doc-src/TutorialI/Recdef/Nested2.thy
changeset 9721 7e51c9f3d5a0
parent 9690 50f22b1b136a
child 9754 a123a64cadeb
equal deleted inserted replaced
9720:3b7b72db57f1 9721:7e51c9f3d5a0
    32 @{term[display,margin=60]"ALL t. t : set ts --> trev (trev t) = t ==> trev (trev (App f ts)) = App f ts"}
    32 @{term[display,margin=60]"ALL t. t : set ts --> trev (trev t) = t ==> trev (trev (App f ts)) = App f ts"}
    33 \end{quote}
    33 \end{quote}
    34 both of which are solved by simplification:
    34 both of which are solved by simplification:
    35 *};
    35 *};
    36 
    36 
    37 by(simp_all del:map_compose add:sym[OF map_compose] rev_map);
    37 by(simp_all add:rev_map sym[OF map_compose]);
    38 
    38 
    39 text{*\noindent
    39 text{*\noindent
    40 If this surprises you, see Datatype/Nested2......
    40 If the proof of the induction step mystifies you, we recommend to go through
       
    41 the chain of simplification steps in detail, probably with the help of
       
    42 \isa{trace\_simp}.
       
    43 %\begin{quote}
       
    44 %{term[display]"trev(trev(App f ts))"}\\
       
    45 %{term[display]"App f (rev(map trev (rev(map trev ts))))"}\\
       
    46 %{term[display]"App f (map trev (rev(rev(map trev ts))))"}\\
       
    47 %{term[display]"App f (map trev (map trev ts))"}\\
       
    48 %{term[display]"App f (map (trev o trev) ts)"}\\
       
    49 %{term[display]"App f (map (%x. x) ts)"}\\
       
    50 %{term[display]"App f ts"}
       
    51 %\end{quote}
    41 
    52 
    42 The above definition of @{term"trev"} is superior to the one in \S\ref{sec:nested-datatype}
    53 The above definition of @{term"trev"} is superior to the one in \S\ref{sec:nested-datatype}
    43 because it brings @{term"rev"} into play, about which already know a lot, in particular
    54 because it brings @{term"rev"} into play, about which already know a lot, in particular
    44 @{prop"rev(rev xs) = xs"}.
    55 @{prop"rev(rev xs) = xs"}.
    45 Thus this proof is a good example of an important principle:
    56 Thus this proof is a good example of an important principle:
    46 \begin{quote}
    57 \begin{quote}
    47 \emph{Chose your definitions carefully\\
    58 \emph{Chose your definitions carefully\\
    48 because they determine the complexity of your proofs.}
    59 because they determine the complexity of your proofs.}
    49 \end{quote}
    60 \end{quote}
    50 
    61 
    51 Let us now return to the question of how \isacommand{recdef} can come up with sensible termination
    62 Let us now return to the question of how \isacommand{recdef} can come up with
    52 conditions in the presence of higher-order functions like @{term"map"}. For a start, if nothing
    63 sensible termination conditions in the presence of higher-order functions
    53 were known about @{term"map"}, @{term"map trev ts"} might apply @{term"trev"} to arbitrary terms,
    64 like @{term"map"}. For a start, if nothing were known about @{term"map"},
    54 and thus \isacommand{recdef} would try to prove the unprovable
    65 @{term"map trev ts"} might apply @{term"trev"} to arbitrary terms, and thus
    55 @{term"size t < Suc (term_size ts)"}, without any assumption about \isa{t}.
    66 \isacommand{recdef} would try to prove the unprovable @{term"size t < Suc
    56 Therefore \isacommand{recdef} has been supplied with the congruence theorem \isa{map\_cong}: 
    67 (term_size ts)"}, without any assumption about \isa{t}.  Therefore
       
    68 \isacommand{recdef} has been supplied with the congruence theorem
       
    69 \isa{map\_cong}:
    57 \begin{quote}
    70 \begin{quote}
    58 @{thm[display,margin=50]"map_cong"[no_vars]}
    71 @{thm[display,margin=50]"map_cong"[no_vars]}
    59 \end{quote}
    72 \end{quote}
    60 Its second premise expresses (indirectly) that the second argument of @{term"map"} is only applied
    73 Its second premise expresses (indirectly) that the second argument of
    61 to elements of its third argument. Congruence rules for other higher-order functions on lists would
    74 @{term"map"} is only applied to elements of its third argument. Congruence
    62 look very similar but have not been proved yet because they were never needed.
    75 rules for other higher-order functions on lists would look very similar but
    63 If you get into a situation where you need to supply \isacommand{recdef} with new congruence
    76 have not been proved yet because they were never needed. If you get into a
       
    77 situation where you need to supply \isacommand{recdef} with new congruence
    64 rules, you can either append the line
    78 rules, you can either append the line
    65 \begin{ttbox}
    79 \begin{ttbox}
    66 congs <congruence rules>
    80 congs <congruence rules>
    67 \end{ttbox}
    81 \end{ttbox}
    68 to the specific occurrence of \isacommand{recdef} or declare them globally:
    82 to the specific occurrence of \isacommand{recdef} or declare them globally: