2 theory Induction = examples + simplification:;
6 Assuming we have defined our function such that Isabelle could prove
7 termination and that the recursion equations (or some suitable derived
8 equations) are simplification rules, we might like to prove something about
9 our function. Since the function is recursive, the natural proof principle is
10 again induction. But this time the structural form of induction that comes
11 with datatypes is unlikely to work well---otherwise we could have defined the
12 function by \isacommand{primrec}. Therefore \isacommand{recdef} automatically
13 proves a suitable induction rule $f$\isa{.induct} that follows the
14 recursion pattern of the particular function $f$. We call this
15 \textbf{recursion induction}. Roughly speaking, it
16 requires you to prove for each \isacommand{recdef} equation that the property
17 you are trying to establish holds for the left-hand side provided it holds
18 for all recursive calls on the right-hand side. Here is a simple example
21 lemma "map f (sep(x,xs)) = sep(f x, map f xs)";
24 involving the predefined \isa{map} functional on lists: \isa{map f xs}
25 is the result of applying \isa{f} to all elements of \isa{xs}. We prove
26 this lemma by recursion induction w.r.t. \isa{sep}:
29 apply(induct_tac x xs rule: sep.induct);
32 The resulting proof state has three subgoals corresponding to the three
33 clauses for \isa{sep}:
35 ~1.~{\isasymAnd}a.~map~f~(sep~(a,~[]))~=~sep~(f~a,~map~f~[])\isanewline
36 ~2.~{\isasymAnd}a~x.~map~f~(sep~(a,~[x]))~=~sep~(f~a,~map~f~[x])\isanewline
37 ~3.~{\isasymAnd}a~x~y~zs.\isanewline
38 ~~~~~~~map~f~(sep~(a,~y~\#~zs))~=~sep~(f~a,~map~f~(y~\#~zs))~{\isasymLongrightarrow}\isanewline
39 ~~~~~~~map~f~(sep~(a,~x~\#~y~\#~zs))~=~sep~(f~a,~map~f~(x~\#~y~\#~zs))%
41 The rest is pure simplification:
47 Try proving the above lemma by structural induction, and you find that you
48 need an additional case distinction. What is worse, the names of variables
49 are invented by Isabelle and have nothing to do with the names in the
50 definition of \isa{sep}.
52 In general, the format of invoking recursion induction is
54 apply(induct_tac \(x@1 \dots x@n\) rule: \(f\).induct)
55 \end{ttbox}\index{*induct_tac}%
56 where $x@1~\dots~x@n$ is a list of free variables in the subgoal and $f$ the
57 name of a function that takes an $n$-tuple. Usually the subgoal will
58 contain the term $f~x@1~\dots~x@n$ but this need not be the case. The
59 induction rules do not mention $f$ at all. For example \isa{sep.induct}
61 {\isasymlbrakk}~{\isasymAnd}a.~P~a~[];\isanewline
62 ~~{\isasymAnd}a~x.~P~a~[x];\isanewline
63 ~~{\isasymAnd}a~x~y~zs.~P~a~(y~\#~zs)~{\isasymLongrightarrow}~P~a~(x~\#~y~\#~zs){\isasymrbrakk}\isanewline
64 {\isasymLongrightarrow}~P~u~v%
66 merely says that in order to prove a property \isa{P} of \isa{u} and
67 \isa{v} you need to prove it for the three cases where \isa{v} is the
68 empty list, the singleton list, and the list with at least two elements
69 (in which case you may assume it holds for the tail of that list).