%!TEX root = root.tex \chapter{The Rules of the Game} \label{chap:rules} This chapter outlines the concepts and techniques that underlie reasoning in Isabelle. Until now, we have proved everything using only induction and simplification, but any serious verification project requires more elaborate forms of inference. The chapter also introduces the fundamentals of predicate logic. The first examples in this chapter will consist of detailed, low-level proof steps. Later, we shall see how to automate such reasoning using the methods \isa{blast}, \isa{auto} and others. Backward or goal-directed proof is our usual style, but the chapter also introduces forward reasoning, where one theorem is transformed to yield another. \section{Natural Deduction} \index{natural deduction|(}% In Isabelle, proofs are constructed using inference rules. The most familiar inference rule is probably \emph{modus ponens}:% \index{modus ponens@\emph{modus ponens}} \[ \infer{Q}{P\imp Q & P} \] This rule says that from $P\imp Q$ and $P$ we may infer~$Q$. \textbf{Natural deduction} is an attempt to formalize logic in a way that mirrors human reasoning patterns. For each logical symbol (say, $\conj$), there are two kinds of rules: \textbf{introduction} and \textbf{elimination} rules. The introduction rules allow us to infer this symbol (say, to infer conjunctions). The elimination rules allow us to deduce consequences from this symbol. Ideally each rule should mention one symbol only. For predicate logic this can be done, but when users define their own concepts they typically have to refer to other symbols as well. It is best not to be dogmatic. Our system is not based on pure natural deduction, but includes elements from the sequent calculus and free-variable tableaux. Natural deduction generally deserves its name. It is easy to use. Each proof step consists of identifying the outermost symbol of a formula and applying the corresponding rule. It creates new subgoals in an obvious way from parts of the chosen formula. Expanding the definitions of constants can blow up the goal enormously. Deriving natural deduction rules for such constants lets us reason in terms of their key properties, which might otherwise be obscured by the technicalities of its definition. Natural deduction rules also lend themselves to automation. Isabelle's \textbf{classical reasoner} accepts any suitable collection of natural deduction rules and uses them to search for proofs automatically. Isabelle is designed around natural deduction and many of its tools use the terminology of introduction and elimination rules.% \index{natural deduction|)} \section{Introduction Rules} \index{introduction rules|(}% An introduction rule tells us when we can infer a formula containing a specific logical symbol. For example, the conjunction introduction rule says that if we have $P$ and if we have $Q$ then we have $P\conj Q$. In a mathematics text, it is typically shown like this: \[ \infer{P\conj Q}{P & Q} \] The rule introduces the conjunction symbol~($\conj$) in its conclusion. In Isabelle proofs we mainly reason backwards. When we apply this rule, the subgoal already has the form of a conjunction; the proof step makes this conjunction symbol disappear. In Isabelle notation, the rule looks like this: \begin{isabelle} \isasymlbrakk?P;\ ?Q\isasymrbrakk\ \isasymLongrightarrow\ ?P\ \isasymand\ ?Q\rulenamedx{conjI} \end{isabelle} Carefully examine the syntax. The premises appear to the left of the arrow and the conclusion to the right. The premises (if more than one) are grouped using the fat brackets. The question marks indicate \textbf{schematic variables} (also called \textbf{unknowns}):\index{unknowns|bold} they may be replaced by arbitrary formulas. If we use the rule backwards, Isabelle tries to unify the current subgoal with the conclusion of the rule, which has the form \isa{?P\ \isasymand\ ?Q}. (Unification is discussed below, {\S}\ref{sec:unification}.) If successful, it yields new subgoals given by the formulas assigned to \isa{?P} and \isa{?Q}. The following trivial proof illustrates how rules work. It also introduces a style of indentation. If a command adds a new subgoal, then the next command's indentation is increased by one space; if it proves a subgoal, then the indentation is reduced. This provides the reader with hints about the subgoal structure. \begin{isabelle} \isacommand{lemma}\ conj_rule:\ "\isasymlbrakk P;\ Q\isasymrbrakk\ \isasymLongrightarrow\ P\ \isasymand\ (Q\ \isasymand\ P)"\isanewline \isacommand{apply}\ (rule\ conjI)\isanewline \ \isacommand{apply}\ assumption\isanewline \isacommand{apply}\ (rule\ conjI)\isanewline \ \isacommand{apply}\ assumption\isanewline \isacommand{apply}\ assumption \end{isabelle} At the start, Isabelle presents us with the assumptions (\isa{P} and~\isa{Q}) and with the goal to be proved, \isa{P\ \isasymand\ (Q\ \isasymand\ P)}. We are working backwards, so when we apply conjunction introduction, the rule removes the outermost occurrence of the \isa{\isasymand} symbol. To apply a rule to a subgoal, we apply the proof method \isa{rule} --- here with \isa{conjI}, the conjunction introduction rule. \begin{isabelle} %\isasymlbrakk P;\ Q\isasymrbrakk\ \isasymLongrightarrow\ P\ \isasymand\ Q\ %\isasymand\ P\isanewline \ 1.\ \isasymlbrakk P;\ Q\isasymrbrakk\ \isasymLongrightarrow\ P\isanewline \ 2.\ \isasymlbrakk P;\ Q\isasymrbrakk\ \isasymLongrightarrow\ Q\ \isasymand\ P \end{isabelle} Isabelle leaves two new subgoals: the two halves of the original conjunction. The first is simply \isa{P}, which is trivial, since \isa{P} is among the assumptions. We can apply the \methdx{assumption} method, which proves a subgoal by finding a matching assumption. \begin{isabelle} \ 1.\ \isasymlbrakk P;\ Q\isasymrbrakk\ \isasymLongrightarrow\ Q\ \isasymand\ P \end{isabelle} We are left with the subgoal of proving \isa{Q\ \isasymand\ P} from the assumptions \isa{P} and~\isa{Q}. We apply \isa{rule conjI} again. \begin{isabelle} \ 1.\ \isasymlbrakk P;\ Q\isasymrbrakk\ \isasymLongrightarrow\ Q\isanewline \ 2.\ \isasymlbrakk P;\ Q\isasymrbrakk\ \isasymLongrightarrow\ P \end{isabelle} We are left with two new subgoals, \isa{Q} and~\isa{P}, each of which can be proved using the \isa{assumption} method.% \index{introduction rules|)} \section{Elimination Rules} \index{elimination rules|(}% Elimination rules work in the opposite direction from introduction rules. In the case of conjunction, there are two such rules. From $P\conj Q$ we infer $P$. also, from $P\conj Q$ we infer $Q$: \[ \infer{P}{P\conj Q} \qquad \infer{Q}{P\conj Q} \] Now consider disjunction. There are two introduction rules, which resemble inverted forms of the conjunction elimination rules: \[ \infer{P\disj Q}{P} \qquad \infer{P\disj Q}{Q} \] What is the disjunction elimination rule? The situation is rather different from conjunction. From $P\disj Q$ we cannot conclude that $P$ is true and we cannot conclude that $Q$ is true; there are no direct elimination rules of the sort that we have seen for conjunction. Instead, there is an elimination rule that works indirectly. If we are trying to prove something else, say $R$, and we know that $P\disj Q$ holds, then we have to consider two cases. We can assume that $P$ is true and prove $R$ and then assume that $Q$ is true and prove $R$ a second time. Here we see a fundamental concept used in natural deduction: that of the \textbf{assumptions}. We have to prove $R$ twice, under different assumptions. The assumptions are local to these subproofs and are visible nowhere else. In a logic text, the disjunction elimination rule might be shown like this: \[ \infer{R}{P\disj Q & \infer*{R}{[P]} & \infer*{R}{[Q]}} \] The assumptions $[P]$ and $[Q]$ are bracketed to emphasize that they are local to their subproofs. In Isabelle notation, the already-familiar \isa{\isasymLongrightarrow} syntax serves the same purpose: \begin{isabelle} \isasymlbrakk?P\ \isasymor\ ?Q;\ ?P\ \isasymLongrightarrow\ ?R;\ ?Q\ \isasymLongrightarrow\ ?R\isasymrbrakk\ \isasymLongrightarrow\ ?R\rulenamedx{disjE} \end{isabelle} When we use this sort of elimination rule backwards, it produces a case split. (We have seen this before, in proofs by induction.) The following proof illustrates the use of disjunction elimination. \begin{isabelle} \isacommand{lemma}\ disj_swap:\ "P\ \isasymor\ Q\ \isasymLongrightarrow\ Q\ \isasymor\ P"\isanewline \isacommand{apply}\ (erule\ disjE)\isanewline \ \isacommand{apply}\ (rule\ disjI2)\isanewline \ \isacommand{apply}\ assumption\isanewline \isacommand{apply}\ (rule\ disjI1)\isanewline \isacommand{apply}\ assumption \end{isabelle} We assume \isa{P\ \isasymor\ Q} and must prove \isa{Q\ \isasymor\ P}\@. Our first step uses the disjunction elimination rule, \isa{disjE}\@. We invoke it using \methdx{erule}, a method designed to work with elimination rules. It looks for an assumption that matches the rule's first premise. It deletes the matching assumption, regards the first premise as proved and returns subgoals corresponding to the remaining premises. When we apply \isa{erule} to \isa{disjE}, only two subgoals result. This is better than applying it using \isa{rule} to get three subgoals, then proving the first by assumption: the other subgoals would have the redundant assumption \hbox{\isa{P\ \isasymor\ Q}}. Most of the time, \isa{erule} is the best way to use elimination rules, since it replaces an assumption by its subformulas; only rarely does the original assumption remain useful. \begin{isabelle} %P\ \isasymor\ Q\ \isasymLongrightarrow\ Q\ \isasymor\ P\isanewline \ 1.\ P\ \isasymLongrightarrow\ Q\ \isasymor\ P\isanewline \ 2.\ Q\ \isasymLongrightarrow\ Q\ \isasymor\ P \end{isabelle} These are the two subgoals returned by \isa{erule}. The first assumes \isa{P} and the second assumes \isa{Q}. Tackling the first subgoal, we need to show \isa{Q\ \isasymor\ P}\@. The second introduction rule (\isa{disjI2}) can reduce this to \isa{P}, which matches the assumption. So, we apply the \isa{rule} method with \isa{disjI2} \ldots \begin{isabelle} \ 1.\ P\ \isasymLongrightarrow\ P\isanewline \ 2.\ Q\ \isasymLongrightarrow\ Q\ \isasymor\ P \end{isabelle} \ldots and finish off with the \isa{assumption} method. We are left with the other subgoal, which assumes \isa{Q}. \begin{isabelle} \ 1.\ Q\ \isasymLongrightarrow\ Q\ \isasymor\ P \end{isabelle} Its proof is similar, using the introduction rule \isa{disjI1}. The result of this proof is a new inference rule \isa{disj_swap}, which is neither an introduction nor an elimination rule, but which might be useful. We can use it to replace any goal of the form $Q\disj P$ by one of the form $P\disj Q$.% \index{elimination rules|)} \section{Destruction Rules: Some Examples} \index{destruction rules|(}% Now let us examine the analogous proof for conjunction. \begin{isabelle} \isacommand{lemma}\ conj_swap:\ "P\ \isasymand\ Q\ \isasymLongrightarrow\ Q\ \isasymand\ P"\isanewline \isacommand{apply}\ (rule\ conjI)\isanewline \ \isacommand{apply}\ (drule\ conjunct2)\isanewline \ \isacommand{apply}\ assumption\isanewline \isacommand{apply}\ (drule\ conjunct1)\isanewline \isacommand{apply}\ assumption \end{isabelle} Recall that the conjunction elimination rules --- whose Isabelle names are \isa{conjunct1} and \isa{conjunct2} --- simply return the first or second half of a conjunction. Rules of this sort (where the conclusion is a subformula of a premise) are called \textbf{destruction} rules because they take apart and destroy a premise.% \footnote{This Isabelle terminology is not used in standard logic texts, although the distinction between the two forms of elimination rule is well known. Girard \cite[page 74]{girard89},\index{Girard, Jean-Yves|fnote} for example, writes ``The elimination rules [for $\disj$ and $\exists$] are very bad. What is catastrophic about them is the parasitic presence of a formula [$R$] which has no structural link with the formula which is eliminated.'' These Isabelle rules are inspired by the sequent calculus.} The first proof step applies conjunction introduction, leaving two subgoals: \begin{isabelle} %P\ \isasymand\ Q\ \isasymLongrightarrow\ Q\ \isasymand\ P\isanewline \ 1.\ P\ \isasymand\ Q\ \isasymLongrightarrow\ Q\isanewline \ 2.\ P\ \isasymand\ Q\ \isasymLongrightarrow\ P \end{isabelle} To invoke the elimination rule, we apply a new method, \isa{drule}. Think of the \isa{d} as standing for \textbf{destruction} (or \textbf{direct}, if you prefer). Applying the second conjunction rule using \isa{drule} replaces the assumption \isa{P\ \isasymand\ Q} by \isa{Q}. \begin{isabelle} \ 1.\ Q\ \isasymLongrightarrow\ Q\isanewline \ 2.\ P\ \isasymand\ Q\ \isasymLongrightarrow\ P \end{isabelle} The resulting subgoal can be proved by applying \isa{assumption}. The other subgoal is similarly proved, using the \isa{conjunct1} rule and the \isa{assumption} method. Choosing among the methods \isa{rule}, \isa{erule} and \isa{drule} is up to you. Isabelle does not attempt to work out whether a rule is an introduction rule or an elimination rule. The method determines how the rule will be interpreted. Many rules can be used in more than one way. For example, \isa{disj_swap} can be applied to assumptions as well as to goals; it replaces any assumption of the form $P\disj Q$ by a one of the form $Q\disj P$. Destruction rules are simpler in form than indirect rules such as \isa{disjE}, but they can be inconvenient. Each of the conjunction rules discards half of the formula, when usually we want to take both parts of the conjunction as new assumptions. The easiest way to do so is by using an alternative conjunction elimination rule that resembles \isa{disjE}\@. It is seldom, if ever, seen in logic books. In Isabelle syntax it looks like this: \begin{isabelle} \isasymlbrakk?P\ \isasymand\ ?Q;\ \isasymlbrakk?P;\ ?Q\isasymrbrakk\ \isasymLongrightarrow\ ?R\isasymrbrakk\ \isasymLongrightarrow\ ?R\rulenamedx{conjE} \end{isabelle} \index{destruction rules|)} \begin{exercise} Use the rule \isa{conjE} to shorten the proof above. \end{exercise} \section{Implication} \index{implication|(}% At the start of this chapter, we saw the rule \emph{modus ponens}. It is, in fact, a destruction rule. The matching introduction rule looks like this in Isabelle: \begin{isabelle} (?P\ \isasymLongrightarrow\ ?Q)\ \isasymLongrightarrow\ ?P\ \isasymlongrightarrow\ ?Q\rulenamedx{impI} \end{isabelle} And this is \emph{modus ponens}\index{modus ponens@\emph{modus ponens}}: \begin{isabelle} \isasymlbrakk?P\ \isasymlongrightarrow\ ?Q;\ ?P\isasymrbrakk\ \isasymLongrightarrow\ ?Q \rulenamedx{mp} \end{isabelle} Here is a proof using the implication rules. This lemma performs a sort of uncurrying, replacing the two antecedents of a nested implication by a conjunction. The proof illustrates how assumptions work. At each proof step, the subgoals inherit the previous assumptions, perhaps with additions or deletions. Rules such as \isa{impI} and \isa{disjE} add assumptions, while applying \isa{erule} or \isa{drule} deletes the matching assumption. \begin{isabelle} \isacommand{lemma}\ imp_uncurry:\ "P\ \isasymlongrightarrow\ (Q\ \isasymlongrightarrow\ R)\ \isasymLongrightarrow\ P\ \isasymand\ Q\ \isasymlongrightarrow\ R"\isanewline \isacommand{apply}\ (rule\ impI)\isanewline \isacommand{apply}\ (erule\ conjE)\isanewline \isacommand{apply}\ (drule\ mp)\isanewline \ \isacommand{apply}\ assumption\isanewline \isacommand{apply}\ (drule\ mp)\isanewline \ \ \isacommand{apply}\ assumption\isanewline \ \isacommand{apply}\ assumption \end{isabelle} First, we state the lemma and apply implication introduction (\isa{rule impI}), which moves the conjunction to the assumptions. \begin{isabelle} %P\ \isasymlongrightarrow\ Q\ \isasymlongrightarrow\ R\ \isasymLongrightarrow\ P\ %\isasymand\ Q\ \isasymlongrightarrow\ R\isanewline \ 1.\ \isasymlbrakk P\ \isasymlongrightarrow\ Q\ \isasymlongrightarrow\ R;\ P\ \isasymand\ Q\isasymrbrakk\ \isasymLongrightarrow\ R \end{isabelle} Next, we apply conjunction elimination (\isa{erule conjE}), which splits this conjunction into two parts. \begin{isabelle} \ 1.\ \isasymlbrakk P\ \isasymlongrightarrow\ Q\ \isasymlongrightarrow\ R;\ P;\ Q\isasymrbrakk\ \isasymLongrightarrow\ R \end{isabelle} Now, we work on the assumption \isa{P\ \isasymlongrightarrow\ (Q\ \isasymlongrightarrow\ R)}, where the parentheses have been inserted for clarity. The nested implication requires two applications of \textit{modus ponens}: \isa{drule mp}. The first use yields the implication \isa{Q\ \isasymlongrightarrow\ R}, but first we must prove the extra subgoal \isa{P}, which we do by assumption. \begin{isabelle} \ 1.\ \isasymlbrakk P;\ Q\isasymrbrakk\ \isasymLongrightarrow\ P\isanewline \ 2.\ \isasymlbrakk P;\ Q;\ Q\ \isasymlongrightarrow\ R\isasymrbrakk\ \isasymLongrightarrow\ R \end{isabelle} Repeating these steps for \isa{Q\ \isasymlongrightarrow\ R} yields the conclusion we seek, namely~\isa{R}. \begin{isabelle} \ 1.\ \isasymlbrakk P;\ Q;\ Q\ \isasymlongrightarrow\ R\isasymrbrakk\ \isasymLongrightarrow\ R \end{isabelle} The symbols \isa{\isasymLongrightarrow} and \isa{\isasymlongrightarrow} both stand for implication, but they differ in many respects. Isabelle uses \isa{\isasymLongrightarrow} to express inference rules; the symbol is built-in and Isabelle's inference mechanisms treat it specially. On the other hand, \isa{\isasymlongrightarrow} is just one of the many connectives available in higher-order logic. We reason about it using inference rules such as \isa{impI} and \isa{mp}, just as we reason about the other connectives. You will have to use \isa{\isasymlongrightarrow} in any context that requires a formula of higher-order logic. Use \isa{\isasymLongrightarrow} to separate a theorem's preconditions from its conclusion.% \index{implication|)} \medskip \index{by@\isacommand{by} (command)|(}% The \isacommand{by} command is useful for proofs like these that use \isa{assumption} heavily. It executes an \isacommand{apply} command, then tries to prove all remaining subgoals using \isa{assumption}. Since (if successful) it ends the proof, it also replaces the \isacommand{done} symbol. For example, the proof above can be shortened: \begin{isabelle} \isacommand{lemma}\ imp_uncurry:\ "P\ \isasymlongrightarrow\ (Q\ \isasymlongrightarrow\ R)\ \isasymLongrightarrow\ P\ \isasymand\ Q\ \isasymlongrightarrow\ R"\isanewline \isacommand{apply}\ (rule\ impI)\isanewline \isacommand{apply}\ (erule\ conjE)\isanewline \isacommand{apply}\ (drule\ mp)\isanewline \ \isacommand{apply}\ assumption\isanewline \isacommand{by}\ (drule\ mp) \end{isabelle} We could use \isacommand{by} to replace the final \isacommand{apply} and \isacommand{done} in any proof, but typically we use it to eliminate calls to \isa{assumption}. It is also a nice way of expressing a one-line proof.% \index{by@\isacommand{by} (command)|)} \section{Negation} \index{negation|(}% Negation causes surprising complexity in proofs. Its natural deduction rules are straightforward, but additional rules seem necessary in order to handle negated assumptions gracefully. This section also illustrates the \isa{intro} method: a convenient way of applying introduction rules. Negation introduction deduces $\lnot P$ if assuming $P$ leads to a contradiction. Negation elimination deduces any formula in the presence of $\lnot P$ together with~$P$: \begin{isabelle} (?P\ \isasymLongrightarrow\ False)\ \isasymLongrightarrow\ \isasymnot\ ?P% \rulenamedx{notI}\isanewline \isasymlbrakk{\isasymnot}\ ?P;\ ?P\isasymrbrakk\ \isasymLongrightarrow\ ?R% \rulenamedx{notE} \end{isabelle} % Classical logic allows us to assume $\lnot P$ when attempting to prove~$P$: \begin{isabelle} (\isasymnot\ ?P\ \isasymLongrightarrow\ ?P)\ \isasymLongrightarrow\ ?P% \rulenamedx{classical} \end{isabelle} \index{contrapositives|(}% The implications $P\imp Q$ and $\lnot Q\imp\lnot P$ are logically equivalent, and each is called the \textbf{contrapositive} of the other. Four further rules support reasoning about contrapositives. They differ in the placement of the negation symbols: \begin{isabelle} \isasymlbrakk?Q;\ \isasymnot\ ?P\ \isasymLongrightarrow\ \isasymnot\ ?Q\isasymrbrakk\ \isasymLongrightarrow\ ?P% \rulename{contrapos_pp}\isanewline \isasymlbrakk?Q;\ ?P\ \isasymLongrightarrow\ \isasymnot\ ?Q\isasymrbrakk\ \isasymLongrightarrow\ \isasymnot\ ?P% \rulename{contrapos_pn}\isanewline \isasymlbrakk{\isasymnot}\ ?Q;\ \isasymnot\ ?P\ \isasymLongrightarrow\ ?Q\isasymrbrakk\ \isasymLongrightarrow\ ?P% \rulename{contrapos_np}\isanewline \isasymlbrakk{\isasymnot}\ ?Q;\ ?P\ \isasymLongrightarrow\ ?Q\isasymrbrakk\ \isasymLongrightarrow\ \isasymnot\ ?P% \rulename{contrapos_nn} \end{isabelle} % These rules are typically applied using the \isa{erule} method, where their effect is to form a contrapositive from an assumption and the goal's conclusion.% \index{contrapositives|)} The most important of these is \isa{contrapos_np}. It is useful for applying introduction rules to negated assumptions. For instance, the assumption $\lnot(P\imp Q)$ is equivalent to the conclusion $P\imp Q$ and we might want to use conjunction introduction on it. Before we can do so, we must move that assumption so that it becomes the conclusion. The following proof demonstrates this technique: \begin{isabelle} \isacommand{lemma}\ "\isasymlbrakk{\isasymnot}(P{\isasymlongrightarrow}Q);\ \isasymnot(R{\isasymlongrightarrow}Q)\isasymrbrakk\ \isasymLongrightarrow\ R"\isanewline \isacommand{apply}\ (erule_tac\ Q = "R{\isasymlongrightarrow}Q"\ \isakeyword{in}\ contrapos_np)\isanewline \isacommand{apply}\ (intro\ impI)\isanewline \isacommand{by}\ (erule\ notE) \end{isabelle} % There are two negated assumptions and we need to exchange the conclusion with the second one. The method \isa{erule contrapos_np} would select the first assumption, which we do not want. So we specify the desired assumption explicitly using a new method, \isa{erule_tac}. This is the resulting subgoal: \begin{isabelle} \ 1.\ \isasymlbrakk{\isasymnot}\ (P\ \isasymlongrightarrow\ Q);\ \isasymnot\ R\isasymrbrakk\ \isasymLongrightarrow\ R\ \isasymlongrightarrow\ Q% \end{isabelle} The former conclusion, namely \isa{R}, now appears negated among the assumptions, while the negated formula \isa{R\ \isasymlongrightarrow\ Q} becomes the new conclusion. We can now apply introduction rules. We use the \methdx{intro} method, which repeatedly applies the given introduction rules. Here its effect is equivalent to \isa{rule impI}. \begin{isabelle} \ 1.\ \isasymlbrakk{\isasymnot}\ (P\ \isasymlongrightarrow\ Q);\ \isasymnot\ R;\ R\isasymrbrakk\ \isasymLongrightarrow\ Q% \end{isabelle} We can see a contradiction in the form of assumptions \isa{\isasymnot\ R} and~\isa{R}, which suggests using negation elimination. If applied on its own, \isa{notE} will select the first negated assumption, which is useless. Instead, we invoke the rule using the \isa{by} command. Now when Isabelle selects the first assumption, it tries to prove \isa{P\ \isasymlongrightarrow\ Q} and fails; it then backtracks, finds the assumption \isa{\isasymnot~R} and finally proves \isa{R} by assumption. That concludes the proof. \medskip The following example may be skipped on a first reading. It involves a peculiar but important rule, a form of disjunction introduction: \begin{isabelle} (\isasymnot \ ?Q\ \isasymLongrightarrow \ ?P)\ \isasymLongrightarrow \ ?P\ \isasymor \ ?Q% \rulenamedx{disjCI} \end{isabelle} This rule combines the effects of \isa{disjI1} and \isa{disjI2}. Its great advantage is that we can remove the disjunction symbol without deciding which disjunction to prove. This treatment of disjunction is standard in sequent and tableau calculi. \begin{isabelle} \isacommand{lemma}\ "(P\ \isasymor\ Q)\ \isasymand\ R\ \isasymLongrightarrow\ P\ \isasymor\ (Q\ \isasymand\ R)"\isanewline \isacommand{apply}\ (rule\ disjCI)\isanewline \isacommand{apply}\ (elim\ conjE\ disjE)\isanewline \ \isacommand{apply}\ assumption \isanewline \isacommand{by}\ (erule\ contrapos_np,\ rule\ conjI) \end{isabelle} % The first proof step to applies the introduction rules \isa{disjCI}. The resulting subgoal has the negative assumption \hbox{\isa{\isasymnot(Q\ \isasymand\ R)}}. \begin{isabelle} \ 1.\ \isasymlbrakk(P\ \isasymor\ Q)\ \isasymand\ R;\ \isasymnot\ (Q\ \isasymand\ R)\isasymrbrakk\ \isasymLongrightarrow\ P% \end{isabelle} Next we apply the \isa{elim} method, which repeatedly applies elimination rules; here, the elimination rules given in the command. One of the subgoals is trivial (\isa{\isacommand{apply} assumption}), leaving us with one other: \begin{isabelle} \ 1.\ \isasymlbrakk{\isasymnot}\ (Q\ \isasymand\ R);\ R;\ Q\isasymrbrakk\ \isasymLongrightarrow\ P% \end{isabelle} % Now we must move the formula \isa{Q\ \isasymand\ R} to be the conclusion. The combination \begin{isabelle} \ \ \ \ \ (erule\ contrapos_np,\ rule\ conjI) \end{isabelle} is robust: the \isa{conjI} forces the \isa{erule} to select a conjunction. The two subgoals are the ones we would expect from applying conjunction introduction to \isa{Q~\isasymand~R}: \begin{isabelle} \ 1.\ \isasymlbrakk R;\ Q;\ \isasymnot\ P\isasymrbrakk\ \isasymLongrightarrow\ Q\isanewline \ 2.\ \isasymlbrakk R;\ Q;\ \isasymnot\ P\isasymrbrakk\ \isasymLongrightarrow\ R% \end{isabelle} They are proved by assumption, which is implicit in the \isacommand{by} command.% \index{negation|)} \section{Interlude: the Basic Methods for Rules} We have seen examples of many tactics that operate on individual rules. It may be helpful to review how they work given an arbitrary rule such as this: \[ \infer{Q}{P@1 & \ldots & P@n} \] Below, we refer to $P@1$ as the \bfindex{major premise}. This concept applies only to elimination and destruction rules. These rules act upon an instance of their major premise, typically to replace it by subformulas of itself. Suppose that the rule above is called~\isa{R}\@. Here are the basic rule methods, most of which we have already seen: \begin{itemize} \item Method \isa{rule\ R} unifies~$Q$ with the current subgoal, replacing it by $n$ new subgoals: instances of $P@1$, \ldots,~$P@n$. This is backward reasoning and is appropriate for introduction rules. \item Method \isa{erule\ R} unifies~$Q$ with the current subgoal and simultaneously unifies $P@1$ with some assumption. The subgoal is replaced by the $n-1$ new subgoals of proving instances of $P@2$, \ldots,~$P@n$, with the matching assumption deleted. It is appropriate for elimination rules. The method \isa{(rule\ R,\ assumption)} is similar, but it does not delete an assumption. \item Method \isa{drule\ R} unifies $P@1$ with some assumption, which it then deletes. The subgoal is replaced by the $n-1$ new subgoals of proving $P@2$, \ldots,~$P@n$; an $n$th subgoal is like the original one but has an additional assumption: an instance of~$Q$. It is appropriate for destruction rules. \item Method \isa{frule\ R} is like \isa{drule\ R} except that the matching assumption is not deleted. (See {\S}\ref{sec:frule} below.) \end{itemize} Other methods apply a rule while constraining some of its variables. The typical form is \begin{isabelle} \ \ \ \ \ \methdx{rule_tac}\ $v@1$ = $t@1$ \isakeyword{and} \ldots \isakeyword{and} $v@k$ = $t@k$ \isakeyword{in} R \end{isabelle} This method behaves like \isa{rule R}, while instantiating the variables $v@1$, \ldots, $v@k$ as specified. We similarly have \methdx{erule_tac}, \methdx{drule_tac} and \methdx{frule_tac}. These methods also let us specify which subgoal to operate on. By default it is the first subgoal, as with nearly all methods, but we can specify that rule \isa{R} should be applied to subgoal number~$i$: \begin{isabelle} \ \ \ \ \ rule_tac\ [$i$] R \end{isabelle} \section{Unification and Substitution}\label{sec:unification} \index{unification|(}% As we have seen, Isabelle rules involve schematic variables, which begin with a question mark and act as placeholders for terms. \textbf{Unification} --- well known to Prolog programmers --- is the act of making two terms identical, possibly replacing their schematic variables by terms. The simplest case is when the two terms are already the same. Next simplest is \textbf{pattern-matching}, which replaces variables in only one of the terms. The \isa{rule} method typically matches the rule's conclusion against the current subgoal. The \isa{assumption} method matches the current subgoal's conclusion against each of its assumptions. Unification can instantiate variables in both terms; the \isa{rule} method can do this if the goal itself contains schematic variables. Other occurrences of the variables in the rule or proof state are updated at the same time. Schematic variables in goals represent unknown terms. Given a goal such as $\exists x.\,P$, they let us proceed with a proof. They can be filled in later, sometimes in stages and often automatically. \begin{pgnote} If unification fails when you think it should succeed, try setting the Proof General flag \pgmenu{Isabelle} $>$ \pgmenu{Settings} $>$ \pgmenu{Trace Unification}, which makes Isabelle show the cause of unification failures (in Proof General's \pgmenu{Trace} buffer). \end{pgnote} \noindent For example, suppose we are trying to prove this subgoal by assumption: \begin{isabelle} \ 1.\ P\ (a,\ f\ (b,\ g\ (e,\ a),\ b),\ a)\ \isasymLongrightarrow \ P\ (a,\ f\ (b,\ g\ (c,\ a),\ b),\ a) \end{isabelle} The \isa{assumption} method having failed, we try again with the flag set: \begin{isabelle} \isacommand{apply} assumption \end{isabelle} In this trivial case, the output clearly shows that \isa{e} clashes with \isa{c}: \begin{isabelle} Clash: e =/= c \end{isabelle} Isabelle uses \textbf{higher-order} unification, which works in the typed $\lambda$-calculus. The procedure requires search and is potentially undecidable. For our purposes, however, the differences from ordinary unification are straightforward. It handles bound variables correctly, avoiding capture. The two terms \isa{{\isasymlambda}x.\ f(x,z)} and \isa{{\isasymlambda}y.\ f(y,z)} are trivially unifiable because they differ only by a bound variable renaming. The two terms \isa{{\isasymlambda}x.\ ?P} and \isa{{\isasymlambda}x.\ t x} are not unifiable; replacing \isa{?P} by \isa{t x} is forbidden because the free occurrence of~\isa{x} would become bound. Unfortunately, even if \isa{trace_unify_fail} is set, Isabelle displays no information about this type of failure. \begin{warn} Higher-order unification sometimes must invent $\lambda$-terms to replace function variables, which can lead to a combinatorial explosion. However, Isabelle proofs tend to involve easy cases where there are few possibilities for the $\lambda$-term being constructed. In the easiest case, the function variable is applied only to bound variables, as when we try to unify \isa{{\isasymlambda}x\ y.\ f(?h x y)} and \isa{{\isasymlambda}x\ y.\ f(x+y+a)}. The only solution is to replace \isa{?h} by \isa{{\isasymlambda}x\ y.\ x+y+a}. Such cases admit at most one unifier, like ordinary unification. A harder case is unifying \isa{?h a} with~\isa{a+b}; it admits two solutions for \isa{?h}, namely \isa{{\isasymlambda}x.~a+b} and \isa{{\isasymlambda}x.~x+b}. Unifying \isa{?h a} with~\isa{a+a+b} admits four solutions; their number is exponential in the number of occurrences of~\isa{a} in the second term. \end{warn} \subsection{Substitution and the {\tt\slshape subst} Method} \label{sec:subst} \index{substitution|(}% Isabelle also uses function variables to express \textbf{substitution}. A typical substitution rule allows us to replace one term by another if we know that two terms are equal. \[ \infer{P[t/x]}{s=t & P[s/x]} \] The rule uses a notation for substitution: $P[t/x]$ is the result of replacing $x$ by~$t$ in~$P$. The rule only substitutes in the positions designated by~$x$. For example, it can derive symmetry of equality from reflexivity. Using $x=s$ for~$P$ replaces just the first $s$ in $s=s$ by~$t$: \[ \infer{t=s}{s=t & \infer{s=s}{}} \] The Isabelle version of the substitution rule looks like this: \begin{isabelle} \isasymlbrakk?t\ =\ ?s;\ ?P\ ?s\isasymrbrakk\ \isasymLongrightarrow\ ?P\ ?t \rulenamedx{ssubst} \end{isabelle} Crucially, \isa{?P} is a function variable. It can be replaced by a $\lambda$-term with one bound variable, whose occurrences identify the places in which $s$ will be replaced by~$t$. The proof above requires \isa{?P} to be replaced by \isa{{\isasymlambda}x.~x=s}; the second premise will then be \isa{s=s} and the conclusion will be \isa{t=s}. The \isa{simp} method also replaces equals by equals, but the substitution rule gives us more control. Consider this proof: \begin{isabelle} \isacommand{lemma}\ "\isasymlbrakk x\ =\ f\ x;\ odd(f\ x)\isasymrbrakk\ \isasymLongrightarrow\ odd\ x"\isanewline \isacommand{by}\ (erule\ ssubst) \end{isabelle} % The assumption \isa{x\ =\ f\ x}, if used for rewriting, would loop, replacing \isa{x} by \isa{f x} and then by \isa{f(f x)} and so forth. (Here \isa{simp} would see the danger and would re-orient the equality, but in more complicated cases it can be fooled.) When we apply the substitution rule, Isabelle replaces every \isa{x} in the subgoal by \isa{f x} just once. It cannot loop. The resulting subgoal is trivial by assumption, so the \isacommand{by} command proves it implicitly. We are using the \isa{erule} method in a novel way. Hitherto, the conclusion of the rule was just a variable such as~\isa{?R}, but it may be any term. The conclusion is unified with the subgoal just as it would be with the \isa{rule} method. At the same time \isa{erule} looks for an assumption that matches the rule's first premise, as usual. With \isa{ssubst} the effect is to find, use and delete an equality assumption. The \methdx{subst} method performs individual substitutions. In simple cases, it closely resembles a use of the substitution rule. Suppose a proof has reached this point: \begin{isabelle} \ 1.\ \isasymlbrakk P\ x\ y\ z;\ Suc\ x\ <\ y\isasymrbrakk \ \isasymLongrightarrow \ f\ z\ =\ x\ *\ y% \end{isabelle} Now we wish to apply a commutative law: \begin{isabelle} ?m\ *\ ?n\ =\ ?n\ *\ ?m% \rulename{mult.commute} \end{isabelle} Isabelle rejects our first attempt: \begin{isabelle} apply (simp add: mult.commute) \end{isabelle} The simplifier notices the danger of looping and refuses to apply the rule.% \footnote{More precisely, it only applies such a rule if the new term is smaller under a specified ordering; here, \isa{x\ *\ y} is already smaller than \isa{y\ *\ x}.} % The \isa{subst} method applies \isa{mult.commute} exactly once. \begin{isabelle} \isacommand{apply}\ (subst\ mult.commute)\isanewline \ 1.\ \isasymlbrakk P\ x\ y\ z;\ Suc\ x\ <\ y\isasymrbrakk \ \isasymLongrightarrow \ f\ z\ =\ y\ *\ x% \end{isabelle} As we wanted, \isa{x\ *\ y} has become \isa{y\ *\ x}. \medskip This use of the \methdx{subst} method has the same effect as the command \begin{isabelle} \isacommand{apply}\ (rule\ mult.commute [THEN ssubst]) \end{isabelle} The attribute \isa{THEN}, which combines two rules, is described in {\S}\ref{sec:THEN} below. The \methdx{subst} method is more powerful than applying the substitution rule. It can perform substitutions in a subgoal's assumptions. Moreover, if the subgoal contains more than one occurrence of the left-hand side of the equality, the \methdx{subst} method lets us specify which occurrence should be replaced. \subsection{Unification and Its Pitfalls} Higher-order unification can be tricky. Here is an example, which you may want to skip on your first reading: \begin{isabelle} \isacommand{lemma}\ "\isasymlbrakk x\ =\ f\ x;\ triple\ (f\ x)\ (f\ x)\ x\isasymrbrakk\ \isasymLongrightarrow\ triple\ x\ x\ x"\isanewline \isacommand{apply}\ (erule\ ssubst)\isanewline \isacommand{back}\isanewline \isacommand{back}\isanewline \isacommand{back}\isanewline \isacommand{back}\isanewline \isacommand{apply}\ assumption\isanewline \isacommand{done} \end{isabelle} % By default, Isabelle tries to substitute for all the occurrences. Applying \isa{erule\ ssubst} yields this subgoal: \begin{isabelle} \ 1.\ triple\ (f\ x)\ (f\ x)\ x\ \isasymLongrightarrow\ triple\ (f\ x)\ (f\ x)\ (f\ x) \end{isabelle} The substitution should have been done in the first two occurrences of~\isa{x} only. Isabelle has gone too far. The \commdx{back} command allows us to reject this possibility and demand a new one: \begin{isabelle} \ 1.\ triple\ (f\ x)\ (f\ x)\ x\ \isasymLongrightarrow\ triple\ x\ (f\ x)\ (f\ x) \end{isabelle} % Now Isabelle has left the first occurrence of~\isa{x} alone. That is promising but it is not the desired combination. So we use \isacommand{back} again: \begin{isabelle} \ 1.\ triple\ (f\ x)\ (f\ x)\ x\ \isasymLongrightarrow\ triple\ (f\ x)\ x\ (f\ x) \end{isabelle} % This also is wrong, so we use \isacommand{back} again: \begin{isabelle} \ 1.\ triple\ (f\ x)\ (f\ x)\ x\ \isasymLongrightarrow\ triple\ x\ x\ (f\ x) \end{isabelle} % And this one is wrong too. Looking carefully at the series of alternatives, we see a binary countdown with reversed bits: 111, 011, 101, 001. Invoke \isacommand{back} again: \begin{isabelle} \ 1.\ triple\ (f\ x)\ (f\ x)\ x\ \isasymLongrightarrow\ triple\ (f\ x)\ (f\ x)\ x% \end{isabelle} At last, we have the right combination! This goal follows by assumption.% \index{unification|)} \medskip This example shows that unification can do strange things with function variables. We were forced to select the right unifier using the \isacommand{back} command. That is all right during exploration, but \isacommand{back} should never appear in the final version of a proof. You can eliminate the need for \isacommand{back} by giving Isabelle less freedom when you apply a rule. One way to constrain the inference is by joining two methods in a \isacommand{apply} command. Isabelle applies the first method and then the second. If the second method fails then Isabelle automatically backtracks. This process continues until the first method produces an output that the second method can use. We get a one-line proof of our example: \begin{isabelle} \isacommand{lemma}\ "\isasymlbrakk x\ =\ f\ x;\ triple\ (f\ x)\ (f\ x)\ x\isasymrbrakk\ \isasymLongrightarrow\ triple\ x\ x\ x"\isanewline \isacommand{apply}\ (erule\ ssubst,\ assumption)\isanewline \isacommand{done} \end{isabelle} \noindent The \isacommand{by} command works too, since it backtracks when proving subgoals by assumption: \begin{isabelle} \isacommand{lemma}\ "\isasymlbrakk x\ =\ f\ x;\ triple\ (f\ x)\ (f\ x)\ x\isasymrbrakk\ \isasymLongrightarrow\ triple\ x\ x\ x"\isanewline \isacommand{by}\ (erule\ ssubst) \end{isabelle} The most general way to constrain unification is by instantiating variables in the rule. The method \isa{rule_tac} is similar to \isa{rule}, but it makes some of the rule's variables denote specified terms. Also available are {\isa{drule_tac}} and \isa{erule_tac}. Here we need \isa{erule_tac} since above we used \isa{erule}. \begin{isabelle} \isacommand{lemma}\ "\isasymlbrakk x\ =\ f\ x;\ triple\ (f\ x)\ (f\ x)\ x\isasymrbrakk\ \isasymLongrightarrow\ triple\ x\ x\ x"\isanewline \isacommand{by}\ (erule_tac\ P = "\isasymlambda u.\ triple\ u\ u\ x"\ \isakeyword{in}\ ssubst) \end{isabelle} % To specify a desired substitution requires instantiating the variable \isa{?P} with a $\lambda$-expression. The bound variable occurrences in \isa{{\isasymlambda}u.\ P\ u\ u\ x} indicate that the first two arguments have to be substituted, leaving the third unchanged. With this instantiation, backtracking is neither necessary nor possible. An alternative to \isa{rule_tac} is to use \isa{rule} with a theorem modified using~\isa{of}, described in {\S}\ref{sec:forward} below. But \isa{rule_tac}, unlike \isa{of}, can express instantiations that refer to \isasymAnd-bound variables in the current subgoal.% \index{substitution|)} \section{Quantifiers} \index{quantifiers!universal|(}% Quantifiers require formalizing syntactic substitution and the notion of arbitrary value. Consider the universal quantifier. In a logic book, its introduction rule looks like this: \[ \infer{\forall x.\,P}{P} \] Typically, a proviso written in English says that $x$ must not occur in the assumptions. This proviso guarantees that $x$ can be regarded as arbitrary, since it has not been assumed to satisfy any special conditions. Isabelle's underlying formalism, called the \bfindex{meta-logic}, eliminates the need for English. It provides its own universal quantifier (\isasymAnd) to express the notion of an arbitrary value. We have already seen another operator of the meta-logic, namely \isa\isasymLongrightarrow, which expresses inference rules and the treatment of assumptions. The only other operator in the meta-logic is \isa\isasymequiv, which can be used to define constants. \subsection{The Universal Introduction Rule} Returning to the universal quantifier, we find that having a similar quantifier as part of the meta-logic makes the introduction rule trivial to express: \begin{isabelle} (\isasymAnd x.\ ?P\ x)\ \isasymLongrightarrow\ {\isasymforall}x.\ ?P\ x\rulenamedx{allI} \end{isabelle} The following trivial proof demonstrates how the universal introduction rule works. \begin{isabelle} \isacommand{lemma}\ "{\isasymforall}x.\ P\ x\ \isasymlongrightarrow\ P\ x"\isanewline \isacommand{apply}\ (rule\ allI)\isanewline \isacommand{by}\ (rule\ impI) \end{isabelle} The first step invokes the rule by applying the method \isa{rule allI}. \begin{isabelle} \ 1.\ \isasymAnd x.\ P\ x\ \isasymlongrightarrow\ P\ x \end{isabelle} Note that the resulting proof state has a bound variable, namely~\isa{x}. The rule has replaced the universal quantifier of higher-order logic by Isabelle's meta-level quantifier. Our goal is to prove \isa{P\ x\ \isasymlongrightarrow\ P\ x} for arbitrary~\isa{x}; it is an implication, so we apply the corresponding introduction rule (\isa{impI}). \begin{isabelle} \ 1.\ \isasymAnd x.\ P\ x\ \isasymLongrightarrow\ P\ x \end{isabelle} This last subgoal is implicitly proved by assumption. \subsection{The Universal Elimination Rule} Now consider universal elimination. In a logic text, the rule looks like this: \[ \infer{P[t/x]}{\forall x.\,P} \] The conclusion is $P$ with $t$ substituted for the variable~$x$. Isabelle expresses substitution using a function variable: \begin{isabelle} {\isasymforall}x.\ ?P\ x\ \isasymLongrightarrow\ ?P\ ?x\rulenamedx{spec} \end{isabelle} This destruction rule takes a universally quantified formula and removes the quantifier, replacing the bound variable \isa{x} by the schematic variable \isa{?x}. Recall that a schematic variable starts with a question mark and acts as a placeholder: it can be replaced by any term. The universal elimination rule is also available in the standard elimination format. Like \isa{conjE}, it never appears in logic books: \begin{isabelle} \isasymlbrakk \isasymforall x.\ ?P\ x;\ ?P\ ?x\ \isasymLongrightarrow \ ?R\isasymrbrakk \ \isasymLongrightarrow \ ?R% \rulenamedx{allE} \end{isabelle} The methods \isa{drule~spec} and \isa{erule~allE} do precisely the same inference. To see how $\forall$-elimination works, let us derive a rule about reducing the scope of a universal quantifier. In mathematical notation we write \[ \infer{P\imp\forall x.\,Q}{\forall x.\,P\imp Q} \] with the proviso ``$x$ not free in~$P$.'' Isabelle's treatment of substitution makes the proviso unnecessary. The conclusion is expressed as \isa{P\ \isasymlongrightarrow\ ({\isasymforall}x.\ Q\ x)}. No substitution for the variable \isa{P} can introduce a dependence upon~\isa{x}: that would be a bound variable capture. Let us walk through the proof. \begin{isabelle} \isacommand{lemma}\ "(\isasymforall x.\ P\ \isasymlongrightarrow \ Q\ x)\ \isasymLongrightarrow \ P\ \isasymlongrightarrow \ (\isasymforall x.\ Q\ x)" \end{isabelle} First we apply implies introduction (\isa{impI}), which moves the \isa{P} from the conclusion to the assumptions. Then we apply universal introduction (\isa{allI}). \begin{isabelle} \isacommand{apply}\ (rule\ impI,\ rule\ allI)\isanewline \ 1.\ \isasymAnd x.\ \isasymlbrakk{\isasymforall}x.\ P\ \isasymlongrightarrow\ Q\ x;\ P\isasymrbrakk\ \isasymLongrightarrow\ Q\ x \end{isabelle} As before, it replaces the HOL quantifier by a meta-level quantifier, producing a subgoal that binds the variable~\isa{x}. The leading bound variables (here \isa{x}) and the assumptions (here \isa{{\isasymforall}x.\ P\ \isasymlongrightarrow\ Q\ x} and \isa{P}) form the \textbf{context} for the conclusion, here \isa{Q\ x}. Subgoals inherit the context, although assumptions can be added or deleted (as we saw earlier), while rules such as \isa{allI} add bound variables. Now, to reason from the universally quantified assumption, we apply the elimination rule using the \isa{drule} method. This rule is called \isa{spec} because it specializes a universal formula to a particular term. \begin{isabelle} \isacommand{apply}\ (drule\ spec)\isanewline \ 1.\ \isasymAnd x.\ \isasymlbrakk P;\ P\ \isasymlongrightarrow\ Q\ (?x2\ x)\isasymrbrakk\ \isasymLongrightarrow\ Q\ x \end{isabelle} Observe how the context has changed. The quantified formula is gone, replaced by a new assumption derived from its body. We have removed the quantifier and replaced the bound variable by the curious term \isa{?x2~x}. This term is a placeholder: it may become any term that can be built from~\isa{x}. (Formally, \isa{?x2} is an unknown of function type, applied to the argument~\isa{x}.) This new assumption is an implication, so we can use \emph{modus ponens} on it, which concludes the proof. \begin{isabelle} \isacommand{by}\ (drule\ mp) \end{isabelle} Let us take a closer look at this last step. \emph{Modus ponens} yields two subgoals: one where we prove the antecedent (in this case \isa{P}) and one where we may assume the consequent. Both of these subgoals are proved by the \isa{assumption} method, which is implicit in the \isacommand{by} command. Replacing the \isacommand{by} command by \isa{\isacommand{apply} (drule\ mp, assumption)} would have left one last subgoal: \begin{isabelle} \ 1.\ \isasymAnd x.\ \isasymlbrakk P;\ Q\ (?x2\ x)\isasymrbrakk\ \isasymLongrightarrow\ Q\ x \end{isabelle} The consequent is \isa{Q} applied to that placeholder. It may be replaced by any term built from~\isa{x}, and here it should simply be~\isa{x}. The assumption need not be identical to the conclusion, provided the two formulas are unifiable.% \index{quantifiers!universal|)} \subsection{The Existential Quantifier} \index{quantifiers!existential|(}% The concepts just presented also apply to the existential quantifier, whose introduction rule looks like this in Isabelle: \begin{isabelle} ?P\ ?x\ \isasymLongrightarrow\ {\isasymexists}x.\ ?P\ x\rulenamedx{exI} \end{isabelle} If we can exhibit some $x$ such that $P(x)$ is true, then $\exists x. P(x)$ is also true. It is a dual of the universal elimination rule, and logic texts present it using the same notation for substitution. The existential elimination rule looks like this in a logic text: \[ \infer{Q}{\exists x.\,P & \infer*{Q}{[P]}} \] % It looks like this in Isabelle: \begin{isabelle} \isasymlbrakk{\isasymexists}x.\ ?P\ x;\ \isasymAnd x.\ ?P\ x\ \isasymLongrightarrow\ ?Q\isasymrbrakk\ \isasymLongrightarrow\ ?Q\rulenamedx{exE} \end{isabelle} % Given an existentially quantified theorem and some formula $Q$ to prove, it creates a new assumption by removing the quantifier. As with the universal introduction rule, the textbook version imposes a proviso on the quantified variable, which Isabelle expresses using its meta-logic. It is enough to have a universal quantifier in the meta-logic; we do not need an existential quantifier to be built in as well. \begin{exercise} Prove the lemma \[ \exists x.\, P\conj Q(x)\Imp P\conj(\exists x.\, Q(x)). \] \emph{Hint}: the proof is similar to the one just above for the universal quantifier. \end{exercise} \index{quantifiers!existential|)} \subsection{Renaming a Bound Variable: {\tt\slshape rename_tac}} \index{assumptions!renaming|(}\index{*rename_tac (method)|(}% When you apply a rule such as \isa{allI}, the quantified variable becomes a new bound variable of the new subgoal. Isabelle tries to avoid changing its name, but sometimes it has to choose a new name in order to avoid a clash. The result may not be ideal: \begin{isabelle} \isacommand{lemma}\ "x\ <\ y\ \isasymLongrightarrow \ \isasymforall x\ y.\ P\ x\ (f\ y)"\isanewline \isacommand{apply}\ (intro allI)\isanewline \ 1.\ \isasymAnd xa\ ya.\ x\ <\ y\ \isasymLongrightarrow \ P\ xa\ (f\ ya) \end{isabelle} % The names \isa{x} and \isa{y} were already in use, so the new bound variables are called \isa{xa} and~\isa{ya}. You can rename them by invoking \isa{rename_tac}: \begin{isabelle} \isacommand{apply}\ (rename_tac\ v\ w)\isanewline \ 1.\ \isasymAnd v\ w.\ x\ <\ y\ \isasymLongrightarrow \ P\ v\ (f\ w) \end{isabelle} Recall that \isa{rule_tac}\index{*rule_tac (method)!and renaming} instantiates a theorem with specified terms. These terms may involve the goal's bound variables, but beware of referring to variables like~\isa{xa}. A future change to your theories could change the set of names produced at top level, so that \isa{xa} changes to~\isa{xb} or reverts to~\isa{x}. It is safer to rename automatically-generated variables before mentioning them. If the subgoal has more bound variables than there are names given to \isa{rename_tac}, the rightmost ones are renamed.% \index{assumptions!renaming|)}\index{*rename_tac (method)|)} \subsection{Reusing an Assumption: {\tt\slshape frule}} \label{sec:frule} \index{assumptions!reusing|(}\index{*frule (method)|(}% Note that \isa{drule spec} removes the universal quantifier and --- as usual with elimination rules --- discards the original formula. Sometimes, a universal formula has to be kept so that it can be used again. Then we use a new method: \isa{frule}. It acts like \isa{drule} but copies rather than replaces the selected assumption. The \isa{f} is for \emph{forward}. In this example, going from \isa{P\ a} to \isa{P(h(h~a))} requires two uses of the quantified assumption, one for each~\isa{h} in~\isa{h(h~a)}. \begin{isabelle} \isacommand{lemma}\ "\isasymlbrakk{\isasymforall}x.\ P\ x\ \isasymlongrightarrow\ P\ (h\ x); \ P\ a\isasymrbrakk\ \isasymLongrightarrow\ P(h\ (h\ a))" \end{isabelle} % Examine the subgoal left by \isa{frule}: \begin{isabelle} \isacommand{apply}\ (frule\ spec)\isanewline \ 1.\ \isasymlbrakk{\isasymforall}x.\ P\ x\ \isasymlongrightarrow\ P\ (h\ x);\ P\ a;\ P\ ?x\ \isasymlongrightarrow\ P\ (h\ ?x)\isasymrbrakk\ \isasymLongrightarrow\ P\ (h\ (h\ a)) \end{isabelle} It is what \isa{drule} would have left except that the quantified assumption is still present. Next we apply \isa{mp} to the implication and the assumption~\isa{P\ a}: \begin{isabelle} \isacommand{apply}\ (drule\ mp,\ assumption)\isanewline \ 1.\ \isasymlbrakk{\isasymforall}x.\ P\ x\ \isasymlongrightarrow\ P\ (h\ x);\ P\ a;\ P\ (h\ a)\isasymrbrakk\ \isasymLongrightarrow\ P\ (h\ (h\ a)) \end{isabelle} % We have created the assumption \isa{P(h\ a)}, which is progress. To continue the proof, we apply \isa{spec} again. We shall not need it again, so we can use \isa{drule}. \begin{isabelle} \isacommand{apply}\ (drule\ spec)\isanewline \ 1.\ \isasymlbrakk P\ a;\ P\ (h\ a);\ P\ ?x2\ \isasymlongrightarrow \ P\ (h\ ?x2)\isasymrbrakk \ \isasymLongrightarrow \ P\ (h\ (h\ a)) \end{isabelle} % The new assumption bridges the gap between \isa{P(h\ a)} and \isa{P(h(h\ a))}. \begin{isabelle} \isacommand{by}\ (drule\ mp) \end{isabelle} \medskip \emph{A final remark}. Replacing this \isacommand{by} command with \begin{isabelle} \isacommand{apply}\ (drule\ mp,\ assumption) \end{isabelle} would not work: it would add a second copy of \isa{P(h~a)} instead of the desired assumption, \isa{P(h(h~a))}. The \isacommand{by} command forces Isabelle to backtrack until it finds the correct one. Alternatively, we could have used the \isacommand{apply} command and bundled the \isa{drule mp} with \emph{two} calls of \isa{assumption}. Or, of course, we could have given the entire proof to \isa{auto}.% \index{assumptions!reusing|)}\index{*frule (method)|)} \subsection{Instantiating a Quantifier Explicitly} \index{quantifiers!instantiating} We can prove a theorem of the form $\exists x.\,P\, x$ by exhibiting a suitable term~$t$ such that $P\,t$ is true. Dually, we can use an assumption of the form $\forall x.\,P\, x$ to generate a new assumption $P\,t$ for a suitable term~$t$. In many cases, Isabelle makes the correct choice automatically, constructing the term by unification. In other cases, the required term is not obvious and we must specify it ourselves. Suitable methods are \isa{rule_tac}, \isa{drule_tac} and \isa{erule_tac}. We have seen (just above, {\S}\ref{sec:frule}) a proof of this lemma: \begin{isabelle} \isacommand{lemma}\ "\isasymlbrakk \isasymforall x.\ P\ x\ \isasymlongrightarrow \ P\ (h\ x);\ P\ a\isasymrbrakk \ \isasymLongrightarrow \ P(h\ (h\ a))" \end{isabelle} We had reached this subgoal: \begin{isabelle} \ 1.\ \isasymlbrakk{\isasymforall}x.\ P\ x\ \isasymlongrightarrow\ P\ (h\ x);\ P\ a;\ P\ (h\ a)\isasymrbrakk\ \isasymLongrightarrow\ P\ (h\ (h\ a)) \end{isabelle} % The proof requires instantiating the quantified assumption with the term~\isa{h~a}. \begin{isabelle} \isacommand{apply}\ (drule_tac\ x\ =\ "h\ a"\ \isakeyword{in}\ spec)\isanewline \ 1.\ \isasymlbrakk P\ a;\ P\ (h\ a);\ P\ (h\ a)\ \isasymlongrightarrow \ P\ (h\ (h\ a))\isasymrbrakk \ \isasymLongrightarrow \ P\ (h\ (h\ a)) \end{isabelle} We have forced the desired instantiation. \medskip Existential formulas can be instantiated too. The next example uses the \textbf{divides} relation\index{divides relation} of number theory: \begin{isabelle} ?m\ dvd\ ?n\ \isasymequiv\ {\isasymexists}k.\ ?n\ =\ ?m\ *\ k \rulename{dvd_def} \end{isabelle} Let us prove that multiplication of natural numbers is monotone with respect to the divides relation: \begin{isabelle} \isacommand{lemma}\ mult_dvd_mono:\ "{\isasymlbrakk}i\ dvd\ m;\ j\ dvd\ n\isasymrbrakk\ \isasymLongrightarrow\ i*j\ dvd\ (m*n\ ::\ nat)"\isanewline \isacommand{apply}\ (simp\ add:\ dvd_def) \end{isabelle} % Unfolding the definition of divides has left this subgoal: \begin{isabelle} \ 1.\ \isasymlbrakk \isasymexists k.\ m\ =\ i\ *\ k;\ \isasymexists k.\ n\ =\ j\ *\ k\isasymrbrakk \ \isasymLongrightarrow \ \isasymexists k.\ m\ *\ n\ =\ i\ *\ j\ *\ k \end{isabelle} % Next, we eliminate the two existential quantifiers in the assumptions: \begin{isabelle} \isacommand{apply}\ (erule\ exE)\isanewline \ 1.\ \isasymAnd k.\ \isasymlbrakk \isasymexists k.\ n\ =\ j\ *\ k;\ m\ =\ i\ *\ k\isasymrbrakk \ \isasymLongrightarrow \ \isasymexists k.\ m\ *\ n\ =\ i\ *\ j\ *\ k% \isanewline \isacommand{apply}\ (erule\ exE) \isanewline \ 1.\ \isasymAnd k\ ka.\ \isasymlbrakk m\ =\ i\ *\ k;\ n\ =\ j\ *\ ka\isasymrbrakk \ \isasymLongrightarrow \ \isasymexists k.\ m\ *\ n\ =\ i\ *\ j\ *\ k \end{isabelle} % The term needed to instantiate the remaining quantifier is~\isa{k*ka}. But \isa{ka} is an automatically-generated name. As noted above, references to such variable names makes a proof less resilient to future changes. So, first we rename the most recent variable to~\isa{l}: \begin{isabelle} \isacommand{apply}\ (rename_tac\ l)\isanewline \ 1.\ \isasymAnd k\ l.\ \isasymlbrakk m\ =\ i\ *\ k;\ n\ =\ j\ *\ l\isasymrbrakk \ \isasymLongrightarrow \ \isasymexists k.\ m\ *\ n\ =\ i\ *\ j\ *\ k% \end{isabelle} We instantiate the quantifier with~\isa{k*l}: \begin{isabelle} \isacommand{apply}\ (rule_tac\ x="k*l"\ \isakeyword{in}\ exI)\ \isanewline \ 1.\ \isasymAnd k\ ka.\ \isasymlbrakk m\ =\ i\ *\ k;\ n\ =\ j\ *\ ka\isasymrbrakk \ \isasymLongrightarrow \ m\ *\ n\ =\ i\ *\ j\ *\ (k\ *\ ka) \end{isabelle} % The rest is automatic, by arithmetic. \begin{isabelle} \isacommand{apply}\ simp\isanewline \isacommand{done}\isanewline \end{isabelle} \section{Description Operators} \label{sec:SOME} \index{description operators|(}% HOL provides two description operators. A \textbf{definite description} formalizes the word ``the,'' as in ``the greatest divisior of~$n$.'' It returns an arbitrary value unless the formula has a unique solution. An \textbf{indefinite description} formalizes the word ``some,'' as in ``some member of~$S$.'' It differs from a definite description in not requiring the solution to be unique: it uses the axiom of choice to pick any solution. \begin{warn} Description operators can be hard to reason about. Novices should try to avoid them. Fortunately, descriptions are seldom required. \end{warn} \subsection{Definite Descriptions} \index{descriptions!definite}% A definite description is traditionally written $\iota x. P(x)$. It denotes the $x$ such that $P(x)$ is true, provided there exists a unique such~$x$; otherwise, it returns an arbitrary value of the expected type. Isabelle uses \sdx{THE} for the Greek letter~$\iota$. %(The traditional notation could be provided, but it is not legible on screen.) We reason using this rule, where \isa{a} is the unique solution: \begin{isabelle} \isasymlbrakk P\ a;\ \isasymAnd x.\ P\ x\ \isasymLongrightarrow \ x\ =\ a\isasymrbrakk \ \isasymLongrightarrow \ (THE\ x.\ P\ x)\ =\ a% \rulenamedx{the_equality} \end{isabelle} For instance, we can define the cardinality of a finite set~$A$ to be that $n$ such that $A$ is in one-to-one correspondence with $\{1,\ldots,n\}$. We can then prove that the cardinality of the empty set is zero (since $n=0$ satisfies the description) and proceed to prove other facts. A more challenging example illustrates how Isabelle/HOL defines the least number operator, which denotes the least \isa{x} satisfying~\isa{P}:% \index{least number operator|see{\protect\isa{LEAST}}} \begin{isabelle} (LEAST\ x.\ P\ x)\ = (THE\ x.\ P\ x\ \isasymand \ (\isasymforall y.\ P\ y\ \isasymlongrightarrow \ x\ \isasymle \ y)) \end{isabelle} % Let us prove the analogue of \isa{the_equality} for \sdx{LEAST}\@. \begin{isabelle} \isacommand{theorem}\ Least_equality:\isanewline \ \ \ \ \ "\isasymlbrakk P\ (k::nat);\ \ \isasymforall x.\ P\ x\ \isasymlongrightarrow \ k\ \isasymle \ x\isasymrbrakk \ \isasymLongrightarrow \ (LEAST\ x.\ P\ x)\ =\ k"\isanewline \isacommand{apply}\ (simp\ add:\ Least_def)\isanewline \isanewline \ 1.\ \isasymlbrakk P\ k;\ \isasymforall x.\ P\ x\ \isasymlongrightarrow \ k\ \isasymle \ x\isasymrbrakk \isanewline \isaindent{\ 1.\ }\isasymLongrightarrow \ (THE\ x.\ P\ x\ \isasymand \ (\isasymforall y.\ P\ y\ \isasymlongrightarrow \ x\ \isasymle \ y))\ =\ k% \end{isabelle} The first step has merely unfolded the definition. \begin{isabelle} \isacommand{apply}\ (rule\ the_equality)\isanewline \isanewline \ 1.\ \isasymlbrakk P\ k;\ \isasymforall x.\ P\ x\ \isasymlongrightarrow \ k\ \isasymle \ x\isasymrbrakk \ \isasymLongrightarrow \ P\ k\ \isasymand \ (\isasymforall y.\ P\ y\ \isasymlongrightarrow \ k\ \isasymle \ y)\isanewline \ 2.\ \isasymAnd x.\ \isasymlbrakk P\ k;\ \isasymforall x.\ P\ x\ \isasymlongrightarrow \ k\ \isasymle \ x;\ P\ x\ \isasymand \ (\isasymforall y.\ P\ y\ \isasymlongrightarrow \ x\ \isasymle \ y)\isasymrbrakk \isanewline \ \ \ \ \ \ \ \ \isasymLongrightarrow \ x\ =\ k% \end{isabelle} As always with \isa{the_equality}, we must show existence and uniqueness of the claimed solution,~\isa{k}. Existence, the first subgoal, is trivial. Uniqueness, the second subgoal, follows by antisymmetry: \begin{isabelle} \isasymlbrakk x\ \isasymle \ y;\ y\ \isasymle \ x\isasymrbrakk \ \isasymLongrightarrow \ x\ =\ y% \rulename{order_antisym} \end{isabelle} The assumptions imply both \isa{k~\isasymle~x} and \isa{x~\isasymle~k}. One call to \isa{auto} does it all: \begin{isabelle} \isacommand{by}\ (auto\ intro:\ order_antisym) \end{isabelle} \subsection{Indefinite Descriptions} \index{Hilbert's $\varepsilon$-operator}% \index{descriptions!indefinite}% An indefinite description is traditionally written $\varepsilon x. P(x)$ and is known as Hilbert's $\varepsilon$-operator. It denotes some $x$ such that $P(x)$ is true, provided one exists. Isabelle uses \sdx{SOME} for the Greek letter~$\varepsilon$. Here is the definition of~\cdx{inv},\footnote{In fact, \isa{inv} is defined via a second constant \isa{inv_into}, which we ignore here.} which expresses inverses of functions: \begin{isabelle} inv\ f\ \isasymequiv \ \isasymlambda y.\ SOME\ x.\ f\ x\ =\ y% \rulename{inv_def} \end{isabelle} Using \isa{SOME} rather than \isa{THE} makes \isa{inv~f} behave well even if \isa{f} is not injective. As it happens, most useful theorems about \isa{inv} do assume the function to be injective. The inverse of \isa{f}, when applied to \isa{y}, returns some~\isa{x} such that \isa{f~x~=~y}. For example, we can prove \isa{inv~Suc} really is the inverse of the \isa{Suc} function \begin{isabelle} \isacommand{lemma}\ "inv\ Suc\ (Suc\ n)\ =\ n"\isanewline \isacommand{by}\ (simp\ add:\ inv_def) \end{isabelle} \noindent The proof is a one-liner: the subgoal simplifies to a degenerate application of \isa{SOME}, which is then erased. In detail, the left-hand side simplifies to \isa{SOME\ x.\ Suc\ x\ =\ Suc\ n}, then to \isa{SOME\ x.\ x\ =\ n} and finally to~\isa{n}. We know nothing about what \isa{inv~Suc} returns when applied to zero. The proof above still treats \isa{SOME} as a definite description, since it only reasons about situations in which the value is described uniquely. Indeed, \isa{SOME} satisfies this rule: \begin{isabelle} \isasymlbrakk P\ a;\ \isasymAnd x.\ P\ x\ \isasymLongrightarrow \ x\ =\ a\isasymrbrakk \ \isasymLongrightarrow \ (SOME\ x.\ P\ x)\ =\ a% \rulenamedx{some_equality} \end{isabelle} To go further is tricky and requires rules such as these: \begin{isabelle} P\ x\ \isasymLongrightarrow \ P\ (SOME\ x.\ P\ x) \rulenamedx{someI}\isanewline \isasymlbrakk P\ a;\ \isasymAnd x.\ P\ x\ \isasymLongrightarrow \ Q\ x\isasymrbrakk \ \isasymLongrightarrow \ Q\ (SOME\ x.\ P\ x) \rulenamedx{someI2} \end{isabelle} Rule \isa{someI} is basic: if anything satisfies \isa{P} then so does \hbox{\isa{SOME\ x.\ P\ x}}. The repetition of~\isa{P} in the conclusion makes it difficult to apply in a backward proof, so the derived rule \isa{someI2} is also provided. \medskip For example, let us prove the \rmindex{axiom of choice}: \begin{isabelle} \isacommand{theorem}\ axiom_of_choice: \ "(\isasymforall x.\ \isasymexists y.\ P\ x\ y)\ \isasymLongrightarrow \ \isasymexists f.\ \isasymforall x.\ P\ x\ (f\ x)"\isanewline \isacommand{apply}\ (rule\ exI,\ rule\ allI)\isanewline \ 1.\ \isasymAnd x.\ \isasymforall x.\ \isasymexists y.\ P\ x\ y\ \isasymLongrightarrow \ P\ x\ (?f\ x) \end{isabelle} % We have applied the introduction rules; now it is time to apply the elimination rules. \begin{isabelle} \isacommand{apply}\ (drule\ spec,\ erule\ exE)\isanewline \ 1.\ \isasymAnd x\ y.\ P\ (?x2\ x)\ y\ \isasymLongrightarrow \ P\ x\ (?f\ x) \end{isabelle} \noindent The rule \isa{someI} automatically instantiates \isa{f} to \hbox{\isa{\isasymlambda x.\ SOME y.\ P\ x\ y}}, which is the choice function. It also instantiates \isa{?x2\ x} to \isa{x}. \begin{isabelle} \isacommand{by}\ (rule\ someI)\isanewline \end{isabelle} \subsubsection{Historical Note} The original purpose of Hilbert's $\varepsilon$-operator was to express an existential destruction rule: \[ \infer{P[(\varepsilon x. P) / \, x]}{\exists x.\,P} \] This rule is seldom used for that purpose --- it can cause exponential blow-up --- but it is occasionally used as an introduction rule for the~$\varepsilon$-operator. Its name in HOL is \tdxbold{someI_ex}.%% \index{description operators|)} \section{Some Proofs That Fail} \index{proofs!examples of failing|(}% Most of the examples in this tutorial involve proving theorems. But not every conjecture is true, and it can be instructive to see how proofs fail. Here we attempt to prove a distributive law involving the existential quantifier and conjunction. \begin{isabelle} \isacommand{lemma}\ "({\isasymexists}x.\ P\ x)\ \isasymand\ ({\isasymexists}x.\ Q\ x)\ \isasymLongrightarrow\ {\isasymexists}x.\ P\ x\ \isasymand\ Q\ x" \end{isabelle} The first steps are routine. We apply conjunction elimination to break the assumption into two existentially quantified assumptions. Applying existential elimination removes one of the quantifiers. \begin{isabelle} \isacommand{apply}\ (erule\ conjE)\isanewline \isacommand{apply}\ (erule\ exE)\isanewline \ 1.\ \isasymAnd x.\ \isasymlbrakk{\isasymexists}x.\ Q\ x;\ P\ x\isasymrbrakk\ \isasymLongrightarrow\ {\isasymexists}x.\ P\ x\ \isasymand\ Q\ x \end{isabelle} % When we remove the other quantifier, we get a different bound variable in the subgoal. (The name \isa{xa} is generated automatically.) \begin{isabelle} \isacommand{apply}\ (erule\ exE)\isanewline \ 1.\ \isasymAnd x\ xa.\ \isasymlbrakk P\ x;\ Q\ xa\isasymrbrakk\ \isasymLongrightarrow\ {\isasymexists}x.\ P\ x\ \isasymand\ Q\ x \end{isabelle} The proviso of the existential elimination rule has forced the variables to differ: we can hardly expect two arbitrary values to be equal! There is no way to prove this subgoal. Removing the conclusion's existential quantifier yields two identical placeholders, which can become any term involving the variables \isa{x} and~\isa{xa}. We need one to become \isa{x} and the other to become~\isa{xa}, but Isabelle requires all instances of a placeholder to be identical. \begin{isabelle} \isacommand{apply}\ (rule\ exI)\isanewline \isacommand{apply}\ (rule\ conjI)\isanewline \ 1.\ \isasymAnd x\ xa.\ \isasymlbrakk P\ x;\ Q\ xa\isasymrbrakk\ \isasymLongrightarrow\ P\ (?x3\ x\ xa)\isanewline \ 2.\ \isasymAnd x\ xa.\ \isasymlbrakk P\ x;\ Q\ xa\isasymrbrakk\ \isasymLongrightarrow\ Q\ (?x3\ x\ xa) \end{isabelle} We can prove either subgoal using the \isa{assumption} method. If we prove the first one, the placeholder changes into~\isa{x}. \begin{isabelle} \ \isacommand{apply}\ assumption\isanewline \ 1.\ \isasymAnd x\ xa.\ \isasymlbrakk P\ x;\ Q\ xa\isasymrbrakk\ \isasymLongrightarrow\ Q\ x \end{isabelle} We are left with a subgoal that cannot be proved. Applying the \isa{assumption} method results in an error message: \begin{isabelle} *** empty result sequence -- proof command failed \end{isabelle} When interacting with Isabelle via the shell interface, you can abandon a proof using the \isacommand{oops} command. \medskip Here is another abortive proof, illustrating the interaction between bound variables and unknowns. If $R$ is a reflexive relation, is there an $x$ such that $R\,x\,y$ holds for all $y$? Let us see what happens when we attempt to prove it. \begin{isabelle} \isacommand{lemma}\ "\isasymforall y.\ R\ y\ y\ \isasymLongrightarrow \ \isasymexists x.\ \isasymforall y.\ R\ x\ y" \end{isabelle} First, we remove the existential quantifier. The new proof state has an unknown, namely~\isa{?x}. \begin{isabelle} \isacommand{apply}\ (rule\ exI)\isanewline \ 1.\ \isasymforall y.\ R\ y\ y\ \isasymLongrightarrow \ \isasymforall y.\ R\ ?x\ y% \end{isabelle} It looks like we can just apply \isa{assumption}, but it fails. Isabelle refuses to substitute \isa{y}, a bound variable, for~\isa{?x}; that would be a bound variable capture. We can still try to finish the proof in some other way. We remove the universal quantifier from the conclusion, moving the bound variable~\isa{y} into the subgoal. But note that it is still bound! \begin{isabelle} \isacommand{apply}\ (rule\ allI)\isanewline \ 1.\ \isasymAnd y.\ \isasymforall y.\ R\ y\ y\ \isasymLongrightarrow \ R\ ?x\ y% \end{isabelle} Finally, we try to apply our reflexivity assumption. We obtain a new assumption whose identical placeholders may be replaced by any term involving~\isa{y}. \begin{isabelle} \isacommand{apply}\ (drule\ spec)\isanewline \ 1.\ \isasymAnd y.\ R\ (?z2\ y)\ (?z2\ y)\ \isasymLongrightarrow\ R\ ?x\ y \end{isabelle} This subgoal can only be proved by putting \isa{y} for all the placeholders, making the assumption and conclusion become \isa{R\ y\ y}. Isabelle can replace \isa{?z2~y} by \isa{y}; this involves instantiating \isa{?z2} to the identity function. But, just as two steps earlier, Isabelle refuses to substitute~\isa{y} for~\isa{?x}. This example is typical of how Isabelle enforces sound quantifier reasoning. \index{proofs!examples of failing|)} \section{Proving Theorems Using the {\tt\slshape blast} Method} \index{*blast (method)|(}% It is hard to prove many theorems using the methods described above. A proof may be hundreds of steps long. You may need to search among different ways of proving certain subgoals. Often a choice that proves one subgoal renders another impossible to prove. There are further complications that we have not discussed, concerning negation and disjunction. Isabelle's \textbf{classical reasoner} is a family of tools that perform such proofs automatically. The most important of these is the \isa{blast} method. In this section, we shall first see how to use the classical reasoner in its default mode and then how to insert additional rules, enabling it to work in new problem domains. We begin with examples from pure predicate logic. The following example is known as Andrew's challenge. Peter Andrews designed it to be hard to prove by automatic means. It is particularly hard for a resolution prover, where converting the nested biconditionals to clause form produces a combinatorial explosion~\cite{pelletier86}. However, the \isa{blast} method proves it in a fraction of a second. \begin{isabelle} \isacommand{lemma}\ "(({\isasymexists}x.\ {\isasymforall}y.\ p(x){=}p(y))\ =\ (({\isasymexists}x.\ q(x))=({\isasymforall}y.\ p(y))))\ \ \ =\ \ \ \ \isanewline \ \ \ \ \ \ \ \ (({\isasymexists}x.\ {\isasymforall}y.\ q(x){=}q(y))\ =\ (({\isasymexists}x.\ p(x))=({\isasymforall}y.\ q(y))))"\isanewline \isacommand{by}\ blast \end{isabelle} The next example is a logic problem composed by Lewis Carroll. The \isa{blast} method finds it trivial. Moreover, it turns out that not all of the assumptions are necessary. We can experiment with variations of this formula and see which ones can be proved. \begin{isabelle} \isacommand{lemma}\ "({\isasymforall}x.\ honest(x)\ \isasymand\ industrious(x)\ \isasymlongrightarrow\ healthy(x))\ \isasymand\ \ \isanewline \ \ \ \ \ \ \ \ \isasymnot\ ({\isasymexists}x.\ grocer(x)\ \isasymand\ healthy(x))\ \isasymand\ \isanewline \ \ \ \ \ \ \ \ ({\isasymforall}x.\ industrious(x)\ \isasymand\ grocer(x)\ \isasymlongrightarrow\ honest(x))\ \isasymand\ \isanewline \ \ \ \ \ \ \ \ ({\isasymforall}x.\ cyclist(x)\ \isasymlongrightarrow\ industrious(x))\ \isasymand\ \isanewline \ \ \ \ \ \ \ \ ({\isasymforall}x.\ {\isasymnot}healthy(x)\ \isasymand\ cyclist(x)\ \isasymlongrightarrow\ {\isasymnot}honest(x))\ \ \isanewline \ \ \ \ \ \ \ \ \isasymlongrightarrow\ ({\isasymforall}x.\ grocer(x)\ \isasymlongrightarrow\ {\isasymnot}cyclist(x))"\isanewline \isacommand{by}\ blast \end{isabelle} The \isa{blast} method is also effective for set theory, which is described in the next chapter. The formula below may look horrible, but the \isa{blast} method proves it in milliseconds. \begin{isabelle} \isacommand{lemma}\ "({\isasymUnion}i{\isasymin}I.\ A(i))\ \isasyminter\ ({\isasymUnion}j{\isasymin}J.\ B(j))\ =\isanewline \ \ \ \ \ \ \ \ ({\isasymUnion}i{\isasymin}I.\ {\isasymUnion}j{\isasymin}J.\ A(i)\ \isasyminter\ B(j))"\isanewline \isacommand{by}\ blast \end{isabelle} Few subgoals are couched purely in predicate logic and set theory. We can extend the scope of the classical reasoner by giving it new rules. Extending it effectively requires understanding the notions of introduction, elimination and destruction rules. Moreover, there is a distinction between safe and unsafe rules. A \textbf{safe}\indexbold{safe rules} rule is one that can be applied backwards without losing information; an \textbf{unsafe}\indexbold{unsafe rules} rule loses information, perhaps transforming the subgoal into one that cannot be proved. The safe/unsafe distinction affects the proof search: if a proof attempt fails, the classical reasoner backtracks to the most recent unsafe rule application and makes another choice. An important special case avoids all these complications. A logical equivalence, which in higher-order logic is an equality between formulas, can be given to the classical reasoner and simplifier by using the attribute \attrdx{iff}. You should do so if the right hand side of the equivalence is simpler than the left-hand side. For example, here is a simple fact about list concatenation. The result of appending two lists is empty if and only if both of the lists are themselves empty. Obviously, applying this equivalence will result in a simpler goal. When stating this lemma, we include the \attrdx{iff} attribute. Once we have proved the lemma, Isabelle will make it known to the classical reasoner (and to the simplifier). \begin{isabelle} \isacommand{lemma}\ [iff]:\ "(xs{\isacharat}ys\ =\ [])\ =\ (xs=[]\ \isasymand\ ys=[])"\isanewline \isacommand{apply}\ (induct_tac\ xs)\isanewline \isacommand{apply}\ (simp_all)\isanewline \isacommand{done} \end{isabelle} % This fact about multiplication is also appropriate for the \attrdx{iff} attribute: \begin{isabelle} (\mbox{?m}\ *\ \mbox{?n}\ =\ 0)\ =\ (\mbox{?m}\ =\ 0\ \isasymor\ \mbox{?n}\ =\ 0) \end{isabelle} A product is zero if and only if one of the factors is zero. The reasoning involves a disjunction. Proving new rules for disjunctive reasoning is hard, but translating to an actual disjunction works: the classical reasoner handles disjunction properly. In more detail, this is how the \attrdx{iff} attribute works. It converts the equivalence $P=Q$ to a pair of rules: the introduction rule $Q\Imp P$ and the destruction rule $P\Imp Q$. It gives both to the classical reasoner as safe rules, ensuring that all occurrences of $P$ in a subgoal are replaced by~$Q$. The simplifier performs the same replacement, since \isa{iff} gives $P=Q$ to the simplifier. Classical reasoning is different from simplification. Simplification is deterministic. It applies rewrite rules repeatedly, as long as possible, transforming a goal into another goal. Classical reasoning uses search and backtracking in order to prove a goal outright.% \index{*blast (method)|)}% \section{Other Classical Reasoning Methods} The \isa{blast} method is our main workhorse for proving theorems automatically. Other components of the classical reasoner interact with the simplifier. Still others perform classical reasoning to a limited extent, giving the user fine control over the proof. Of the latter methods, the most useful is \methdx{clarify}. It performs all obvious reasoning steps without splitting the goal into multiple parts. It does not apply unsafe rules that could render the goal unprovable. By performing the obvious steps, \isa{clarify} lays bare the difficult parts of the problem, where human intervention is necessary. For example, the following conjecture is false: \begin{isabelle} \isacommand{lemma}\ "({\isasymforall}x.\ P\ x)\ \isasymand\ ({\isasymexists}x.\ Q\ x)\ \isasymlongrightarrow\ ({\isasymforall}x.\ P\ x\ \isasymand\ Q\ x)"\isanewline \isacommand{apply}\ clarify \end{isabelle} The \isa{blast} method would simply fail, but \isa{clarify} presents a subgoal that helps us see why we cannot continue the proof. \begin{isabelle} \ 1.\ \isasymAnd x\ xa.\ \isasymlbrakk{\isasymforall}x.\ P\ x;\ Q\ xa\isasymrbrakk\ \isasymLongrightarrow\ P\ x\ \isasymand\ Q\ x \end{isabelle} The proof must fail because the assumption \isa{Q\ xa} and conclusion \isa{Q\ x} refer to distinct bound variables. To reach this state, \isa{clarify} applied the introduction rules for \isa{\isasymlongrightarrow} and \isa{\isasymforall} and the elimination rule for \isa{\isasymand}. It did not apply the introduction rule for \isa{\isasymand} because of its policy never to split goals. Also available is \methdx{clarsimp}, a method that interleaves \isa{clarify} and \isa{simp}. Also there is \methdx{safe}, which like \isa{clarify} performs obvious steps but even applies those that split goals. The \methdx{force} method applies the classical reasoner and simplifier to one goal. Unless it can prove the goal, it fails. Contrast that with the \isa{auto} method, which also combines classical reasoning with simplification. The latter's purpose is to prove all the easy subgoals and parts of subgoals. Unfortunately, it can produce large numbers of new subgoals; also, since it proves some subgoals and splits others, it obscures the structure of the proof tree. The \isa{force} method does not have these drawbacks. Another difference: \isa{force} tries harder than {\isa{auto}} to prove its goal, so it can take much longer to terminate. Older components of the classical reasoner have largely been superseded by \isa{blast}, but they still have niche applications. Most important among these are \isa{fast} and \isa{best}. While \isa{blast} searches for proofs using a built-in first-order reasoner, these earlier methods search for proofs using standard Isabelle inference. That makes them slower but enables them to work in the presence of the more unusual features of Isabelle rules, such as type classes and function unknowns. For example, recall the introduction rule for Hilbert's $\varepsilon$-operator: \begin{isabelle} ?P\ ?x\ \isasymLongrightarrow\ ?P\ (SOME\ x.\ ?P x) \rulename{someI} \end{isabelle} % The repeated occurrence of the variable \isa{?P} makes this rule tricky to apply. Consider this contrived example: \begin{isabelle} \isacommand{lemma}\ "\isasymlbrakk Q\ a;\ P\ a\isasymrbrakk\isanewline \ \ \ \ \ \ \ \ \,\isasymLongrightarrow\ P\ (SOME\ x.\ P\ x\ \isasymand\ Q\ x)\ \isasymand\ Q\ (SOME\ x.\ P\ x\ \isasymand\ Q\ x)"\isanewline \isacommand{apply}\ (rule\ someI) \end{isabelle} % We can apply rule \isa{someI} explicitly. It yields the following subgoal: \begin{isabelle} \ 1.\ \isasymlbrakk Q\ a;\ P\ a\isasymrbrakk\ \isasymLongrightarrow\ P\ ?x\ \isasymand\ Q\ ?x% \end{isabelle} The proof from this point is trivial. Could we have proved the theorem with a single command? Not using \isa{blast}: it cannot perform the higher-order unification needed here. The \methdx{fast} method succeeds: \begin{isabelle} \isacommand{apply}\ (fast\ intro!:\ someI) \end{isabelle} The \methdx{best} method is similar to \isa{fast} but it uses a best-first search instead of depth-first search. Accordingly, it is slower but is less susceptible to divergence. Transitivity rules usually cause \isa{fast} to loop where \isa{best} can often manage. Here is a summary of the classical reasoning methods: \begin{itemize} \item \methdx{blast} works automatically and is the fastest \item \methdx{clarify} and \methdx{clarsimp} perform obvious steps without splitting the goal; \methdx{safe} even splits goals \item \methdx{force} uses classical reasoning and simplification to prove a goal; \methdx{auto} is similar but leaves what it cannot prove \item \methdx{fast} and \methdx{best} are legacy methods that work well with rules involving unusual features \end{itemize} A table illustrates the relationships among four of these methods. \begin{center} \begin{tabular}{r|l|l|} & no split & split \\ \hline no simp & \methdx{clarify} & \methdx{safe} \\ \hline simp & \methdx{clarsimp} & \methdx{auto} \\ \hline \end{tabular} \end{center} \section{Finding More Theorems} \label{sec:find2} \input{find2.tex} \section{Forward Proof: Transforming Theorems}\label{sec:forward} \index{forward proof|(}% Forward proof means deriving new facts from old ones. It is the most fundamental type of proof. Backward proof, by working from goals to subgoals, can help us find a difficult proof. But it is not always the best way of presenting the proof thus found. Forward proof is particularly good for reasoning from the general to the specific. For example, consider this distributive law for the greatest common divisor: \[ k\times\gcd(m,n) = \gcd(k\times m,k\times n)\] Putting $m=1$ we get (since $\gcd(1,n)=1$ and $k\times1=k$) \[ k = \gcd(k,k\times n)\] We have derived a new fact; if re-oriented, it might be useful for simplification. After re-orienting it and putting $n=1$, we derive another useful law: \[ \gcd(k,k)=k \] Substituting values for variables --- instantiation --- is a forward step. Re-orientation works by applying the symmetry of equality to an equation, so it too is a forward step. \subsection{Modifying a Theorem using {\tt\slshape of}, {\tt\slshape where} and {\tt\slshape THEN}} \label{sec:THEN} Let us reproduce our examples in Isabelle. Recall that in {\S}\ref{sec:fun-simplification} we declared the recursive function \isa{gcd}:\index{*gcd (constant)|(} \begin{isabelle} \isacommand{fun}\ gcd\ ::\ "nat\ \isasymRightarrow \ nat\ \isasymRightarrow \ nat"\ \isakeyword{where}\isanewline \ \ "gcd\ m\ n\ =\ (if\ n=0\ then\ m\ else\ gcd\ n\ (m\ mod\ n))" \end{isabelle} % From this definition, it is possible to prove the distributive law. That takes us to the starting point for our example. \begin{isabelle} ?k\ *\ gcd\ ?m\ ?n\ =\ gcd\ (?k\ *\ ?m)\ (?k\ *\ ?n) \rulename{gcd_mult_distrib2} \end{isabelle} % The first step in our derivation is to replace \isa{?m} by~1. We instantiate the theorem using~\attrdx{of}, which identifies variables in order of their appearance from left to right. In this case, the variables are \isa{?k}, \isa{?m} and~\isa{?n}. So, the expression \hbox{\texttt{[of k 1]}} replaces \isa{?k} by~\isa{k} and \isa{?m} by~\isa{1}. \begin{isabelle} \isacommand{lemmas}\ gcd_mult_0\ =\ gcd_mult_distrib2\ [of\ k\ 1] \end{isabelle} % The keyword \commdx{lemmas} declares a new theorem, which can be derived from an existing one using attributes such as \isa{[of~k~1]}. The command \isa{thm gcd_mult_0} displays the result: \begin{isabelle} \ \ \ \ \ k\ *\ gcd\ 1\ ?n\ =\ gcd\ (k\ *\ 1)\ (k\ *\ ?n) \end{isabelle} Something is odd: \isa{k} is an ordinary variable, while \isa{?n} is schematic. We did not specify an instantiation for \isa{?n}. In its present form, the theorem does not allow substitution for \isa{k}. One solution is to avoid giving an instantiation for \isa{?k}: instead of a term we can put an underscore~(\isa{_}). For example, \begin{isabelle} \ \ \ \ \ gcd_mult_distrib2\ [of\ _\ 1] \end{isabelle} replaces \isa{?m} by~\isa{1} but leaves \isa{?k} unchanged. An equivalent solution is to use the attribute \isa{where}. \begin{isabelle} \ \ \ \ \ gcd\_mult\_distrib2\ [where\ m=1] \end{isabelle} While \isa{of} refers to variables by their position, \isa{where} refers to variables by name. Multiple instantiations are separated by~\isa{and}, as in this example: \begin{isabelle} \ \ \ \ \ gcd\_mult\_distrib2\ [where\ m=1\ and\ k=1] \end{isabelle} We now continue the present example with the version of \isa{gcd_mult_0} shown above, which has \isa{k} instead of \isa{?k}. Once we have replaced \isa{?m} by~1, we must next simplify the theorem \isa{gcd_mult_0}, performing the steps $\gcd(1,n)=1$ and $k\times1=k$. The \attrdx{simplified} attribute takes a theorem and returns the result of simplifying it, with respect to the default simplification rules: \begin{isabelle} \isacommand{lemmas}\ gcd_mult_1\ =\ gcd_mult_0\ [simplified]% \end{isabelle} % Again, we display the resulting theorem: \begin{isabelle} \ \ \ \ \ k\ =\ gcd\ k\ (k\ *\ ?n) \end{isabelle} % To re-orient the equation requires the symmetry rule: \begin{isabelle} ?s\ =\ ?t\ \isasymLongrightarrow\ ?t\ =\ ?s% \rulenamedx{sym} \end{isabelle} The following declaration gives our equation to \isa{sym}: \begin{isabelle} \ \ \ \isacommand{lemmas}\ gcd_mult\ =\ gcd_mult_1\ [THEN\ sym] \end{isabelle} % Here is the result: \begin{isabelle} \ \ \ \ \ gcd\ k\ (k\ *\ ?n)\ =\ k% \end{isabelle} \isa{THEN~sym}\indexbold{*THEN (attribute)} gives the current theorem to the rule \isa{sym} and returns the resulting conclusion. The effect is to exchange the two operands of the equality. Typically \isa{THEN} is used with destruction rules. Also useful is \isa{THEN~spec}, which removes the quantifier from a theorem of the form $\forall x.\,P$, and \isa{THEN~mp}, which converts the implication $P\imp Q$ into the rule $\vcenter{\infer{Q}{P}}$. Similar to \isa{mp} are the following two rules, which extract the two directions of reasoning about a boolean equivalence: \begin{isabelle} \isasymlbrakk?Q\ =\ ?P;\ ?Q\isasymrbrakk\ \isasymLongrightarrow\ ?P% \rulenamedx{iffD1}% \isanewline \isasymlbrakk?P\ =\ ?Q;\ ?Q\isasymrbrakk\ \isasymLongrightarrow\ ?P% \rulenamedx{iffD2} \end{isabelle} % Normally we would never name the intermediate theorems such as \isa{gcd_mult_0} and \isa{gcd_mult_1} but would combine the three forward steps: \begin{isabelle} \isacommand{lemmas}\ gcd_mult\ =\ gcd_mult_distrib2\ [of\ k\ 1,\ simplified,\ THEN\ sym]% \end{isabelle} The directives, or attributes, are processed from left to right. This declaration of \isa{gcd_mult} is equivalent to the previous one. Such declarations can make the proof script hard to read. Better is to state the new lemma explicitly and to prove it using a single \isa{rule} method whose operand is expressed using forward reasoning: \begin{isabelle} \isacommand{lemma}\ gcd\_mult\ [simp]:\ "gcd\ k\ (k*n)\ =\ k"\isanewline \isacommand{by}\ (rule\ gcd_mult_distrib2\ [of\ k\ 1,\ simplified,\ THEN\ sym]) \end{isabelle} Compared with the previous proof of \isa{gcd_mult}, this version shows the reader what has been proved. Also, the result will be processed in the normal way. In particular, Isabelle generalizes over all variables: the resulting theorem will have {\isa{?k}} instead of {\isa{k}}. At the start of this section, we also saw a proof of $\gcd(k,k)=k$. Here is the Isabelle version:\index{*gcd (constant)|)} \begin{isabelle} \isacommand{lemma}\ gcd\_self\ [simp]:\ "gcd\ k\ k\ =\ k"\isanewline \isacommand{by}\ (rule\ gcd_mult\ [of\ k\ 1,\ simplified]) \end{isabelle} \begin{warn} To give~\isa{of} a nonatomic term, enclose it in quotation marks, as in \isa{[of "k*m"]}. The term must not contain unknowns: an attribute such as \isa{[of "?k*m"]} will be rejected. \end{warn} %Answer is now included in that section! Is a modified version of this % exercise worth including? E.g. find a difference between the two ways % of substituting. %\begin{exercise} %In {\S}\ref{sec:subst} the method \isa{subst\ mult.commute} was applied. How %can we achieve the same effect using \isa{THEN} with the rule \isa{ssubst}? %% answer rule (mult.commute [THEN ssubst]) %\end{exercise} \subsection{Modifying a Theorem using {\tt\slshape OF}} \index{*OF (attribute)|(}% Recall that \isa{of} generates an instance of a rule by specifying values for its variables. Analogous is \isa{OF}, which generates an instance of a rule by specifying facts for its premises. We again need the divides relation\index{divides relation} of number theory, which as we recall is defined by \begin{isabelle} ?m\ dvd\ ?n\ \isasymequiv\ {\isasymexists}k.\ ?n\ =\ ?m\ *\ k \rulename{dvd_def} \end{isabelle} % Suppose, for example, that we have proved the following rule. It states that if $k$ and $n$ are relatively prime and if $k$ divides $m\times n$ then $k$ divides $m$. \begin{isabelle} \isasymlbrakk gcd ?k ?n {=} 1;\ ?k\ dvd\ ?m * ?n\isasymrbrakk\ \isasymLongrightarrow\ ?k\ dvd\ ?m \rulename{relprime_dvd_mult} \end{isabelle} We can use \isa{OF} to create an instance of this rule. First, we prove an instance of its first premise: \begin{isabelle} \isacommand{lemma}\ relprime\_20\_81:\ "gcd\ 20\ 81\ =\ 1"\isanewline \isacommand{by}\ (simp\ add:\ gcd.simps) \end{isabelle} We have evaluated an application of the \isa{gcd} function by simplification. Expression evaluation involving recursive functions is not guaranteed to terminate, and it can be slow; Isabelle performs arithmetic by rewriting symbolic bit strings. Here, however, the simplification takes less than one second. We can give this new lemma to \isa{OF}. The expression \begin{isabelle} \ \ \ \ \ relprime_dvd_mult [OF relprime_20_81] \end{isabelle} yields the theorem \begin{isabelle} \ \ \ \ \ 20\ dvd\ (?m\ *\ 81)\ \isasymLongrightarrow\ 20\ dvd\ ?m% \end{isabelle} % \isa{OF} takes any number of operands. Consider the following facts about the divides relation: \begin{isabelle} \isasymlbrakk?k\ dvd\ ?m;\ ?k\ dvd\ ?n\isasymrbrakk\ \isasymLongrightarrow\ ?k\ dvd\ ?m\ +\ ?n \rulename{dvd_add}\isanewline ?m\ dvd\ ?m% \rulename{dvd_refl} \end{isabelle} Let us supply \isa{dvd_refl} for each of the premises of \isa{dvd_add}: \begin{isabelle} \ \ \ \ \ dvd_add [OF dvd_refl dvd_refl] \end{isabelle} Here is the theorem that we have expressed: \begin{isabelle} \ \ \ \ \ ?k\ dvd\ (?k\ +\ ?k) \end{isabelle} As with \isa{of}, we can use the \isa{_} symbol to leave some positions unspecified: \begin{isabelle} \ \ \ \ \ dvd_add [OF _ dvd_refl] \end{isabelle} The result is \begin{isabelle} \ \ \ \ \ ?k\ dvd\ ?m\ \isasymLongrightarrow\ ?k\ dvd\ ?m\ +\ ?k \end{isabelle} You may have noticed that \isa{THEN} and \isa{OF} are based on the same idea, namely to combine two rules. They differ in the order of the combination and thus in their effect. We use \isa{THEN} typically with a destruction rule to extract a subformula of the current theorem. We use \isa{OF} with a list of facts to generate an instance of the current theorem.% \index{*OF (attribute)|)} Here is a summary of some primitives for forward reasoning: \begin{itemize} \item \attrdx{of} instantiates the variables of a rule to a list of terms \item \attrdx{OF} applies a rule to a list of theorems \item \attrdx{THEN} gives a theorem to a named rule and returns the conclusion %\item \attrdx{rule_format} puts a theorem into standard form % by removing \isa{\isasymlongrightarrow} and~\isa{\isasymforall} \item \attrdx{simplified} applies the simplifier to a theorem \item \isacommand{lemmas} assigns a name to the theorem produced by the attributes above \end{itemize} \section{Forward Reasoning in a Backward Proof} We have seen that the forward proof directives work well within a backward proof. There are many ways to achieve a forward style using our existing proof methods. We shall also meet some new methods that perform forward reasoning. The methods \isa{drule}, \isa{frule}, \isa{drule_tac}, etc., reason forward from a subgoal. We have seen them already, using rules such as \isa{mp} and \isa{spec} to operate on formulae. They can also operate on terms, using rules such as these: \begin{isabelle} x\ =\ y\ \isasymLongrightarrow \ f\ x\ =\ f\ y% \rulenamedx{arg_cong}\isanewline i\ \isasymle \ j\ \isasymLongrightarrow \ i\ *\ k\ \isasymle \ j\ *\ k% \rulename{mult_le_mono1} \end{isabelle} For example, let us prove a fact about divisibility in the natural numbers: \begin{isabelle} \isacommand{lemma}\ "2\ \isasymle \ u\ \isasymLongrightarrow \ u*m\ \isasymnoteq \ Suc(u*n)"\isanewline \isacommand{apply}\ (intro\ notI)\isanewline \ 1.\ \isasymlbrakk 2\ \isasymle \ u;\ u\ *\ m\ =\ Suc\ (u\ *\ n)\isasymrbrakk \ \isasymLongrightarrow \ False% \end{isabelle} % The key step is to apply the function \ldots\isa{mod\ u} to both sides of the equation \isa{u*m\ =\ Suc(u*n)}:\index{*drule_tac (method)} \begin{isabelle} \isacommand{apply}\ (drule_tac\ f="\isasymlambda x.\ x\ mod\ u"\ \isakeyword{in}\ arg_cong)\isanewline \ 1.\ \isasymlbrakk 2\ \isasymle \ u;\ u\ *\ m\ mod\ u\ =\ Suc\ (u\ *\ n)\ mod\ u\isasymrbrakk \ \isasymLongrightarrow \ False \end{isabelle} % Simplification reduces the left side to 0 and the right side to~1, yielding the required contradiction. \begin{isabelle} \isacommand{apply}\ (simp\ add:\ mod_Suc)\isanewline \isacommand{done} \end{isabelle} Our proof has used a fact about remainder: \begin{isabelle} Suc\ m\ mod\ n\ =\isanewline (if\ Suc\ (m\ mod\ n)\ =\ n\ then\ 0\ else\ Suc\ (m\ mod\ n)) \rulename{mod_Suc} \end{isabelle} \subsection{The Method {\tt\slshape insert}} \index{*insert (method)|(}% The \isa{insert} method inserts a given theorem as a new assumption of all subgoals. This already is a forward step; moreover, we may (as always when using a theorem) apply \isa{of}, \isa{THEN} and other directives. The new assumption can then be used to help prove the subgoals. For example, consider this theorem about the divides relation. The first proof step inserts the distributive law for \isa{gcd}. We specify its variables as shown. \begin{isabelle} \isacommand{lemma}\ relprime\_dvd\_mult:\ \isanewline \ \ \ \ \ \ "\isasymlbrakk \ gcd\ k\ n\ =\ 1;\ k\ dvd\ m*n\ \isasymrbrakk \ \isasymLongrightarrow \ k\ dvd\ m"\isanewline \isacommand{apply}\ (insert\ gcd_mult_distrib2\ [of\ m\ k\ n]) \end{isabelle} In the resulting subgoal, note how the equation has been inserted: \begin{isabelle} \ 1.\ \isasymlbrakk gcd\ k\ n\ =\ 1;\ k\ dvd\ m\ *\ n;\ m\ *\ gcd\ k\ n\ =\ gcd\ (m\ *\ k)\ (m\ *\ n)\isasymrbrakk \isanewline \isaindent{\ 1.\ }\isasymLongrightarrow \ k\ dvd\ m% \end{isabelle} The next proof step utilizes the assumption \isa{gcd\ k\ n\ =\ 1} (note that \isa{Suc\ 0} is another expression for 1): \begin{isabelle} \isacommand{apply}(simp)\isanewline \ 1.\ \isasymlbrakk gcd\ k\ n\ =\ Suc\ 0;\ k\ dvd\ m\ *\ n;\ m\ =\ gcd\ (m\ *\ k)\ (m\ *\ n)\isasymrbrakk \isanewline \isaindent{\ 1.\ }\isasymLongrightarrow \ k\ dvd\ m% \end{isabelle} Simplification has yielded an equation for~\isa{m}. The rest of the proof is omitted. \medskip Here is another demonstration of \isa{insert}. Division and remainder obey a well-known law: \begin{isabelle} (?m\ div\ ?n)\ *\ ?n\ +\ ?m\ mod\ ?n\ =\ ?m \rulename{div_mult_mod_eq} \end{isabelle} We refer to this law explicitly in the following proof: \begin{isabelle} \isacommand{lemma}\ div_mult_self_is_m:\ \isanewline \ \ \ \ \ \ "0{\isacharless}n\ \isasymLongrightarrow\ (m*n)\ div\ n\ =\ (m::nat)"\isanewline \isacommand{apply}\ (insert\ div_mult_mod_eq\ [of\ "m*n"\ n])\isanewline \isacommand{apply}\ (simp)\isanewline \isacommand{done} \end{isabelle} The first step inserts the law, specifying \isa{m*n} and \isa{n} for its variables. Notice that non-trivial expressions must be enclosed in quotation marks. Here is the resulting subgoal, with its new assumption: \begin{isabelle} %0\ \isacharless\ n\ \isasymLongrightarrow\ (m\ %*\ n)\ div\ n\ =\ m\isanewline \ 1.\ \isasymlbrakk0\ \isacharless\ n;\ \ (m\ *\ n)\ div\ n\ *\ n\ +\ (m\ *\ n)\ mod\ n\ =\ m\ *\ n\isasymrbrakk\isanewline \ \ \ \ \isasymLongrightarrow\ (m\ *\ n)\ div\ n\ =\ m \end{isabelle} Simplification reduces \isa{(m\ *\ n)\ mod\ n} to zero. Then it cancels the factor~\isa{n} on both sides of the equation \isa{(m\ *\ n)\ div\ n\ *\ n\ =\ m\ *\ n}, proving the theorem. \begin{warn} Any unknowns in the theorem given to \methdx{insert} will be universally quantified in the new assumption. \end{warn}% \index{*insert (method)|)} \subsection{The Method {\tt\slshape subgoal_tac}} \index{*subgoal_tac (method)}% A related method is \isa{subgoal_tac}, but instead of inserting a theorem as an assumption, it inserts an arbitrary formula. This formula must be proved later as a separate subgoal. The idea is to claim that the formula holds on the basis of the current assumptions, to use this claim to complete the proof, and finally to justify the claim. It gives the proof some structure. If you find yourself generating a complex assumption by a long series of forward steps, consider using \isa{subgoal_tac} instead: you can state the formula you are aiming for, and perhaps prove it automatically. Look at the following example. \begin{isabelle} \isacommand{lemma}\ "\isasymlbrakk(z::int)\ <\ 37;\ 66\ <\ 2*z;\ z*z\ \isasymnoteq\ 1225;\ Q(34);\ Q(36)\isasymrbrakk\isanewline \ \ \ \ \ \ \ \ \,\isasymLongrightarrow\ Q(z)"\isanewline \isacommand{apply}\ (subgoal_tac\ "z\ =\ 34\ \isasymor\ z\ =\ 36")\isanewline \isacommand{apply}\ blast\isanewline \isacommand{apply}\ (subgoal_tac\ "z\ \isasymnoteq\ 35")\isanewline \isacommand{apply}\ arith\isanewline \isacommand{apply}\ force\isanewline \isacommand{done} \end{isabelle} The first assumption tells us that \isa{z} is no greater than~36. The second tells us that \isa{z} is at least~34. The third assumption tells us that \isa{z} cannot be 35, since $35\times35=1225$. So \isa{z} is either 34 or~36, and since \isa{Q} holds for both of those values, we have the conclusion. The Isabelle proof closely follows this reasoning. The first step is to claim that \isa{z} is either 34 or 36. The resulting proof state gives us two subgoals: \begin{isabelle} %\isasymlbrakk z\ <\ 37;\ 66\ <\ 2\ *\ z;\ z\ *\ z\ \isasymnoteq\ 1225;\ %Q\ 34;\ Q\ 36\isasymrbrakk\ \isasymLongrightarrow\ Q\ z\isanewline \ 1.\ \isasymlbrakk z\ <\ 37;\ 66\ <\ 2\ *\ z;\ z\ *\ z\ \isasymnoteq\ 1225;\ Q\ 34;\ Q\ 36;\isanewline \ \ \ \ \ z\ =\ 34\ \isasymor\ z\ =\ 36\isasymrbrakk\isanewline \ \ \ \ \isasymLongrightarrow\ Q\ z\isanewline \ 2.\ \isasymlbrakk z\ <\ 37;\ 66\ <\ 2\ *\ z;\ z\ *\ z\ \isasymnoteq\ 1225;\ Q\ 34;\ Q\ 36\isasymrbrakk\isanewline \ \ \ \ \isasymLongrightarrow\ z\ =\ 34\ \isasymor\ z\ =\ 36 \end{isabelle} The first subgoal is trivial (\isa{blast}), but for the second Isabelle needs help to eliminate the case \isa{z}=35. The second invocation of {\isa{subgoal_tac}} leaves two subgoals: \begin{isabelle} \ 1.\ \isasymlbrakk z\ <\ 37;\ 66\ <\ 2\ *\ z;\ z\ *\ z\ \isasymnoteq\ 1225;\ Q\ 34;\ Q\ 36;\isanewline \ \ \ \ \ z\ \isasymnoteq\ 35\isasymrbrakk\isanewline \ \ \ \ \isasymLongrightarrow\ z\ =\ 34\ \isasymor\ z\ =\ 36\isanewline \ 2.\ \isasymlbrakk z\ <\ 37;\ 66\ <\ 2\ *\ z;\ z\ *\ z\ \isasymnoteq\ 1225;\ Q\ 34;\ Q\ 36\isasymrbrakk\isanewline \ \ \ \ \isasymLongrightarrow\ z\ \isasymnoteq\ 35 \end{isabelle} Assuming that \isa{z} is not 35, the first subgoal follows by linear arithmetic (\isa{arith}). For the second subgoal we apply the method \isa{force}, which proceeds by assuming that \isa{z}=35 and arriving at a contradiction. \medskip Summary of these methods: \begin{itemize} \item \methdx{insert} adds a theorem as a new assumption \item \methdx{subgoal_tac} adds a formula as a new assumption and leaves the subgoal of proving that formula \end{itemize} \index{forward proof|)} \section{Managing Large Proofs} Naturally you should try to divide proofs into manageable parts. Look for lemmas that can be proved separately. Sometimes you will observe that they are instances of much simpler facts. On other occasions, no lemmas suggest themselves and you are forced to cope with a long proof involving many subgoals. \subsection{Tacticals, or Control Structures} \index{tacticals|(}% If the proof is long, perhaps it at least has some regularity. Then you can express it more concisely using \textbf{tacticals}, which provide control structures. Here is a proof (it would be a one-liner using \isa{blast}, but forget that) that contains a series of repeated commands: % \begin{isabelle} \isacommand{lemma}\ "\isasymlbrakk P\isasymlongrightarrow Q;\ Q\isasymlongrightarrow R;\ R\isasymlongrightarrow S;\ P\isasymrbrakk \ \isasymLongrightarrow \ S"\isanewline \isacommand{apply}\ (drule\ mp,\ assumption)\isanewline \isacommand{apply}\ (drule\ mp,\ assumption)\isanewline \isacommand{apply}\ (drule\ mp,\ assumption)\isanewline \isacommand{apply}\ (assumption)\isanewline \isacommand{done} \end{isabelle} % Each of the three identical commands finds an implication and proves its antecedent by assumption. The first one finds \isa{P\isasymlongrightarrow Q} and \isa{P}, concluding~\isa{Q}; the second one concludes~\isa{R} and the third one concludes~\isa{S}. The final step matches the assumption \isa{S} with the goal to be proved. Suffixing a method with a plus sign~(\isa+)\index{*"+ (tactical)} expresses one or more repetitions: \begin{isabelle} \isacommand{lemma}\ "\isasymlbrakk P\isasymlongrightarrow Q;\ Q\isasymlongrightarrow R;\ R\isasymlongrightarrow S;\ P\isasymrbrakk \ \isasymLongrightarrow \ S"\isanewline \isacommand{by}\ (drule\ mp,\ assumption)+ \end{isabelle} % Using \isacommand{by} takes care of the final use of \isa{assumption}. The new proof is more concise. It is also more general: the repetitive method works for a chain of implications having any length, not just three. Choice is another control structure. Separating two methods by a vertical % we must use ?? rather than "| as the sorting item because somehow the presence % of | (even quoted) stops hyperref from putting |hyperpage at the end of the index % entry. bar~(\isa|)\index{??@\texttt{"|} (tactical)} gives the effect of applying the first method, and if that fails, trying the second. It can be combined with repetition, when the choice must be made over and over again. Here is a chain of implications in which most of the antecedents are proved by assumption, but one is proved by arithmetic: \begin{isabelle} \isacommand{lemma}\ "\isasymlbrakk Q\isasymlongrightarrow R;\ P\isasymlongrightarrow Q;\ x<5\isasymlongrightarrow P;\ Suc\ x\ <\ 5\isasymrbrakk \ \isasymLongrightarrow \ R"\ \isanewline \isacommand{by}\ (drule\ mp,\ (assumption|arith))+ \end{isabelle} The \isa{arith} method can prove $x<5$ from $x+1<5$, but it cannot duplicate the effect of \isa{assumption}. Therefore, we combine these methods using the choice operator. A postfixed question mark~(\isa?)\index{*"? (tactical)} expresses zero or one repetitions of a method. It can also be viewed as the choice between executing a method and doing nothing. It is useless at top level but can be valuable within other control structures; for example, \isa{($m$+)?} performs zero or more repetitions of method~$m$.% \index{tacticals|)} \subsection{Subgoal Numbering} Another problem in large proofs is contending with huge subgoals or many subgoals. Induction can produce a proof state that looks like this: \begin{isabelle} \ 1.\ bigsubgoal1\isanewline \ 2.\ bigsubgoal2\isanewline \ 3.\ bigsubgoal3\isanewline \ 4.\ bigsubgoal4\isanewline \ 5.\ bigsubgoal5\isanewline \ 6.\ bigsubgoal6 \end{isabelle} If each \isa{bigsubgoal} is 15 lines or so, the proof state will be too big to scroll through. By default, Isabelle displays at most 10 subgoals. The \commdx{pr} command lets you change this limit: \begin{isabelle} \isacommand{pr}\ 2\isanewline \ 1.\ bigsubgoal1\isanewline \ 2.\ bigsubgoal2\isanewline A total of 6 subgoals... \end{isabelle} \medskip All methods apply to the first subgoal. Sometimes, not only in a large proof, you may want to focus on some other subgoal. Then you should try the commands \isacommand{defer} or \isacommand{prefer}. In the following example, the first subgoal looks hard, while the others look as if \isa{blast} alone could prove them: \begin{isabelle} \ 1.\ hard\isanewline \ 2.\ \isasymnot \ \isasymnot \ P\ \isasymLongrightarrow \ P\isanewline \ 3.\ Q\ \isasymLongrightarrow \ Q% \end{isabelle} % The \commdx{defer} command moves the first subgoal into the last position. \begin{isabelle} \isacommand{defer}\ 1\isanewline \ 1.\ \isasymnot \ \isasymnot \ P\ \isasymLongrightarrow \ P\isanewline \ 2.\ Q\ \isasymLongrightarrow \ Q\isanewline \ 3.\ hard% \end{isabelle} % Now we apply \isa{blast} repeatedly to the easy subgoals: \begin{isabelle} \isacommand{apply}\ blast+\isanewline \ 1.\ hard% \end{isabelle} Using \isacommand{defer}, we have cleared away the trivial parts of the proof so that we can devote attention to the difficult part. \medskip The \commdx{prefer} command moves the specified subgoal into the first position. For example, if you suspect that one of your subgoals is invalid (not a theorem), then you should investigate that subgoal first. If it cannot be proved, then there is no point in proving the other subgoals. \begin{isabelle} \ 1.\ ok1\isanewline \ 2.\ ok2\isanewline \ 3.\ doubtful% \end{isabelle} % We decide to work on the third subgoal. \begin{isabelle} \isacommand{prefer}\ 3\isanewline \ 1.\ doubtful\isanewline \ 2.\ ok1\isanewline \ 3.\ ok2 \end{isabelle} If we manage to prove \isa{doubtful}, then we can work on the other subgoals, confident that we are not wasting our time. Finally we revise the proof script to remove the \isacommand{prefer} command, since we needed it only to focus our exploration. The previous example is different: its use of \isacommand{defer} stops trivial subgoals from cluttering the rest of the proof. Even there, we should consider proving \isa{hard} as a preliminary lemma. Always seek ways to streamline your proofs. \medskip Summary: \begin{itemize} \item the control structures \isa+, \isa? and \isa| help express complicated proofs \item the \isacommand{pr} command can limit the number of subgoals to display \item the \isacommand{defer} and \isacommand{prefer} commands move a subgoal to the last or first position \end{itemize} \begin{exercise} Explain the use of \isa? and \isa+ in this proof. \begin{isabelle} \isacommand{lemma}\ "\isasymlbrakk P\isasymand Q\isasymlongrightarrow R;\ P\isasymlongrightarrow Q;\ P\isasymrbrakk \ \isasymLongrightarrow \ R"\isanewline \isacommand{by}\ (drule\ mp,\ (intro conjI)?,\ assumption+)+ \end{isabelle} \end{exercise} \section{Proving the Correctness of Euclid's Algorithm} \label{sec:proving-euclid} \index{Euclid's algorithm|(}\index{*gcd (constant)|(}\index{divides relation|(}% A brief development will demonstrate the techniques of this chapter, including \isa{blast} applied with additional rules. We shall also see \isa{case_tac} used to perform a Boolean case split. Let us prove that \isa{gcd} computes the greatest common divisor of its two arguments. % We use induction: \isa{gcd.induct} is the induction rule returned by \isa{fun}. We simplify using rules proved in {\S}\ref{sec:fun-simplification}, since rewriting by the definition of \isa{gcd} can loop. \begin{isabelle} \isacommand{lemma}\ gcd\_dvd\_both:\ "(gcd\ m\ n\ dvd\ m)\ \isasymand \ (gcd\ m\ n\ dvd\ n)" \end{isabelle} The induction formula must be a conjunction. In the inductive step, each conjunct establishes the other. \begin{isabelle} \ 1.\ \isasymAnd m\ n.\ (n\ \isasymnoteq \ 0\ \isasymLongrightarrow \isanewline \isaindent{\ 1.\ \isasymAnd m\ n.\ (}gcd\ n\ (m\ mod\ n)\ dvd\ n\ \isasymand \isanewline \isaindent{\ 1.\ \isasymAnd m\ n.\ (}gcd\ n\ (m\ mod\ n)\ dvd\ m\ mod\ n)\ \isasymLongrightarrow \isanewline \isaindent{\ 1.\ \isasymAnd m\ n.\ }gcd\ m\ n\ dvd\ m\ \isasymand \ gcd\ m\ n\ dvd\ n% \end{isabelle} The conditional induction hypothesis suggests doing a case analysis on \isa{n=0}. We apply \methdx{case_tac} with type \isa{bool} --- and not with a datatype, as we have done until now. Since \isa{nat} is a datatype, we could have written \isa{case_tac~n} instead of \isa{case_tac~"n=0"}. However, the definition of \isa{gcd} makes a Boolean decision: \begin{isabelle} \ \ \ \ "gcd\ m\ n\ =\ (if\ n=0\ then\ m\ else\ gcd\ n\ (m\ mod\ n))" \end{isabelle} Proofs about a function frequently follow the function's definition, so we perform case analysis over the same formula. \begin{isabelle} \isacommand{apply}\ (case_tac\ "n=0")\isanewline \ 1.\ \isasymAnd m\ n.\ \isasymlbrakk n\ \isasymnoteq \ 0\ \isasymLongrightarrow \isanewline \isaindent{\ 1.\ \isasymAnd m\ n.\ \isasymlbrakk }gcd\ n\ (m\ mod\ n)\ dvd\ n\ \isasymand \ gcd\ n\ (m\ mod\ n)\ dvd\ m\ mod\ n;\isanewline \isaindent{\ 1.\ \isasymAnd m\ n.\ \ }n\ =\ 0\isasymrbrakk \isanewline \isaindent{\ 1.\ \isasymAnd m\ n.\ }\isasymLongrightarrow \ gcd\ m\ n\ dvd\ m\ \isasymand \ gcd\ m\ n\ dvd\ n\isanewline \ 2.\ \isasymAnd m\ n.\ \isasymlbrakk n\ \isasymnoteq \ 0\ \isasymLongrightarrow \isanewline \isaindent{\ 2.\ \isasymAnd m\ n.\ \isasymlbrakk }gcd\ n\ (m\ mod\ n)\ dvd\ n\ \isasymand \ gcd\ n\ (m\ mod\ n)\ dvd\ m\ mod\ n;\isanewline \isaindent{\ 2.\ \isasymAnd m\ n.\ \ }n\ \isasymnoteq \ 0\isasymrbrakk \isanewline \isaindent{\ 2.\ \isasymAnd m\ n.\ }\isasymLongrightarrow \ gcd\ m\ n\ dvd\ m\ \isasymand \ gcd\ m\ n\ dvd\ n% \end{isabelle} % Simplification leaves one subgoal: \begin{isabelle} \isacommand{apply}\ (simp_all)\isanewline \ 1.\ \isasymAnd m\ n.\ \isasymlbrakk gcd\ n\ (m\ mod\ n)\ dvd\ n\ \isasymand \ gcd\ n\ (m\ mod\ n)\ dvd\ m\ mod\ n;\isanewline \isaindent{\ 1.\ \isasymAnd m\ n.\ \ }0\ <\ n\isasymrbrakk \isanewline \isaindent{\ 1.\ \isasymAnd m\ n.\ }\isasymLongrightarrow \ gcd\ n\ (m\ mod\ n)\ dvd\ m% \end{isabelle} % Here, we can use \isa{blast}. One of the assumptions, the induction hypothesis, is a conjunction. The two divides relationships it asserts are enough to prove the conclusion, for we have the following theorem at our disposal: \begin{isabelle} \isasymlbrakk?k\ dvd\ (?m\ mod\ ?n){;}\ ?k\ dvd\ ?n\isasymrbrakk\ \isasymLongrightarrow\ ?k\ dvd\ ?m% \rulename{dvd_mod_imp_dvd} \end{isabelle} % This theorem can be applied in various ways. As an introduction rule, it would cause backward chaining from the conclusion (namely \isa{?k~dvd~?m}) to the two premises, which also involve the divides relation. This process does not look promising and could easily loop. More sensible is to apply the rule in the forward direction; each step would eliminate an occurrence of the \isa{mod} symbol, so the process must terminate. \begin{isabelle} \isacommand{apply}\ (blast\ dest:\ dvd_mod_imp_dvd)\isanewline \isacommand{done} \end{isabelle} Attaching the \attrdx{dest} attribute to \isa{dvd_mod_imp_dvd} tells \isa{blast} to use it as destruction rule; that is, in the forward direction. \medskip We have proved a conjunction. Now, let us give names to each of the two halves: \begin{isabelle} \isacommand{lemmas}\ gcd_dvd1\ [iff]\ =\ gcd_dvd_both\ [THEN\ conjunct1]\isanewline \isacommand{lemmas}\ gcd_dvd2\ [iff]\ =\ gcd_dvd_both\ [THEN\ conjunct2]% \end{isabelle} Here we see \commdx{lemmas} used with the \attrdx{iff} attribute, which supplies the new theorems to the classical reasoner and the simplifier. Recall that \attrdx{THEN} is frequently used with destruction rules; \isa{THEN conjunct1} extracts the first half of a conjunctive theorem. Given \isa{gcd_dvd_both} it yields \begin{isabelle} \ \ \ \ \ gcd\ ?m1\ ?n1\ dvd\ ?m1 \end{isabelle} The variable names \isa{?m1} and \isa{?n1} arise because Isabelle renames schematic variables to prevent clashes. The second \isacommand{lemmas} declaration yields \begin{isabelle} \ \ \ \ \ gcd\ ?m1\ ?n1\ dvd\ ?n1 \end{isabelle} \medskip To complete the verification of the \isa{gcd} function, we must prove that it returns the greatest of all the common divisors of its arguments. The proof is by induction, case analysis and simplification. \begin{isabelle} \isacommand{lemma}\ gcd\_greatest\ [rule\_format]:\isanewline \ \ \ \ \ \ "k\ dvd\ m\ \isasymlongrightarrow \ k\ dvd\ n\ \isasymlongrightarrow \ k\ dvd\ gcd\ m\ n" \end{isabelle} % The goal is expressed using HOL implication, \isa{\isasymlongrightarrow}, because the induction affects the two preconditions. The directive \isa{rule_format} tells Isabelle to replace each \isa{\isasymlongrightarrow} by \isa{\isasymLongrightarrow} before storing the eventual theorem. This directive can also remove outer universal quantifiers, converting the theorem into the usual format for inference rules. It can replace any series of applications of \isa{THEN} to the rules \isa{mp} and \isa{spec}. We did not have to write this: \begin{isabelle} \isacommand{lemma}\ gcd_greatest\ [THEN mp, THEN mp]:\isanewline \ \ \ \ \ \ "k\ dvd\ m\ \isasymlongrightarrow \ k\ dvd\ n\ \isasymlongrightarrow \ k\ dvd\ gcd\ m\ n" \end{isabelle} Because we are again reasoning about \isa{gcd}, we perform the same induction and case analysis as in the previous proof: \begingroup\samepage \begin{isabelle} \ 1.\ \isasymAnd m\ n.\ \isasymlbrakk n\ \isasymnoteq \ 0\ \isasymLongrightarrow \isanewline \isaindent{\ 1.\ \isasymAnd m\ n.\ \isasymlbrakk }k\ dvd\ n\ \isasymlongrightarrow \ k\ dvd\ m\ mod\ n\ \isasymlongrightarrow \ k\ dvd\ gcd\ n\ (m\ mod\ n);\isanewline \isaindent{\ 1.\ \isasymAnd m\ n.\ \ }n\ =\ 0\isasymrbrakk \isanewline \isaindent{\ 1.\ \isasymAnd m\ n.\ }\isasymLongrightarrow \ k\ dvd\ m\ \isasymlongrightarrow \ k\ dvd\ n\ \isasymlongrightarrow \ k\ dvd\ gcd\ m\ n\isanewline \ 2.\ \isasymAnd m\ n.\ \isasymlbrakk n\ \isasymnoteq \ 0\ \isasymLongrightarrow \isanewline \isaindent{\ 2.\ \isasymAnd m\ n.\ \isasymlbrakk }k\ dvd\ n\ \isasymlongrightarrow \ k\ dvd\ m\ mod\ n\ \isasymlongrightarrow \ k\ dvd\ gcd\ n\ (m\ mod\ n);\isanewline \isaindent{\ 2.\ \isasymAnd m\ n.\ \ }n\ \isasymnoteq \ 0\isasymrbrakk \isanewline \isaindent{\ 2.\ \isasymAnd m\ n.\ }\isasymLongrightarrow \ k\ dvd\ m\ \isasymlongrightarrow \ k\ dvd\ n\ \isasymlongrightarrow \ k\ dvd\ gcd\ m\ n% \end{isabelle} \endgroup \noindent Simplification proves both subgoals. \begin{isabelle} \isacommand{apply}\ (simp_all\ add:\ dvd_mod)\isanewline \isacommand{done} \end{isabelle} In the first, where \isa{n=0}, the implication becomes trivial: \isa{k\ dvd\ gcd\ m\ n} goes to~\isa{k\ dvd\ m}. The second subgoal is proved by an unfolding of \isa{gcd}, using this rule about divides: \begin{isabelle} \isasymlbrakk ?f\ dvd\ ?m;\ ?f\ dvd\ ?n\isasymrbrakk \ \isasymLongrightarrow \ ?f\ dvd\ ?m\ mod\ ?n% \rulename{dvd_mod} \end{isabelle} \medskip The facts proved above can be summarized as a single logical equivalence. This step gives us a chance to see another application of \isa{blast}. \begin{isabelle} \isacommand{theorem}\ gcd\_greatest\_iff\ [iff]:\ \isanewline \ \ \ \ \ \ \ \ "(k\ dvd\ gcd\ m\ n)\ =\ (k\ dvd\ m\ \isasymand \ k\ dvd\ n)"\isanewline \isacommand{by}\ (blast\ intro!:\ gcd_greatest\ intro:\ dvd_trans) \end{isabelle} This theorem concisely expresses the correctness of the \isa{gcd} function. We state it with the \isa{iff} attribute so that Isabelle can use it to remove some occurrences of \isa{gcd}. The theorem has a one-line proof using \isa{blast} supplied with two additional introduction rules. The exclamation mark ({\isa{intro}}{\isa{!}})\ signifies safe rules, which are applied aggressively. Rules given without the exclamation mark are applied reluctantly and their uses can be undone if the search backtracks. Here the unsafe rule expresses transitivity of the divides relation: \begin{isabelle} \isasymlbrakk?m\ dvd\ ?n;\ ?n\ dvd\ ?p\isasymrbrakk\ \isasymLongrightarrow\ ?m\ dvd\ ?p% \rulename{dvd_trans} \end{isabelle} Applying \isa{dvd_trans} as an introduction rule entails a risk of looping, for it multiplies occurrences of the divides symbol. However, this proof relies on transitivity reasoning. The rule {\isa{gcd\_greatest}} is safe to apply aggressively because it yields simpler subgoals. The proof implicitly uses \isa{gcd_dvd1} and \isa{gcd_dvd2} as safe rules, because they were declared using \isa{iff}.% \index{Euclid's algorithm|)}\index{*gcd (constant)|)}\index{divides relation|)}