
diff -r 22fa8b16c3ae -r 13b32661dde4 doc-src/TutorialI/Recdef/simplification.thy
--- /dev/null	Thu Jan 01 00:00:00 1970 +0000
+++ b/doc-src/TutorialI/Recdef/simplification.thy	Wed Apr 19 11:56:31 2000 +0200
@@ -0,0 +1,105 @@
+(*<*)
+theory simplification = Main:;
+(*>*)
+
+text{*
+Once we have succeeded in proving all termination conditions, the recursion
+equations become simplification rules, just as with
+\isacommand{primrec}. In most cases this works fine, but there is a subtle
+problem that must be mentioned: simplification may not
+terminate because of automatic splitting of \isa{if}.
+Let us look at an example:
+*}
+
+consts gcd :: "nat*nat \\<Rightarrow> nat";
+recdef gcd "measure (\\<lambda>(m,n).n)"
+  "gcd (m, n) = (if n=0 then m else gcd(n, m mod n))";
+
+text{*\noindent
+According to the measure function, the second argument should decrease with
+each recursive call. The resulting termination condition
+*}
+
+(*<*)term(*>*) "n \\<noteq> 0 \\<Longrightarrow> m mod n < n";
+
+text{*\noindent
+is provded automatically because it is already present as a lemma in the
+arithmetic library. Thus the recursion equation becomes a simplification
+rule. Of course the equation is nonterminating if we are allowed to unfold
+the recursive call inside the \isa{else} branch, which is why programming
+languages and our simplifier don't do that. Unfortunately the simplifier does
+something else which leads to the same problem: it splits \isa{if}s if the
+condition simplifies to neither \isa{True} nor \isa{False}. For
+example, simplification reduces
+*}
+
+(*<*)term(*>*) "gcd(m,n) = k";
+
+text{*\noindent
+in one step to
+*}
+
+(*<*)term(*>*) "(if n=0 then m else gcd(n, m mod n)) = k";
+
+text{*\noindent
+where the condition cannot be reduced further, and splitting leads to
+*}
+
+(*<*)term(*>*) "(n=0 \\<longrightarrow> m=k) \\<and> (n\\<noteq>0 \\<longrightarrow> gcd(n, m mod n)=k)";
+
+text{*\noindent
+Since the recursive call \isa{gcd(n, m mod n)} is no longer protected by
+an \isa{if}, this leads to an infinite chain of simplification steps.
+Fortunately, this problem can be avoided in many different ways.
+
+Of course the most radical solution is to disable the offending
+\isa{split_if} as shown in the section on case splits in
+\S\ref{sec:SimpFeatures}.
+However, we do not recommend this because it means you will often have to
+invoke the rule explicitly when \isa{if} is involved.
+
+If possible, the definition should be given by pattern matching on the left
+rather than \isa{if} on the right. In the case of \isa{gcd} the
+following alternative definition suggests itself:
+*}
+
+consts gcd1 :: "nat*nat \\<Rightarrow> nat";
+recdef gcd1 "measure (\\<lambda>(m,n).n)"
+  "gcd1 (m, 0) = m"
+  "gcd1 (m, n) = gcd1(n, m mod n)";
+
+
+text{*\noindent
+Note that the order of equations is important and hides the side condition
+\isa{n \isasymnoteq\ 0}. Unfortunately, in general the case distinction
+may not be expressible by pattern matching.
+
+A very simple alternative is to replace \isa{if} by \isa{case}, which
+is also available for \isa{bool} but is not split automatically:
+*}
+
+consts gcd2 :: "nat*nat \\<Rightarrow> nat";
+recdef gcd2 "measure (\\<lambda>(m,n).n)"
+  "gcd2(m,n) = (case n=0 of True \\<Rightarrow> m | False \\<Rightarrow> gcd2(n,m mod n))";
+
+text{*\noindent
+In fact, this is probably the neatest solution next to pattern matching.
+
+A final alternative is to replace the offending simplification rules by
+derived conditional ones. For \isa{gcd} it means we have to prove
+*}
+
+lemma [simp]: "gcd (m, 0) = m";
+apply(simp).;
+lemma [simp]: "n \\<noteq> 0 \\<Longrightarrow> gcd(m, n) = gcd(n, m mod n)";
+apply(simp).;
+
+text{*\noindent
+after which we can disable the original simplification rule:
+*}
+
+lemmas [simp del] = gcd.simps;
+
+(*<*)
+end
+(*>*)




