From 35de8baadf52a3f6488fbd0463cdc331a4ea02e3 Mon Sep 17 00:00:00 2001
From: chris
Date: Sun, 15 Mar 2015 13:26:22 -0400
Subject: [PATCH]
---
exercises/_assignment6.mdwn | 6 +++---
1 file changed, 3 insertions(+), 3 deletions(-)
diff --git a/exercises/_assignment6.mdwn b/exercises/_assignment6.mdwn
index 50e89ac8..b6a00ded 100644
--- a/exercises/_assignment6.mdwn
+++ b/exercises/_assignment6.mdwn
@@ -134,8 +134,8 @@ piece, which we can think of as a function from a type to a type.
Call this type function M, and let P, Q, R, and S be variables over types.
Recall that a monad requires a singleton function 1:P-> MP, and a
-composition operator >=>: (P->MQ) -> (Q->MR) -> (P->MR) [type type for
-the composition operator corrects a "type"-o from the class handout]
+composition operator >=>: (P->MQ) -> (Q->MR) -> (P->MR) [the type for
+the composition operator given here corrects a "type"-o from the class handout]
that obey the following laws:
1 >=> k = k
@@ -170,7 +170,7 @@ Show your composition operator obeys the monad laws.
'a, let the boxed type be a list of objects of type 'a. The singleton
is `\p.[p]`, and the composition operator is
- >=> (first:P->[Q]) (second:Q->[R]) :(P->[R]) = fun p -> [r | q <- first p, r <- second q]
+ >=> (first:P->[Q]) (second:Q->[R]) :(P->[R]) = fun p -> [r | q <- first p, r <- second q]
Sanity check:
--
2.11.0