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diff --git a/doc/tutorial/Tutorial.tex b/doc/tutorial/Tutorial.tex deleted file mode 100644 index 77ce8574f..000000000 --- a/doc/tutorial/Tutorial.tex +++ /dev/null @@ -1,1575 +0,0 @@ -\documentclass[11pt,a4paper]{book} -\usepackage[T1]{fontenc} -\usepackage[utf8]{inputenc} -\usepackage{textcomp} -\usepackage{pslatex} -\usepackage{hyperref} - -\input{../common/version.tex} -\input{../common/macros.tex} -\input{../common/title.tex} - -%\makeindex - -\begin{document} -\coverpage{A Tutorial}{Gérard Huet, Gilles Kahn and Christine Paulin-Mohring}{} - -%\tableofcontents - -\chapter*{Getting started} - -\Coq{} is a Proof Assistant for a Logical Framework known as the Calculus -of Inductive Constructions. It allows the interactive construction of -formal proofs, and also the manipulation of functional programs -consistently with their specifications. It runs as a computer program -on many architectures. - -It is available with a variety of user interfaces. The present -document does not attempt to present a comprehensive view of all the -possibilities of \Coq, but rather to present in the most elementary -manner a tutorial on the basic specification language, called Gallina, -in which formal axiomatisations may be developed, and on the main -proof tools. For more advanced information, the reader could refer to -the \Coq{} Reference Manual or the \textit{Coq'Art}, a book by Y. -Bertot and P. Castéran on practical uses of the \Coq{} system. - -Instructions on installation procedures, as well as more comprehensive -documentation, may be found in the standard distribution of \Coq, -which may be obtained from \Coq{} web site -\url{https://coq.inria.fr/}\footnote{You can report any bug you find -while using \Coq{} at \url{https://coq.inria.fr/bugs}. Make sure to -always provide a way to reproduce it and to specify the exact version -you used. You can get this information by running \texttt{coqc -v}}. -\Coq{} is distributed together with a graphical user interface called -CoqIDE. Alternative interfaces exist such as -Proof General\footnote{See \url{https://proofgeneral.github.io/}.}. - -In the following examples, lines preceded by the prompt \verb:Coq < : -represent user input, terminated by a period. -The following lines usually show \Coq's answer. -When used from a graphical user interface such as -CoqIDE, the prompt is not displayed: user input is given in one window -and \Coq's answers are displayed in a different window. - -\chapter{Basic Predicate Calculus} - -\section{An overview of the specification language Gallina} - -A formal development in Gallina consists in a sequence of {\sl declarations} -and {\sl definitions}. - -\subsection{Declarations} - -A declaration associates a {\sl name} with a {\sl specification}. -A name corresponds roughly to an identifier in a programming -language, i.e. to a string of letters, digits, and a few ASCII symbols like -underscore (\verb"_") and prime (\verb"'"), starting with a letter. -We use case distinction, so that the names \verb"A" and \verb"a" are distinct. -Certain strings are reserved as key-words of \Coq, and thus are forbidden -as user identifiers. - -A specification is a formal expression which classifies the notion which is -being declared. There are basically three kinds of specifications: -{\sl logical propositions}, {\sl mathematical collections}, and -{\sl abstract types}. They are classified by the three basic sorts -of the system, called respectively \verb:Prop:, \verb:Set:, and -\verb:Type:, which are themselves atomic abstract types. - -Every valid expression $e$ in Gallina is associated with a specification, -itself a valid expression, called its {\sl type} $\tau(E)$. We write -$e:\tau(E)$ for the judgment that $e$ is of type $E$. -You may request \Coq{} to return to you the type of a valid expression by using -the command \verb:Check:: - -\begin{coq_eval} -Set Printing Width 60. -\end{coq_eval} - -\begin{coq_example} -Check O. -\end{coq_example} - -Thus we know that the identifier \verb:O: (the name `O', not to be -confused with the numeral `0' which is not a proper identifier!) is -known in the current context, and that its type is the specification -\verb:nat:. This specification is itself classified as a mathematical -collection, as we may readily check: - -\begin{coq_example} -Check nat. -\end{coq_example} - -The specification \verb:Set: is an abstract type, one of the basic -sorts of the Gallina language, whereas the notions $nat$ and $O$ are -notions which are defined in the arithmetic prelude, -automatically loaded when running the \Coq{} system. - -We start by introducing a so-called section name. The role of sections -is to structure the modelisation by limiting the scope of parameters, -hypotheses and definitions. It will also give a convenient way to -reset part of the development. - -\begin{coq_example} -Section Declaration. -\end{coq_example} -With what we already know, we may now enter in the system a declaration, -corresponding to the informal mathematics {\sl let n be a natural - number}. - -\begin{coq_example} -Variable n : nat. -\end{coq_example} - -If we want to translate a more precise statement, such as -{\sl let n be a positive natural number}, -we have to add another declaration, which will declare explicitly the -hypothesis \verb:Pos_n:, with specification the proper logical -proposition: -\begin{coq_example} -Hypothesis Pos_n : (gt n 0). -\end{coq_example} - -Indeed we may check that the relation \verb:gt: is known with the right type -in the current context: - -\begin{coq_example} -Check gt. -\end{coq_example} - -which tells us that \texttt{gt} is a function expecting two arguments of -type \texttt{nat} in order to build a logical proposition. -What happens here is similar to what we are used to in a functional -programming language: we may compose the (specification) type \texttt{nat} -with the (abstract) type \texttt{Prop} of logical propositions through the -arrow function constructor, in order to get a functional type -\texttt{nat -> Prop}: -\begin{coq_example} -Check (nat -> Prop). -\end{coq_example} -which may be composed once more with \verb:nat: in order to obtain the -type \texttt{nat -> nat -> Prop} of binary relations over natural numbers. -Actually the type \texttt{nat -> nat -> Prop} is an abbreviation for -\texttt{nat -> (nat -> Prop)}. - -Functional notions may be composed in the usual way. An expression $f$ -of type $A\ra B$ may be applied to an expression $e$ of type $A$ in order -to form the expression $(f~e)$ of type $B$. Here we get that -the expression \verb:(gt n): is well-formed of type \texttt{nat -> Prop}, -and thus that the expression \verb:(gt n O):, which abbreviates -\verb:((gt n) O):, is a well-formed proposition. -\begin{coq_example} -Check gt n O. -\end{coq_example} - -\subsection{Definitions} - -The initial prelude contains a few arithmetic definitions: -\texttt{nat} is defined as a mathematical collection (type \texttt{Set}), -constants \texttt{O}, \texttt{S}, \texttt{plus}, are defined as objects of -types respectively \texttt{nat}, \texttt{nat -> nat}, and \texttt{nat -> -nat -> nat}. -You may introduce new definitions, which link a name to a well-typed value. -For instance, we may introduce the constant \texttt{one} as being defined -to be equal to the successor of zero: -\begin{coq_example} -Definition one := (S O). -\end{coq_example} -We may optionally indicate the required type: -\begin{coq_example} -Definition two : nat := S one. -\end{coq_example} - -Actually \Coq{} allows several possible syntaxes: -\begin{coq_example} -Definition three := S two : nat. -\end{coq_example} - -Here is a way to define the doubling function, which expects an -argument \verb:m: of type \verb:nat: in order to build its result as -\verb:(plus m m):: - -\begin{coq_example} -Definition double (m : nat) := plus m m. -\end{coq_example} -This introduces the constant \texttt{double} defined as the -expression \texttt{fun m : nat => plus m m}. -The abstraction introduced by \texttt{fun} is explained as follows. -The expression \texttt{fun x : A => e} is well formed of type -\texttt{A -> B} in a context whenever the expression \texttt{e} is -well-formed of type \texttt{B} in the given context to which we add the -declaration that \texttt{x} is of type \texttt{A}. Here \texttt{x} is a -bound, or dummy variable in the expression \texttt{fun x : A => e}. -For instance we could as well have defined \texttt{double} as -\texttt{fun n : nat => (plus n n)}. - -Bound (local) variables and free (global) variables may be mixed. -For instance, we may define the function which adds the constant \verb:n: -to its argument as -\begin{coq_example} -Definition add_n (m:nat) := plus m n. -\end{coq_example} -However, note that here we may not rename the formal argument $m$ into $n$ -without capturing the free occurrence of $n$, and thus changing the meaning -of the defined notion. - -Binding operations are well known for instance in logic, where they -are called quantifiers. Thus we may universally quantify a -proposition such as $m>0$ in order to get a universal proposition -$\forall m\cdot m>0$. Indeed this operator is available in \Coq, with -the following syntax: \texttt{forall m : nat, gt m O}. Similarly to the -case of the functional abstraction binding, we are obliged to declare -explicitly the type of the quantified variable. We check: -\begin{coq_example} -Check (forall m : nat, gt m 0). -\end{coq_example} - -\begin{coq_eval} -Reset Initial. -Set Printing Width 60. -Set Printing Compact Contexts. -\end{coq_eval} - -\section{Introduction to the proof engine: Minimal Logic} - -In the following, we are going to consider various propositions, built -from atomic propositions $A, B, C$. This may be done easily, by -introducing these atoms as global variables declared of type \verb:Prop:. -It is easy to declare several names with the same specification: -\begin{coq_example} -Section Minimal_Logic. -Variables A B C : Prop. -\end{coq_example} - -We shall consider simple implications, such as $A\ra B$, read as -``$A$ implies $B$''. Note that we overload the arrow symbol, which -has been used above as the functionality type constructor, and which -may be used as well as propositional connective: -\begin{coq_example} -Check (A -> B). -\end{coq_example} - -Let us now embark on a simple proof. We want to prove the easy tautology -$((A\ra (B\ra C))\ra (A\ra B)\ra (A\ra C)$. -We enter the proof engine by the command -\verb:Goal:, followed by the conjecture we want to verify: -\begin{coq_example} -Goal (A -> B -> C) -> (A -> B) -> A -> C. -\end{coq_example} - -The system displays the current goal below a double line, local hypotheses -(there are none initially) being displayed above the line. We call -the combination of local hypotheses with a goal a {\sl judgment}. -We are now in an inner -loop of the system, in proof mode. -New commands are available in this -mode, such as {\sl tactics}, which are proof combining primitives. -A tactic operates on the current goal by attempting to construct a proof -of the corresponding judgment, possibly from proofs of some -hypothetical judgments, which are then added to the current -list of conjectured judgments. -For instance, the \verb:intro: tactic is applicable to any judgment -whose goal is an implication, by moving the proposition to the left -of the application to the list of local hypotheses: -\begin{coq_example} -intro H. -\end{coq_example} - -Several introductions may be done in one step: -\begin{coq_example} -intros H' HA. -\end{coq_example} - -We notice that $C$, the current goal, may be obtained from hypothesis -\verb:H:, provided the truth of $A$ and $B$ are established. -The tactic \verb:apply: implements this piece of reasoning: -\begin{coq_example} -apply H. -\end{coq_example} - -We are now in the situation where we have two judgments as conjectures -that remain to be proved. Only the first is listed in full, for the -others the system displays only the corresponding subgoal, without its -local hypotheses list. Note that \verb:apply: has kept the local -hypotheses of its father judgment, which are still available for -the judgments it generated. - -In order to solve the current goal, we just have to notice that it is -exactly available as hypothesis $HA$: -\begin{coq_example} -exact HA. -\end{coq_example} - -Now $H'$ applies: -\begin{coq_example} -apply H'. -\end{coq_example} - -And we may now conclude the proof as before, with \verb:exact HA.: -Actually, we may not bother with the name \verb:HA:, and just state that -the current goal is solvable from the current local assumptions: -\begin{coq_example} -assumption. -\end{coq_example} - -The proof is now finished. We are now going to ask \Coq{}'s kernel -to check and save the proof. -\begin{coq_example} -Qed. -\end{coq_example} - -Let us redo the same proof with a few variations. First of all we may name -the initial goal as a conjectured lemma: -\begin{coq_example} -Lemma distr_impl : (A -> B -> C) -> (A -> B) -> A -> C. -\end{coq_example} - -Next, we may omit the names of local assumptions created by the introduction -tactics, they can be automatically created by the proof engine as new -non-clashing names. -\begin{coq_example} -intros. -\end{coq_example} - -The \verb:intros: tactic, with no arguments, effects as many individual -applications of \verb:intro: as is legal. - -Then, we may compose several tactics together in sequence, or in parallel, -through {\sl tacticals}, that is tactic combinators. The main constructions -are the following: -\begin{itemize} -\item $T_1 ; T_2$ (read $T_1$ then $T_2$) applies tactic $T_1$ to the current -goal, and then tactic $T_2$ to all the subgoals generated by $T_1$. -\item $T; [T_1 | T_2 | ... | T_n]$ applies tactic $T$ to the current -goal, and then tactic $T_1$ to the first newly generated subgoal, -..., $T_n$ to the nth. -\end{itemize} - -We may thus complete the proof of \verb:distr_impl: with one composite tactic: -\begin{coq_example} -apply H; [ assumption | apply H0; assumption ]. -\end{coq_example} - -You should be aware however that relying on automatically generated names is -not robust to slight updates to this proof script. Consequently, it is -discouraged in finished proof scripts. As for the composition of tactics with -\texttt{:} it may hinder the readability of the proof script and it is also -harder to see what's going on when replaying the proof because composed -tactics are evaluated in one go. - -Actually, such an easy combination of tactics \verb:intro:, \verb:apply: -and \verb:assumption: may be found completely automatically by an automatic -tactic, called \verb:auto:, without user guidance: - -\begin{coq_eval} -Abort. -\end{coq_eval} -\begin{coq_example} -Lemma distr_impl : (A -> B -> C) -> (A -> B) -> A -> C. -auto. -\end{coq_example} - -Let us now save lemma \verb:distr_impl:: -\begin{coq_example} -Qed. -\end{coq_example} - -\section{Propositional Calculus} - -\subsection{Conjunction} - -We have seen how \verb:intro: and \verb:apply: tactics could be combined -in order to prove implicational statements. More generally, \Coq{} favors a style -of reasoning, called {\sl Natural Deduction}, which decomposes reasoning into -so called {\sl introduction rules}, which tell how to prove a goal whose main -operator is a given propositional connective, and {\sl elimination rules}, -which tell how to use an hypothesis whose main operator is the propositional -connective. Let us show how to use these ideas for the propositional connectives -\verb:/\: and \verb:\/:. - -\begin{coq_example} -Lemma and_commutative : A /\ B -> B /\ A. -intro H. -\end{coq_example} - -We make use of the conjunctive hypothesis \verb:H: with the \verb:elim: tactic, -which breaks it into its components: -\begin{coq_example} -elim H. -\end{coq_example} - -We now use the conjunction introduction tactic \verb:split:, which splits the -conjunctive goal into the two subgoals: -\begin{coq_example} -split. -\end{coq_example} -and the proof is now trivial. Indeed, the whole proof is obtainable as follows: -\begin{coq_eval} -Abort. -\end{coq_eval} -\begin{coq_example} -Lemma and_commutative : A /\ B -> B /\ A. -intro H; elim H; auto. -Qed. -\end{coq_example} - -The tactic \verb:auto: succeeded here because it knows as a hint the -conjunction introduction operator \verb+conj+ -\begin{coq_example} -Check conj. -\end{coq_example} - -Actually, the tactic \verb+split+ is just an abbreviation for \verb+apply conj.+ - -What we have just seen is that the \verb:auto: tactic is more powerful than -just a simple application of local hypotheses; it tries to apply as well -lemmas which have been specified as hints. A -\verb:Hint Resolve: command registers a -lemma as a hint to be used from now on by the \verb:auto: tactic, whose power -may thus be incrementally augmented. - -\subsection{Disjunction} - -In a similar fashion, let us consider disjunction: - -\begin{coq_example} -Lemma or_commutative : A \/ B -> B \/ A. -intro H; elim H. -\end{coq_example} - -Let us prove the first subgoal in detail. We use \verb:intro: in order to -be left to prove \verb:B\/A: from \verb:A:: - -\begin{coq_example} -intro HA. -\end{coq_example} - -Here the hypothesis \verb:H: is not needed anymore. We could choose to -actually erase it with the tactic \verb:clear:; in this simple proof it -does not really matter, but in bigger proof developments it is useful to -clear away unnecessary hypotheses which may clutter your screen. -\begin{coq_example} -clear H. -\end{coq_example} - -The tactic \verb:destruct: combines the effects of \verb:elim:, \verb:intros:, -and \verb:clear:: - -\begin{coq_eval} -Abort. -\end{coq_eval} -\begin{coq_example} -Lemma or_commutative : A \/ B -> B \/ A. -intros H; destruct H. -\end{coq_example} - -The disjunction connective has two introduction rules, since \verb:P\/Q: -may be obtained from \verb:P: or from \verb:Q:; the two corresponding -proof constructors are called respectively \verb:or_introl: and -\verb:or_intror:; they are applied to the current goal by tactics -\verb:left: and \verb:right: respectively. For instance: -\begin{coq_example} -right. -trivial. -\end{coq_example} -The tactic \verb:trivial: works like \verb:auto: with the hints -database, but it only tries those tactics that can solve the goal in one -step. - -As before, all these tedious elementary steps may be performed automatically, -as shown for the second symmetric case: - -\begin{coq_example} -auto. -\end{coq_example} - -However, \verb:auto: alone does not succeed in proving the full lemma, because -it does not try any elimination step. -It is a bit disappointing that \verb:auto: is not able to prove automatically -such a simple tautology. The reason is that we want to keep -\verb:auto: efficient, so that it is always effective to use. - -\subsection{Tauto} - -A complete tactic for propositional -tautologies is indeed available in \Coq{} as the \verb:tauto: tactic. -\begin{coq_eval} -Abort. -\end{coq_eval} -\begin{coq_example} -Lemma or_commutative : A \/ B -> B \/ A. -tauto. -Qed. -\end{coq_example} - -It is possible to inspect the actual proof tree constructed by \verb:tauto:, -using a standard command of the system, which prints the value of any notion -currently defined in the context: -\begin{coq_example} -Print or_commutative. -\end{coq_example} - -It is not easy to understand the notation for proof terms without some -explanations. The \texttt{fun} prefix, such as \verb+fun H : A\/B =>+, -corresponds -to \verb:intro H:, whereas a subterm such as -\verb:(or_intror: \verb:B H0): -corresponds to the sequence of tactics \verb:apply or_intror; exact H0:. -The generic combinator \verb:or_intror: needs to be instantiated by -the two properties \verb:B: and \verb:A:. Because \verb:A: can be -deduced from the type of \verb:H0:, only \verb:B: is printed. -The two instantiations are effected automatically by the tactic -\verb:apply: when pattern-matching a goal. The specialist will of course -recognize our proof term as a $\lambda$-term, used as notation for the -natural deduction proof term through the Curry-Howard isomorphism. The -naive user of \Coq{} may safely ignore these formal details. - -Let us exercise the \verb:tauto: tactic on a more complex example: -\begin{coq_example} -Lemma distr_and : A -> B /\ C -> (A -> B) /\ (A -> C). -tauto. -Qed. -\end{coq_example} - -\subsection{Classical reasoning} - -The tactic \verb:tauto: always comes back with an answer. Here is an example where it -fails: -\begin{coq_example} -Lemma Peirce : ((A -> B) -> A) -> A. -try tauto. -\end{coq_example} - -Note the use of the \verb:try: tactical, which does nothing if its tactic -argument fails. - -This may come as a surprise to someone familiar with classical reasoning. -Peirce's lemma is true in Boolean logic, i.e. it evaluates to \verb:true: for -every truth-assignment to \verb:A: and \verb:B:. Indeed the double negation -of Peirce's law may be proved in \Coq{} using \verb:tauto:: -\begin{coq_eval} -Abort. -\end{coq_eval} -\begin{coq_example} -Lemma NNPeirce : ~ ~ (((A -> B) -> A) -> A). -tauto. -Qed. -\end{coq_example} - -In classical logic, the double negation of a proposition is equivalent to this -proposition, but in the constructive logic of \Coq{} this is not so. If you -want to use classical logic in \Coq, you have to import explicitly the -\verb:Classical: module, which will declare the axiom \verb:classic: -of excluded middle, and classical tautologies such as de Morgan's laws. -The \verb:Require: command is used to import a module from \Coq's library: -\begin{coq_example} -Require Import Classical. -Check NNPP. -\end{coq_example} - -and it is now easy (although admittedly not the most direct way) to prove -a classical law such as Peirce's: -\begin{coq_example} -Lemma Peirce : ((A -> B) -> A) -> A. -apply NNPP; tauto. -Qed. -\end{coq_example} - -Here is one more example of propositional reasoning, in the shape of -a Scottish puzzle. A private club has the following rules: -\begin{enumerate} -\item Every non-scottish member wears red socks -\item Every member wears a kilt or doesn't wear red socks -\item The married members don't go out on Sunday -\item A member goes out on Sunday if and only if he is Scottish -\item Every member who wears a kilt is Scottish and married -\item Every scottish member wears a kilt -\end{enumerate} -Now, we show that these rules are so strict that no one can be accepted. -\begin{coq_example} -Section club. -Variables Scottish RedSocks WearKilt Married GoOutSunday : Prop. -Hypothesis rule1 : ~ Scottish -> RedSocks. -Hypothesis rule2 : WearKilt \/ ~ RedSocks. -Hypothesis rule3 : Married -> ~ GoOutSunday. -Hypothesis rule4 : GoOutSunday <-> Scottish. -Hypothesis rule5 : WearKilt -> Scottish /\ Married. -Hypothesis rule6 : Scottish -> WearKilt. -Lemma NoMember : False. -tauto. -Qed. -\end{coq_example} -At that point \verb:NoMember: is a proof of the absurdity depending on -hypotheses. -We may end the section, in that case, the variables and hypotheses -will be discharged, and the type of \verb:NoMember: will be -generalised. - -\begin{coq_example} -End club. -Check NoMember. -\end{coq_example} - -\section{Predicate Calculus} - -Let us now move into predicate logic, and first of all into first-order -predicate calculus. The essence of predicate calculus is that to try to prove -theorems in the most abstract possible way, without using the definitions of -the mathematical notions, but by formal manipulations of uninterpreted -function and predicate symbols. - -\subsection{Sections and signatures} - -Usually one works in some domain of discourse, over which range the individual -variables and function symbols. In \Coq{}, we speak in a language with a rich -variety of types, so we may mix several domains of discourse, in our -multi-sorted language. For the moment, we just do a few exercises, over a -domain of discourse \verb:D: axiomatised as a \verb:Set:, and we consider two -predicate symbols \verb:P: and \verb:R: over \verb:D:, of arities -1 and 2, respectively. - -\begin{coq_eval} -Reset Initial. -Set Printing Width 60. -Set Printing Compact Contexts. -\end{coq_eval} - -We start by assuming a domain of -discourse \verb:D:, and a binary relation \verb:R: over \verb:D:: -\begin{coq_example} -Section Predicate_calculus. -Variable D : Set. -Variable R : D -> D -> Prop. -\end{coq_example} - -As a simple example of predicate calculus reasoning, let us assume -that relation \verb:R: is symmetric and transitive, and let us show that -\verb:R: is reflexive in any point \verb:x: which has an \verb:R: successor. -Since we do not want to make the assumptions about \verb:R: global axioms of -a theory, but rather local hypotheses to a theorem, we open a specific -section to this effect. -\begin{coq_example} -Section R_sym_trans. -Hypothesis R_symmetric : forall x y : D, R x y -> R y x. -Hypothesis R_transitive : - forall x y z : D, R x y -> R y z -> R x z. -\end{coq_example} - -Note the syntax \verb+forall x : D,+ which stands for universal quantification -$\forall x : D$. - -\subsection{Existential quantification} - -We now state our lemma, and enter proof mode. -\begin{coq_example} -Lemma refl_if : forall x : D, (exists y, R x y) -> R x x. -\end{coq_example} - -The hypotheses that are local to the currently opened sections -are listed as local hypotheses to the current goals. -That is because these hypotheses are going to be discharged, as -we shall see, when we shall close the corresponding sections. - -Note the functional syntax for existential quantification. The existential -quantifier is built from the operator \verb:ex:, which expects a -predicate as argument: -\begin{coq_example} -Check ex. -\end{coq_example} -and the notation \verb+(exists x : D, P x)+ is just concrete syntax for -the expression \verb+(ex D (fun x : D => P x))+. -Existential quantification is handled in \Coq{} in a similar -fashion to the connectives \verb:/\: and \verb:\/:: it is introduced by -the proof combinator \verb:ex_intro:, which is invoked by the specific -tactic \verb:exists:, and its elimination provides a witness \verb+a : D+ to -\verb:P:, together with an assumption \verb+h : (P a)+ that indeed \verb+a+ -verifies \verb:P:. Let us see how this works on this simple example. -\begin{coq_example} -intros x x_Rlinked. -\end{coq_example} - -Note that \verb:intros: treats universal quantification in the same way -as the premises of implications. Renaming of bound variables occurs -when it is needed; for instance, had we started with \verb:intro y:, -we would have obtained the goal: -\begin{coq_eval} -Undo. -\end{coq_eval} -\begin{coq_example} -intro y. -\end{coq_example} -\begin{coq_eval} -Undo. -intros x x_Rlinked. -\end{coq_eval} - -Let us now use the existential hypothesis \verb:x_Rlinked: to -exhibit an R-successor y of x. This is done in two steps, first with -\verb:elim:, then with \verb:intros: - -\begin{coq_example} -elim x_Rlinked. -intros y Rxy. -\end{coq_example} - -Now we want to use \verb:R_transitive:. The \verb:apply: tactic will know -how to match \verb:x: with \verb:x:, and \verb:z: with \verb:x:, but needs -help on how to instantiate \verb:y:, which appear in the hypotheses of -\verb:R_transitive:, but not in its conclusion. We give the proper hint -to \verb:apply: in a \verb:with: clause, as follows: -\begin{coq_example} -apply R_transitive with y. -\end{coq_example} - -The rest of the proof is routine: -\begin{coq_example} -assumption. -apply R_symmetric; assumption. -\end{coq_example} -\begin{coq_example*} -Qed. -\end{coq_example*} - -Let us now close the current section. -\begin{coq_example} -End R_sym_trans. -\end{coq_example} - -All the local hypotheses have been -discharged in the statement of \verb:refl_if:, which now becomes a general -theorem in the first-order language declared in section -\verb:Predicate_calculus:. In this particular example, section -\verb:R_sym_trans: has not been really useful, since we could have -instead stated theorem \verb:refl_if: in its general form, and done -basically the same proof, obtaining \verb:R_symmetric: and -\verb:R_transitive: as local hypotheses by initial \verb:intros: rather -than as global hypotheses in the context. But if we had pursued the -theory by proving more theorems about relation \verb:R:, -we would have obtained all general statements at the closing of the section, -with minimal dependencies on the hypotheses of symmetry and transitivity. - -\subsection{Paradoxes of classical predicate calculus} - -Let us illustrate this feature by pursuing our \verb:Predicate_calculus: -section with an enrichment of our language: we declare a unary predicate -\verb:P: and a constant \verb:d:: -\begin{coq_example} -Variable P : D -> Prop. -Variable d : D. -\end{coq_example} - -We shall now prove a well-known fact from first-order logic: a universal -predicate is non-empty, or in other terms existential quantification -follows from universal quantification. -\begin{coq_example} -Lemma weird : (forall x:D, P x) -> exists a, P a. - intro UnivP. -\end{coq_example} - -First of all, notice the pair of parentheses around -\verb+forall x : D, P x+ in -the statement of lemma \verb:weird:. -If we had omitted them, \Coq's parser would have interpreted the -statement as a truly trivial fact, since we would -postulate an \verb:x: verifying \verb:(P x):. Here the situation is indeed -more problematic. If we have some element in \verb:Set: \verb:D:, we may -apply \verb:UnivP: to it and conclude, otherwise we are stuck. Indeed -such an element \verb:d: exists, but this is just by virtue of our -new signature. This points out a subtle difference between standard -predicate calculus and \Coq. In standard first-order logic, -the equivalent of lemma \verb:weird: always holds, -because such a rule is wired in the inference rules for quantifiers, the -semantic justification being that the interpretation domain is assumed to -be non-empty. Whereas in \Coq, where types are not assumed to be -systematically inhabited, lemma \verb:weird: only holds in signatures -which allow the explicit construction of an element in the domain of -the predicate. - -Let us conclude the proof, in order to show the use of the \verb:exists: -tactic: -\begin{coq_example} -exists d; trivial. -Qed. -\end{coq_example} - -Another fact which illustrates the sometimes disconcerting rules of -classical -predicate calculus is Smullyan's drinkers' paradox: ``In any non-empty -bar, there is a person such that if she drinks, then everyone drinks''. -We modelize the bar by Set \verb:D:, drinking by predicate \verb:P:. -We shall need classical reasoning. Instead of loading the \verb:Classical: -module as we did above, we just state the law of excluded middle as a -local hypothesis schema at this point: -\begin{coq_example} -Hypothesis EM : forall A : Prop, A \/ ~ A. -Lemma drinker : exists x : D, P x -> forall x : D, P x. -\end{coq_example} -The proof goes by cases on whether or not -there is someone who does not drink. Such reasoning by cases proceeds -by invoking the excluded middle principle, via \verb:elim: of the -proper instance of \verb:EM:: -\begin{coq_example} -elim (EM (exists x, ~ P x)). -\end{coq_example} - -We first look at the first case. Let Tom be the non-drinker. -The following combines at once the effect of \verb:intros: and -\verb:destruct:: -\begin{coq_example} -intros (Tom, Tom_does_not_drink). -\end{coq_example} - -We conclude in that case by considering Tom, since his drinking leads to -a contradiction: -\begin{coq_example} -exists Tom; intro Tom_drinks. -\end{coq_example} - -There are several ways in which we may eliminate a contradictory case; -in this case, we use \verb:contradiction: to let \Coq{} find out the -two contradictory hypotheses: -\begin{coq_example} -contradiction. -\end{coq_example} - -We now proceed with the second case, in which actually any person will do; -such a John Doe is given by the non-emptiness witness \verb:d:: -\begin{coq_example} -intro No_nondrinker; exists d; intro d_drinks. -\end{coq_example} - -Now we consider any Dick in the bar, and reason by cases according to its -drinking or not: -\begin{coq_example} -intro Dick; elim (EM (P Dick)); trivial. -\end{coq_example} - -The only non-trivial case is again treated by contradiction: -\begin{coq_example} -intro Dick_does_not_drink; absurd (exists x, ~ P x); trivial. -exists Dick; trivial. -Qed. -\end{coq_example} - -Now, let us close the main section and look at the complete statements -we proved: -\begin{coq_example} -End Predicate_calculus. -Check refl_if. -Check weird. -Check drinker. -\end{coq_example} - -Note how the three theorems are completely generic in the most general -fashion; -the domain \verb:D: is discharged in all of them, \verb:R: is discharged in -\verb:refl_if: only, \verb:P: is discharged only in \verb:weird: and -\verb:drinker:, along with the hypothesis that \verb:D: is inhabited. -Finally, the excluded middle hypothesis is discharged only in -\verb:drinker:. - -Note, too, that the name \verb:d: has vanished from -the statements of \verb:weird: and \verb:drinker:, -since \Coq's pretty-printer replaces -systematically a quantification such as \texttt{forall d : D, E}, -where \texttt{d} does not occur in \texttt{E}, -by the functional notation \texttt{D -> E}. -Similarly the name \texttt{EM} does not appear in \texttt{drinker}. - -Actually, universal quantification, implication, -as well as function formation, are -all special cases of one general construct of type theory called -{\sl dependent product}. This is the mathematical construction -corresponding to an indexed family of functions. A function -$f\in \Pi x:D\cdot Cx$ maps an element $x$ of its domain $D$ to its -(indexed) codomain $Cx$. Thus a proof of $\forall x:D\cdot Px$ is -a function mapping an element $x$ of $D$ to a proof of proposition $Px$. - - -\subsection{Flexible use of local assumptions} - -Very often during the course of a proof we want to retrieve a local -assumption and reintroduce it explicitly in the goal, for instance -in order to get a more general induction hypothesis. The tactic -\verb:generalize: is what is needed here: - -\begin{coq_example} -Section Predicate_Calculus. -Variables P Q : nat -> Prop. -Variable R : nat -> nat -> Prop. -Lemma PQR : - forall x y:nat, (R x x -> P x -> Q x) -> P x -> R x y -> Q x. -intros. -generalize H0. -\end{coq_example} - -Sometimes it may be convenient to state an intermediate fact. -The tactic \verb:assert: does this and introduces a new subgoal -for this fact to be proved first. The tactic \verb:enough: does -the same while keeping this goal for later. -\begin{coq_example} -enough (R x x) by auto. -\end{coq_example} -We clean the goal by doing an \verb:Abort: command. -\begin{coq_example*} -Abort. -\end{coq_example*} - - -\subsection{Equality} - -The basic equality provided in \Coq{} is Leibniz equality, noted infix like -\texttt{x = y}, when \texttt{x} and \texttt{y} are two expressions of -type the same Set. The replacement of \texttt{x} by \texttt{y} in any -term is effected by a variety of tactics, such as \texttt{rewrite} -and \texttt{replace}. - -Let us give a few examples of equality replacement. Let us assume that -some arithmetic function \verb:f: is null in zero: -\begin{coq_example} -Variable f : nat -> nat. -Hypothesis foo : f 0 = 0. -\end{coq_example} - -We want to prove the following conditional equality: -\begin{coq_example*} -Lemma L1 : forall k:nat, k = 0 -> f k = k. -\end{coq_example*} - -As usual, we first get rid of local assumptions with \verb:intro:: -\begin{coq_example} -intros k E. -\end{coq_example} - -Let us now use equation \verb:E: as a left-to-right rewriting: -\begin{coq_example} -rewrite E. -\end{coq_example} -This replaced both occurrences of \verb:k: by \verb:O:. - -Now \verb:apply foo: will finish the proof: - -\begin{coq_example} -apply foo. -Qed. -\end{coq_example} - -When one wants to rewrite an equality in a right to left fashion, we should -use \verb:rewrite <- E: rather than \verb:rewrite E: or the equivalent -\verb:rewrite -> E:. -Let us now illustrate the tactic \verb:replace:. -\begin{coq_example} -Hypothesis f10 : f 1 = f 0. -Lemma L2 : f (f 1) = 0. -replace (f 1) with 0. -\end{coq_example} -What happened here is that the replacement left the first subgoal to be -proved, but another proof obligation was generated by the \verb:replace: -tactic, as the second subgoal. The first subgoal is solved immediately -by applying lemma \verb:foo:; the second one transitivity and then -symmetry of equality, for instance with tactics \verb:transitivity: and -\verb:symmetry:: -\begin{coq_example} -apply foo. -transitivity (f 0); symmetry; trivial. -\end{coq_example} -In case the equality $t=u$ generated by \verb:replace: $u$ \verb:with: -$t$ is an assumption -(possibly modulo symmetry), it will be automatically proved and the -corresponding goal will not appear. For instance: - -\begin{coq_eval} -Restart. -\end{coq_eval} -\begin{coq_example} -Lemma L2 : f (f 1) = 0. -replace (f 1) with (f 0). -replace (f 0) with 0; trivial. -Qed. -\end{coq_example} - -\section{Using definitions} - -The development of mathematics does not simply proceed by logical -argumentation from first principles: definitions are used in an essential way. -A formal development proceeds by a dual process of abstraction, where one -proves abstract statements in predicate calculus, and use of definitions, -which in the contrary one instantiates general statements with particular -notions in order to use the structure of mathematical values for the proof of -more specialised properties. - -\subsection{Unfolding definitions} - -Assume that we want to develop the theory of sets represented as characteristic -predicates over some universe \verb:U:. For instance: -\begin{coq_example} -Variable U : Type. -Definition set := U -> Prop. -Definition element (x : U) (S : set) := S x. -Definition subset (A B : set) := - forall x : U, element x A -> element x B. -\end{coq_example} - -Now, assume that we have loaded a module of general properties about -relations over some abstract type \verb:T:, such as transitivity: - -\begin{coq_example} -Definition transitive (T : Type) (R : T -> T -> Prop) := - forall x y z : T, R x y -> R y z -> R x z. -\end{coq_example} - -We want to prove that \verb:subset: is a \verb:transitive: -relation. -\begin{coq_example} -Lemma subset_transitive : transitive set subset. -\end{coq_example} - -In order to make any progress, one needs to use the definition of -\verb:transitive:. The \verb:unfold: tactic, which replaces all -occurrences of a defined notion by its definition in the current goal, -may be used here. -\begin{coq_example} -unfold transitive. -\end{coq_example} - -Now, we must unfold \verb:subset:: -\begin{coq_example} -unfold subset. -\end{coq_example} -Now, unfolding \verb:element: would be a mistake, because indeed a simple proof -can be found by \verb:auto:, keeping \verb:element: an abstract predicate: -\begin{coq_example} -auto. -\end{coq_example} - -Many variations on \verb:unfold: are provided in \Coq. For instance, -instead of unfolding all occurrences of \verb:subset:, we may want to -unfold only one designated occurrence: -\begin{coq_eval} -Undo 2. -\end{coq_eval} -\begin{coq_example} -unfold subset at 2. -\end{coq_example} - -One may also unfold a definition in a given local hypothesis, using the -\verb:in: notation: -\begin{coq_example} -intros. -unfold subset in H. -\end{coq_example} - -Finally, the tactic \verb:red: does only unfolding of the head occurrence -of the current goal: -\begin{coq_example} -red. -auto. -Qed. -\end{coq_example} - - -\subsection{Principle of proof irrelevance} - -Even though in principle the proof term associated with a verified lemma -corresponds to a defined value of the corresponding specification, such -definitions cannot be unfolded in \Coq: a lemma is considered an {\sl opaque} -definition. This conforms to the mathematical tradition of {\sl proof -irrelevance}: the proof of a logical proposition does not matter, and the -mathematical justification of a logical development relies only on -{\sl provability} of the lemmas used in the formal proof. - -Conversely, ordinary mathematical definitions can be unfolded at will, they -are {\sl transparent}. - -\chapter{Induction} - -\begin{coq_eval} -Reset Initial. -Set Printing Width 60. -Set Printing Compact Contexts. -\end{coq_eval} - -\section{Data Types as Inductively Defined Mathematical Collections} - -All the notions which were studied until now pertain to traditional -mathematical logic. Specifications of objects were abstract properties -used in reasoning more or less constructively; we are now entering -the realm of inductive types, which specify the existence of concrete -mathematical constructions. - -\subsection{Booleans} - -Let us start with the collection of booleans, as they are specified -in the \Coq's \verb:Prelude: module: -\begin{coq_example} -Inductive bool : Set := true | false. -\end{coq_example} - -Such a declaration defines several objects at once. First, a new -\verb:Set: is declared, with name \verb:bool:. Then the {\sl constructors} -of this \verb:Set: are declared, called \verb:true: and \verb:false:. -Those are analogous to introduction rules of the new Set \verb:bool:. -Finally, a specific elimination rule for \verb:bool: is now available, which -permits to reason by cases on \verb:bool: values. Three instances are -indeed defined as new combinators in the global context: \verb:bool_ind:, -a proof combinator corresponding to reasoning by cases, -\verb:bool_rec:, an if-then-else programming construct, -and \verb:bool_rect:, a similar combinator at the level of types. -Indeed: -\begin{coq_example} -Check bool_ind. -Check bool_rec. -Check bool_rect. -\end{coq_example} - -Let us for instance prove that every Boolean is true or false. -\begin{coq_example} -Lemma duality : forall b:bool, b = true \/ b = false. -intro b. -\end{coq_example} - -We use the knowledge that \verb:b: is a \verb:bool: by calling tactic -\verb:elim:, which is this case will appeal to combinator \verb:bool_ind: -in order to split the proof according to the two cases: -\begin{coq_example} -elim b. -\end{coq_example} - -It is easy to conclude in each case: -\begin{coq_example} -left; trivial. -right; trivial. -\end{coq_example} - -Indeed, the whole proof can be done with the combination of the - \verb:destruct:, which combines \verb:intro: and \verb:elim:, -with good old \verb:auto:: -\begin{coq_eval} -Abort. -\end{coq_eval} -\begin{coq_example} -Lemma duality : forall b:bool, b = true \/ b = false. -destruct b; auto. -Qed. -\end{coq_example} - -\subsection{Natural numbers} - -Similarly to Booleans, natural numbers are defined in the \verb:Prelude: -module with constructors \verb:S: and \verb:O:: -\begin{coq_example} -Inductive nat : Set := - | O : nat - | S : nat -> nat. -\end{coq_example} - -The elimination principles which are automatically generated are Peano's -induction principle, and a recursion operator: -\begin{coq_example} -Check nat_ind. -Check nat_rec. -\end{coq_example} - -Let us start by showing how to program the standard primitive recursion -operator \verb:prim_rec: from the more general \verb:nat_rec:: -\begin{coq_example} -Definition prim_rec := nat_rec (fun i : nat => nat). -\end{coq_example} - -That is, instead of computing for natural \verb:i: an element of the indexed -\verb:Set: \verb:(P i):, \verb:prim_rec: computes uniformly an element of -\verb:nat:. Let us check the type of \verb:prim_rec:: -\begin{coq_example} -About prim_rec. -\end{coq_example} - -Oops! Instead of the expected type \verb+nat->(nat->nat->nat)->nat->nat+ we -get an apparently more complicated expression. -In fact, the two types are convertible and one way of having the proper -type would be to do some computation before actually defining \verb:prim_rec: -as such: - -\begin{coq_eval} -Reset Initial. -Set Printing Width 60. -Set Printing Compact Contexts. -\end{coq_eval} - -\begin{coq_example} -Definition prim_rec := - Eval compute in nat_rec (fun i : nat => nat). -About prim_rec. -\end{coq_example} - -Let us now show how to program addition with primitive recursion: -\begin{coq_example} -Definition addition (n m:nat) := - prim_rec m (fun p rec : nat => S rec) n. -\end{coq_example} - -That is, we specify that \verb+(addition n m)+ computes by cases on \verb:n: -according to its main constructor; when \verb:n = O:, we get \verb:m:; - when \verb:n = S p:, we get \verb:(S rec):, where \verb:rec: is the result -of the recursive computation \verb+(addition p m)+. Let us verify it by -asking \Coq{} to compute for us say $2+3$: -\begin{coq_example} -Eval compute in (addition (S (S O)) (S (S (S O)))). -\end{coq_example} - -Actually, we do not have to do all explicitly. {\Coq} provides a -special syntax {\tt Fixpoint/match} for generic primitive recursion, -and we could thus have defined directly addition as: - -\begin{coq_example} -Fixpoint plus (n m:nat) {struct n} : nat := - match n with - | O => m - | S p => S (plus p m) - end. -\end{coq_example} - -\begin{coq_eval} -\begin{coq_example} -Reset Initial. -Set Printing Width 60. -Set Printing Compact Contexts. -\end{coq_eval} - -\subsection{Simple proofs by induction} - -Let us now show how to do proofs by structural induction. We start with easy -properties of the \verb:plus: function we just defined. Let us first -show that $n=n+0$. -\begin{coq_example} -Lemma plus_n_O : forall n : nat, n = n + 0. -intro n; elim n. -\end{coq_example} - -What happened was that \texttt{elim n}, in order to construct a \texttt{Prop} -(the initial goal) from a \texttt{nat} (i.e. \texttt{n}), appealed to the -corresponding induction principle \texttt{nat\_ind} which we saw was indeed -exactly Peano's induction scheme. Pattern-matching instantiated the -corresponding predicate \texttt{P} to \texttt{fun n : nat => n = n + 0}, -and we get as subgoals the corresponding instantiations of the base case -\texttt{(P O)}, and of the inductive step -\texttt{forall y : nat, P y -> P (S y)}. -In each case we get an instance of function \texttt{plus} in which its second -argument starts with a constructor, and is thus amenable to simplification -by primitive recursion. The \Coq{} tactic \texttt{simpl} can be used for -this purpose: -\begin{coq_example} -simpl. -auto. -\end{coq_example} - -We proceed in the same way for the base step: -\begin{coq_example} -simpl; auto. -Qed. -\end{coq_example} - -Here \verb:auto: succeeded, because it used as a hint lemma \verb:eq_S:, -which say that successor preserves equality: -\begin{coq_example} -Check eq_S. -\end{coq_example} - -Actually, let us see how to declare our lemma \verb:plus_n_O: as a hint -to be used by \verb:auto:: -\begin{coq_example} -Hint Resolve plus_n_O . -\end{coq_example} - -We now proceed to the similar property concerning the other constructor -\verb:S:: -\begin{coq_example} -Lemma plus_n_S : forall n m:nat, S (n + m) = n + S m. -\end{coq_example} - -We now go faster, using the tactic \verb:induction:, which does the -necessary \verb:intros: before applying \verb:elim:. Factoring simplification -and automation in both cases thanks to tactic composition, we prove this -lemma in one line: -\begin{coq_example} -induction n; simpl; auto. -Qed. -Hint Resolve plus_n_S . -\end{coq_example} - -Let us end this exercise with the commutativity of \verb:plus:: - -\begin{coq_example} -Lemma plus_com : forall n m:nat, n + m = m + n. -\end{coq_example} - -Here we have a choice on doing an induction on \verb:n: or on \verb:m:, the -situation being symmetric. For instance: -\begin{coq_example} -induction m as [ | m IHm ]; simpl; auto. -\end{coq_example} - -Here \verb:auto: succeeded on the base case, thanks to our hint -\verb:plus_n_O:, but the induction step requires rewriting, which -\verb:auto: does not handle: - -\begin{coq_example} -rewrite <- IHm; auto. -Qed. -\end{coq_example} - -\subsection{Discriminate} - -It is also possible to define new propositions by primitive recursion. -Let us for instance define the predicate which discriminates between -the constructors \verb:O: and \verb:S:: it computes to \verb:False: -when its argument is \verb:O:, and to \verb:True: when its argument is -of the form \verb:(S n):: -\begin{coq_example} -Definition Is_S (n : nat) := match n with - | O => False - | S p => True - end. -\end{coq_example} - -Now we may use the computational power of \verb:Is_S: to prove -trivially that \verb:(Is_S (S n)):: -\begin{coq_example} -Lemma S_Is_S : forall n:nat, Is_S (S n). -simpl; trivial. -Qed. -\end{coq_example} - -But we may also use it to transform a \verb:False: goal into -\verb:(Is_S O):. Let us show a particularly important use of this feature; -we want to prove that \verb:O: and \verb:S: construct different values, one -of Peano's axioms: -\begin{coq_example} -Lemma no_confusion : forall n:nat, 0 <> S n. -\end{coq_example} - -First of all, we replace negation by its definition, by reducing the -goal with tactic \verb:red:; then we get contradiction by successive -\verb:intros:: -\begin{coq_example} -red; intros n H. -\end{coq_example} - -Now we use our trick: -\begin{coq_example} -change (Is_S 0). -\end{coq_example} - -Now we use equality in order to get a subgoal which computes out to -\verb:True:, which finishes the proof: -\begin{coq_example} -rewrite H; trivial. -simpl; trivial. -\end{coq_example} - -Actually, a specific tactic \verb:discriminate: is provided -to produce mechanically such proofs, without the need for the user to define -explicitly the relevant discrimination predicates: - -\begin{coq_eval} -Abort. -\end{coq_eval} -\begin{coq_example} -Lemma no_confusion : forall n:nat, 0 <> S n. -intro n; discriminate. -Qed. -\end{coq_example} - - -\section{Logic programming} - -In the same way as we defined standard data-types above, we -may define inductive families, and for instance inductive predicates. -Here is the definition of predicate $\le$ over type \verb:nat:, as -given in \Coq's \verb:Prelude: module: -\begin{coq_example*} -Inductive le (n : nat) : nat -> Prop := - | le_n : le n n - | le_S : forall m : nat, le n m -> le n (S m). -\end{coq_example*} - -This definition introduces a new predicate -\verb+le : nat -> nat -> Prop+, -and the two constructors \verb:le_n: and \verb:le_S:, which are the -defining clauses of \verb:le:. That is, we get not only the ``axioms'' -\verb:le_n: and \verb:le_S:, but also the converse property, that -\verb:(le n m): if and only if this statement can be obtained as a -consequence of these defining clauses; that is, \verb:le: is the -minimal predicate verifying clauses \verb:le_n: and \verb:le_S:. This is -insured, as in the case of inductive data types, by an elimination principle, -which here amounts to an induction principle \verb:le_ind:, stating this -minimality property: -\begin{coq_example} -Check le. -Check le_ind. -\end{coq_example} - -Let us show how proofs may be conducted with this principle. -First we show that $n\le m \Rightarrow n+1\le m+1$: -\begin{coq_example} -Lemma le_n_S : forall n m : nat, le n m -> le (S n) (S m). -intros n m n_le_m. -elim n_le_m. -\end{coq_example} - -What happens here is similar to the behaviour of \verb:elim: on natural -numbers: it appeals to the relevant induction principle, here \verb:le_ind:, -which generates the two subgoals, which may then be solved easily -with the help of the defining clauses of \verb:le:. -\begin{coq_example} -apply le_n; trivial. -intros; apply le_S; trivial. -\end{coq_example} - -Now we know that it is a good idea to give the defining clauses as hints, -so that the proof may proceed with a simple combination of -\verb:induction: and \verb:auto:. \verb:Hint Constructors le: -is just an abbreviation for \verb:Hint Resolve le_n le_S:. -\begin{coq_eval} -Abort. -\end{coq_eval} -\begin{coq_example} -Hint Constructors le. -Lemma le_n_S : forall n m : nat, le n m -> le (S n) (S m). -\end{coq_example} - -We have a slight problem however. We want to say ``Do an induction on -hypothesis \verb:(le n m):'', but we have no explicit name for it. What we -do in this case is to say ``Do an induction on the first unnamed hypothesis'', -as follows. -\begin{coq_example} -induction 1; auto. -Qed. -\end{coq_example} - -Here is a more tricky problem. Assume we want to show that -$n\le 0 \Rightarrow n=0$. This reasoning ought to follow simply from the -fact that only the first defining clause of \verb:le: applies. -\begin{coq_example} -Lemma tricky : forall n:nat, le n 0 -> n = 0. -\end{coq_example} - -However, here trying something like \verb:induction 1: would lead -nowhere (try it and see what happens). -An induction on \verb:n: would not be convenient either. -What we must do here is analyse the definition of \verb"le" in order -to match hypothesis \verb:(le n O): with the defining clauses, to find -that only \verb:le_n: applies, whence the result. -This analysis may be performed by the ``inversion'' tactic -\verb:inversion_clear: as follows: -\begin{coq_example} -intros n H; inversion_clear H. -trivial. -Qed. -\end{coq_example} - -\chapter{Modules} - -\begin{coq_eval} -Reset Initial. -Set Printing Width 60. -Set Printing Compact Contexts. -\end{coq_eval} - -\section{Opening library modules} - -When you start \Coq{} without further requirements in the command line, -you get a bare system with few libraries loaded. As we saw, a standard -prelude module provides the standard logic connectives, and a few -arithmetic notions. If you want to load and open other modules from -the library, you have to use the \verb"Require" command, as we saw for -classical logic above. For instance, if you want more arithmetic -constructions, you should request: -\begin{coq_example*} -Require Import Arith. -\end{coq_example*} - -Such a command looks for a (compiled) module file \verb:Arith.vo: in -the libraries registered by \Coq. Libraries inherit the structure of -the file system of the operating system and are registered with the -command \verb:Add LoadPath:. Physical directories are mapped to -logical directories. Especially the standard library of \Coq{} is -pre-registered as a library of name \verb=Coq=. Modules have absolute -unique names denoting their place in \Coq{} libraries. An absolute -name is a sequence of single identifiers separated by dots. E.g. the -module \verb=Arith= has full name \verb=Coq.Arith.Arith= and because -it resides in eponym subdirectory \verb=Arith= of the standard -library, it can be as well required by the command - -\begin{coq_example*} -Require Import Coq.Arith.Arith. -\end{coq_example*} - -This may be useful to avoid ambiguities if somewhere, in another branch -of the libraries known by Coq, another module is also called -\verb=Arith=. Notice that by default, when a library is registered, -all its contents, and all the contents of its subdirectories recursively are -visible and accessible by a short (relative) name as \verb=Arith=. -Notice also that modules or definitions not explicitly registered in -a library are put in a default library called \verb=Top=. - -The loading of a compiled file is quick, because the corresponding -development is not type-checked again. - -\section{Creating your own modules} - -You may create your own module files, by writing {\Coq} commands in a file, -say \verb:my_module.v:. Such a module may be simply loaded in the current -context, with command \verb:Load my_module:. It may also be compiled, -in ``batch'' mode, using the UNIX command -\verb:coqc:. Compiling the module \verb:my_module.v: creates a -file \verb:my_module.vo:{} that can be reloaded with command -\verb:Require: \verb:Import: \verb:my_module:. - -If a required module depends on other modules then the latters are -automatically required beforehand. However their contents is not -automatically visible. If you want a module \verb=M= required in a -module \verb=N= to be automatically visible when \verb=N= is required, -you should use \verb:Require Export M: in your module \verb:N:. - -\section{Managing the context} - -It is often difficult to remember the names of all lemmas and -definitions available in the current context, especially if large -libraries have been loaded. A convenient \verb:Search: command -is available to lookup all known facts -concerning a given predicate. For instance, if you want to know all the -known lemmas about both the successor and the less or equal relation, just ask: -\begin{coq_eval} -Reset Initial. -Set Printing Width 60. -Set Printing Compact Contexts. -\end{coq_eval} -\begin{coq_example} -Search S le. -\end{coq_example} -Another command \verb:SearchHead: displays only lemmas where the searched -predicate appears at the head position in the conclusion. -\begin{coq_example} -SearchHead le. -\end{coq_example} - -The \verb:Search: commands also allows finding the theorems -containing a given pattern, where \verb:_: can be used in -place of an arbitrary term. As shown in this example, \Coq{} -provides usual infix notations for arithmetic operators. - -\begin{coq_example} -Search (_ + _ = _). -\end{coq_example} - -\section{Now you are on your own} - -This tutorial is necessarily incomplete. If you wish to pursue serious -proving in \Coq, you should now get your hands on \Coq's Reference Manual, -which contains a complete description of all the tactics we saw, -plus many more. -You also should look in the library of developed theories which is distributed -with \Coq, in order to acquaint yourself with various proof techniques. - - -\end{document} - |