function in the Lambda Calculus. We said "What we really want to do is something like this:
+Recall our initial, abortive attempt above to define the `length` function in the Lambda Calculus. We said:
+
+>   What we really want to do is something like this:

 
 

-    \xs. (empty? xs) 0 (succ (... (tail xs)))
+>       \xs. (empty? xs) 0 (succ ((...) (tail xs)))

 
 

-where this very same formula occupies the `...` position."
+>   where this very same formula occupies the `...` position...

 
 Imagine replacing the `...` with some expression `LENGTH` that computes the
 length function. Then we have
 
 Imagine replacing the `...` with some expression `LENGTH` that computes the
 length function. Then we have


@@ -260,14 +258,14 @@ symbol `LENGTH`.  Technically, it has the status of an unbound
 variable.
 
 Imagine now binding the mysterious variable, and calling the resulting
 variable.
 
 Imagine now binding the mysterious variable, and calling the resulting

-function `h`:
+term `h`:

 
     h ≡ \length \xs. (empty? xs) 0 (succ (length (tail xs)))
 
 Now we have no unbound variables, and we have complete non-recursive
 definitions of each of the other symbols (`empty?`, `0`, `succ`, and `tail`).
 
 
     h ≡ \length \xs. (empty? xs) 0 (succ (length (tail xs)))
 
 Now we have no unbound variables, and we have complete non-recursive
 definitions of each of the other symbols (`empty?`, `0`, `succ`, and `tail`).
 

-So `h` takes an argument, and returns a function that accurately
+So `h` takes a `length` argument, and returns a function that accurately

 computes the length of a list --- as long as the argument we supply is
 already the length function we are trying to define.  (Dehydrated
 water: to reconstitute, just add water!)
 computes the length of a list --- as long as the argument we supply is
 already the length function we are trying to define.  (Dehydrated
 water: to reconstitute, just add water!)


@@ -275,7 +273,7 @@ water: to reconstitute, just add water!)
 Here is where the discussion of fixed points becomes relevant.  Saying
 that `h` is looking for an argument (call it `LENGTH`) that has the same
 behavior as the result of applying `h` to `LENGTH` is just another way of
 Here is where the discussion of fixed points becomes relevant.  Saying
 that `h` is looking for an argument (call it `LENGTH`) that has the same
 behavior as the result of applying `h` to `LENGTH` is just another way of

-saying that we are looking for a fixed point for `h`.
+saying that we are looking for a fixed point for `h`:

 
     h LENGTH <~~> LENGTH
 
 
     h LENGTH <~~> LENGTH
 


@@ -306,73 +304,75 @@ applied `h` to a list, but `h` expects as its first argument the
 length function.
 
 So let's adjust `h`, calling the adjusted function `H`. (We'll use `u` as the variable
 length function.
 
 So let's adjust `h`, calling the adjusted function `H`. (We'll use `u` as the variable

-that expects to be bound to `H`'s fixed point, rather than `length`. This will make it easier
+that expects to be bound to the as-yet-*unknown* argument, rather than `length`. This will make it easier

 to discuss generalizations of this strategy.)
 
 to discuss generalizations of this strategy.)
 

-    H ≡ \u \xs. (empty? xs) 0 (succ ((u u) (tail xs)))
+    h ≡ \length \xs. (empty? xs) 0 (succ (length (tail xs)))
+    H ≡ \u      \xs. (empty? xs) 0 (succ ((u u)  (tail xs)))

 
 

-This is the key creative step.  Instead of applying `u` to a list, we
-apply it first to itself.  After applying `u` to an argument, it's
-ready to apply to a list, so we've solved the problem noted above with `h (tail list)`.
-We're not done yet, of course; we don't yet know what argument to give
+This is the key creative step.  Instead of applying `u` to a list, as happened
+when we self-applied `h`, `H` applies its argument `u` first to *itself*: `u u`.
+After `u` gets an argument, the *result* is ready to apply to a list, so we've solved the problem noted above with `h (tail list)`.
+We're not done yet, of course; we don't yet know what argument `u` to give

 to `H` that will behave in the desired way.
 
 to `H` that will behave in the desired way.
 

-So let's reason about `H`.  What exactly is H expecting as its first
+So let's reason about `H`.  What exactly is `H` expecting as its first

 argument?  Based on the excerpt `(u u) (tail xs)`, it appears that
 `H`'s argument, `u`, should be a function that is ready to take itself
 as an argument, and that returns a function that takes a list as an
 argument.  `H` itself fits the bill:
 
     H H <~~> (\u \xs. (empty? xs) 0 (succ ((u u) (tail xs)))) H
 argument?  Based on the excerpt `(u u) (tail xs)`, it appears that
 `H`'s argument, `u`, should be a function that is ready to take itself
 as an argument, and that returns a function that takes a list as an
 argument.  `H` itself fits the bill:
 
     H H <~~> (\u \xs. (empty? xs) 0 (succ ((u u) (tail xs)))) H

-        <~~> \xs. (empty? xs) 0 (succ ((H H) (tail xs)))
-           ≡ \xs. (empty? xs) 0 (succ (
-                    (\xs. (empty? xs) 0 (succ ((H H) (tail xs))))
-                                             (tail xs) ))
-        <~~> \xs. (empty? xs) 0 (succ (
-                    (empty? (tail xs)) 0 (succ ((H H) (tail (tail xs)))) ))
+        <~~>     \xs. (empty? xs) 0 (succ ((H H) (tail xs)))
+        <~~>     \xs. (empty? xs) 0 (succ ((
+                 \xs. (empty? xs) 0 (succ ((H H) (tail xs)))
+                                               ) (tail xs)))
+        <~~>     \xs. (empty? xs) 0 (succ (
+                      (empty? (tail xs)) 0 (succ ((H H) (tail (tail xs)))) ))

 
 We're in business!
 
 How does the recursion work?
 We've defined `H` in such a way that `H H` turns out to be the length function.
 
 We're in business!
 
 How does the recursion work?
 We've defined `H` in such a way that `H H` turns out to be the length function.

+That is, `H H` is the `LENGTH` we were looking for.

 In order to evaluate `H H`, we substitute `H` into the body of the
 In order to evaluate `H H`, we substitute `H` into the body of the

-lambda term.  Inside the lambda term, once the substitution has
+lambda term `H`.  Inside that lambda term, once the substitution has

 occurred, we are once again faced with evaluating `H H`.  And so on.
 
 occurred, we are once again faced with evaluating `H H`.  And so on.
 

-We've got the potentially infinite regress we desired, defined in terms of a
+We've got the (potentially) infinite regress we desired, defined in terms of a

 finite lambda term with no undefined symbols.
 
 Since `H H` turns out to be the length function, we can think of `H`
 finite lambda term with no undefined symbols.
 
 Since `H H` turns out to be the length function, we can think of `H`

-by itself as half of the length function (which is why we called it
-`H`, of course).  Can you think up a recursion strategy that involves
+by itself as *half* of the length function (which is why we called it
+`H`, of course).  (Thought exercise: Can you think up a recursion strategy that involves

 "dividing" the recursive function into equal thirds `T`, such that the
 "dividing" the recursive function into equal thirds `T`, such that the

-length function <~~> `T T T`?
+length function <~~> `T T T`?)

 
 We've starting with a particular recursive definition, and arrived at
 a fixed point for that definition.
 What's the general recipe?
 
 
 We've starting with a particular recursive definition, and arrived at
 a fixed point for that definition.
 What's the general recipe?
 

-1.   Start with any recursive definition `h` that takes itself as an arg: `h ≡ \self ... self ...`
-2.   Next, define `H ≡ \u . h (u u)`
-3.   Then compute `H H ≡ ((\u . h (u u)) (\u . h (u u)))`
-4.   That's the fixed point, the recursive function we're trying to define
+1.   Start with a formula `h` that takes the recursive function you're seeking as an argument: `h ≡ \length. ... length ...`
+2.   Next, define `H ≡ \u. h (u u)`
+3.   Then compute `H H ≡ ((\u. h (u u)) (\u. h (u u)))`
+4.   That's the fixed point of `h`, the recursive function you're seeking.

 
 

-So here is a general method for taking an arbitrary `h`-style recursive function
-and returning a fixed point for that function:
+Expressed in terms of a single formula, here is this method for taking an arbitrary `h`-style term and returning
+the recursive function that term expects as an argument, which as we've seen will be the `h`-term's fixed point:

 
      Y ≡ \h. (\u. h (u u)) (\u. h (u u))
 
 
      Y ≡ \h. (\u. h (u u)) (\u. h (u u))
 

-Test:
+Let's test that `Y h` will indeed be `h`'s fixed point:

 
     Y h  ≡ (\h. (\u. h (u u)) (\u. h (u u))) h
 
     Y h  ≡ (\h. (\u. h (u u)) (\u. h (u u))) h

-       ~~> (\u. h (u u)) (\u. h (u u))
-       ~~> h ((\u. h (u u)) (\u. h (u u)))
+       ~~>      (\u. h (u u)) (\u. h (u u))
+       ~~>           h ((\u. h (u u)) (\u. h (u u)))

 
 

-But the argument of `h` in the last line is just the same as the second line, which `<~~> Y h`. So the last line `<~~> h (Y h)`. In other words, `Y h <~~> h (Y h)`. So by definition, `Y h` is a fixed point for `h`.
+But the argument of `h` in the last line is just the same as the second line, which <\~~> `Y h`. So the last line <\~~> `h (Y h)`. In other words, `Y h <~~> h (Y h)`. So by definition, `Y h` is a fixed point for `h`.

 
 Works!
 
 
 Works!
 

-##Coming at it another way##
+###Coming at it another way###

 
 TODO
 
 
 TODO
 


@@ -383,16 +383,17 @@ 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
 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`. The term `K`
+be like.  We're looking for a fixed point for `K`, i.e., `\x y. x`. The term `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
 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
+also have `KX ≡ (\x y. 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`
 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 as many arguments as you give it.  
+must be a function that ignores as many arguments as you give it.
+

 Our expectation, then, is that our recipe for finding fixed points
 will build us a term that somehow manages to ignore arbitrarily many arguments.
 
 Our expectation, then, is that our recipe for finding fixed points
 will build us a term that somehow manages to ignore arbitrarily many arguments.
 


@@ -400,9 +401,9 @@ will build us a term that somehow manages to ignore arbitrarily many arguments.
     H ≡ \u.h(uu) ≡ \u.(\xy.x)(uu) ~~> \uy.uu
     H H ≡ (\uy.uu)(\uy.uu) ~~> \y.(\uy.uu)(\uy.uu)
 
     H ≡ \u.h(uu) ≡ \u.(\xy.x)(uu) ~~> \uy.uu
     H H ≡ (\uy.uu)(\uy.uu) ~~> \y.(\uy.uu)(\uy.uu)
 

-Let's check that it is in fact a fixed point:
+Let's check that it is in fact a fixed point for `K`:

 
 

-    K(H H) ≡ (\xy.x)((\uy.uu)(\uy.uu)
+    K(H H) ≡ (\xy.x)((\uy.uu)(\uy.uu))

            ~~> \y.(\uy.uu)(\uy.uu)
 
 Yep, `H H` and `K(H H)` both reduce to the same term.  
            ~~> \y.(\uy.uu)(\uy.uu)
 
 Yep, `H H` and `K(H H)` both reduce to the same term.  


@@ -414,15 +415,15 @@ To see what this fixed point does, let's reduce it a bit more:
         ~~> \yy.(\uy.uu)(\uy.uu)
         ~~> \yyy.(\uy.uu)(\uy.uu)
     
         ~~> \yy.(\uy.uu)(\uy.uu)
         ~~> \yyy.(\uy.uu)(\uy.uu)
     

-Sure enough, this fixed point ignores an endless, infinite series of
+Sure enough, this fixed point ignores an endless, arbitrarily-long 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,
 
 arguments.  It's a write-only memory, a black hole.
 
 Now that we have one fixed point, we can find others, for instance,
 

-    (\uy.uuu)(\uy.uuu) 
-    ~~> \y.(\uy.uuu)(\uy.uuu)(\uy.uuu)
-    ~~> \yy.(\uy.uuu)(\uy.uuu)(\uy.uuu)(\uy.uuu)
-    ~~> \yyy.(\uy.uuu)(\uy.uuu)(\uy.uuu)(\uy.uuu)(\uy.uuu)
+    (\uy.[uu]u) (\uy.uuu) 
+    ~~> \y. [(\uy.uuu) (\uy.uuu)] (\uy.uuu)
+    ~~> \y. [\y. (\uy.uuu) (\uy.uuu) (\uy.uuu)] (\uy.uuu)
+    ~~> \yyy. (\uy.uuu) (\uy.uuu) (\uy.uuu) (\uy.uuu) (\uy.uuu)

 
 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
 
 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


@@ -433,24 +434,24 @@ series of arguments.
 As we've seen, the recipe just given for finding a fixed point worked
 great for our `h`, which we wrote as a definition for the length
 function.  But the recipe doesn't make any assumptions about the
 As we've seen, the recipe just given for finding a fixed point worked
 great for our `h`, which we wrote as a definition for the length
 function.  But the recipe doesn't make any assumptions about the

-internal structure of the function it works with.  That means it can
-find a fixed point for literally any function whatsoever.
+internal structure of the term it works with.  That means it can
+find a fixed point for literally any lambda term whatsoever.

 
 

-In particular, what could the fixed point for the
+In particular, what could the fixed point for (our encoding of) the

 successor function possibly be like?
 
 successor function possibly be like?
 

-Well, you might think, only some of the formulas that we might give to the `succ` as arguments would really represent numbers. If we said something like:
+Well, you might think, only some of the formulas that we might give to `succ` as arguments would really represent numbers. If we said something like:

 
     succ pair
 
 who knows what we'd get back? Perhaps there's some non-number-representing formula such that when we feed it to `succ` as an argument, we get the same formula back.
 
 
     succ pair
 
 who knows what we'd get back? Perhaps there's some non-number-representing formula such that when we feed it to `succ` as an argument, we get the same formula back.
 

-Yes! That's exactly right. And which formula this is will depend on the particular way you've implemented the succ function.
+Yes! That's exactly right. And which formula this is will depend on the particular way you've encoded the successor function.

 
 One (by now obvious) upshot is that the recipes that enable us to name
 
 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 `ξ` such that
-neither `ξ` nor `f ξ` have normal forms. (Indeed, what they give us
+fixed points for any given formula `h` aren't *guaranteed* to give us
+*terminating, normalizing* fixed points. They might give us formulas `ξ` such that
+neither `ξ` nor `h ξ` have normal forms. (Indeed, what they give us

 for the `square` function isn't any of the Church numerals, but is
 rather an expression with no normal form.) However, if we take care we
 can ensure that we *do* get terminating fixed points. And this gives
 for the `square` function isn't any of the Church numerals, but is
 rather an expression with no normal form.) However, if we take care we
 can ensure that we *do* get terminating fixed points. And this gives


@@ -464,7 +465,7 @@ wasn't always clear how to force the computation to "keep going."
 
 OK, so how do we make use of this?
 
 
 OK, so how do we make use of this?
 

-Many fixed-point combinators have been discovered. (And some
+Many fixed-point combinators have been discovered. (And as we've seen, some

 fixed-point combinators give us models for building infinitely many
 more, non-equivalent fixed-point combinators.)
 
 fixed-point combinators give us models for building infinitely many
 more, non-equivalent fixed-point combinators.)
 


@@ -482,33 +483,33 @@ Indeed you can, getting the simpler:
     Θ ≡ (\u h. h (u u h)) (\u h. h (u u h))
     Y ≡ \h. (\u. h (u u)) (\u. h (u u))
 
     Θ ≡ (\u h. h (u u h)) (\u h. h (u u h))
     Y ≡ \h. (\u. h (u u)) (\u. h (u u))
 

-I stated the more complex formulas for the following reason: in a language whose evaluation order is *call-by-value*, the evaluation of `Θ (\self. BODY)` and `Y (\self. BODY)` will in general not terminate. But evaluation of the eta-unreduced primed versions will.
+We stated the more complex formulas for the following reason: in a language whose evaluation order is *call-by-value*, the evaluation of `Θ (\body. BODY)` and `Y (\body. BODY)` will in general not terminate. But evaluation of the eta-unreduced primed versions will.

 
 

-Of course, if you define your `\self. BODY` stupidly, your formula will never terminate. For example, it doesn't matter what fixed point combinator you use for `Ψ` in:
+Of course, if you define your `\body. BODY` stupidly, your formula will never terminate. For example, it doesn't matter what fixed point combinator you use for `Ψ` in:

 
 

-    Ψ (\self. \n. self n)
+    Ψ (\body. \n. body n)

 
 When you try to evaluate the application of that to some argument `M`, it's going to try to give you back:
 
 
 When you try to evaluate the application of that to some argument `M`, it's going to try to give you back:
 

-    (\n. self n) M
+    (\n. body n) M

 
 

-where `self` is equivalent to the very formula `\n. self n` that contains it. So the evaluation will proceed:
+where `body` is equivalent to the very formula `\n. body n` that contains it. So the evaluation will proceed:

 
 

-    (\n. self n) M ~~>
-    self M <~~>
-    (\n. self n) M ~~>
-    self M <~~>
+    (\n. body n) M ~~>
+    body M <~~>
+    (\n. body n) M ~~>
+    body M <~~>

     ...
 
 You've written an infinite loop!
 
 However, when we evaluate the application of our:
 
     ...
 
 You've written an infinite loop!
 
 However, when we evaluate the application of our:
 

-    Ψ (\self (\xs. (empty? xs) 0 (succ (self (tail xs))) ))
+    Ψ (\body. (\xs. (empty? xs) 0 (succ (body (tail xs))) ))

 
 to some list, we're not going to go into an infinite evaluation loop of that sort. At each cycle, we're going to be evaluating the application of:
 
 
 to some list, we're not going to go into an infinite evaluation loop of that sort. At each cycle, we're going to be evaluating the application of:
 

-    \xs. (empty? xs) 0 (succ (self (tail xs)))
+    \xs. (empty? xs) 0 (succ (body (tail xs)))

 
 to *the tail* of the list we were evaluating its application to at the previous stage. Assuming our lists are finite (and the encodings we've been using so far don't permit otherwise), at some point one will get a list whose tail is empty, and then the evaluation of that formula to that tail will return `0`. So the recursion eventually bottoms out in a base value.
 
 
 to *the tail* of the list we were evaluating its application to at the previous stage. Assuming our lists are finite (and the encodings we've been using so far don't permit otherwise), at some point one will get a list whose tail is empty, and then the evaluation of that formula to that tail will return `0`. So the recursion eventually bottoms out in a base value.
 


@@ -567,7 +568,7 @@ Now we try the next most complex example:
     3. (\h. (\u. h [u u]) (\u. h (u u))) (\sb.bsI) true false
     4. (\u. (\sb.bsI) [u u]) (\u. (\sb.bsI) (u u)) true false
     5. (\sb.bsI) [(\u. (\sb.bsI) (u u)) (\u. (\sb.bsI) (u u))] true false
     3. (\h. (\u. h [u u]) (\u. h (u u))) (\sb.bsI) true false
     4. (\u. (\sb.bsI) [u u]) (\u. (\sb.bsI) (u u)) true false
     5. (\sb.bsI) [(\u. (\sb.bsI) (u u)) (\u. (\sb.bsI) (u u))] true false

-    6. (\b.b [(\u. (\sb.bsI) (u u)) (\u. (\sb.bsI) (u u))] I) true false
+    6. (\b. b [(\u. (\sb.bsI) (u u)) (\u. (\sb.bsI) (u u))] I) true false

     7. true [(\u. (\sb.bsI) (u u)) (\u. (\sb.bsI) (u u))] I false
     8. [(\u. (\sb.bsI) (u u)) (\u. (\sb.bsI) (u u))] false
 
     7. true [(\u. (\sb.bsI) (u u)) (\u. (\sb.bsI) (u u))] I false
     8. [(\u. (\sb.bsI) (u u)) (\u. (\sb.bsI) (u u))] false
 


@@ -575,7 +576,7 @@ We've now arrived at line (4) of the first computation, so the result
 is again `I`.
 
 You should be able to see that `sink` will consume as many `true`s as
 is again `I`.
 
 You should be able to see that `sink` will consume as many `true`s as

-we throw at it, then turn into the identity function after it
+we throw at it, then turn into the identity function when it

 encounters the first `false`.
 
 The key to the recursion is that, thanks to `Y`, the definition of
 encounters the first `false`.
 
 The key to the recursion is that, thanks to `Y`, the definition of


@@ -587,6 +588,8 @@ will be discarded, and the recursion will stop.
 
 That's about as simple as recursion gets.
 
 
 That's about as simple as recursion gets.
 

+<!-- TODO Perhaps move rest to new document? -->
+

 ##Application to the truth teller/liar paradoxes##
 
 ###Base cases, and their lack###
 ##Application to the truth teller/liar paradoxes##
 
 ###Base cases, and their lack###


@@ -606,7 +609,7 @@ factorial of `n-1`.  But if we leave out the base case, we get
 That's why it's crucial to declare that `0!` = `1`, in which case the
 recursive rule does not apply.  In our terms,
 
 That's why it's crucial to declare that `0!` = `1`, in which case the
 recursive rule does not apply.  In our terms,
 

-    fact ≡ Y (\fact n. zero? n 1 (fact (predecessor n)))
+    fact ≡ Y (\fact n. (zero? n) 1 (fact (pred n)))

 
 If `n` is `0`, `fact` reduces to `1`, without computing the recursive case.
 
 
 If `n` is `0`, `fact` reduces to `1`, without computing the recursive case.
 


@@ -665,8 +668,8 @@ What about the liar paradox?
 (2)  This sentence is false.
 
 Used in a context in which *this sentence* refers to the utterance of
 (2)  This sentence is false.
 
 Used in a context in which *this sentence* refers to the utterance of

-(2) in which it occurs, (2) will denote a fixed point for `\f. neg f`,
-or `\f l r. f r l`, which is the `C` combinator.  So in such a
+(2) in which that noun phrase occurs, (2) will denote a fixed point for `\f. neg f`,
+or `\f y n. f n y`, which is the `C` combinator.  So in such a

 context, (2) might denote
 
      Y C
 context, (2) might denote
 
      Y C


@@ -705,7 +708,7 @@ for any choice of `X` whatsoever.
 So the `Y` combinator is only guaranteed to give us one fixed point out
 of infinitely many --- and not always the intuitively most useful
 one. (For instance, the squaring function `\x. mul x x` has `0` as a fixed point,
 So the `Y` combinator is only guaranteed to give us one fixed point out
 of infinitely many --- and not always the intuitively most useful
 one. (For instance, the squaring function `\x. mul x x` has `0` as a fixed point,

-since `0 * 0 = 0`, and `1` as a fixed point, since `1 * 1 = 1`, but `Y
+since `square 0 <~~> 0`, and `1` as a fixed point, since `square 1 <~~> 1`, but `Y

 (\x. mul x x)` doesn't give us `0` or `1`.) So with respect to the
 truth-teller paradox, why in the reasoning we've
 just gone through should we be reaching for just this fixed point at
 (\x. mul x x)` doesn't give us `0` or `1`.) So with respect to the
 truth-teller paradox, why in the reasoning we've
 just gone through should we be reaching for just this fixed point at


@@ -729,6 +732,8 @@ gloss on pronouns such as *it*.  In the system of
 [Jacobson 1999](http://www.springerlink.com/content/j706674r4w217jj5/),
 pronouns denote (you guessed it!) identity functions...
 
 [Jacobson 1999](http://www.springerlink.com/content/j706674r4w217jj5/),
 pronouns denote (you guessed it!) identity functions...
 

+<!-- Jim says: haven't made clear how we got from the self-referential (3) to I. -->
+

 Ultimately, in the context of this course, these paradoxes are more
 useful as a way of gaining leverage on the concepts of fixed points
 and recursion, rather than the other way around.
 Ultimately, in the context of this course, these paradoxes are more
 useful as a way of gaining leverage on the concepts of fixed points
 and recursion, rather than the other way around.