author wenzelm
Sat Nov 01 14:20:38 2014 +0100 (2014-11-01)
changeset 58860 fee7cfa69c50
parent 48985 5386df44a037
child 67406 23307fd33906
permissions -rw-r--r--
eliminated spurious semicolons;
     1 (*<*)
     2 theory Itrev
     3 imports Main
     4 begin
     5 declare [[names_unique = false]]
     6 (*>*)
     8 section{*Induction Heuristics*}
    10 text{*\label{sec:InductionHeuristics}
    11 \index{induction heuristics|(}%
    12 The purpose of this section is to illustrate some simple heuristics for
    13 inductive proofs. The first one we have already mentioned in our initial
    14 example:
    15 \begin{quote}
    16 \emph{Theorems about recursive functions are proved by induction.}
    17 \end{quote}
    18 In case the function has more than one argument
    19 \begin{quote}
    20 \emph{Do induction on argument number $i$ if the function is defined by
    21 recursion in argument number $i$.}
    22 \end{quote}
    23 When we look at the proof of @{text"(xs@ys) @ zs = xs @ (ys@zs)"}
    24 in \S\ref{sec:intro-proof} we find
    25 \begin{itemize}
    26 \item @{text"@"} is recursive in
    27 the first argument
    28 \item @{term xs}  occurs only as the first argument of
    29 @{text"@"}
    30 \item both @{term ys} and @{term zs} occur at least once as
    31 the second argument of @{text"@"}
    32 \end{itemize}
    33 Hence it is natural to perform induction on~@{term xs}.
    35 The key heuristic, and the main point of this section, is to
    36 \emph{generalize the goal before induction}.
    37 The reason is simple: if the goal is
    38 too specific, the induction hypothesis is too weak to allow the induction
    39 step to go through. Let us illustrate the idea with an example.
    41 Function \cdx{rev} has quadratic worst-case running time
    42 because it calls function @{text"@"} for each element of the list and
    43 @{text"@"} is linear in its first argument.  A linear time version of
    44 @{term"rev"} reqires an extra argument where the result is accumulated
    45 gradually, using only~@{text"#"}:
    46 *}
    48 primrec itrev :: "'a list \<Rightarrow> 'a list \<Rightarrow> 'a list" where
    49 "itrev []     ys = ys" |
    50 "itrev (x#xs) ys = itrev xs (x#ys)"
    52 text{*\noindent
    53 The behaviour of \cdx{itrev} is simple: it reverses
    54 its first argument by stacking its elements onto the second argument,
    55 and returning that second argument when the first one becomes
    56 empty. Note that @{term"itrev"} is tail-recursive: it can be
    57 compiled into a loop.
    59 Naturally, we would like to show that @{term"itrev"} does indeed reverse
    60 its first argument provided the second one is empty:
    61 *}
    63 lemma "itrev xs [] = rev xs"
    65 txt{*\noindent
    66 There is no choice as to the induction variable, and we immediately simplify:
    67 *}
    69 apply(induct_tac xs, simp_all)
    71 txt{*\noindent
    72 Unfortunately, this attempt does not prove
    73 the induction step:
    74 @{subgoals[display,indent=0,margin=70]}
    75 The induction hypothesis is too weak.  The fixed
    76 argument,~@{term"[]"}, prevents it from rewriting the conclusion.  
    77 This example suggests a heuristic:
    78 \begin{quote}\index{generalizing induction formulae}%
    79 \emph{Generalize goals for induction by replacing constants by variables.}
    80 \end{quote}
    81 Of course one cannot do this na\"{\i}vely: @{term"itrev xs ys = rev xs"} is
    82 just not true.  The correct generalization is
    83 *}
    84 (*<*)oops(*>*)
    85 lemma "itrev xs ys = rev xs @ ys"
    86 (*<*)apply(induct_tac xs, simp_all)(*>*)
    87 txt{*\noindent
    88 If @{term"ys"} is replaced by @{term"[]"}, the right-hand side simplifies to
    89 @{term"rev xs"}, as required.
    91 In this instance it was easy to guess the right generalization.
    92 Other situations can require a good deal of creativity.  
    94 Although we now have two variables, only @{term"xs"} is suitable for
    95 induction, and we repeat our proof attempt. Unfortunately, we are still
    96 not there:
    97 @{subgoals[display,indent=0,goals_limit=1]}
    98 The induction hypothesis is still too weak, but this time it takes no
    99 intuition to generalize: the problem is that @{term"ys"} is fixed throughout
   100 the subgoal, but the induction hypothesis needs to be applied with
   101 @{term"a # ys"} instead of @{term"ys"}. Hence we prove the theorem
   102 for all @{term"ys"} instead of a fixed one:
   103 *}
   104 (*<*)oops(*>*)
   105 lemma "\<forall>ys. itrev xs ys = rev xs @ ys"
   106 (*<*)
   107 by(induct_tac xs, simp_all)
   108 (*>*)
   110 text{*\noindent
   111 This time induction on @{term"xs"} followed by simplification succeeds. This
   112 leads to another heuristic for generalization:
   113 \begin{quote}
   114 \emph{Generalize goals for induction by universally quantifying all free
   115 variables {\em(except the induction variable itself!)}.}
   116 \end{quote}
   117 This prevents trivial failures like the one above and does not affect the
   118 validity of the goal.  However, this heuristic should not be applied blindly.
   119 It is not always required, and the additional quantifiers can complicate
   120 matters in some cases. The variables that should be quantified are typically
   121 those that change in recursive calls.
   123 A final point worth mentioning is the orientation of the equation we just
   124 proved: the more complex notion (@{const itrev}) is on the left-hand
   125 side, the simpler one (@{term rev}) on the right-hand side. This constitutes
   126 another, albeit weak heuristic that is not restricted to induction:
   127 \begin{quote}
   128   \emph{The right-hand side of an equation should (in some sense) be simpler
   129     than the left-hand side.}
   130 \end{quote}
   131 This heuristic is tricky to apply because it is not obvious that
   132 @{term"rev xs @ ys"} is simpler than @{term"itrev xs ys"}. But see what
   133 happens if you try to prove @{prop"rev xs @ ys = itrev xs ys"}!
   135 If you have tried these heuristics and still find your
   136 induction does not go through, and no obvious lemma suggests itself, you may
   137 need to generalize your proposition even further. This requires insight into
   138 the problem at hand and is beyond simple rules of thumb.  
   139 Additionally, you can read \S\ref{sec:advanced-ind}
   140 to learn about some advanced techniques for inductive proofs.%
   141 \index{induction heuristics|)}
   142 *}
   143 (*<*)
   144 declare [[names_unique = true]]
   145 end
   146 (*>*)