From: Camil Staps <info@camilstaps.nl>
Date: Tue, 17 Mar 2015 21:29:26 +0000 (+0100)
Subject: w6 camil
X-Git-Url: https://git.martlubbers.net/?a=commitdiff_plain;h=509520ebc8b75075575671dc96dc6ae3733f51a6;p=fp1415.git

w6 camil
---

diff --git a/week6/camil/9.4.1 b/week6/camil/9.4.1
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index 0000000..0f279ca
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+++ b/week6/camil/9.4.1
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+9.4.1 - proof by induction over as
+
+Induction base:
+    Suppose as = []. Then we have:
+    map f (as ++ bs) = map f ([] ++ bs) = map f bs = [] ++ (map f bs) = (map f []) ++ (map f bs) = (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. Then we have:
+    map f ([a:as] ++ bs) = map f [a:as ++ bs] = [f a : map f (as ++ bs)] = [f a : (map f as) ++ (map f bs)] = [f a : map f as] ++ (map f bs) = (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.
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diff --git a/week6/camil/9.4.2 b/week6/camil/9.4.2
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+9.4.2 - proof by induction over xs
+
+Induction base:
+    Suppose xs = []. Then we have:
+    flatten (map (map f) xs) = flatten (map (map f) []) = flatten [] = [] = map f [] = map f (flatten []) = map f (flatten xs).
+
+Induction step:
+    Suppose flatten (map (map f) xs) = map f (flatten xs) for certain xs of finite length. Then we have:
+    flatten (map (map f) [x:xs]) = flatten [map f x : map (map f) xs] = (map f x) ++ flatten (map (map f) xs) = (map f x) ++ (map f (flatten xs)) =(9.4.1) map f (x ++ (flatten xs)) = map f (flatten [x:xs]).
+
+By the principle of induction we have now proven that flatten (map (map f) xs) = map f (flatten xs) for any list of finite length xs.
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