From ad4eef901340bd0411469aebe125de740864f9b6 Mon Sep 17 00:00:00 2001
From: Chris Barker
Date: Mon, 17 Jan 2011 23:14:53 -0500
Subject: [PATCH] edits
---
hints/assignment_4_hint_3_alternate_2.mdwn | 8 +++++---
1 file changed, 5 insertions(+), 3 deletions(-)
diff --git a/hints/assignment_4_hint_3_alternate_2.mdwn b/hints/assignment_4_hint_3_alternate_2.mdwn
index b2ddef82..a0ed78dc 100644
--- a/hints/assignment_4_hint_3_alternate_2.mdwn
+++ b/hints/assignment_4_hint_3_alternate_2.mdwn
@@ -83,7 +83,9 @@ In the original, `h` was somehow *half* of the fixed point, so that `h
h` computed the fixed point. In the schema here, `h1` had better be a
function which, when you give it suitable arguments, computes the
first fixed point `X1` (likewise for `h2` wrt the second fixed point
-`X2`). Then we can arrange for our definition to return the desired
+`X2`).
+
+Then we can arrange for our definition to return the desired
fixed point like this:
let Y1 = \pe po . (\h1 h2 . pe (h1 [blah])(h2 [blah]))
@@ -91,7 +93,7 @@ fixed point like this:
(\h1 h2 . ...)
The term in the middle line is going in for `h1`, so it had better
-be the kind of thing which, when you give it suitable arguments,
+also be the kind of thing which, when you give it suitable arguments,
computes a fixed point for `pe`:
let Y1 = \pe po . (\h1 h2 . pe (h1 [blah]) (h2 [blah]))
@@ -102,7 +104,7 @@ But the third line must help compute a fixed point for `po`.
All we need to do is figure out what the arguments to `h1` and `h2`
ought to be. Final guess: in the original, `h` took one argument (a
-copy of itself), so once again, we'll need two arguments. Here's
+copy of itself), so once again, here we'll need two arguments. Here's
where the mutual recursion comes in: the two arguments to `h1` are a
copy of itself, and a copy of `h2` (symmetrically for `h2`). So the
complete definition is
--
2.11.0