src/Doc/Tutorial/Recdef/Nested2.thy
author wenzelm
Tue, 28 Aug 2012 18:57:32 +0200
changeset 48985 5386df44a037
parent 40878 doc-src/TutorialI/Recdef/Nested2.thy@7695e4de4d86
child 56846 9df717fef2bb
permissions -rw-r--r--
renamed doc-src to src/Doc; renamed TutorialI to Tutorial;
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(*<*)
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theory Nested2 imports Nested0 begin
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(*>*)
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lemma [simp]: "t \<in> set ts \<longrightarrow> size t < Suc(term_list_size ts)"
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by(induct_tac ts, auto)
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(*<*)
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recdef trev "measure size"
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 "trev (Var x) = Var x"
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 "trev (App f ts) = App f (rev(map trev ts))"
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(*>*)
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text{*\noindent
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By making this theorem a simplification rule, \isacommand{recdef}
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applies it automatically and the definition of @{term"trev"}
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succeeds now. As a reward for our effort, we can now prove the desired
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lemma directly.  We no longer need the verbose
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induction schema for type @{text"term"} and can use the simpler one arising from
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@{term"trev"}:
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*}
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lemma "trev(trev t) = t"
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apply(induct_tac t rule: trev.induct)
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txt{*
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@{subgoals[display,indent=0]}
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Both the base case and the induction step fall to simplification:
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*}
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by(simp_all add: rev_map sym[OF map_compose] cong: map_cong)
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text{*\noindent
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If the proof of the induction step mystifies you, we recommend that you go through
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the chain of simplification steps in detail; you will probably need the help of
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@{text"simp_trace"}. Theorem @{thm[source]map_cong} is discussed below.
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%\begin{quote}
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%{term[display]"trev(trev(App f ts))"}\\
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%{term[display]"App f (rev(map trev (rev(map trev ts))))"}\\
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%{term[display]"App f (map trev (rev(rev(map trev ts))))"}\\
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%{term[display]"App f (map trev (map trev ts))"}\\
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%{term[display]"App f (map (trev o trev) ts)"}\\
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%{term[display]"App f (map (%x. x) ts)"}\\
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%{term[display]"App f ts"}
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%\end{quote}
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The definition of @{term"trev"} above is superior to the one in
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\S\ref{sec:nested-datatype} because it uses @{term"rev"}
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and lets us use existing facts such as \hbox{@{prop"rev(rev xs) = xs"}}.
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Thus this proof is a good example of an important principle:
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\begin{quote}
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\emph{Chose your definitions carefully\\
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because they determine the complexity of your proofs.}
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\end{quote}
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Let us now return to the question of how \isacommand{recdef} can come up with
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sensible termination conditions in the presence of higher-order functions
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like @{term"map"}. For a start, if nothing were known about @{term"map"}, then
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@{term"map trev ts"} might apply @{term"trev"} to arbitrary terms, and thus
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\isacommand{recdef} would try to prove the unprovable @{term"size t < Suc
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(term_list_size ts)"}, without any assumption about @{term"t"}.  Therefore
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\isacommand{recdef} has been supplied with the congruence theorem
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@{thm[source]map_cong}:
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@{thm[display,margin=50]"map_cong"[no_vars]}
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Its second premise expresses that in @{term"map f xs"},
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function @{term"f"} is only applied to elements of list @{term"xs"}.  Congruence
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rules for other higher-order functions on lists are similar.  If you get
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into a situation where you need to supply \isacommand{recdef} with new
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congruence rules, you can append a hint after the end of
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the recursion equations:\cmmdx{hints}
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*}
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(*<*)
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consts dummy :: "nat => nat"
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recdef dummy "{}"
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"dummy n = n"
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(*>*)
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(hints recdef_cong: map_cong)
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text{*\noindent
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Or you can declare them globally
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by giving them the \attrdx{recdef_cong} attribute:
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*}
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declare map_cong[recdef_cong]
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text{*
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The @{text cong} and @{text recdef_cong} attributes are
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intentionally kept apart because they control different activities, namely
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simplification and making recursive definitions.
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%The simplifier's congruence rules cannot be used by recdef.
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%For example the weak congruence rules for if and case would prevent
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%recdef from generating sensible termination conditions.
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*}
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(*<*)end(*>*)