# HG changeset patch # User paulson # Date 972377164 -7200 # Node ID 5b36035e4dff598b2ff51e2d13bea9afeaa55b7c # Parent 51e830bb7abeeee8b6088ca953181c381758073d even numbers example diff -r 51e830bb7abe -r 5b36035e4dff doc-src/TutorialI/Inductive/Even.thy --- /dev/null Thu Jan 01 00:00:00 1970 +0000 +++ b/doc-src/TutorialI/Inductive/Even.thy Tue Oct 24 10:46:04 2000 +0200 @@ -0,0 +1,130 @@ +theory Even = Main: + +text{*We begin with a simple example: the set of even numbers. Obviously this +concept can be expressed already using the divides relation (dvd). We shall +prove below that the two formulations coincide. + +First, we declare the constant \isa{even} to be a set of natural numbers. +Then, an inductive declaration gives it the desired properties. +*} + + +consts even :: "nat set" +inductive even +intros +zero[intro!]: "0 \ even" +step[intro!]: "n \ even \ (Suc (Suc n)) \ even" + +text{*An inductive definition consists of introduction rules. The first one +above states that 0 is even; the second states that if $n$ is even, so is +$n+2$. Given this declaration, Isabelle generates a fixed point definition +for \isa{even} and proves many theorems about it. These theorems include the +introduction rules specified in the declaration, an elimination rule for case +analysis and an induction rule. We can refer to these theorems by +automatically-generated names: + +@{thm[display] even.step} +\rulename{even.step} + +@{thm[display] even.induct} +\rulename{even.induct} + +Attributes can be given to the introduction rules. Here both rules are +specified as \isa{intro!}, which will cause them to be applied aggressively. +Obviously, regarding 0 as even is always safe. The second rule is also safe +because $n+2$ is even if and only if $n$ is even. We prove this equivalence +later.*} + +text{*Our first lemma states that numbers of the form $2\times k$ are even. +Introduction rules are used to show that given values belong to the inductive +set. Often, as here, the proof involves some other sort of induction.*} +lemma two_times_even[intro!]: "#2*k \ even" +apply (induct "k") + apply auto +done + +text{* The first step is induction on the natural number \isa{k}, which leaves +two subgoals: + +pr(latex xsymbols symbols); +proof\ {\isacharparenleft}prove{\isacharparenright}{\isacharcolon}\ step\ {\isadigit{1}}\isanewline +\isanewline +goal\ {\isacharparenleft}lemma\ two{\isacharunderscore}times{\isacharunderscore}even{\isacharparenright}{\isacharcolon}\isanewline +{\isacharhash}{\isadigit{2}}\ {\isacharasterisk}\ k\ {\isasymin}\ even\isanewline +\ {\isadigit{1}}{\isachardot}\ {\isacharhash}{\isadigit{2}}\ {\isacharasterisk}\ {\isadigit{0}}\ {\isasymin}\ even\isanewline +\ {\isadigit{2}}{\isachardot}\ {\isasymAnd}n{\isachardot}\ {\isacharhash}{\isadigit{2}}\ {\isacharasterisk}\ n\ {\isasymin}\ even\ {\isasymLongrightarrow}\ {\isacharhash}{\isadigit{2}}\ {\isacharasterisk}\ Suc\ n\ {\isasymin}\ even + +Here \isa{auto} simplifies both subgoals so that they match the introduction +rules, which then are applied automatically. *} + +text{*Our goal is to prove the equivalence between the traditional definition +of even (using the divides relation) and our inductive definition. Half of +this equivalence is trivial using the lemma just proved, whose \isa{intro!} +attribute ensures it will be applied automatically. *} + +lemma dvd_imp_even: "#2 dvd n \ n \ even" +apply (auto simp add: dvd_def) +done + +text{*our first rule induction!*} +lemma even_imp_dvd: "n \ even \ #2 dvd n" +apply (erule even.induct) + apply simp +apply (simp add: dvd_def) +apply clarify +apply (rule_tac x = "Suc k" in exI) +apply simp +done +text{* +proof\ {\isacharparenleft}prove{\isacharparenright}{\isacharcolon}\ step\ {\isadigit{1}}\isanewline +\isanewline +goal\ {\isacharparenleft}lemma\ even{\isacharunderscore}imp{\isacharunderscore}dvd{\isacharparenright}{\isacharcolon}\isanewline +n\ {\isasymin}\ even\ {\isasymLongrightarrow}\ {\isacharhash}{\isadigit{2}}\ dvd\ n\isanewline +\ {\isadigit{1}}{\isachardot}\ {\isacharhash}{\isadigit{2}}\ dvd\ {\isadigit{0}}\isanewline +\ {\isadigit{2}}{\isachardot}\ {\isasymAnd}n{\isachardot}\ {\isasymlbrakk}n\ {\isasymin}\ even{\isacharsemicolon}\ {\isacharhash}{\isadigit{2}}\ dvd\ n{\isasymrbrakk}\ {\isasymLongrightarrow}\ {\isacharhash}{\isadigit{2}}\ dvd\ Suc\ {\isacharparenleft}Suc\ n{\isacharparenright} + +proof\ {\isacharparenleft}prove{\isacharparenright}{\isacharcolon}\ step\ {\isadigit{4}}\isanewline +\isanewline +goal\ {\isacharparenleft}lemma\ even{\isacharunderscore}imp{\isacharunderscore}dvd{\isacharparenright}{\isacharcolon}\isanewline +n\ {\isasymin}\ even\ {\isasymLongrightarrow}\ {\isacharhash}{\isadigit{2}}\ dvd\ n\isanewline +\ {\isadigit{1}}{\isachardot}\ {\isasymAnd}n\ k{\isachardot}\ {\isacharhash}{\isadigit{2}}\ {\isacharasterisk}\ k\ {\isasymin}\ even\ {\isasymLongrightarrow}\ {\isasymexists}ka{\isachardot}\ Suc\ {\isacharparenleft}Suc\ {\isacharparenleft}{\isacharhash}{\isadigit{2}}\ {\isacharasterisk}\ k{\isacharparenright}{\isacharparenright}\ {\isacharequal}\ {\isacharhash}{\isadigit{2}}\ {\isacharasterisk}\ ka +*} + + +text{*no iff-attribute because we don't always want to use it*} +theorem even_iff_dvd: "(n \ even) = (#2 dvd n)" +apply (blast intro: dvd_imp_even even_imp_dvd) +done + +text{*this result ISN'T inductive...*} +lemma Suc_Suc_even_imp_even: "Suc (Suc n) \ even \ n \ even" +apply (erule even.induct) +oops +text{* +proof\ {\isacharparenleft}prove{\isacharparenright}{\isacharcolon}\ step\ {\isadigit{1}}\isanewline +\isanewline +goal\ {\isacharparenleft}lemma\ Suc{\isacharunderscore}Suc{\isacharunderscore}even{\isacharunderscore}imp{\isacharunderscore}even{\isacharparenright}{\isacharcolon}\isanewline +Suc\ {\isacharparenleft}Suc\ n{\isacharparenright}\ {\isasymin}\ even\ {\isasymLongrightarrow}\ n\ {\isasymin}\ even\isanewline +\ {\isadigit{1}}{\isachardot}\ n\ {\isasymin}\ even\isanewline +\ {\isadigit{2}}{\isachardot}\ {\isasymAnd}na{\isachardot}\ {\isasymlbrakk}na\ {\isasymin}\ even{\isacharsemicolon}\ n\ {\isasymin}\ even{\isasymrbrakk}\ {\isasymLongrightarrow}\ n\ {\isasymin}\ even +*} + + +text{*...so we need an inductive lemma...*} +lemma even_imp_even_minus_2: "n \ even \ n-#2 \ even" +apply (erule even.induct) +apply auto +done + +text{*...and prove it in a separate step*} +lemma Suc_Suc_even_imp_even: "Suc (Suc n) \ even \ n \ even" +apply (drule even_imp_even_minus_2) +apply simp +done + +lemma Suc_Suc_even_iff [iff]: "((Suc (Suc n)) \ even) = (n \ even)" +apply (blast dest: Suc_Suc_even_imp_even) +done + +end +