doc-src/TutorialI/Inductive/Mutual.thy
author nipkow
Tue, 02 Jan 2001 12:04:33 +0100
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(*<*)theory Mutual = Main:(*>*)
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subsection{*Mutual inductive definitions*}
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text{*
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Just as there are datatypes defined by mutual recursion, there are sets defined
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by mutual induction. As a trivial example we consider the even and odd natural numbers:
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*}
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consts even :: "nat set"
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       odd  :: "nat set"
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inductive even odd
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intros
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zero:  "0 \<in> even"
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evenI: "n \<in> odd \<Longrightarrow> Suc n \<in> even"
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oddI:  "n \<in> even \<Longrightarrow> Suc n \<in> odd"
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text{*\noindent
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The simultaneous inductive definition of multiple sets is no different from that
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of a single set, except for induction: just as for mutually recursive datatypes,
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induction needs to involve all the simultaneously defined sets. In the above case,
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the induction rule is called @{thm[source]even_odd.induct} (simply concenate the names
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of the sets involved) and has the conclusion
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@{text[display]"(?x \<in> even \<longrightarrow> ?P ?x) \<and> (?y \<in> odd \<longrightarrow> ?Q ?y)"}
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If we want to prove that all even numbers are divisible by 2, we have to generalize
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the statement as follows:
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*}
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lemma "(m \<in> even \<longrightarrow> 2 dvd m) \<and> (n \<in> odd \<longrightarrow> 2 dvd (Suc n))"
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txt{*\noindent
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The proof is by rule induction. Because of the form of the induction theorem, it is
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applied by @{text rule} rather than @{text erule} as for ordinary inductive definitions:
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*}
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apply(rule even_odd.induct)
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txt{*
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@{subgoals[display,indent=0]}
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The first two subgoals are proved by simplification and the final one can be
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proved in the same manner as in \S\ref{sec:rule-induction}
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where the same subgoal was encountered before.
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We do not show the proof script.
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*}
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(*<*)
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  apply simp
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 apply simp
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apply(simp add:dvd_def)
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apply(clarify)
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apply(rule_tac x = "Suc k" in exI)
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apply simp
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done
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(*>*)
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(*
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Exercise: 1 : odd
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*)
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(*<*)end(*>*)