author nipkow Fri, 18 Aug 2000 11:14:23 +0200 changeset 9645 20ae97cd2a16 parent 9644 6b0b6b471855 child 9646 aa3b82085e07
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+++ b/doc-src/TutorialI/Misc/AdvancedInd.thy	Fri Aug 18 11:14:23 2000 +0200
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+(*<*)
+(*>*)
+
+text{*\noindent
+Now that we have learned about rules and logic, we take another look at the
+finer points of induction. The two questions we answer are: what to do if the
+proposition to be proved is not directly amenable to induction, and how to
+utilize and even derive new induction schemas.
+*}
+
+subsection{*Massaging the proposition\label{sec:ind-var-in-prems}*}
+
+text{*
+\noindent
+So far we have assumed that the theorem we want to prove is already in a form
+that is amenable to induction, but this is not always the case:
+*}
+
+lemma "xs \\<noteq> [] \\<Longrightarrow> hd(rev xs) = last xs";
+apply(induct_tac xs);
+
+txt{*\noindent
+(where \isa{hd} and \isa{last} return the first and last element of a
+non-empty list)
+produces the warning
+\begin{quote}\tt
+Induction variable occurs also among premises!
+\end{quote}
+and leads to the base case
+\begin{isabellepar}%
+\ 1.\ xs\ {\isasymnoteq}\ []\ {\isasymLongrightarrow}\ hd\ (rev\ [])\ =\ last\ []
+\end{isabellepar}%
+which, after simplification, becomes
+\begin{isabellepar}%
+\ 1.\ xs\ {\isasymnoteq}\ []\ {\isasymLongrightarrow}\ hd\ []\ =\ last\ []
+\end{isabellepar}%
+We cannot prove this equality because we do not know what \isa{hd} and
+\isa{last} return when applied to \isa{[]}.
+
+The point is that we have violated the above warning. Because the induction
+formula is only the conclusion, the occurrence of \isa{xs} in the premises is
+not modified by induction. Thus the case that should have been trivial
+becomes unprovable. Fortunately, the solution is easy:
+\begin{quote}
+\emph{Pull all occurrences of the induction variable into the conclusion
+using \isa{\isasymlongrightarrow}.}
+\end{quote}
+This means we should prove
+*}
+(*<*)oops(*>*)
+lemma hd_rev: "xs \\<noteq> [] \\<longrightarrow> hd(rev xs) = last xs";
+(*<*)
+by(induct_tac xs, auto)
+(*>*)
+
+text{*\noindent
+This time, induction leaves us with the following base case
+\begin{isabellepar}%
+\ 1.\ []\ {\isasymnoteq}\ []\ {\isasymlongrightarrow}\ hd\ (rev\ [])\ =\ last\ []
+\end{isabellepar}%
+which is trivial, and \isa{auto} finishes the whole proof.
+
+If \isa{hd\_rev} is meant to be simplification rule, you are done. But if you
+really need the \isa{\isasymLongrightarrow}-version of \isa{hd\_rev}, for
+example because you want to apply it as an introduction rule, you need to
+derive it separately, by combining it with modus ponens:
+*}
+
+lemmas hd_revI = hd_rev[THEN mp]
+
+text{*\noindent
+which yields the lemma we originally set out to prove.
+
+In case there are multiple premises $A@1$, \dots, $A@n$ containing the
+induction variable, you should turn the conclusion $C$ into
+$A@1 \longrightarrow \cdots A@n \longrightarrow C$
+(see the remark?? in \S\ref{??}).
+Additionally, you may also have to universally quantify some other variables,
+which can yield a fairly complex conclusion.
+Here is a simple example (which is proved by \isa{blast}):
+*}
+
+lemma simple: "\\<forall> y. A y \\<longrightarrow> B y \<longrightarrow> B y & A y"
+(*<*)by blast(*>*)
+
+text{*\noindent
+You can get the desired lemma by explicit
+application of modus ponens and \isa{spec}:
+*}
+
+lemmas myrule = simple[THEN spec, THEN mp, THEN mp]
+
+text{*\noindent
+or the wholesale stripping of \isa{\isasymforall} and
+\isa{\isasymlongrightarrow} in the conclusion via \isa{rulify}
+*}
+
+lemmas myrule = simple[rulify]
+
+text{*\noindent
+yielding @{thm"myrule"}.
+You can go one step further and include these derivations already in the
+statement of your original lemma, thus avoiding the intermediate step:
+*}
+
+lemma myrule[rulify]:  "\\<forall> y. A y \\<longrightarrow> B y \<longrightarrow> B y & A y"
+(*<*)
+by blast
+(*>*)
+
+text{*
+\bigskip
+
+A second reason why your proposition may not be amenable to induction is that
+you want to induct on a whole term, rather than an individual variable. In
+general, when inducting on some term $t$ you must rephrase the conclusion as
+$\forall y@1 \dots y@n.~ x = t \longrightarrow C$ where $y@1 \dots y@n$
+are the free variables in $t$ and $x$ is new, and perform induction on $x$
+afterwards. An example appears below.
+
+*}
+
+subsection{*Beyond structural induction*}
+
+text{*
+So far, inductive proofs where by structural induction for
+primitive recursive functions and recursion induction for total recursive
+functions. But sometimes structural induction is awkward and there is no
+recursive function in sight either that could furnish a more appropriate
+induction schema. In such cases some existing standard induction schema can
+be helpful. We show how to apply such induction schemas by an example.
+
+Structural induction on \isa{nat} is
+usually known as mathematical induction''. There is also complete
+induction'', where you must prove $P(n)$ under the assumption that $P(m)$
+holds for all $m<n$. In Isabelle, this is the theorem \isa{less\_induct}:
+\begin{quote}
+@{thm[display]"less_induct"}
+\end{quote}
+Here is an example of its application.
+*}
+
+consts f :: "nat => nat"
+axioms f_ax: "f(f(n)) < f(Suc(n))"
+
+text{*\noindent
+From the above axiom\footnote{In general, the use of axioms is strongly
+discouraged, because of the danger of inconsistencies. The above axiom does
+not introduce an inconsistency because, for example, the identity function
+satisfies it.}
+for \isa{f} it follows that @{term"n <= f n"}, which can
+be proved by induction on @{term"f n"}. Following the recipy outlined
+above, we have to phrase the proposition as follows to allow induction:
+*}
+
+lemma f_incr_lem: "\\<forall>i. k = f i \\<longrightarrow> i \\<le> f i"
+
+txt{*\noindent
+To perform induction on \isa{k} using \isa{less\_induct}, we use the same
+general induction method as for recursion induction (see
+\S\ref{sec:recdef-induction}):
+*}
+
+apply(induct_tac k rule:less_induct)
+(*<*)
+apply(rule allI)
+apply(case_tac i);
+ apply(simp);
+(*>*)
+txt{*\noindent
+which leaves us with the following proof state:
+\begin{isabellepar}%
+\ 1.\ {\isasymAnd}\mbox{n}.\ {\isasymforall}\mbox{m}.\ \mbox{m}\ <\ \mbox{n}\ {\isasymlongrightarrow}\ ({\isasymforall}\mbox{i}.\ \mbox{m}\ =\ f\ \mbox{i}\ {\isasymlongrightarrow}\ \mbox{i}\ {\isasymle}\ f\ \mbox{i})\isanewline
+\ \ \ \ \ \ \ {\isasymLongrightarrow}\ {\isasymforall}\mbox{i}.\ \mbox{n}\ =\ f\ \mbox{i}\ {\isasymlongrightarrow}\ \mbox{i}\ {\isasymle}\ f\ \mbox{i}
+\end{isabellepar}%
+After stripping the \isa{\isasymforall i}, the proof continues with a case
+distinction on \isa{i}. The case \isa{i = 0} is trivial and we focus on the
+other case:
+\begin{isabellepar}%
+\ 1.\ {\isasymAnd}\mbox{n}\ \mbox{i}\ \mbox{nat}.\isanewline
+\ \ \ \ \ \ \ {\isasymlbrakk}{\isasymforall}\mbox{m}.\ \mbox{m}\ <\ \mbox{n}\ {\isasymlongrightarrow}\ ({\isasymforall}\mbox{i}.\ \mbox{m}\ =\ f\ \mbox{i}\ {\isasymlongrightarrow}\ \mbox{i}\ {\isasymle}\ f\ \mbox{i});\ \mbox{i}\ =\ Suc\ \mbox{nat}{\isasymrbrakk}\isanewline
+\ \ \ \ \ \ \ {\isasymLongrightarrow}\ \mbox{n}\ =\ f\ \mbox{i}\ {\isasymlongrightarrow}\ \mbox{i}\ {\isasymle}\ f\ \mbox{i}
+\end{isabellepar}%
+*}
+
+by(blast intro!: f_ax Suc_leI intro:le_less_trans);
+
+text{*\noindent
+It is not surprising if you find the last step puzzling.
+The proof goes like this (writing \isa{j} instead of \isa{nat}).
+Since @{term"i = Suc j"} it suffices to show
+@{term"j < f(Suc j)"} (by \isa{Suc\_leI}: @{thm"Suc_leI"}). This is
+proved as follows. From \isa{f\_ax} we have @{term"f (f j) < f (Suc j)"}
+(1) which implies @{term"f j <= f (f j)"} (by the induction hypothesis).
+Using (1) once more we obtain @{term"f j < f(Suc j)"} (2) by transitivity
+(\isa{le_less_trans}: @{thm"le_less_trans"}).
+Using the induction hypothesis once more we obtain @{term"j <= f j"}
+which, together with (2) yields @{term"j < f (Suc j)"} (again by
+\isa{le_less_trans}).
+
+This last step shows both the power and the danger of automatic proofs: they
+will usually not tell you how the proof goes, because it can be very hard to
+translate the internal proof into a human-readable format. Therefore
+\S\ref{sec:part2?} introduces a language for writing readable yet concise
+proofs.
+
+We can now derive the desired @{term"i <= f i"} from \isa{f\_incr}:
+*}
+
+lemmas f_incr = f_incr_lem[rulify, OF refl];
+
+text{*
+The final \isa{refl} gets rid of the premise \isa{?k = f ?i}. Again, we could
+have included this derivation in the original statement of the lemma:
+*}
+
+lemma f_incr[rulify, OF refl]: "\\<forall>i. k = f i \\<longrightarrow> i \\<le> f i"
+(*<*)oops(*>*)
+
+text{*
+\begin{exercise}
+From the above axiom and lemma for \isa{f} show that \isa{f} is the identity.
+\end{exercise}
+
+In general, \isa{induct\_tac} can be applied with any rule \isa{r}
+whose conclusion is of the form \isa{?P ?x1 \dots ?xn}, in which case the
+format is
+\begin{ttbox}
+apply(induct_tac y1 ... yn rule: r)
+\end{ttbox}\index{*induct_tac}%
+where \isa{y1}, \dots, \isa{yn} are variables in the first subgoal.
+In fact, \isa{induct\_tac} even allows the conclusion of
+\isa{r} to be an (iterated) conjunction of formulae of the above form, in
+which case the application is
+\begin{ttbox}
+apply(induct_tac y1 ... yn and ... and z1 ... zm rule: r)
+\end{ttbox}
+
+Finally we should mention that HOL already provides the mother of all
+inductions, \emph{wellfounded induction} (\isa{wf\_induct}):
+\begin{quote}
+@{thm[display]"wf_induct"}
+\end{quote}
+For details see the library.
+*}
+
+(*<*)
+end
+(*>*)
--- /dev/null	Thu Jan 01 00:00:00 1970 +0000
+++ b/doc-src/TutorialI/Recdef/Nested0.thy	Fri Aug 18 11:14:23 2000 +0200
@@ -0,0 +1,25 @@
+(*<*)
+theory Nested0 = Main:
+(*>*)
+
+text{*
+In \S\ref{sec:nested-datatype} we defined the datatype of terms
+*}
+
+datatype ('a,'b)"term" = Var 'a | App 'b "('a,'b)term list"
+
+text{*\noindent
+and closed with the observation that the associated schema for the definition
+of primitive recursive functions leads to overly verbose definitions. Moreover,
+if you have worked exercise~\ref{ex:trev-trev} you will have noticed that
+you needed to reprove many lemmas reminiscent of similar lemmas about
+@{term"rev"}. We will now show you how \isacommand{recdef} can simplify
+definitions and proofs about nested recursive datatypes. As an example we
+chose exercise~\ref{ex:trev-trev}:
+
+FIXME: declare trev now!
+*}
+(* consts trev  :: "('a,'b)term => ('a,'b)term" *)
+(*<*)
+end
+(*>*)
--- /dev/null	Thu Jan 01 00:00:00 1970 +0000
+++ b/doc-src/TutorialI/Recdef/Nested1.thy	Fri Aug 18 11:14:23 2000 +0200
@@ -0,0 +1,42 @@
+(*<*)
+theory Nested1 = Nested0:
+(*>*)
+consts trev  :: "('a,'b)term => ('a,'b)term"
+
+text{*\noindent
+Although the definition of @{term"trev"} is quite natural, we will have
+overcome a minor difficulty in convincing Isabelle of is termination.
+It is precisely this difficulty that is the \textit{rasion d'\^etre} of
+this subsection.
+
+Defining @{term"trev"} by \isacommand{recdef} rather than \isacommand{primrec}
+simplifies matters because we are now free to use the recursion equation
+suggested at the end of \S\ref{sec:nested-datatype}:
+*}
+ML"Prim.print_recdef_congs(Context.the_context())";
+
+recdef trev "measure size"
+ "trev (Var x) = Var x"
+ "trev (App f ts) = App f (rev(map trev ts))";
+ML"Prim.congs []";
+congs map_cong;
+thm trev.simps;
+(*<*)ML"Pretty.setmargin 60"(*>*)
+text{*
+FIXME: recdef should complain and generate unprovable termination condition!
+
+The point is that the above termination condition is unprovable because it
+ignores some crucial information: the recursive call of @{term"trev"} will
+not involve arbitrary terms but only those in @{term"ts"}. This is expressed
+by theorem \isa{map\_cong}:
+\begin{quote}
+@{thm[display]"map_cong"}
+\end{quote}
+*}
+
+(*<*)
+end
+(*>*)
+
+