
diff -r ff6787d730d5 -r 260090b54ef9 doc-src/Exercises/0304/a2/a2.thy
--- a/doc-src/Exercises/0304/a2/a2.thy	Fri Apr 29 18:13:28 2005 +0200
+++ /dev/null	Thu Jan 01 00:00:00 1970 +0000
@@ -1,129 +0,0 @@
-
-(*<*) theory a2 = Main: (*>*)
-
-subsection {* Folding Lists and Trees *}
-
-subsubsection {* Some more list functions *}
-
-text{* Recall the summation function *}
-
-(*<*) consts (*>*)
-  sum :: "nat list \<Rightarrow> nat"
-primrec
-  "sum [] = 0"
-  "sum (x # xs) = x + sum xs"
-
-
-text{* In the Isabelle library, you will find in theory
-\texttt{List.thy} the functions @{text foldr} and @{text foldl}, which
-allow you to define some list functions, among them @{text sum} and
-@{text length}. Show the following: *}
-
-
-lemma sum_foldr: "sum xs = foldr (op +) xs 0"
-(*<*) oops (*>*)
-
-lemma length_foldr: "length xs = foldr (\<lambda> x res. 1 + res) xs 0"
-(*<*) oops (*>*)
-
-text {* Repeated application of @{text foldr} and @{text map} has the
-disadvantage that a list is traversed several times. A single traversal is sufficient, as
-illustrated by the following example: *}
-
-lemma "sum (map (\<lambda> x. x + 3) xs) = foldr h xs b"
-(*<*) oops (*>*)
-
-text {* Find terms @{text h} and @{text b} which solve this
-equation. Generalize this result, i.e. show for appropriate @{text h}
-and @{text b}: *}
-
-
-lemma "foldr g (map f xs) a = foldr h xs b"
-(*<*) oops (*>*)
-
-text {* Hint: Isabelle can help you find the solution if you use the
-equalities arising during a proof attempt. *}
-
-text {* The following function @{text rev_acc} reverses a list in linear time: *}
-
-
-consts
-  rev_acc :: "['a list, 'a list] \<Rightarrow> 'a list"
-primrec
-  "rev_acc [] ys = ys"
-  "rev_acc (x#xs) ys = (rev_acc xs (x#ys))"
-
-text{* Show that @{text rev_acc} can be defined by means of @{text foldl}. *}
-
-lemma rev_acc_foldl: "rev_acc xs a = foldl (\<lambda> ys x. x # ys) a xs"
-(*<*) oops (*>*)
-
-text {* On the first exercise sheet, we have shown: *}
-
-lemma sum_append [simp]: "sum (xs @ ys) = sum xs + sum ys"
-  by (induct xs) auto
-
-text {* Prove a similar distributivity property for @{text foldr},
-i.e. something like @{text "foldr f (xs @ ys) a = f (foldr f xs a)
-(foldr f ys a)"}. However, you will have to strengthen the premisses
-by taking into account algebraic properties of @{text f} and @{text
-a}. *}
-
-
-lemma foldr_append: "foldr f (xs @ ys) a = f (foldr f xs a) (foldr f ys a)"
-(*<*) oops (*>*)
-
-text {* Now, define the function @{text prod}, which computes the product of all list elements: *}
-(*<*) consts (*>*)
-  prod :: "nat list \<Rightarrow> nat"
-
-text {* direcly with the aid of  a fold and prove the following: *}
-lemma "prod (xs @ ys) = prod xs * prod ys"
-(*<*) oops (*>*)
-
-
-subsubsection {* Functions on Trees *}
-
-text {* Consider the following type of binary trees: *}
-datatype 'a tree = Tip | Node "'a tree" 'a "'a tree"
-
-text {* Define functions which convert a tree into a list by traversing it in pre- resp. postorder: *}
-(*<*) consts (*>*)
-  preorder :: "'a tree \<Rightarrow> 'a list"
-  postorder :: "'a tree \<Rightarrow> 'a list"
-
-text {* You have certainly realized that computation of postorder traversal can be efficiently realized with an accumulator, in analogy to  @{text rev_acc}: *} 
-
-consts
-  postorder_acc :: "['a tree, 'a list] \<Rightarrow> 'a list"
-
-text {* Define this function and show: *}
-lemma "postorder_acc t xs = (postorder t) @ xs"
-(*<*) oops (*>*)
-
-
-text {* @{text postorder_acc} is the instance of a function
-@{text foldl_tree}, which is similar to @{text foldl}. *}
-
-consts
-  foldl_tree :: "('b => 'a => 'b) \<Rightarrow> 'b \<Rightarrow> 'a tree \<Rightarrow> 'b"
-
-text {* Show the following: *}
-
-lemma "\<forall> a. postorder_acc t a = foldl_tree (\<lambda> xs x. Cons x xs) a t"
-(*<*) oops (*>*)
-
-text {* Define a function @{text tree_sum} that computes the sum of
-the elements of a tree of natural numbers: *}
-
-consts
-  tree_sum :: "nat tree \<Rightarrow> nat"
-
-text {* and show that this function satisfies *}
-
-lemma "tree_sum t = sum (preorder t)"
-(*<*) oops (*>*)
-
-
-(*<*) end (*>*)
-




