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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
16 \emph{Theorems about recursive functions are proved by induction.}
18 In case the function has more than one argument
20 \emph{Do induction on argument number $i$ if the function is defined by
21 recursion in argument number $i$.}
23 When we look at the proof of @{text"(xs@ys) @ zs = xs @ (ys@zs)"}
24 in \S\ref{sec:intro-proof} we find
26 \item @{text"@"} is recursive in
28 \item @{term xs} occurs only as the first argument of
30 \item both @{term ys} and @{term zs} occur at least once as
31 the second argument of @{text"@"}
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"#"}:
48 primrec itrev :: "'a list \<Rightarrow> 'a list \<Rightarrow> 'a list" where
50 "itrev (x#xs) ys = itrev xs (x#ys)"
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
59 Naturally, we would like to show that @{term"itrev"} does indeed reverse
60 its first argument provided the second one is empty:
63 lemma "itrev xs [] = rev xs";
66 There is no choice as to the induction variable, and we immediately simplify:
69 apply(induct_tac xs, simp_all);
72 Unfortunately, this attempt does not prove
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.}
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
85 lemma "itrev xs ys = rev xs @ ys";
86 (*<*)apply(induct_tac xs, simp_all)(*>*)
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
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:
105 lemma "\<forall>ys. itrev xs ys = rev xs @ ys";
107 by(induct_tac xs, simp_all);
111 This time induction on @{term"xs"} followed by simplification succeeds. This
112 leads to another heuristic for generalization:
114 \emph{Generalize goals for induction by universally quantifying all free
115 variables {\em(except the induction variable itself!)}.}
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:
128 \emph{The right-hand side of an equation should (in some sense) be simpler
129 than the left-hand side.}
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|)}
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