doc-src/TutorialI/Recdef/Nested2.thy

--- /dev/null	Thu Jan 01 00:00:00 1970 +0000
+++ b/doc-src/TutorialI/Recdef/Nested2.thy	Mon Aug 28 10:16:58 2000 +0200
@@ -0,0 +1,78 @@
+(*<*)
+theory Nested2 = Nested0:;
+consts trev  :: "('a,'b)term => ('a,'b)term";
+(*>*)
+
+text{*\noindent
+The termintion condition is easily proved by induction:
+*};
+
+lemma [simp]: "t \\<in> set ts \\<longrightarrow> size t < Suc(term_size ts)";
+by(induct_tac ts, auto);
+(*<*)
+recdef trev "measure size"
+ "trev (Var x) = Var x"
+ "trev (App f ts) = App f (rev(map trev ts))";
+(*>*)
+text{*\noindent
+By making this theorem a simplification rule, \isacommand{recdef}
+applies it automatically and the above definition of @{term"trev"}
+succeeds now. As a reward for our effort, we can now prove the desired
+lemma directly. The key is the fact that we no longer need the verbose
+induction schema for type \isa{term} but the simpler one arising from
+@{term"trev"}:
+*};
+
+lemmas [cong] = map_cong;
+lemma "trev(trev t) = t";
+apply(induct_tac t rule:trev.induct);
+txt{*\noindent
+This leaves us with a trivial base case @{term"trev (trev (Var x)) = Var x"} and the step case
+\begin{quote}
+@{term[display,margin=60]"ALL t. t : set ts --> trev (trev t) = t ==> trev (trev (App f ts)) = App f ts"}
+\end{quote}
+both of which are solved by simplification:
+*};
+
+by(simp_all del:map_compose add:sym[OF map_compose] rev_map);
+
+text{*\noindent
+If this surprises you, see Datatype/Nested2......
+
+The above definition of @{term"trev"} is superior to the one in \S\ref{sec:nested-datatype}
+because it brings @{term"rev"} into play, about which already know a lot, in particular
+@{prop"rev(rev xs) = xs"}.
+Thus this proof is a good example of an important principle:
+\begin{quote}
+\emph{Chose your definitions carefully\\
+because they determine the complexity of your proofs.}
+\end{quote}
+
+Let us now return to the question of how \isacommand{recdef} can come up with sensible termination
+conditions in the presence of higher-order functions like @{term"map"}. For a start, if nothing
+were known about @{term"map"}, @{term"map trev ts"} might apply @{term"trev"} to arbitrary terms,
+and thus \isacommand{recdef} would try to prove the unprovable
+@{term"size t < Suc (term_size ts)"}, without any assumption about \isa{t}.
+Therefore \isacommand{recdef} has been supplied with the congruence theorem \isa{map\_cong}: 
+\begin{quote}
+@{thm[display,margin=50]"map_cong"[no_vars]}
+\end{quote}
+Its second premise expresses (indirectly) that the second argument of @{term"map"} is only applied
+to elements of its third argument. Congruence rules for other higher-order functions on lists would
+look very similar but have not been proved yet because they were never needed.
+If you get into a situation where you need to supply \isacommand{recdef} with new congruence
+rules, you can either append the line
+\begin{ttbox}
+congs <congruence rules>
+\end{ttbox}
+to the specific occurrence of \isacommand{recdef} or declare them globally:
+\begin{ttbox}
+lemmas [????????] = <congruence rules>
+\end{ttbox}
+
+Note that \isacommand{recdef} feeds on exactly the same \emph{kind} of
+congruence rules as the simplifier (\S\ref{sec:simp-cong}) but that
+declaring a congruence rule for the simplifier does not make it
+available to \isacommand{recdef}, and vice versa. This is intentional.
+*};
+(*<*)end;(*>*)