(*<*)
theory Nested2 = Nested0:;
(*>*)
text{*\noindent
The termintion condition is easily proved by induction:
*};
lemma [simp]: "t \<in> set ts \<longrightarrow> size t < Suc(term_list_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 @{text"term"} but the simpler one arising from
@{term"trev"}:
*};
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
@{term[display,margin=60]"ALL t. t : set ts --> trev (trev t) = t ==> trev (trev (App f ts)) = App f ts"}
both of which are solved by simplification:
*};
by(simp_all add:rev_map sym[OF map_compose] cong:map_cong);
text{*\noindent
If the proof of the induction step mystifies you, we recommend to go through
the chain of simplification steps in detail; you will probably need the help of
@{text"trace_simp"}. Theorem @{thm[source]map_cong} is discussed below.
%\begin{quote}
%{term[display]"trev(trev(App f ts))"}\\
%{term[display]"App f (rev(map trev (rev(map trev ts))))"}\\
%{term[display]"App f (map trev (rev(rev(map trev ts))))"}\\
%{term[display]"App f (map trev (map trev ts))"}\\
%{term[display]"App f (map (trev o trev) ts)"}\\
%{term[display]"App f (map (%x. x) ts)"}\\
%{term[display]"App f ts"}
%\end{quote}
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_list_size ts)"}, without any assumption about @{term"t"}. Therefore
\isacommand{recdef} has been supplied with the congruence theorem
@{thm[source]map_cong}:
@{thm[display,margin=50]"map_cong"[no_vars]}
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 look very similar. If you get
into a situation where you need to supply \isacommand{recdef} with new
congruence rules, you can either append a hint locally
to the specific occurrence of \isacommand{recdef}
*}
(*<*)
consts dummy :: "nat => nat"
recdef dummy "{}"
"dummy n = n"
(*>*)
(hints recdef_cong: map_cong)
text{*\noindent
or declare them globally
by giving them the @{text recdef_cong} attribute as in
*}
declare map_cong[recdef_cong];
text{*
Note that the @{text cong} and @{text recdef_cong} attributes are
intentionally kept apart because they control different activities, namely
simplification and making recursive definitions.
% The local @{text cong} in
% the hints section of \isacommand{recdef} is merely short for @{text recdef_cong}.
%The simplifier's congruence rules cannot be used by recdef.
%For example the weak congruence rules for if and case would prevent
%recdef from generating sensible termination conditions.
*};
(*<*)end;(*>*)