From 6f604b19d3f5966e5c1d7c4fdf3703bd6ff0861c Mon Sep 17 00:00:00 2001
From: Mart Lubbers
Date: Thu, 16 Apr 2015 21:22:20 +0200
Subject: update to fp2 yay, public and licence

---
 week6/mart/BewijsMapFlatten.icl | 97 -----------------------------------------
 1 file changed, 97 deletions(-)
 delete mode 100644 week6/mart/BewijsMapFlatten.icl

(limited to 'week6/mart')

diff --git a/week6/mart/BewijsMapFlatten.icl b/week6/mart/BewijsMapFlatten.icl
deleted file mode 100644
index 59fa5bc..0000000
--- a/week6/mart/BewijsMapFlatten.icl
+++ /dev/null
@@ -1,97 +0,0 @@
-Zij gegeven:
-
-(++) :: [a] [a] -> [a]
-(++) []     xs = xs                (1)
-(++) [y:ys] xs = [y : ys ++ xs]    (2)
-
-map :: (a -> b) [a] -> [b]
-map f []       = []                (3)
-map f [x:xs]   = [f x : map f xs]  (4)
-
-flatten :: [[a]] -> [a]
-flatten []     = []                (5)
-flatten [x:xs] = x ++ (flatten xs) (6)
-
-1. 
-Te bewijzen: 
-	voor iedere functie f, eindige lijst as en bs:
-		
-		map f (as ++ bs) = (map f as) ++ (map f bs)
-
-Bewijs:
-	met inductie naar de lengte van as.
-
-	Basis:
-	aanname: as = [].
-
-	     map f ([] ++ bs) = (map f []) ++ (map f bs)                  basis
-	           ********
-	  => map f (bs) = (map f []) ++ (map f bs)                        (1)
-	                   ********
-	  => map f (bs) = [] ++ (map f bs)                                (3)
-	                  ****************
-	  => map f bs = map f bs                                          (1)
-
-	Inductiestap:
-	aanname: stelling geldt voor zekere as, ofwel:
-		map f (as ++ bs) = (map f as) ++ (map f bs) (IH)
-
-	Te bewijzen: stelling geldt ook voor as, ofwel:
-		map f ([a:as] ++ bs) = (map f [a:as]) ++ (map f bs)
-
-	     map f ([a:as] ++ bs) = (map f [a:as]) ++ (map f bs)          basis
-	  	 ************
-	  => map f [a:as ++ bs] = (map f [a:as]) ++ (map f bs)            (2)
-	    ******************
-	  => [f a : map f (as ++ bs)] = (map f [a:as]) ++ (map f bs)      (4)
-	                                ************
-	  => [f a : map f (as ++ bs)] = [f a : map f as] ++ (map f bs)    (4)
-	                                ****************************
-	  => [f a : map f (as ++ bs)] = [f a : (map f as) ++ (map f bs)]  (4)
-	           ****************           ************************
-	  => map f (as ++ bs) = (map f as) ++ (map f bs)                  (IH)
-
-	Dus: basis + inductiestap => stelling bewezen.
-
-2. 
-Te bewijzen:
-	voor iedere functie f, voor iedere eindige lijst xs:
-	
-		flatten (map (map f) xs) = map f (flatten xs)
-
-Bewijs:
-	met inductio van de lengte van xs
-	
-	Basis:
-	aanname: xs = [].
-
-	    flatten (map (map f) []) = map f (flatten [])                                basis
-	             **************
-	  = flatten [] = map f (flatten [])                                              (3)
-	                        **********
-	  = flatten [] = map f []                                                        (5)
-	    **********
-	  = [] = map f []                                                                (5)
-	         *****
-	  = [] = []                                                                      (3)
-
-	Inductiestap:
-	aanname: stelling geldt voor zekere xs, ofwel:
-		flatten (map (map f) xs) = map f (flatten xs)
-
-	Te bewijzen: stelling geldt ook voor xs, ofwel:
-		flatten (map (map f) [x:xs]) = map f (flatten [x:xs])
-
-	     flatten (map (map f) [x:xs]) = map f (flatten [x:xs])                       basis
-	              ******************
-	  => flatten [map f x: map (map f) xs] = map f (flatten [x:xs])                  (4)
-	     *********************************
-	  => (map f x) ++ (flatten (map (map f) xs)) = map f (flatten [x:xs])            (6)
-	                                                    **************
-	  => (map f x) ++ (flatten (map (map f) xs)) = map f (x ++ (flatten xs))         (6)
-	                                               *************************
-	  => (map f x) ++ (flatten (map (map f) xs)) = (map f x) ++ (map f (flatten xs)) (9.4.1)
-	                   ************************                  ****************
-	  => flatten (map (map f) xs) = map f (flatten xs)                               (IH)
-
-	Dus: basis + inductiestap => stelling  bewezen.
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