doc-src/TutorialI/Inductive/Star.thy
author nipkow
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rule inversion
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(*<*)theory Star = Main:(*>*)
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section{*The reflexive transitive closure*}
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text{*\label{sec:rtc}
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{\bf Say something about inductive relations as opposed to sets? Or has that
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been said already? If not, explain induction!}
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A perfect example of an inductive definition is the reflexive transitive
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closure of a relation. This concept was already introduced in
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\S\ref{sec:Relations}, where the operator @{text"^*"} was
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defined as a least fixed point because
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inductive definitions were not yet available. But now they are:
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*}
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consts rtc :: "('a \<times> 'a)set \<Rightarrow> ('a \<times> 'a)set"   ("_*" [1000] 999)
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inductive "r*"
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intros
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rtc_refl[iff]:  "(x,x) \<in> r*"
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rtc_step:       "\<lbrakk> (x,y) \<in> r; (y,z) \<in> r* \<rbrakk> \<Longrightarrow> (x,z) \<in> r*"
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text{*\noindent
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The function @{term rtc} is annotated with concrete syntax: instead of
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@{text"rtc r"} we can read and write {term"r*"}. The actual definition
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consists of two rules. Reflexivity is obvious and is immediately declared an
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equivalence rule.  Thus the automatic tools will apply it automatically. The
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second rule, @{thm[source]rtc_step}, says that we can always add one more
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@{term r}-step to the left. Although we could make @{thm[source]rtc_step} an
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introduction rule, this is dangerous: the recursion slows down and may
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even kill the automatic tactics.
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The above definition of the concept of reflexive transitive closure may
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be sufficiently intuitive but it is certainly not the only possible one:
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for a start, it does not even mention transitivity explicitly.
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The rest of this section is devoted to proving that it is equivalent to
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the ``standard'' definition. We start with a simple lemma:
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*}
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lemma [intro]: "(x,y) : r \<Longrightarrow> (x,y) \<in> r*"
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by(blast intro: rtc_step);
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text{*\noindent
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Although the lemma itself is an unremarkable consequence of the basic rules,
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it has the advantage that it can be declared an introduction rule without the
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danger of killing the automatic tactics because @{term"r*"} occurs only in
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the conclusion and not in the premise. Thus some proofs that would otherwise
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need @{thm[source]rtc_step} can now be found automatically. The proof also
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shows that @{term blast} is quite able to handle @{thm[source]rtc_step}. But
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some of the other automatic tactics are more sensitive, and even @{text
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blast} can be lead astray in the presence of large numbers of rules.
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Let us now turn to transitivity. It should be a consequence of the definition.
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*}
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lemma rtc_trans:
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  "\<lbrakk> (x,y) \<in> r*; (y,z) \<in> r* \<rbrakk> \<Longrightarrow> (x,z) \<in> r*"
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txt{*\noindent
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The proof starts canonically by rule induction:
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*}
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apply(erule rtc.induct)
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txt{*\noindent
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However, even the resulting base case is a problem
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@{subgoals[display,indent=0,goals_limit=1]}
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and maybe not what you had expected. We have to abandon this proof attempt.
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To understand what is going on,
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let us look at the induction rule @{thm[source]rtc.induct}:
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\[ \frac{(x,y) \in r^* \qquad \bigwedge x.~P~x~x \quad \dots}{P~x~y} \]
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When applying this rule, $x$ becomes @{term x}, $y$ becomes
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@{term y} and $P~x~y$ becomes @{prop"(x,z) : r*"}, thus
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yielding the above subgoal. So what went wrong?
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When looking at the instantiation of $P~x~y$ we see
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that $P$ does not depend on its second parameter at
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all. The reason is that in our original goal, of the pair @{term"(x,y)"} only
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@{term x} appears also in the conclusion, but not @{term y}. Thus our
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induction statement is too weak. Fortunately, it can easily be strengthened:
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transfer the additional premise @{prop"(y,z):r*"} into the conclusion:*}
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(*<*)oops(*>*)
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lemma rtc_trans[rule_format]:
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  "(x,y) \<in> r* \<Longrightarrow> (y,z) \<in> r* \<longrightarrow> (x,z) \<in> r*"
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txt{*\noindent
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This is not an obscure trick but a generally applicable heuristic:
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\begin{quote}\em
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Whe proving a statement by rule induction on $(x@1,\dots,x@n) \in R$,
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pull all other premises containing any of the $x@i$ into the conclusion
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using $\longrightarrow$.
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\end{quote}
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A similar heuristic for other kinds of inductions is formulated in
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\S\ref{sec:ind-var-in-prems}. The @{text rule_format} directive turns
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@{text"\<longrightarrow>"} back into @{text"\<Longrightarrow>"}. Thus in the end we obtain the original
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statement of our lemma.
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*}
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apply(erule rtc.induct)
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txt{*\noindent
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Now induction produces two subgoals which are both proved automatically:
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@{subgoals[display,indent=0]}
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*}
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 apply(blast);
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apply(blast intro: rtc_step);
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done
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text{*
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Let us now prove that @{term"r*"} is really the reflexive transitive closure
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of @{term r}, i.e.\ the least reflexive and transitive
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relation containing @{term r}. The latter is easily formalized
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*}
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consts rtc2 :: "('a \<times> 'a)set \<Rightarrow> ('a \<times> 'a)set"
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inductive "rtc2 r"
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intros
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"(x,y) \<in> r \<Longrightarrow> (x,y) \<in> rtc2 r"
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"(x,x) \<in> rtc2 r"
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"\<lbrakk> (x,y) \<in> rtc2 r; (y,z) \<in> rtc2 r \<rbrakk> \<Longrightarrow> (x,z) \<in> rtc2 r"
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text{*\noindent
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and the equivalence of the two definitions is easily shown by the obvious rule
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inductions:
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*}
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lemma "(x,y) \<in> rtc2 r \<Longrightarrow> (x,y) \<in> r*"
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apply(erule rtc2.induct);
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  apply(blast);
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 apply(blast);
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apply(blast intro: rtc_trans);
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done
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lemma "(x,y) \<in> r* \<Longrightarrow> (x,y) \<in> rtc2 r"
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apply(erule rtc.induct);
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 apply(blast intro: rtc2.intros);
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apply(blast intro: rtc2.intros);
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done
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text{*
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So why did we start with the first definition? Because it is simpler. It
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contains only two rules, and the single step rule is simpler than
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transitivity.  As a consequence, @{thm[source]rtc.induct} is simpler than
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@{thm[source]rtc2.induct}. Since inductive proofs are hard enough, we should
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certainly pick the simplest induction schema available for any concept.
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Hence @{term rtc} is the definition of choice.
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\begin{exercise}
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Show that the converse of @{thm[source]rtc_step} also holds:
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@{prop[display]"[| (x,y) : r*; (y,z) : r |] ==> (x,z) : r*"}
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\end{exercise}
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*}
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(*<*)
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lemma rtc_step2[rule_format]: "(x,y) : r* \<Longrightarrow> (y,z) : r --> (x,z) : r*"
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apply(erule rtc.induct);
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 apply blast;
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apply(blast intro:rtc_step)
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done
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end
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(*>*)