From e038bd97c96eef24418d63dd236116073205ad56 Mon Sep 17 00:00:00 2001
From: Camil Staps
Date: Wed, 18 Mar 2015 13:05:33 +0100
Subject: Put w6 in format

---
 week6/camil/9.4.1 | 23 -----------------------
 1 file changed, 23 deletions(-)
 delete mode 100644 week6/camil/9.4.1

(limited to 'week6/camil/9.4.1')

diff --git a/week6/camil/9.4.1 b/week6/camil/9.4.1
deleted file mode 100644
index d9b2608..0000000
--- a/week6/camil/9.4.1
+++ /dev/null
@@ -1,23 +0,0 @@
-9.4.1 - proof by induction over as
-
-Induction base:
-    Suppose as = []. Then we have:
-
-    map f (as ++ bs)                        // assumption as = []
-    = map f ([] ++ bs)                      // definition of ++, rule 1
-    = map f bs                              // definition of ++, rule 1
-    = [] ++ (map f bs)                      // definition of map, rule 3
-    = (map f []) ++ (map f bs)              // assumption as = []
-    = (map f as) ++ (map f bs).
-
-Induction step:
-    Suppose map f (as ++ bs) = (map f as) ++ (map f bs) for certain as and any bs (induction hypothesis). Then we have:
-    
-    map f ([a:as] ++ bs)                    // definition of ++, rule 2
-    = map f [a:as ++ bs]                    // definition of map, rule 4
-    = [f a : map f (as ++ bs)]              // induction hypothesis: assumption map f (as ++ bs) = (map f as) ++ (map f bs)
-    = [f a : (map f as) ++ (map f bs)]      // rewriting list
-    = [f a : map f as] ++ (map f bs)        // definition of map, rule 4
-    = (map f [a:as]) ++ (map f bs).
-
-By the principle of induction we have now proven that map f (as ++ bs) = (map f as) ++ (map f bs) for any finite lists as, bs.
\ No newline at end of file
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