5 section{*Induction Heuristics*}
7 text{*\label{sec:InductionHeuristics}
8 The purpose of this section is to illustrate some simple heuristics for
9 inductive proofs. The first one we have already mentioned in our initial
12 \emph{Theorems about recursive functions are proved by induction.}
14 In case the function has more than one argument
16 \emph{Do induction on argument number $i$ if the function is defined by
17 recursion in argument number $i$.}
19 When we look at the proof of @{term[source]"(xs @ ys) @ zs = xs @ (ys @ zs)"}
20 in \S\ref{sec:intro-proof} we find (a) @{text"@"} is recursive in
21 the first argument, (b) @{term xs} occurs only as the first argument of
22 @{text"@"}, and (c) both @{term ys} and @{term zs} occur at least once as
23 the second argument of @{text"@"}. Hence it is natural to perform induction
26 The key heuristic, and the main point of this section, is to
27 generalize the goal before induction. The reason is simple: if the goal is
28 too specific, the induction hypothesis is too weak to allow the induction
29 step to go through. Let us now illustrate the idea with an example.
31 Function @{term"rev"} has quadratic worst-case running time
32 because it calls function @{text"@"} for each element of the list and
33 @{text"@"} is linear in its first argument. A linear time version of
34 @{term"rev"} reqires an extra argument where the result is accumulated
35 gradually, using only @{text"#"}:
38 consts itrev :: "'a list \<Rightarrow> 'a list \<Rightarrow> 'a list";
41 "itrev (x#xs) ys = itrev xs (x#ys)";
44 The behaviour of @{term"itrev"} is simple: it reverses
45 its first argument by stacking its elements onto the second argument,
46 and returning that second argument when the first one becomes
47 empty. Note that @{term"itrev"} is tail-recursive, i.e.\ it can be
50 Naturally, we would like to show that @{term"itrev"} does indeed reverse
51 its first argument provided the second one is empty:
54 lemma "itrev xs [] = rev xs";
57 There is no choice as to the induction variable, and we immediately simplify:
60 apply(induct_tac xs, simp_all);
63 Unfortunately, this is not a complete success:
64 @{subgoals[display,indent=0,margin=65]}
65 Just as predicted above, the overall goal, and hence the induction
66 hypothesis, is too weak to solve the induction step because of the fixed
67 argument, @{term"[]"}. This suggests a heuristic:
69 \emph{Generalize goals for induction by replacing constants by variables.}
71 Of course one cannot do this na\"{\i}vely: @{term"itrev xs ys = rev xs"} is
72 just not true---the correct generalization is
75 lemma "itrev xs ys = rev xs @ ys";
76 (*<*)apply(induct_tac xs, simp_all)(*>*)
78 If @{term"ys"} is replaced by @{term"[]"}, the right-hand side simplifies to
79 @{term"rev xs"}, just as required.
81 In this particular instance it was easy to guess the right generalization,
82 but in more complex situations a good deal of creativity is needed. This is
83 the main source of complications in inductive proofs.
85 Although we now have two variables, only @{term"xs"} is suitable for
86 induction, and we repeat our above proof attempt. Unfortunately, we are still
88 @{subgoals[display,indent=0,goals_limit=1]}
89 The induction hypothesis is still too weak, but this time it takes no
90 intuition to generalize: the problem is that @{term"ys"} is fixed throughout
91 the subgoal, but the induction hypothesis needs to be applied with
92 @{term"a # ys"} instead of @{term"ys"}. Hence we prove the theorem
93 for all @{term"ys"} instead of a fixed one:
96 lemma "\<forall>ys. itrev xs ys = rev xs @ ys";
98 by(induct_tac xs, simp_all);
102 This time induction on @{term"xs"} followed by simplification succeeds. This
103 leads to another heuristic for generalization:
105 \emph{Generalize goals for induction by universally quantifying all free
106 variables {\em(except the induction variable itself!)}.}
108 This prevents trivial failures like the above and does not change the
109 provability of the goal. Because it is not always required, and may even
110 complicate matters in some cases, this heuristic is often not
112 The variables that require generalization are typically those that
113 change in recursive calls.
115 In general, if you have tried the above heuristics and still find your
116 induction does not go through, and no obvious lemma suggests itself, you may
117 need to generalize your proposition even further. This requires insight into
118 the problem at hand and is beyond simple rules of thumb. You
119 will need to be creative. Additionally, you can read \S\ref{sec:advanced-ind}
120 to learn about some advanced techniques for inductive proofs.
122 A final point worth mentioning is the orientation of the equation we just
123 proved: the more complex notion (@{term itrev}) is on the left-hand
124 side, the simpler one (@{term rev}) on the right-hand side. This constitutes
125 another, albeit weak heuristic that is not restricted to induction:
127 \emph{The right-hand side of an equation should (in some sense) be simpler
128 than the left-hand side.}
130 This heuristic is tricky to apply because it is not obvious that
131 @{term"rev xs @ ys"} is simpler than @{term"itrev xs ys"}. But see what
132 happens if you try to prove @{prop"rev xs @ ys = itrev xs ys"}!