
diff -r 22fa8b16c3ae -r 13b32661dde4 doc-src/TutorialI/Recdef/Induction.thy
--- /dev/null	Thu Jan 01 00:00:00 1970 +0000
+++ b/doc-src/TutorialI/Recdef/Induction.thy	Wed Apr 19 11:56:31 2000 +0200
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+(*<*)
+theory Induction = examples + simplification:;
+(*>*)
+
+text{*
+Assuming we have defined our function such that Isabelle could prove
+termination and that the recursion equations (or some suitable derived
+equations) are simplification rules, we might like to prove something about
+our function. Since the function is recursive, the natural proof principle is
+again induction. But this time the structural form of induction that comes
+with datatypes is unlikely to work well---otherwise we could have defined the
+function by \isacommand{primrec}. Therefore \isacommand{recdef} automatically
+proves a suitable induction rule $f$\isa{.induct} that follows the
+recursion pattern of the particular function $f$. We call this
+\textbf{recursion induction}. Roughly speaking, it
+requires you to prove for each \isacommand{recdef} equation that the property
+you are trying to establish holds for the left-hand side provided it holds
+for all recursive calls on the right-hand side. Here is a simple example:
+*}
+
+lemma "map f (sep(x,xs)) = sep(f x, map f xs)";
+
+txt{*\noindent
+involving the predefined \isa{map} functional on lists: \isa{map f xs}
+is the result of applying \isa{f} to all elements of \isa{xs}. We prove
+this lemma by recursion induction w.r.t. \isa{sep}:
+*}
+
+apply(induct_tac x xs rule: sep.induct);
+
+txt{*\noindent
+The resulting proof state has three subgoals corresponding to the three
+clauses for \isa{sep}:
+\begin{isabellepar}%
+~1.~{\isasymAnd}a.~map~f~(sep~(a,~[]))~=~sep~(f~a,~map~f~[])\isanewline
+~2.~{\isasymAnd}a~x.~map~f~(sep~(a,~[x]))~=~sep~(f~a,~map~f~[x])\isanewline
+~3.~{\isasymAnd}a~x~y~zs.\isanewline
+~~~~~~~map~f~(sep~(a,~y~\#~zs))~=~sep~(f~a,~map~f~(y~\#~zs))~{\isasymLongrightarrow}\isanewline
+~~~~~~~map~f~(sep~(a,~x~\#~y~\#~zs))~=~sep~(f~a,~map~f~(x~\#~y~\#~zs))%
+\end{isabellepar}%
+The rest is pure simplification:
+*}
+
+apply auto.;
+
+text{*
+Try proving the above lemma by structural induction, and you find that you
+need an additional case distinction. What is worse, the names of variables
+are invented by Isabelle and have nothing to do with the names in the
+definition of \isa{sep}.
+
+In general, the format of invoking recursion induction is
+\begin{ttbox}
+apply(induct_tac \(x@1 \dots x@n\) rule: \(f\).induct)
+\end{ttbox}\index{*induct_tac}%
+where $x@1~\dots~x@n$ is a list of free variables in the subgoal and $f$ the
+name of a function that takes an $n$-tuple. Usually the subgoal will
+contain the term $f~x@1~\dots~x@n$ but this need not be the case. The
+induction rules do not mention $f$ at all. For example \isa{sep.induct}
+\begin{isabellepar}%
+{\isasymlbrakk}~{\isasymAnd}a.~?P~a~[];\isanewline
+~~{\isasymAnd}a~x.~?P~a~[x];\isanewline
+~~{\isasymAnd}a~x~y~zs.~?P~a~(y~\#~zs)~{\isasymLongrightarrow}~?P~a~(x~\#~y~\#~zs){\isasymrbrakk}\isanewline
+{\isasymLongrightarrow}~?P~?u~?v%
+\end{isabellepar}%
+merely says that in order to prove a property \isa{?P} of \isa{?u} and
+\isa{?v} you need to prove it for the three cases where \isa{?v} is the
+empty list, the singleton list, and the list with at least two elements
+(in which case you may assume it holds for the tail of that list).
+*}
+
+(*<*)
+end
+(*>*)




