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From: Stephane Glondu <steph@glondu.net>
Date: Sat, 10 Jul 2010 15:57:24 +0100
Subject: Remove non-DFSG contents

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-\documentclass[11pt,a4paper]{book}
-\usepackage[T1]{fontenc}
-\usepackage[latin1]{inputenc}
-\usepackage{pslatex}
-
-\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.
-%, and mainly on Unix machines. 
-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 new book by Y.
-Bertot and P. Castéran on practical uses of the \Coq{} system.
-
-Coq can be used from a standard teletype-like shell window but
-preferably through the graphical user interface
-CoqIde\footnote{Alternative graphical interfaces exist: Proof General
-and Pcoq.}.
-
-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 \texttt{http://coq.inria.fr}.
-
-In the following, we assume that \Coq~ is called from a standard
-teletype-like shell window. All examples preceded by the prompting
-sequence \verb:Coq < : represent user input, terminated by a
-period. 
-
-The following lines usually show \Coq's answer as it appears on the
-users screen. 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.
-
-The sequence of such examples is a valid \Coq~
-session, unless otherwise specified. This version of the tutorial has
-been prepared on a PC workstation running Linux.  The standard
-invocation of \Coq\ delivers a message such as:
-
-\begin{small}
-\begin{flushleft}
-\begin{verbatim}
-unix:~> coqtop
-Welcome to Coq 8.3 (October 2010)
-
-Coq < 
-\end{verbatim}
-\end{flushleft}
-\end{small}
-
-The first line gives a banner stating the precise version of \Coq~
-used. You should always return this banner when you report an anomaly
-to our bug-tracking system
-\verb|http://coq.inria.fr/bugs|
-
-\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}. You may also send \Coq~ {\sl commands} which are
-not really part of the formal development, but correspond to information
-requests, or service routine invocations. For instance, the command:
-\begin{verbatim}
-Coq < Quit.
-\end{verbatim}
-terminates the current session.
-
-\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 \verb:gt: is a function expecting two arguments of
-type \verb: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 \verb:nat:
-with the (abstract) type \verb:Prop: of logical propositions through the
-arrow function constructor, in order to get a functional type
-\verb:nat->Prop::
-\begin{coq_example}
-Check (nat -> Prop).
-\end{coq_example}
-which may be composed one more times with \verb:nat: in order to obtain the
-type \verb:nat->nat->Prop: of binary relations over natural numbers.
-Actually the type \verb:nat->nat->Prop: is an abbreviation for 
-\verb: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 \verb: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:
-\verb:nat: is defined as a mathematical collection (type \verb:Set:), constants
-\verb:O:, \verb:S:, \verb:plus:, are defined as objects of types
-respectively \verb:nat:, \verb:nat->nat:, and \verb:nat->nat->nat:.
-You may introduce new definitions, which link a name to a well-typed value.
-For instance, we may introduce the constant \verb: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 : nat := S two.
-\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
-\verb+fun x:A => e+ is well formed of type \verb+A->B+ in a context
-whenever the expression \verb+e+ is well-formed of type \verb+B+ in 
-the given context to which we add the declaration that \verb+x+
-is of type \verb+A+. Here \verb+x+ is a bound, or dummy variable in
-the expression \verb+fun x:A => e+. For instance we could as well have
-defined \verb:double: as \verb+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: \verb+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}
-We may clean-up the development by removing the contents of the
-current section:
-\begin{coq_example}
-Reset Declaration.
-\end{coq_example}
-
-\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 $B$''. Remark 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. Remark 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 may either discard it, by using the
-command \verb:Abort: which returns to the standard \Coq~ toplevel loop
-without further ado, or else save it as a lemma in the current context,
-under name say \verb:trivial_lemma::
-\begin{coq_example}
-Save trivial_lemma.
-\end{coq_example}
-
-As a comment, the system shows the proof script listing all tactic
-commands used in the proof. 
-
-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}
-
-Let us now save lemma \verb:distr_impl::
-\begin{coq_example}
-Save.
-\end{coq_example}
-
-Here \verb:Save: needs no argument, since we gave the name \verb:distr_impl: 
-in advance;
-it is however possible to override the given name by giving a different 
-argument to command \verb:Save:.
-
-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_example}
-Lemma distr_imp : (A -> B -> C) -> (A -> B) -> A -> C.
-auto.
-\end{coq_example}
-
-This time, we do not save the proof, we just discard it with the \verb:Abort: 
-command:
-
-\begin{coq_example}
-Abort.
-\end{coq_example}
-
-At any point during a proof, we may use \verb:Abort: to exit the proof mode
-and go back to Coq's main loop. We may also use \verb:Restart: to restart
-from scratch the proof of the same lemma. We may also use \verb:Undo: to
-backtrack one step, and more generally \verb:Undo n: to
-backtrack n steps.
-
-We end this section by showing a useful command, \verb:Inspect n.:,
-which inspects the global \Coq~ environment, showing the last \verb:n: declared
-notions: 
-\begin{coq_example}
-Inspect 3.
-\end{coq_example}
-
-The declarations, whether global parameters or axioms, are shown preceded by 
-\verb:***:; definitions and lemmas are stated with their specification, but
-their value (or proof-term) is omitted.
-
-\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.
-\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_example}
-Restart.
-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 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_example}
-Restart.
-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 a few
-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_example}
-Abort.
-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 me 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 
-respectively 1 and 2. Such abstract entities may be entered in the context
-as global variables. But we must be careful about the pollution of our
-global environment by such declarations. For instance, we have already 
-polluted our \Coq~ session by declaring the variables
-\verb:n:, \verb:Pos_n:, \verb:A:, \verb:B:, and \verb:C:. If we want to revert to the clean state of
-our initial session, we may use the \Coq~ \verb:Reset: command, which returns
-to the state just prior the given global notion as we did before to
-remove a section, or we may return to the initial state using~:
-\begin{coq_example}
-Reset Initial.
-\end{coq_example}
-\begin{coq_eval}
-Set Printing Width 60.
-\end{coq_eval}
-
-We shall now declare a new \verb:Section:, which will allow us to define
-notions local to a well-delimited scope. 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}
-
-Remark 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}
-
-Remark that the hypotheses which are local to the currently opened sections
-are listed as local hypotheses to the current goals.
-The rationale is that 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}
-
-Remark 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}
-
-Here \Coq's printout is a warning that all 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, the use of section
-\verb:R_sym_trans: has not been really significant, 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:
-\begin{coq_example}
-intro Non_drinker; elim Non_drinker;
-  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;
-a simple one is to use the \verb:absurd: tactic as follows:
-\begin{coq_example}
-absurd (P Tom); trivial.
-\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}
-
-Remark 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 that the name \verb:d: has vanished as well from
-the statements of \verb:weird: and \verb:drinker:, 
-since \Coq's pretty-printer replaces
-systematically a quantification such as \verb+forall d:D, E+, where \verb:d:
-does not occur in \verb:E:, by the functional notation \verb:D->E:. 
-Similarly the name \verb:EM: does not appear in \verb: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 use a lemma, although we do not have
-a direct way to appeal to such an already proven fact. The tactic \verb:cut:
-permits to use the lemma at this point, keeping the corresponding proof
-obligation as a new subgoal:
-\begin{coq_example}
-cut (R x x); trivial.
-\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
-\verb+x=y+, when \verb:x: and \verb:y: are two expressions of
-type the same Set. The replacement of \verb:x: by \verb:y: in any
-term is effected by a variety of tactics, such as \verb:rewrite:
-and \verb: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_example}
-Restart.
-replace (f 0) with 0.
-rewrite f10; rewrite foo; 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}
-
-Now, assume that 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,
-we may selectively unfold one designated occurrence:
-\begin{coq_example}
-Undo 2.
-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}
-
-\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:simple: \verb:induction:, which combines \verb:intro: and \verb:elim:,
-with good old \verb:auto::
-\begin{coq_example}
-Restart.
-simple induction 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}
-Check 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. Indeed the type of
-\verb:prim_rec: is equivalent by rule $\beta$ to its expected type; this may
-be checked in \Coq~ by command \verb:Eval Cbv Beta:, which $\beta$-reduces
-an expression to its {\sl normal form}:
-\begin{coq_example}
-Eval cbv beta in
-  ((fun _:nat => nat) O ->
-   (forall y:nat, 
-      (fun _:nat => nat) y -> (fun _:nat => nat) (S y)) ->
-   forall n:nat, (fun _:nat => nat) n).
-\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}
-
-For the rest of the session, we shall clean up what we did so far with 
-types \verb:bool: and \verb:nat:, in order to use the initial definitions
-given in \Coq's \verb:Prelude: module, and not to get confusing error
-messages due to our redefinitions. We thus revert to the state before
-our definition of \verb:bool: with the \verb:Reset: command:
-\begin{coq_example}
-Reset bool.
-\end{coq_example}
-
-
-\subsection{Simple proofs by induction}
-
-\begin{coq_eval}
-Reset Initial.
-\end{coq_eval}
-\begin{coq_eval}
-Set Printing Width 60.
-\end{coq_eval}
-
-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 \verb:elim n:, in order to construct a \verb:Prop:
-(the initial goal) from a \verb:nat: (i.e. \verb:n:), appealed to the
-corresponding induction principle \verb:nat_ind: which we saw was indeed
-exactly Peano's induction scheme. Pattern-matching instantiated the 
-corresponding predicate \verb:P: to \verb+fun n:nat => n = n + 0+, and we get
-as subgoals the corresponding instantiations of the base case \verb:(P O): ,
-and of the inductive step \verb+forall y:nat, P y -> P (S y)+.
-In each case we get an instance of function \verb:plus: in which its second
-argument starts with a constructor, and is thus amenable to simplification
-by primitive recursion. The \Coq~tactic \verb: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, remembering that tactic \verb:simple induction: 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}
-simple 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}
-simple induction m; 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}
-intros m' E; rewrite <- E; 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: in order 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_example}
-Restart.
-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:.
-\begin{coq_example}
-Restart.
-Hint Resolve le_n le_S .
-\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}
-simple 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}
-
-\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:SearchAbout: command
-is available to lookup all known facts 
-concerning a given predicate. For instance, if you want to know all the
-known lemmas about the less or equal relation, just ask:
-\begin{coq_example}
-SearchAbout le.
-\end{coq_example}
-Another command \verb:Search: displays only lemmas where the searched
-predicate appears at the head position in the conclusion.
-\begin{coq_example}
-Search le.
-\end{coq_example}
-
-A new and more convenient search tool is \textsf{SearchPattern}
-developed by Yves Bertot. It allows to find the theorems with a
-conclusion matching a given pattern, where \verb:\_: can be used in
-place of an arbitrary term. We remark in this example, that \Coq{}
-provides usual infix notations for arithmetic operators.
-
-\begin{coq_example}
-SearchPattern (_ + _ = _).
-\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}
-
-% $Id: Tutorial.tex 13548 2010-10-14 12:35:43Z notin $ 
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