doc-src/TutorialI/Trie/Trie.thy

--- a/doc-src/TutorialI/Trie/Trie.thy	Fri Jul 28 13:04:59 2000 +0200
+++ b/doc-src/TutorialI/Trie/Trie.thy	Fri Jul 28 16:02:51 2000 +0200
@@ -7,7 +7,7 @@
 representation where the subtries are held in an association list, i.e.\ a
 list of (letter,trie) pairs.  Abstracting over the alphabet \isa{'a} and the
 values \isa{'v} we define a trie as follows:
-*}
+*};
 
 datatype ('a,'v)trie = Trie  "'v option"  "('a * ('a,'v)trie)list";
 
@@ -16,7 +16,7 @@
 association list of subtries.  This is an example of nested recursion involving products,
 which is fine because products are datatypes as well.
 We define two selector functions:
-*}
+*};
 
 consts value :: "('a,'v)trie \\<Rightarrow> 'v option"
        alist :: "('a,'v)trie \\<Rightarrow> ('a * ('a,'v)trie)list";
@@ -25,7 +25,7 @@
 
 text{*\noindent
 Association lists come with a generic lookup function:
-*}
+*};
 
 consts   assoc :: "('key * 'val)list \\<Rightarrow> 'key \\<Rightarrow> 'val option";
 primrec "assoc [] x = None"
@@ -37,7 +37,7 @@
 examining the letters of the search string one by one. As
 recursion on lists is simpler than on tries, let us express this as primitive
 recursion on the search string argument:
-*}
+*};
 
 consts   lookup :: "('a,'v)trie \\<Rightarrow> 'a list \\<Rightarrow> 'v option";
 primrec "lookup t [] = value t"
@@ -49,16 +49,16 @@
 As a first simple property we prove that looking up a string in the empty
 trie \isa{Trie~None~[]} always returns \isa{None}. The proof merely
 distinguishes the two cases whether the search string is empty or not:
-*}
+*};
 
 lemma [simp]: "lookup (Trie None []) as = None";
-apply(case_tac as, auto).;
+by(case_tac as, auto);
 
 text{*
 Things begin to get interesting with the definition of an update function
 that adds a new (string,value) pair to a trie, overwriting the old value
 associated with that string:
-*}
+*};
 
 consts update :: "('a,'v)trie \\<Rightarrow> 'a list \\<Rightarrow> 'v \\<Rightarrow> ('a,'v)trie";
 primrec
@@ -79,7 +79,7 @@
 Before we start on any proofs about \isa{update} we tell the simplifier to
 expand all \isa{let}s and to split all \isa{case}-constructs over
 options:
-*}
+*};
 
 theorems [simp] = Let_def;
 theorems [split] = option.split;
@@ -91,7 +91,7 @@
 
 Our main goal is to prove the correct interaction of \isa{update} and
 \isa{lookup}:
-*}
+*};
 
 theorem "\\<forall>t v bs. lookup (update t as v) bs =
                     (if as=bs then Some v else lookup t bs)";
@@ -104,7 +104,7 @@
 if \isa{update} has already been simplified, which can only happen if
 \isa{as} is instantiated.
 The start of the proof is completely conventional:
-*}
+*};
 
 apply(induct_tac as, auto);
 
@@ -118,10 +118,10 @@
 Clearly, if we want to make headway we have to instantiate \isa{bs} as
 well now. It turns out that instead of induction, case distinction
 suffices:
-*}
+*};
 
 apply(case_tac[!] bs);
-apply(auto).;
+by(auto);
 
 text{*\noindent
 Both \isaindex{case_tac} and \isaindex{induct_tac}
@@ -136,8 +136,8 @@
 goals up in such a way that case distinction on \isa{bs} makes sense and
 solves the proof. Part~\ref{Isar} shows you how to write readable and stable
 proofs.
-*}
+*};
 
 (*<*)
-end
+end;
 (*>*)