doc-src/TutorialI/Recdef/simplification.thy

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(*<*)
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 @{name"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
\begin{quote}
@{term[display]"n ~= 0 ==> m mod n < n"}
\end{quote}
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 @{name"if"} 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 @{name"if"}s if the
condition simplifies to neither @{term"True"} nor @{term"False"}. For
example, simplification reduces
\begin{quote}
@{term[display]"gcd(m,n) = k"}
\end{quote}
in one step to
\begin{quote}
@{term[display]"(if n=0 then m else gcd(n, m mod n)) = k"}
\end{quote}
where the condition cannot be reduced further, and splitting leads to
\begin{quote}
@{term[display]"(n=0 --> m=k) & (n ~= 0 --> gcd(n, m mod n)=k)"}
\end{quote}
Since the recursive call @{term"gcd(n, m mod n)"} is no longer protected by
an @{name"if"}, it is unfolded again, which leads to an infinite chain of
simplification steps. Fortunately, this problem can be avoided in many
different ways.

The most radical solution is to disable the offending \@{name"split_if"} as
shown in the section on case splits in \S\ref{sec:Simplification}.  However,
we do not recommend this because it means you will often have to invoke the
rule explicitly when @{name"if"} is involved.

If possible, the definition should be given by pattern matching on the left
rather than @{name"if"} on the right. In the case of @{term"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
@{prop"n ~= 0"}. Unfortunately, in general the case distinction
may not be expressible by pattern matching.

A very simple alternative is to replace @{name"if"} by @{name"case"}, which
is also available for @{typ"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 @{term"gcd"} it means we have to prove
*}

lemma [simp]: "gcd (m, 0) = m";
by(simp);
lemma [simp]: "n \\<noteq> 0 \\<Longrightarrow> gcd(m, n) = gcd(n, m mod n)";
by(simp);

text{*\noindent
after which we can disable the original simplification rule:
*}

lemmas [simp del] = gcd.simps;

(*<*)
end
(*>*)