diff -r 751fde5307e4 -r 50f22b1b136a 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 \\ set ts \\ 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 +\end{ttbox} +to the specific occurrence of \isacommand{recdef} or declare them globally: +\begin{ttbox} +lemmas [????????] = +\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;(*>*)