| author | wenzelm |
| Mon, 17 Dec 2012 14:07:34 +0100 | |
| changeset 50575 | ae1da46022d1 |
| parent 48985 | 5386df44a037 |
| child 56059 | 2390391584c2 |
| permissions | -rw-r--r-- |
(*<*)theory Even imports Main begin ML_file "../../antiquote_setup.ML" setup Antiquote_Setup.setup (*>*) section{* The Set of Even Numbers *} text {* \index{even numbers!defining inductively|(}% The set of even numbers can be inductively defined as the least set containing 0 and closed under the operation $+2$. Obviously, \emph{even} can also be expressed using the divides relation (@{text dvd}). We shall prove below that the two formulations coincide. On the way we shall examine the primary means of reasoning about inductively defined sets: rule induction. *} subsection{* Making an Inductive Definition *} text {* Using \commdx{inductive\protect\_set}, we declare the constant @{text even} to be a set of natural numbers with the desired properties. *} inductive_set even :: "nat set" where zero[intro!]: "0 \<in> even" | step[intro!]: "n \<in> even \<Longrightarrow> (Suc (Suc n)) \<in> 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, then so is~$n+2$. Given this declaration, Isabelle generates a fixed point definition for @{term even} and proves theorems about it, thus following the definitional approach (see {\S}\ref{sec:definitional}). 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. Here are two examples: @{named_thms[display,indent=0] even.zero[no_vars] (even.zero) even.step[no_vars] (even.step)} The introduction rules can be given attributes. Here both rules are specified as \isa{intro!},% \index{intro"!@\isa {intro"!} (attribute)} directing the classical reasoner to apply them aggressively. Obviously, regarding 0 as even is safe. The @{text step} rule is also safe because $n+2$ is even if and only if $n$ is even. We prove this equivalence later. *} subsection{*Using Introduction Rules*} text {* Our first lemma states that numbers of the form $2\times k$ are even. Introduction rules are used to show that specific values belong to the inductive set. Such proofs typically involve induction, perhaps over some other inductive set. *} lemma two_times_even[intro!]: "2*k \<in> even" apply (induct_tac k) apply auto done (*<*) lemma "2*k \<in> even" apply (induct_tac k) (*>*) txt {* \noindent The first step is induction on the natural number @{text k}, which leaves two subgoals: @{subgoals[display,indent=0,margin=65]} Here @{text auto} simplifies both subgoals so that they match the introduction rules, which are then applied automatically. Our ultimate goal is to prove the equivalence between the traditional definition of @{text even} (using the divides relation) and our inductive definition. One direction of this equivalence is immediate by the lemma just proved, whose @{text "intro!"} attribute ensures it is applied automatically. *} (*<*)oops(*>*) lemma dvd_imp_even: "2 dvd n \<Longrightarrow> n \<in> even" by (auto simp add: dvd_def) subsection{* Rule Induction \label{sec:rule-induction} *} text {* \index{rule induction|(}% From the definition of the set @{term even}, Isabelle has generated an induction rule: @{named_thms [display,indent=0,margin=40] even.induct [no_vars] (even.induct)} A property @{term P} holds for every even number provided it holds for~@{text 0} and is closed under the operation \isa{Suc(Suc \(\cdot\))}. Then @{term P} is closed under the introduction rules for @{term even}, which is the least set closed under those rules. This type of inductive argument is called \textbf{rule induction}. Apart from the double application of @{term Suc}, the induction rule above resembles the familiar mathematical induction, which indeed is an instance of rule induction; the natural numbers can be defined inductively to be the least set containing @{text 0} and closed under~@{term Suc}. Induction is the usual way of proving a property of the elements of an inductively defined set. Let us prove that all members of the set @{term even} are multiples of two. *} lemma even_imp_dvd: "n \<in> even \<Longrightarrow> 2 dvd n" txt {* We begin by applying induction. Note that @{text even.induct} has the form of an elimination rule, so we use the method @{text erule}. We get two subgoals: *} apply (erule even.induct) txt {* @{subgoals[display,indent=0]} We unfold the definition of @{text dvd} in both subgoals, proving the first one and simplifying the second: *} apply (simp_all add: dvd_def) txt {* @{subgoals[display,indent=0]} The next command eliminates the existential quantifier from the assumption and replaces @{text n} by @{text "2 * k"}. *} apply clarify txt {* @{subgoals[display,indent=0]} To conclude, we tell Isabelle that the desired value is @{term "Suc k"}. With this hint, the subgoal falls to @{text simp}. *} apply (rule_tac x = "Suc k" in exI, simp) (*<*)done(*>*) text {* Combining the previous two results yields our objective, the equivalence relating @{term even} and @{text dvd}. % %we don't want [iff]: discuss? *} theorem even_iff_dvd: "(n \<in> even) = (2 dvd n)" by (blast intro: dvd_imp_even even_imp_dvd) subsection{* Generalization and Rule Induction \label{sec:gen-rule-induction} *} text {* \index{generalizing for induction}% Before applying induction, we typically must generalize the induction formula. With rule induction, the required generalization can be hard to find and sometimes requires a complete reformulation of the problem. In this example, our first attempt uses the obvious statement of the result. It fails: *} lemma "Suc (Suc n) \<in> even \<Longrightarrow> n \<in> even" apply (erule even.induct) oops (*<*) lemma "Suc (Suc n) \<in> even \<Longrightarrow> n \<in> even" apply (erule even.induct) (*>*) txt {* Rule induction finds no occurrences of @{term "Suc(Suc n)"} in the conclusion, which it therefore leaves unchanged. (Look at @{text even.induct} to see why this happens.) We have these subgoals: @{subgoals[display,indent=0]} The first one is hopeless. Rule induction on a non-variable term discards information, and usually fails. How to deal with such situations in general is described in {\S}\ref{sec:ind-var-in-prems} below. In the current case the solution is easy because we have the necessary inverse, subtraction: *} (*<*)oops(*>*) lemma even_imp_even_minus_2: "n \<in> even \<Longrightarrow> n - 2 \<in> even" apply (erule even.induct) apply auto done (*<*) lemma "n \<in> even \<Longrightarrow> n - 2 \<in> even" apply (erule even.induct) (*>*) txt {* This lemma is trivially inductive. Here are the subgoals: @{subgoals[display,indent=0]} The first is trivial because @{text "0 - 2"} simplifies to @{text 0}, which is even. The second is trivial too: @{term "Suc (Suc n) - 2"} simplifies to @{term n}, matching the assumption.% \index{rule induction|)} %the sequel isn't really about induction \medskip Using our lemma, we can easily prove the result we originally wanted: *} (*<*)oops(*>*) lemma Suc_Suc_even_imp_even: "Suc (Suc n) \<in> even \<Longrightarrow> n \<in> even" by (drule even_imp_even_minus_2, simp) text {* We have just proved the converse of the introduction rule @{text even.step}. This suggests proving the following equivalence. We give it the \attrdx{iff} attribute because of its obvious value for simplification. *} lemma [iff]: "((Suc (Suc n)) \<in> even) = (n \<in> even)" by (blast dest: Suc_Suc_even_imp_even) subsection{* Rule Inversion \label{sec:rule-inversion} *} text {* \index{rule inversion|(}% Case analysis on an inductive definition is called \textbf{rule inversion}. It is frequently used in proofs about operational semantics. It can be highly effective when it is applied automatically. Let us look at how rule inversion is done in Isabelle/HOL\@. Recall that @{term even} is the minimal set closed under these two rules: @{thm [display,indent=0] even.intros [no_vars]} Minimality means that @{term even} contains only the elements that these rules force it to contain. If we are told that @{term a} belongs to @{term even} then there are only two possibilities. Either @{term a} is @{text 0} or else @{term a} has the form @{term "Suc(Suc n)"}, for some suitable @{term n} that belongs to @{term even}. That is the gist of the @{term cases} rule, which Isabelle proves for us when it accepts an inductive definition: @{named_thms [display,indent=0,margin=40] even.cases [no_vars] (even.cases)} This general rule is less useful than instances of it for specific patterns. For example, if @{term a} has the form @{term "Suc(Suc n)"} then the first case becomes irrelevant, while the second case tells us that @{term n} belongs to @{term even}. Isabelle will generate this instance for us: *} inductive_cases Suc_Suc_cases [elim!]: "Suc(Suc n) \<in> even" text {* The \commdx{inductive\protect\_cases} command generates an instance of the @{text cases} rule for the supplied pattern and gives it the supplied name: @{named_thms [display,indent=0] Suc_Suc_cases [no_vars] (Suc_Suc_cases)} Applying this as an elimination rule yields one case where @{text even.cases} would yield two. Rule inversion works well when the conclusions of the introduction rules involve datatype constructors like @{term Suc} and @{text "#"} (list ``cons''); freeness reasoning discards all but one or two cases. In the \isacommand{inductive\_cases} command we supplied an attribute, @{text "elim!"}, \index{elim"!@\isa {elim"!} (attribute)}% indicating that this elimination rule can be applied aggressively. The original @{term cases} rule would loop if used in that manner because the pattern~@{term a} matches everything. The rule @{text Suc_Suc_cases} is equivalent to the following implication: @{term [display,indent=0] "Suc (Suc n) \<in> even \<Longrightarrow> n \<in> even"} Just above we devoted some effort to reaching precisely this result. Yet we could have obtained it by a one-line declaration, dispensing with the lemma @{text even_imp_even_minus_2}. This example also justifies the terminology \textbf{rule inversion}: the new rule inverts the introduction rule @{text even.step}. In general, a rule can be inverted when the set of elements it introduces is disjoint from those of the other introduction rules. For one-off applications of rule inversion, use the \methdx{ind_cases} method. Here is an example: *} (*<*)lemma "Suc(Suc n) \<in> even \<Longrightarrow> P"(*>*) apply (ind_cases "Suc(Suc n) \<in> even") (*<*)oops(*>*) text {* The specified instance of the @{text cases} rule is generated, then applied as an elimination rule. To summarize, every inductive definition produces a @{text cases} rule. The \commdx{inductive\protect\_cases} command stores an instance of the @{text cases} rule for a given pattern. Within a proof, the @{text ind_cases} method applies an instance of the @{text cases} rule. The even numbers example has shown how inductive definitions can be used. Later examples will show that they are actually worth using.% \index{rule inversion|)}% \index{even numbers!defining inductively|)} *} (*<*)end(*>*)