edits

diff --git a/week4.mdwn b/week4.mdwn
index 8d89a5a..edf9552 100644 (file)
--- a/week4.mdwn
+++ b/week4.mdwn
@@ -18,7 +18,7 @@ X = WW = (\x.T(xx))W = T(WW) = TX
 Please slow down and make sure that you understand what justified each
 of the equalities in the last line.
 
 Please slow down and make sure that you understand what justified each
 of the equalities in the last line.
 

-Q: How do you know that for any term T, YT is a fixed point of T?
+Q: How do you know that for any term `T`, `YT` is a fixed point of `T`?

 
 A: Note that in the proof given in the previous answer, we chose `T`
 and then set `X = WW = (\x.T(xx))(\x.T(xx))`.  If we abstract over
 
 A: Note that in the proof given in the previous answer, we chose `T`
 and then set `X = WW = (\x.T(xx))(\x.T(xx))`.  If we abstract over


@@ -26,7 +26,7 @@ and then set `X = WW = (\x.T(xx))(\x.T(xx))`.  If we abstract over
 what argument `T` we feed Y, it returns some `X` that is a fixed point
 of `T`, by the reasoning in the previous answer.
 
 what argument `T` we feed Y, it returns some `X` that is a fixed point
 of `T`, by the reasoning in the previous answer.
 

-Q: So if every term has a fixed point, even Y has fixed point.
+Q: So if every term has a fixed point, even `Y` has fixed point.

 
 A: Right:
 
 
 A: Right:
 


@@ -42,7 +42,7 @@ Q: Ouch!  Stop hurting my brain.
 A: Let's come at it from the direction of arithmetic.  Recall that we
 claimed that even `succ`---the function that added one to any
 number---had a fixed point.  How could there be an X such that X = X+1?
 A: Let's come at it from the direction of arithmetic.  Recall that we
 claimed that even `succ`---the function that added one to any
 number---had a fixed point.  How could there be an X such that X = X+1?

-Then
+That would imply that

 
     X = succ X = succ (succ X) = succ (succ (succ (X))) = succ (... (succ X)...)
 
 
     X = succ X = succ (succ X) = succ (succ (succ (X))) = succ (... (succ X)...)
 


@@ -58,7 +58,7 @@ successor.  Let's just check that `X = succ X`:
 
 You should see the close similarity with YY here.
 
 
 You should see the close similarity with YY here.
 

-Q. So `Y` applied to `succ` returns infinity!
+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:
 
 A. Yes!  Let's see why it makes sense to think of `Y succ` as a Church
 numeral:


@@ -82,8 +82,10 @@ succ)(Y succ)`?  What would you expect infinity minus infinity to be?
 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
 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:
+represents arithmetic infinity.  
+
+It's important to bear in mind the simplest term in question is not
+infinite:

 
      Y succ = (\f.(\x.f(xx))(\x.f(xx)))(\n s z. s (n s z))
 
 
      Y succ = (\f.(\x.f(xx))(\x.f(xx)))(\n s z. s (n s z))
 


@@ -122,12 +124,13 @@ endless reduction:
 
 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`,
 
 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
 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
+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. 
 
 Q.  You claimed that the Ackerman function couldn't be implemented
 `prefac` if we are forced to. 
 
 Q.  You claimed that the Ackerman function couldn't be implemented


@@ -160,5 +163,5 @@ For instance,
 A 1 x is to A 0 x as addition is to the successor function;
 A 2 x is to A 1 x as multiplication is to addition;
 A 3 x is to A 2 x as exponentiation is to multiplication---
 A 1 x is to A 0 x as addition is to the successor function;
 A 2 x is to A 1 x as multiplication is to addition;
 A 3 x is to A 2 x as exponentiation is to multiplication---

-so A 4 x is to A 3 x as super-exponentiation is to exponentiation...
+so A 4 x is to A 3 x as hyper-exponentiation is to exponentiation...