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diff --git a/doc/faq/FAQ.tex b/doc/faq/FAQ.tex deleted file mode 100644 index 2b5d898f..00000000 --- a/doc/faq/FAQ.tex +++ /dev/null @@ -1,2481 +0,0 @@ -\RequirePackage{ifpdf} -\ifpdf % si on est en pdflatex -\documentclass[a4paper,pdftex]{article} -\else -\documentclass[a4paper]{article} -\fi -\pagestyle{plain} - -% yay les symboles -\usepackage{stmaryrd} -\usepackage{amssymb} -\usepackage{url} -%\usepackage{multicol} -\usepackage{hevea} -\usepackage{fullpage} -\usepackage[latin1]{inputenc} -\usepackage[english]{babel} - -\ifpdf % si on est en pdflatex - \usepackage[pdftex]{graphicx} -\else - \usepackage[dvips]{graphicx} -\fi - -%\input{../macros.tex} - -% Making hevea happy -%HEVEA \renewcommand{\textbar}{|} -%HEVEA \renewcommand{\textunderscore}{\_} - -\def\Question#1{\stepcounter{question}\subsubsection{#1}} - -% version et date -\def\faqversion{0.1} - -% les macros d'amour -\def\Coq{\textsc{Coq}} -\def\Why{\textsc{Why}} -\def\Caduceus{\textsc{Caduceus}} -\def\Krakatoa{\textsc{Krakatoa}} -\def\Ltac{\textsc{Ltac}} -\def\CoqIde{\textsc{CoqIde}} - -\newcommand{\coqtt}[1]{{\tt #1}} -\newcommand{\coqimp}{{\mbox{\tt ->}}} -\newcommand{\coqequiv}{{\mbox{\tt <->}}} - - -% macro pour les tactics -\def\split{{\tt split}} -\def\assumption{{\tt assumption}} -\def\auto{{\tt auto}} -\def\trivial{{\tt trivial}} -\def\tauto{{\tt tauto}} -\def\left{{\tt left}} -\def\right{{\tt right}} -\def\decompose{{\tt decompose}} -\def\intro{{\tt intro}} -\def\intros{{\tt intros}} -\def\field{{\tt field}} -\def\ring{{\tt ring}} -\def\apply{{\tt apply}} -\def\exact{{\tt exact}} -\def\cut{{\tt cut}} -\def\assert{{\tt assert}} -\def\solve{{\tt solve}} -\def\idtac{{\tt idtac}} -\def\fail{{\tt fail}} -\def\existstac{{\tt exists}} -\def\firstorder{{\tt firstorder}} -\def\congruence{{\tt congruence}} -\def\gb{{\tt gb}} -\def\generalize{{\tt generalize}} -\def\abstracttac{{\tt abstract}} -\def\eapply{{\tt eapply}} -\def\unfold{{\tt unfold}} -\def\rewrite{{\tt rewrite}} -\def\replace{{\tt replace}} -\def\simpl{{\tt simpl}} -\def\elim{{\tt elim}} -\def\set{{\tt set}} -\def\pose{{\tt pose}} -\def\case{{\tt case}} -\def\destruct{{\tt destruct}} -\def\reflexivity{{\tt reflexivity}} -\def\transitivity{{\tt transitivity}} -\def\symmetry{{\tt symmetry}} -\def\Focus{{\tt Focus}} -\def\discriminate{{\tt discriminate}} -\def\contradiction{{\tt contradiction}} -\def\intuition{{\tt intuition}} -\def\try{{\tt try}} -\def\repeat{{\tt repeat}} -\def\eauto{{\tt eauto}} -\def\subst{{\tt subst}} -\def\symmetryin{{\tt symmetryin}} -\def\instantiate{{\tt instantiate}} -\def\inversion{{\tt inversion}} -\def\Defined{{\tt Defined}} -\def\Qed{{\tt Qed}} -\def\pattern{{\tt pattern}} -\def\Type{{\tt Type}} -\def\Prop{{\tt Prop}} -\def\Set{{\tt Set}} - - -\newcommand\vfile[2]{\ahref{#1}{\tt {#2}.v}} -\urldef{\InitWf}{\url} - {http://coq.inria.fr/library/Coq.Init.Wf.html} -\urldef{\LogicBerardi}{\url} - {http://coq.inria.fr/library/Coq.Logic.Berardi.html} -\urldef{\LogicClassical}{\url} - {http://coq.inria.fr/library/Coq.Logic.Classical.html} -\urldef{\LogicClassicalFacts}{\url} - {http://coq.inria.fr/library/Coq.Logic.ClassicalFacts.html} -\urldef{\LogicClassicalDescription}{\url} - {http://coq.inria.fr/library/Coq.Logic.ClassicalDescription.html} -\urldef{\LogicProofIrrelevance}{\url} - {http://coq.inria.fr/library/Coq.Logic.ProofIrrelevance.html} -\urldef{\LogicEqdep}{\url} - {http://coq.inria.fr/library/Coq.Logic.Eqdep.html} -\urldef{\LogicEqdepDec}{\url} - {http://coq.inria.fr/library/Coq.Logic.Eqdep_dec.html} - - - - -\begin{document} -\bibliographystyle{plain} -\newcounter{question} -\renewcommand{\thesubsubsection}{\arabic{question}} - -%%%%%%% Coq pour les nuls %%%%%%% - -\title{Coq Version 8.0 for the Clueless\\ - \large(\protect\ref{lastquestion} - \ Hints) -} -\author{Pierre Castéran \and Hugo Herbelin \and Florent Kirchner \and Benjamin Monate \and Julien Narboux} -\maketitle - -%%%%%%% - -\begin{abstract} -This note intends to provide an easy way to get acquainted with the -{\Coq} theorem prover. It tries to formulate appropriate answers -to some of the questions any newcomers will face, and to give -pointers to other references when possible. -\end{abstract} - -%%%%%%% - -%\begin{multicols}{2} -\tableofcontents -%\end{multicols} - -%%%%%%% - -\newpage - -\section{Introduction} -This FAQ is the sum of the questions that came to mind as we developed -proofs in \Coq. Since we are singularly short-minded, we wrote the -answers we found on bits of papers to have them at hand whenever the -situation occurs again. This is pretty much the result of that: a -collection of tips one can refer to when proofs become intricate. Yes, -this means we won't take the blame for the shortcomings of this -FAQ. But if you want to contribute and send in your own question and -answers, feel free to write to us\ldots - -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% - -\section{Presentation} - -\Question{What is {\Coq}?}\label{whatiscoq} -The {\Coq} tool is a formal proof management system: a proof done with {\Coq} is mechanically checked by the machine. -In particular, {\Coq} allows: -\begin{itemize} - \item the definition of mathematical objects and programming objects, - \item to state mathematical theorems and software specifications, - \item to interactively develop formal proofs of these theorems, - \item to check these proofs by a small certification ``kernel''. -\end{itemize} -{\Coq} is based on a logical framework called ``Calculus of Inductive -Constructions'' extended by a modular development system for theories. - -\Question{Did you really need to name it like that?} -Some French computer scientists have a tradition of naming their -software as animal species: Caml, Elan, Foc or Phox are examples -of this tacit convention. In French, ``coq'' means rooster, and it -sounds like the initials of the Calculus of Constructions CoC on which -it is based. - -\Question{Is {\Coq} a theorem prover?} - -{\Coq} comes with decision and semi-decision procedures ( -propositional calculus, Presburger's arithmetic, ring and field -simplification, resolution, ...) but the main style for proving -theorems is interactively by using LCF-style tactics. - - -\Question{What are the other theorem provers?} -Many other theorem provers are available for use nowadays. -Isabelle, HOL, HOL Light, Lego, Nuprl, PVS are examples of provers that are fairly similar -to {\Coq} by the way they interact with the user. Other relatives of -{\Coq} are ACL2, Agda/Alfa, Twelf, Kiv, Mizar, NqThm, -\begin{htmlonly}% -Omega\ldots -\end{htmlonly} -\begin{latexonly}% -{$\Omega$}mega\ldots -\end{latexonly} - -\Question{What do I have to trust when I see a proof checked by Coq?} - -You have to trust: - -\begin{description} -\item[The theory behind Coq] The theory of {\Coq} version 8.0 is -generally admitted to be consistent wrt Zermelo-Fraenkel set theory + -inaccessible cardinals. Proofs of consistency of subsystems of the -theory of Coq can be found in the literature. -\item[The Coq kernel implementation] You have to trust that the -implementation of the {\Coq} kernel mirrors the theory behind {\Coq}. The -kernel is intentionally small to limit the risk of conceptual or -accidental implementation bugs. -\item[The Objective Caml compiler] The {\Coq} kernel is written using the -Objective Caml language but it uses only the most standard features -(no object, no label ...), so that it is highly unprobable that an -Objective Caml bug breaks the consistency of {\Coq} without breaking all -other kinds of features of {\Coq} or of other software compiled with -Objective Caml. -\item[Your hardware] In theory, if your hardware does not work -properly, it can accidentally be the case that False becomes -provable. But it is more likely the case that the whole {\Coq} system -will be unusable. You can check your proof using different computers -if you feel the need to. -\item[Your axioms] Your axioms must be consistent with the theory -behind {\Coq}. -\end{description} - - -\Question{Where can I find information about the theory behind {\Coq}?} -\begin{description} -\item[The Calculus of Inductive Constructions] The -\ahref{http://coq.inria.fr/doc/Reference-Manual006.html}{corresponding} -chapter and the chapter on -\ahref{http://coq.inria.fr/doc/Reference-Manual007.html}{modules} in -the {\Coq} Reference Manual. -\item[Type theory] A book~\cite{ProofsTypes} or some lecture -notes~\cite{Types:Dowek}. -\item[Inductive types] -Christine Paulin-Mohring's habilitation thesis~\cite{Pau96b}. -\item[Co-Inductive types] -Eduardo Giménez' thesis~\cite{EGThese}. -\item[Miscellaneous] A -\ahref{http://coq.inria.fr/doc/biblio.html}{bibliography} about Coq -\end{description} - - -\Question{How can I use {\Coq} to prove programs?} - -You can either extract a program from a proof by using the extraction -mechanism or use dedicated tools, such as -\ahref{http://why.lri.fr}{\Why}, -\ahref{http://krakatoa.lri.fr}{\Krakatoa}, -\ahref{http://why.lri.fr/caduceus/index.en.html}{\Caduceus}, to prove -annotated programs written in other languages. - -%\Question{How many {\Coq} users are there?} -% -%An estimation is about 100 regular users. - -\Question{How old is {\Coq}?} - -The first implementation is from 1985 (it was named {\sf CoC} which is -the acronym of the name of the logic it implemented: the Calculus of -Constructions). The first official release of {\Coq} (version 4.10) -was distributed in 1989. - -\Question{What are the \Coq-related tools?} - -There are graphical user interfaces: -\begin{description} -\item[Coqide] A GTK based GUI for \Coq. -\item[Pcoq] A GUI for {\Coq} with proof by pointing and pretty printing. -\item[coqwc] A tool similar to {\tt wc} to count lines in {\Coq} files. -\item[Proof General] A emacs mode for {\Coq} and many other proof assistants. -\end{description} - -There are documentation and browsing tools: - -\begin{description} -\item[Helm/Mowgli] A rendering, searching and publishing tool. -\item[coq-tex] A tool to insert {\Coq} examples within .tex files. -\item[coqdoc] A documentation tool for \Coq. -\end{description} - -There are front-ends for specific languages: - -\begin{description} -\item[Why] A back-end generator of verification conditions. -\item[Krakatoa] A Java code certification tool that uses both {\Coq} and {\Why} to verify the soundness of implementations with regards to the specifications. -\item[Caduceus] A C code certification tool that uses both {\Coq} and \Why. -\item[Zenon] A first-order theorem prover. -\item[Focal] The \ahref{http://focal.inria.fr}{Focal} project aims at building an environment to develop certified computer algebra libraries. -\end{description} - -\Question{What are the high-level tactics of \Coq} - -\begin{itemize} -\item Decision of quantifier-free Presburger's Arithmetic -\item Simplification of expressions on rings and fields -\item Decision of closed systems of equations -\item Semi-decision of first-order logic -\item Prolog-style proof search, possibly involving equalities -\end{itemize} - -\Question{What are the main libraries available for \Coq} - -\begin{itemize} -\item Basic Peano's arithmetic, binary integer numbers, rational numbers, -\item Real analysis, -\item Libraries for lists, boolean, maps, floating-point numbers, -\item Libraries for relations, sets and constructive algebra, -\item Geometry -\end{itemize} - - -\Question{What are the mathematical applications for {\Coq}?} - -{\Coq} is used for formalizing mathematical theories, for teaching, -and for proving properties of algorithms or programs libraries. - -The largest mathematical formalization has been done at the University -of Nijmegen (see the -\ahref{http://vacuumcleaner.cs.kun.nl/c-corn}{Constructive Coq -Repository at Nijmegen}). - -A symbolic step has also been obtained by formalizing in full a proof -of the Four Color Theorem. - -\Question{What are the industrial applications for {\Coq}?} - -{\Coq} is used e.g. to prove properties of the JavaCard system -(especially by Schlumberger and Trusted Logic). It has -also been used to formalize the semantics of the Lucid-Synchrone -data-flow synchronous calculus used by Esterel-Technologies. - -\iffalse -todo christine compilo lustre? -\fi - -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% - -\section{Documentation} - -\Question{Where can I find documentation about {\Coq}?} -All the documentation about \Coq, from the reference manual~\cite{Coq:manual} to -friendly tutorials~\cite{Coq:Tutorial} and documentation of the standard library, is available -\ahref{http://coq.inria.fr/doc-eng.html}{online}. -All these documents are viewable either in browsable HTML, or as -downloadable postscripts. - -\Question{Where can I find this FAQ on the web?} - -This FAQ is available online at \ahref{http://coq.inria.fr/doc/faq.html}{\url{http://coq.inria.fr/doc/faq.html}}. - -\Question{How can I submit suggestions / improvements / additions for this FAQ?} - -This FAQ is unfinished (in the sense that there are some obvious -sections that are missing). Please send contributions to \texttt{Florent.Kirchner at lix.polytechnique.fr} and \texttt{Julien.Narboux at inria.fr}. - -\Question{Is there any mailing list about {\Coq}?} -The main {\Coq} mailing list is \url{coq-club@pauillac.inria.fr}, which -broadcasts questions and suggestions about the implementation, the -logical formalism or proof developments. See -\ahref{http://coq.inria.fr/mailman/listinfo/coq-club}{\url{http://pauillac.inria.fr/mailman/listinfo/coq-club}} for -subscription. For bugs reports see question \ref{coqbug}. - -\Question{Where can I find an archive of the list?} -The archives of the {\Coq} mailing list are available at -\ahref{http://pauillac.inria.fr/pipermail/coq-club}{\url{http://coq.inria.fr/pipermail/coq-club}}. - - -\Question{How can I be kept informed of new releases of {\Coq}?} - -New versions of {\Coq} are announced on the coq-club mailing list. If you only want to receive information about new releases, you can subscribe to {\Coq} on \ahref{http://freshmeat.net/projects/coq/}{\url{http://freshmeat.net/projects/coq/}}. - - -\Question{Is there any book about {\Coq}?} - -The first book on \Coq, Yves Bertot and Pierre Castéran's Coq'Art has been published by Springer-Verlag in 2004: -\begin{quote} -``This book provides a pragmatic introduction to the development of -proofs and certified programs using \Coq. With its large collection of -examples and exercises it is an invaluable tool for researchers, -students, and engineers interested in formal methods and the -development of zero-default software.'' -\end{quote} - -\Question{Where can I find some {\Coq} examples?} - -There are examples in the manual~\cite{Coq:manual} and in the -Coq'Art~\cite{Coq:coqart} exercises \ahref{\url{http://www.labri.fr/Perso/~casteran/CoqArt/index.html}}{\url{http://www.labri.fr/Perso/~casteran/CoqArt/index.html}}. -You can also find large developments using -{\Coq} in the {\Coq} user contributions: -\ahref{http://coq.inria.fr/contrib-eng.html}{\url{http://coq.inria.fr/contrib-eng.html}}. - -\Question{How can I report a bug?}\label{coqbug} - -You can use the web interface accessible at \ahref{http://coq.inria.fr}{\url{http://coq.inria.fr}}, link ``contacts''. - - -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% - -\section{Installation} - -\Question{What is the license of {\Coq}?} -{\Coq} is distributed under the GNU Lesser General License -(LGPL). - -\Question{Where can I find the sources of {\Coq}?} -The sources of {\Coq} can be found online in the tar.gz'ed packages -(\ahref{http://coq.inria.fr}{\url{http://coq.inria.fr}}, link -``download''). Development sources can be accessed at -\ahref{http://coq.gforge.inria.fr/}{\url{http://coq.gforge.inria.fr/}} - -\Question{On which platform is {\Coq} available?} -Compiled binaries are available for Linux, MacOS X, and Windows. The -sources can be easily compiled on all platforms supporting Objective -Caml. - -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% - -\section{The logic of {\Coq}} - -\subsection{General} - -\Question{What is the logic of \Coq?} - -{\Coq} is based on an axiom-free type theory called -the Calculus of Inductive Constructions (see Coquand \cite{CoHu86}, -Luo~\cite{Luo90} -and Coquand--Paulin-Mohring \cite{CoPa89}). It includes higher-order -functions and predicates, inductive and co-inductive datatypes and -predicates, and a stratified hierarchy of sets. - -\Question{Is \Coq's logic intuitionistic or classical?} - -{\Coq}'s logic is modular. The core logic is intuitionistic -(i.e. excluded-middle $A\vee\neg A$ is not granted by default). It can -be extended to classical logic on demand by requiring an -optional module stating $A\vee\neg A$. - -\Question{Can I define non-terminating programs in \Coq?} - -All programs in {\Coq} are terminating. Especially, loops -must come with an evidence of their termination. - -Non-terminating programs can be simulated by passing around a -bound on how long the program is allowed to run before dying. - -\Question{How is equational reasoning working in {\Coq}?} - - {\Coq} comes with an internal notion of computation called -{\em conversion} (e.g. $(x+1)+y$ is internally equivalent to -$(x+y)+1$; similarly applying argument $a$ to a function mapping $x$ -to some expression $t$ converts to the expression $t$ where $x$ is -replaced by $a$). This notion of conversion (which is decidable -because {\Coq} programs are terminating) covers a certain part of -equational reasoning but is limited to sequential evaluation of -expressions of (not necessarily closed) programs. Besides conversion, -equations have to be treated by hand or using specialised tactics. - -\subsection{Axioms} - -\Question{What axioms can be safely added to {\Coq}?} - -There are a few typical useful axioms that are independent from the -Calculus of Inductive Constructions and that can be safely added to -{\Coq}. These axioms are stated in the directory {\tt Logic} of the -standard library of {\Coq}. The most interesting ones are - -\begin{itemize} -\item Excluded-middle: $\forall A:Prop, A \vee \neg A$ -\item Proof-irrelevance: $\forall A:Prop \forall p_1 p_2:A, p_1=p_2$ -\item Unicity of equality proofs (or equivalently Streicher's axiom $K$): -$\forall A \forall x y:A \forall p_1 p_2:x=y, p_1=p_2$ -\item The axiom of unique choice: $\forall x \exists! y R(x,y) \rightarrow \exists f \forall x R(x,f(x))$ -\item The functional axiom of choice: $\forall x \exists y R(x,y) \rightarrow \exists f \forall x R(x,f(x))$ -\item Extensionality of predicates: $\forall P Q:A\rightarrow Prop, (\forall x, P(x) \leftrightarrow Q(x)) \rightarrow P=Q$ -\item Extensionality of functions: $\forall f g:A\rightarrow B, (\forall x, f(x)=g(x)) \rightarrow f=g$ -\end{itemize} - -Here is a summary of the relative strength of these axioms, most -proofs can be found in directory {\tt Logic} of the standard library. -The justification of their validity relies on the interpretability in -set theory. - -%HEVEA\imgsrc{axioms.png} -%BEGIN LATEX -\ifpdf % si on est en pdflatex -\includegraphics[width=1.0\textwidth]{axioms.png} -\else -\includegraphics[width=1.0\textwidth]{axioms.eps} -\fi -%END LATEX - -\Question{What standard axioms are inconsistent with {\Coq}?} - -The axiom of unique choice together with classical logic -(e.g. excluded-middle) are inconsistent in the variant of the Calculus -of Inductive Constructions where {\Set} is impredicative. - -As a consequence, the functional form of the axiom of choice and -excluded-middle, or any form of the axiom of choice together with -predicate extensionality are inconsistent in the {\Set}-impredicative -version of the Calculus of Inductive Constructions. - -The main purpose of the \Set-predicative restriction of the Calculus -of Inductive Constructions is precisely to accommodate these axioms -which are quite standard in mathematical usage. - -The $\Set$-predicative system is commonly considered consistent by -interpreting it in a standard set-theoretic boolean model, even with -classical logic, axiom of choice and predicate extensionality added. - -\Question{What is Streicher's axiom $K$} -\label{Streicher} - -Streicher's axiom $K$~\cite{HofStr98} is an axiom that asserts -dependent elimination of reflexive equality proofs. - -\begin{coq_example*} -Axiom Streicher_K : - forall (A:Type) (x:A) (P: x=x -> Prop), - P (refl_equal x) -> forall p: x=x, P p. -\end{coq_example*} - -In the general case, axiom $K$ is an independent statement of the -Calculus of Inductive Constructions. However, it is true on decidable -domains (see file \vfile{\LogicEqdepDec}{Eqdep\_dec}). It is also -trivially a consequence of proof-irrelevance (see -\ref{proof-irrelevance}) hence of classical logic. - -Axiom $K$ is equivalent to {\em Uniqueness of Identity Proofs} \cite{HofStr98} - -\begin{coq_example*} -Axiom UIP : forall (A:Set) (x y:A) (p1 p2: x=y), p1 = p2. -\end{coq_example*} - -Axiom $K$ is also equivalent to {\em Uniqueness of Reflexive Identity Proofs} \cite{HofStr98} - -\begin{coq_example*} -Axiom UIP_refl : forall (A:Set) (x:A) (p: x=x), p = refl_equal x. -\end{coq_example*} - -Axiom $K$ is also equivalent to - -\begin{coq_example*} -Axiom - eq_rec_eq : - forall (A:Set) (x:A) (P: A->Set) (p:P x) (h: x=x), - p = eq_rect x P p x h. -\end{coq_example*} - -It is also equivalent to the injectivity of dependent equality (dependent equality is itself equivalent to equality of dependent pairs). - -\begin{coq_example*} -Inductive eq_dep (U:Set) (P:U -> Set) (p:U) (x:P p) : -forall q:U, P q -> Prop := - eq_dep_intro : eq_dep U P p x p x. -Axiom - eq_dep_eq : - forall (U:Set) (u:U) (P:U -> Set) (p1 p2:P u), - eq_dep U P u p1 u p2 -> p1 = p2. -\end{coq_example*} - -\Question{What is proof-irrelevance} -\label{proof-irrelevance} - -A specificity of the Calculus of Inductive Constructions is to permit -statements about proofs. This leads to the question of comparing two -proofs of the same proposition. Identifying all proofs of the same -proposition is called {\em proof-irrelevance}: -$$ -\forall A:\Prop, \forall p q:A, p=q -$$ - -Proof-irrelevance (in {\Prop}) can be assumed without contradiction in -{\Coq}. It expresses that only provability matters, whatever the exact -form of the proof is. This is in harmony with the common purely -logical interpretation of {\Prop}. Contrastingly, proof-irrelevance is -inconsistent in {\Set} since there are types in {\Set}, such as the -type of booleans, that are provably more than 2 elements. - -Proof-irrelevance (in {\Prop}) is a consequence of classical logic -(see proofs in file \vfile{\LogicClassical}{Classical} and -\vfile{\LogicBerardi}{Berardi}). Proof-irrelevance is also a -consequence of propositional extensionality (i.e. \coqtt{(A {\coqequiv} B) -{\coqimp} A=B}, see the proof in file -\vfile{\LogicClassicalFacts}{ClassicalFacts}). - -Proof-irrelevance directly implies Streicher's axiom $K$. - -\Question{What about functional extensionality?} - -Extensionality of functions is admittedly consistent with the -Set-predicative Calculus of Inductive Constructions. - -%\begin{coq_example*} -% Axiom extensionality : (A,B:Set)(f,g:(A->B))(x:A)(f x)=(g x)->f=g. -%\end{coq_example*} - -Let {\tt A}, {\tt B} be types. To deal with extensionality on -\verb=A->B= without relying on a general extensionality axiom, -a possible approach is to define one's own extensional equality on -\verb=A->B=. - -\begin{coq_eval} -Variables A B : Set. -\end{coq_eval} - -\begin{coq_example*} -Definition ext_eq (f g: A->B) := forall x:A, f x = g x. -\end{coq_example*} - -and to reason on \verb=A->B= as a setoid (see the Chapter on -Setoids in the Reference Manual). - -\Question{Is {\Prop} impredicative?} - -Yes, the sort {\Prop} of propositions is {\em -impredicative}. Otherwise said, a statement of the form $\forall -A:Prop, P(A)$ can be instantiated by itself: if $\forall A:\Prop, P(A)$ -is provable, then $P(\forall A:\Prop, P(A))$ is. - -\Question{Is {\Set} impredicative?} - -No, the sort {\Set} lying at the bottom of the hierarchy of -computational types is {\em predicative} in the basic {\Coq} system. -This means that a family of types in {\Set}, e.g. $\forall A:\Set, A -\rightarrow A$, is not a type in {\Set} and it cannot be applied on -itself. - -However, the sort {\Set} was impredicative in the original versions of -{\Coq}. For backward compatibility, or for experiments by -knowledgeable users, the logic of {\Coq} can be set impredicative for -{\Set} by calling {\Coq} with the option {\tt -impredicative-set}. - -{\Set} has been made predicative from version 8.0 of {\Coq}. The main -reason is to interact smoothly with a classical mathematical world -where both excluded-middle and the axiom of description are valid (see -file \vfile{\LogicClassicalDescription}{ClassicalDescription} for a -proof that excluded-middle and description implies the double negation -of excluded-middle in {\Set} and file {\tt Hurkens\_Set.v} from the -user contribution {\tt Rocq/PARADOXES} for a proof that -impredicativity of {\Set} implies the simple negation of -excluded-middle in {\Set}). - -\Question{Is {\Type} impredicative?} - -No, {\Type} is stratified. This is hidden for the -user, but {\Coq} internally maintains a set of constraints ensuring -stratification. - -If {\Type} were impredicative then it would be possible to encode -Girard's systems $U-$ and $U$ in {\Coq} and it is known from Girard, -Coquand, Hurkens and Miquel that systems $U-$ and $U$ are inconsistent -[Girard 1972, Coquand 1991, Hurkens 1993, Miquel 2001]. This encoding -can be found in file {\tt Logic/Hurkens.v} of {\Coq} standard library. - -For instance, when the user see {\tt $\forall$ X:Type, X->X : Type}, each -occurrence of {\Type} is implicitly bound to a different level, say -$\alpha$ and $\beta$ and the actual statement is {\tt -forall X:Type($\alpha$), X->X : Type($\beta$)} with the constraint -$\alpha<\beta$. - -When a statement violates a constraint, the message {\tt Universe -inconsistency} appears. Example: {\tt fun (x:Type) (y:$\forall$ X:Type, X -{\coqimp} X) => y x x}. - -\Question{I have two proofs of the same proposition. Can I prove they are equal?} - -In the base {\Coq} system, the answer is generally no. However, if -classical logic is set, the answer is yes for propositions in {\Prop}. -The answer is also yes if proof irrelevance holds (see question -\ref{proof-irrelevance}). - -There are also ``simple enough'' propositions for which you can prove -the equality without requiring any extra axioms. This is typically -the case for propositions defined deterministically as a first-order -inductive predicate on decidable sets. See for instance in question -\ref{le-uniqueness} an axiom-free proof of the unicity of the proofs of -the proposition {\tt le m n} (less or equal on {\tt nat}). - -% It is an ongoing work of research to natively include proof -% irrelevance in {\Coq}. - -\Question{I have two proofs of an equality statement. Can I prove they are -equal?} - - Yes, if equality is decidable on the domain considered (which -is the case for {\tt nat}, {\tt bool}, etc): see {\Coq} file -\verb=Eqdep_dec.v=). No otherwise, unless -assuming Streicher's axiom $K$ (see \cite{HofStr98}) or a more general -assumption such as proof-irrelevance (see \ref{proof-irrelevance}) or -classical logic. - -All of these statements can be found in file \vfile{\LogicEqdep}{Eqdep}. - -\Question{Can I prove that the second components of equal dependent -pairs are equal?} - - The answer is the same as for proofs of equality -statements. It is provable if equality on the domain of the first -component is decidable (look at \verb=inj_right_pair= from file -\vfile{\LogicEqdepDec}{Eqdep\_dec}), but not provable in the general -case. However, it is consistent (with the Calculus of Constructions) -to assume it is true. The file \vfile{\LogicEqdep}{Eqdep} actually -provides an axiom (equivalent to Streicher's axiom $K$) which entails -the result (look at \verb=inj_pair2= in \vfile{\LogicEqdep}{Eqdep}). - -\subsection{Impredicativity} - -\Question{Why {\tt injection} does not work on impredicative {\tt Set}?} - - E.g. in this case (this occurs only in the {\tt Set}-impredicative - variant of \Coq): - -\begin{coq_eval} -Reset Initial. -\end{coq_eval} - -\begin{coq_example*} -Inductive I : Type := - intro : forall k:Set, k -> I. -Lemma eq_jdef : - forall x y:nat, intro _ x = intro _ y -> x = y. -Proof. - intros x y H; injection H. -\end{coq_example*} - - Injectivity of constructors is restricted to predicative types. If -injectivity on large inductive types were not restricted, we would be -allowed to derive an inconsistency (e.g. following the lines of -Burali-Forti paradox). The question remains open whether injectivity -is consistent on some large inductive types not expressive enough to -encode known paradoxes (such as type I above). - - -\Question{What is a ``large inductive definition''?} - -An inductive definition in {\Prop} or {\Set} is called large -if its constructors embed sets or propositions. As an example, here is -a large inductive type: - -\begin{coq_example*} -Inductive sigST (P:Set -> Set) : Type := - existST : forall X:Set, P X -> sigST P. -\end{coq_example*} - -In the {\tt Set} impredicative variant of {\Coq}, large inductive -definitions in {\tt Set} have restricted elimination schemes to -prevent inconsistencies. Especially, projecting the set or the -proposition content of a large inductive definition is forbidden. If -it were allowed, it would be possible to encode e.g. Burali-Forti -paradox \cite{Gir70,Coq85}. - - -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -\section{Talkin' with the Rooster} - - -%%%%%%% -\subsection{My goal is ..., how can I prove it?} - - -\Question{My goal is a conjunction, how can I prove it?} - -Use some theorem or assumption or use the {\split} tactic. -\begin{coq_example} -Goal forall A B:Prop, A->B-> A/\B. -intros. -split. -assumption. -assumption. -Qed. -\end{coq_example} - -\Question{My goal contains a conjunction as an hypothesis, how can I use it?} - -If you want to decompose your hypothesis into other hypothesis you can use the {\decompose} tactic: - -\begin{coq_example} -Goal forall A B:Prop, A/\B-> B. -intros. -decompose [and] H. -assumption. -Qed. -\end{coq_example} - - -\Question{My goal is a disjunction, how can I prove it?} - -You can prove the left part or the right part of the disjunction using -{\left} or {\right} tactics. If you want to do a classical -reasoning step, use the {\tt classic} axiom to prove the right part with the assumption -that the left part of the disjunction is false. - -\begin{coq_example} -Goal forall A B:Prop, A-> A\/B. -intros. -left. -assumption. -Qed. -\end{coq_example} - -An example using classical reasoning: - -\begin{coq_example} -Require Import Classical. - -Ltac classical_right := -match goal with -| _:_ |-?X1 \/ _ => (elim (classic X1);intro;[left;trivial|right]) -end. - -Ltac classical_left := -match goal with -| _:_ |- _ \/?X1 => (elim (classic X1);intro;[right;trivial|left]) -end. - - -Goal forall A B:Prop, (~A -> B) -> A\/B. -intros. -classical_right. -auto. -Qed. -\end{coq_example} - -\Question{My goal is an universally quantified statement, how can I prove it?} - -Use some theorem or assumption or introduce the quantified variable in -the context using the {\intro} tactic. If there are several -variables you can use the {\intros} tactic. A good habit is to -provide names for these variables: {\Coq} will do it anyway, but such -automatic naming decreases legibility and robustness. - - -\Question{My goal is an existential, how can I prove it?} - -Use some theorem or assumption or exhibit the witness using the {\existstac} tactic. -\begin{coq_example} -Goal exists x:nat, forall y, x+y=y. -exists 0. -intros. -auto. -Qed. -\end{coq_example} - - -\Question{My goal is solvable by some lemma, how can I prove it?} - -Just use the {\apply} tactic. - -\begin{coq_eval} -Reset Initial. -\end{coq_eval} - -\begin{coq_example} -Lemma mylemma : forall x, x+0 = x. -auto. -Qed. - -Goal 3+0 = 3. -apply mylemma. -Qed. -\end{coq_example} - - - -\Question{My goal contains False as an hypothesis, how can I prove it?} - -You can use the {\contradiction} or {\intuition} tactics. - - -\Question{My goal is an equality of two convertible terms, how can I prove it?} - -Just use the {\reflexivity} tactic. - -\begin{coq_example} -Goal forall x, 0+x = x. -intros. -reflexivity. -Qed. -\end{coq_example} - -\Question{My goal is a {\tt let x := a in ...}, how can I prove it?} - -Just use the {\intro} tactic. - - -\Question{My goal is a {\tt let (a, ..., b) := c in}, how can I prove it?} - -Just use the {\destruct} c as (a,...,b) tactic. - - -\Question{My goal contains some existential hypotheses, how can I use it?} - -You can use the tactic {\elim} with you hypotheses as an argument. - -\Question{My goal contains some existential hypotheses, how can I use it and decompose my knowledge about this new thing into different hypotheses?} - -\begin{verbatim} -Ltac DecompEx H P := elim H;intro P;intro TO;decompose [and] TO;clear TO;clear H. -\end{verbatim} - - -\Question{My goal is an equality, how can I swap the left and right hand terms?} - -Just use the {\symmetry} tactic. -\begin{coq_example} -Goal forall x y : nat, x=y -> y=x. -intros. -symmetry. -assumption. -Qed. -\end{coq_example} - -\Question{My hypothesis is an equality, how can I swap the left and right hand terms?} - -Just use the {\symmetryin} tactic. - -\begin{coq_example} -Goal forall x y : nat, x=y -> y=x. -intros. -symmetry in H. -assumption. -Qed. -\end{coq_example} - - -\Question{My goal is an equality, how can I prove it by transitivity?} - -Just use the {\transitivity} tactic. -\begin{coq_example} -Goal forall x y z : nat, x=y -> y=z -> x=z. -intros. -transitivity y. -assumption. -assumption. -Qed. -\end{coq_example} - - -\Question{My goal would be solvable using {\tt apply;assumption} if it would not create meta-variables, how can I prove it?} - -You can use {\tt eapply yourtheorem;eauto} but it won't work in all cases ! (for example if more than one hypothesis match one of the subgoals generated by \eapply) so you should rather use {\tt try solve [eapply yourtheorem;eauto]}, otherwise some metavariables may be incorrectly instantiated. - -\begin{coq_example} -Lemma trans : forall x y z : nat, x=y -> y=z -> x=z. -intros. -transitivity y;assumption. -Qed. - -Goal forall x y z : nat, x=y -> y=z -> x=z. -intros. -eapply trans;eauto. -Qed. - -Goal forall x y z t : nat, x=y -> x=t -> y=z -> x=z. -intros. -eapply trans;eauto. -Undo. -eapply trans. -apply H. -auto. -Qed. - -Goal forall x y z t : nat, x=y -> x=t -> y=z -> x=z. -intros. -eapply trans;eauto. -Undo. -try solve [eapply trans;eauto]. -eapply trans. -apply H. -auto. -Qed. - -\end{coq_example} - -\Question{My goal is solvable by some lemma within a set of lemmas and I don't want to remember which one, how can I prove it?} - -You can use a what is called a hints' base. - -\begin{coq_example} -Require Import ZArith. -Require Ring. -Open Local Scope Z_scope. -Lemma toto1 : 1+1 = 2. -ring. -Qed. -Lemma toto2 : 2+2 = 4. -ring. -Qed. -Lemma toto3 : 2+1 = 3. -ring. -Qed. - -Hint Resolve toto1 toto2 toto3 : mybase. - -Goal 2+(1+1)=4. -auto with mybase. -Qed. -\end{coq_example} - - -\Question{My goal is one of the hypotheses, how can I prove it?} - -Use the {\assumption} tactic. - -\begin{coq_example} -Goal 1=1 -> 1=1. -intro. -assumption. -Qed. -\end{coq_example} - - -\Question{My goal appears twice in the hypotheses and I want to choose which one is used, how can I do it?} - -Use the {\exact} tactic. -\begin{coq_example} -Goal 1=1 -> 1=1 -> 1=1. -intros. -exact H0. -Qed. -\end{coq_example} - -\Question{What can be the difference between applying one hypothesis or another in the context of the last question?} - -From a proof point of view it is equivalent but if you want to extract -a program from your proof, the two hypotheses can lead to different -programs. - - -\Question{My goal is a propositional tautology, how can I prove it?} - -Just use the {\tauto} tactic. -\begin{coq_example} -Goal forall A B:Prop, A-> (A\/B) /\ A. -intros. -tauto. -Qed. -\end{coq_example} - -\Question{My goal is a first order formula, how can I prove it?} - -Just use the semi-decision tactic: \firstorder. - -\iffalse -todo: demander un exemple à Pierre -\fi - -\Question{My goal is solvable by a sequence of rewrites, how can I prove it?} - -Just use the {\congruence} tactic. -\begin{coq_example} -Goal forall a b c d e, a=d -> b=e -> c+b=d -> c+e=a. -intros. -congruence. -Qed. -\end{coq_example} - - -\Question{My goal is a disequality solvable by a sequence of rewrites, how can I prove it?} - -Just use the {\congruence} tactic. - -\begin{coq_example} -Goal forall a b c d, a<>d -> b=a -> d=c+b -> b<>c+b. -intros. -congruence. -Qed. -\end{coq_example} - - -\Question{My goal is an equality on some ring (e.g. natural numbers), how can I prove it?} - -Just use the {\ring} tactic. - -\begin{coq_example} -Require Import ZArith. -Require Ring. -Open Local Scope Z_scope. -Goal forall a b : Z, (a+b)*(a+b) = a*a + 2*a*b + b*b. -intros. -ring. -Qed. -\end{coq_example} - -\Question{My goal is an equality on some field (e.g. real numbers), how can I prove it?} - -Just use the {\field} tactic. - -\begin{coq_example} -Require Import Reals. -Require Ring. -Open Local Scope R_scope. -Goal forall a b : R, b*a<>0 -> (a/b) * (b/a) = 1. -intros. -field. -assumption. -Qed. -\end{coq_example} - - -\Question{My goal is an inequality on integers in Presburger's arithmetic (an expression build from +,-,constants and variables), how can I prove it?} - - -\begin{coq_example} -Require Import ZArith. -Require Omega. -Open Local Scope Z_scope. -Goal forall a : Z, a>0 -> a+a > a. -intros. -omega. -Qed. -\end{coq_example} - - -\Question{My goal is an equation solvable using equational hypothesis on some ring (e.g. natural numbers), how can I prove it?} - -You need the {\gb} tactic (see Loïc Pottier's homepage). - -\subsection{Tactics usage} - -\Question{I want to state a fact that I will use later as an hypothesis, how can I do it?} - -If you want to use forward reasoning (first proving the fact and then -using it) you just need to use the {\assert} tactic. If you want to use -backward reasoning (proving your goal using an assumption and then -proving the assumption) use the {\cut} tactic. - -\begin{coq_example} -Goal forall A B C D : Prop, (A -> B) -> (B->C) -> A -> C. -intros. -assert (A->C). -intro;apply H0;apply H;assumption. -apply H2. -assumption. -Qed. - -Goal forall A B C D : Prop, (A -> B) -> (B->C) -> A -> C. -intros. -cut (A->C). -intro. -apply H2;assumption. -intro;apply H0;apply H;assumption. -Qed. -\end{coq_example} - - - - -\Question{I want to state a fact that I will use later as an hypothesis and prove it later, how can I do it?} - -You can use {\cut} followed by {\intro} or you can use the following {\Ltac} command: -\begin{verbatim} -Ltac assert_later t := cut t;[intro|idtac]. -\end{verbatim} - -\Question{What is the difference between {\Qed} and {\Defined}?} - -These two commands perform type checking, but when {\Defined} is used the new definition is set as transparent, otherwise it is defined as opaque (see \ref{opaque}). - - -\Question{How can I know what a tactic does?} - -You can use the {\tt info} command. - - - -\Question{Why {\auto} does not work? How can I fix it?} - -You can increase the depth of the proof search or add some lemmas in the base of hints. -Perhaps you may need to use \eauto. - -\Question{What is {\eauto}?} - -This is the same tactic as \auto, but it relies on {\eapply} instead of \apply. - -\iffalse -todo les espaces -\fi - -\Question{How can I speed up {\auto}?} - -You can use \texttt{info }\auto to replace {\auto} by the tactics it generates. -You can split your hint bases into smaller ones. - - -\Question{What is the equivalent of {\tauto} for classical logic?} - -Currently there are no equivalent tactic for classical logic. You can use Gödel's ``not not'' translation. - - -\Question{I want to replace some term with another in the goal, how can I do it?} - -If one of your hypothesis (say {\tt H}) states that the terms are equal you can use the {\rewrite} tactic. Otherwise you can use the {\replace} {\tt with} tactic. - -\Question{I want to replace some term with another in an hypothesis, how can I do it?} - -You can use the {\rewrite} {\tt in} tactic. - -\Question{I want to replace some symbol with its definition, how can I do it?} - -You can use the {\unfold} tactic. - -\Question{How can I reduce some term?} - -You can use the {\simpl} tactic. - -\Question{How can I declare a shortcut for some term?} - -You can use the {\set} or {\pose} tactics. - -\Question{How can I perform case analysis?} - -You can use the {\case} or {\destruct} tactics. - - -\Question{Why should I name my intros?} - -When you use the {\intro} tactic you don't have to give a name to your -hypothesis. If you do so the name will be generated by {\Coq} but your -scripts may be less robust. If you add some hypothesis to your theorem -(or change their order), you will have to change your proof to adapt -to the new names. - -\Question{How can I automatize the naming?} - -You can use the {\tt Show Intro.} or {\tt Show Intros.} commands to generate the names and use your editor to generate a fully named {\intro} tactic. -This can be automatized within {\tt xemacs}. - -\begin{coq_example} -Goal forall A B C : Prop, A -> B -> C -> A/\B/\C. -Show Intros. -(* -A B C H H0 -H1 -*) -intros A B C H H0 H1. -repeat split;assumption. -Qed. -\end{coq_example} - -\Question{I want to automatize the use of some tactic, how can I do it?} - -You need to use the {\tt proof with T} command and add {\ldots} at the -end of your sentences. - -For instance: -\begin{coq_example} -Goal forall A B C : Prop, A -> B/\C -> A/\B/\C. -Proof with assumption. -intros. -split... -Qed. -\end{coq_example} - -\Question{I want to execute the {\texttt proof with} tactic only if it solves the goal, how can I do it?} - -You need to use the {\try} and {\solve} tactics. For instance: -\begin{coq_example} -Require Import ZArith. -Require Ring. -Open Local Scope Z_scope. -Goal forall a b c : Z, a+b=b+a. -Proof with try solve [ring]. -intros... -Qed. -\end{coq_example} - -\Question{How can I do the opposite of the {\intro} tactic?} - -You can use the {\generalize} tactic. - -\begin{coq_example} -Goal forall A B : Prop, A->B-> A/\B. -intros. -generalize H. -intro. -auto. -Qed. -\end{coq_example} - -\Question{One of the hypothesis is an equality between a variable and some term, I want to get rid of this variable, how can I do it?} - -You can use the {\subst} tactic. This will rewrite the equality everywhere and clear the assumption. - -\Question{What can I do if I get ``{\tt generated subgoal term has metavariables in it }''?} - -You should use the {\eapply} tactic, this will generate some goals containing metavariables. - -\Question{How can I instantiate some metavariable?} - -Just use the {\instantiate} tactic. - - -\Question{What is the use of the {\pattern} tactic?} - -The {\pattern} tactic transforms the current goal, performing -beta-expansion on all the applications featuring this tactic's -argument. For instance, if the current goal includes a subterm {\tt -phi(t)}, then {\tt pattern t} transforms the subterm into {\tt (fun -x:A => phi(x)) t}. This can be useful when {\apply} fails on matching, -to abstract the appropriate terms. - -\Question{What is the difference between assert, cut and generalize?} - -PS: Notice for people that are interested in proof rendering that \assert -and {\pose} (and \cut) are not rendered the same as {\generalize} (see the -HELM experimental rendering tool at \ahref{http://helm.cs.unibo.it/library.html}{\url{http://helm.cs.unibo.it}}, link -HELM, link COQ Online). Indeed {\generalize} builds a beta-expanded term -while \assert, {\pose} and {\cut} uses a let-in. - -\begin{verbatim} - (* Goal is T *) - generalize (H1 H2). - (* Goal is A->T *) - ... a proof of A->T ... -\end{verbatim} - -is rendered into something like -\begin{verbatim} - (h) ... the proof of A->T ... - we proved A->T - (h0) by (H1 H2) we proved A - by (h h0) we proved T -\end{verbatim} -while -\begin{verbatim} - (* Goal is T *) - assert q := (H1 H2). - (* Goal is A *) - ... a proof of A ... - (* Goal is A |- T *) - ... a proof of T ... -\end{verbatim} -is rendered into something like -\begin{verbatim} - (q) ... the proof of A ... - we proved A - ... the proof of T ... - we proved T -\end{verbatim} -Otherwise said, {\generalize} is not rendered in a forward-reasoning way, -while {\assert} is. - -\Question{What can I do if \Coq can not infer some implicit argument ?} - -You can state explicitely what this implicit argument is. See \ref{implicit}. - -\Question{How can I explicit some implicit argument ?}\label{implicit} - -Just use \texttt{A:=term} where \texttt{A} is the argument. - -For instance if you want to use the existence of ``nil'' on nat*nat lists: -\begin{verbatim} -exists (nil (A:=(nat*nat))). -\end{verbatim} - -\iffalse -\Question{Is there anyway to do pattern matching with dependent types?} - -todo -\fi - -\subsection{Proof management} - - -\Question{How can I change the order of the subgoals?} - -You can use the {\Focus} command to concentrate on some goal. When the goal is proved you will see the remaining goals. - -\Question{How can I change the order of the hypothesis?} - -You can use the {\tt Move ... after} command. - -\Question{How can I change the name of an hypothesis?} - -You can use the {\tt Rename ... into} command. - -\Question{How can I delete some hypothesis?} - -You can use the {\tt Clear} command. - -\Question{How can use a proof which is not finished?} - -You can use the {\tt Admitted} command to state your current proof as an axiom. - -\Question{How can I state a conjecture?} - -You can use the {\tt Admitted} command to state your current proof as an axiom. - -\Question{What is the difference between a lemma, a fact and a theorem?} - -From {\Coq} point of view there are no difference. But some tools can -have a different behavior when you use a lemma rather than a -theorem. For instance {\tt coqdoc} will not generate documentation for -the lemmas within your development. - -\Question{How can I organize my proofs?} - -You can organize your proofs using the section mechanism of \Coq. Have -a look at the manual for further information. - - -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -\section{Inductive and Co-inductive types} - -\subsection{General} - -\Question{How can I prove that two constructors are different?} - -You can use the {\discriminate} tactic. - -\begin{coq_example} -Inductive toto : Set := | C1 : toto | C2 : toto. -Goal C1 <> C2. -discriminate. -Qed. -\end{coq_example} - -\Question{During an inductive proof, how to get rid of impossible cases of an inductive definition?} - -Use the {\inversion} tactic. - - -\Question{How can I prove that 2 terms in an inductive set are equal? Or different?} - -Have a look at \coqtt{decide equality} and \coqtt{discriminate} in the \ahref{http://coq.inria.fr/doc/main.html}{Reference Manual}. - -\Question{Why is the proof of \coqtt{0+n=n} on natural numbers -trivial but the proof of \coqtt{n+0=n} is not?} - - Since \coqtt{+} (\coqtt{plus}) on natural numbers is defined by analysis on its first argument - -\begin{coq_example} -Print plus. -\end{coq_example} - -{\noindent} The expression \coqtt{0+n} evaluates to \coqtt{n}. As {\Coq} reasons -modulo evaluation of expressions, \coqtt{0+n} and \coqtt{n} are -considered equal and the theorem \coqtt{0+n=n} is an instance of the -reflexivity of equality. On the other side, \coqtt{n+0} does not -evaluate to \coqtt{n} and a proof by induction on \coqtt{n} is -necessary to trigger the evaluation of \coqtt{+}. - -\Question{Why is dependent elimination in Prop not -available by default?} - - -This is just because most of the time it is not needed. To derive a -dependent elimination principle in {\tt Prop}, use the command {\tt Scheme} and -apply the elimination scheme using the \verb=using= option of -\verb=elim=, \verb=destruct= or \verb=induction=. - - -\Question{Argh! I cannot write expressions like ``~{\tt if n <= p then p else n}~'', as in any programming language} -\label{minmax} - -The short answer : You should use {\texttt le\_lt\_dec n p} instead.\\ - -That's right, you can't. -If you type for instance the following ``definition'': -\begin{coq_eval} -Reset Initial. -\end{coq_eval} -\begin{coq_example} -Definition max (n p : nat) := if n <= p then p else n. -\end{coq_example} - -As \Coq~ says, the term ``~\texttt{n <= p}~'' is a proposition, i.e. a -statement that belongs to the mathematical world. There are many ways to -prove such a proposition, either by some computation, or using some already -proven theoremas. For instance, proving $3-2 \leq 2^{45503}$ is very easy, -using some theorems on arithmetical operations. If you compute both numbers -before comparing them, you risk to use a lot of time and space. - - -On the contrary, a function for computing the greatest of two natural numbers -is an algorithm which, called on two natural numbers -$n$ and $p$, determines wether $n\leq p$ or $p < n$. -Such a function is a \emph{decision procedure} for the inequality of - \texttt{nat}. The possibility of writing such a procedure comes -directly from de decidability of the order $\leq$ on natural numbers. - - -When you write a piece of code like -``~\texttt{if n <= p then \dots{} else \dots}~'' -in a -programming language like \emph{ML} or \emph{Java}, a call to such a -decision procedure is generated. The decision procedure is in general -a primitive function, written in a low-level language, in the correctness -of which you have to trust. - -The standard Library of the system \emph{Coq} contains a -(constructive) proof of decidability of the order $\leq$ on -\texttt{nat} : the function \texttt{le\_lt\_dec} of -the module \texttt{Compare\_dec} of library \texttt{Arith}. - -The following code shows how to define correctly \texttt{min} and -\texttt{max}, and prove some properties of these functions. - -\begin{coq_example} -Require Import Compare_dec. - -Definition max (n p : nat) := if le_lt_dec n p then p else n. - -Definition min (n p : nat) := if le_lt_dec n p then n else p. - -Eval compute in (min 4 7). - -Theorem min_plus_max : forall n p, min n p + max n p = n + p. -Proof. - intros n p; - unfold min, max; - case (le_lt_dec n p); - simpl; auto with arith. -Qed. - -Theorem max_equiv : forall n p, max n p = p <-> n <= p. -Proof. - unfold max; intros n p; case (le_lt_dec n p);simpl; auto. - intuition auto with arith. - split. - intro e; rewrite e; auto with arith. - intro H; absurd (p < p); eauto with arith. -Qed. -\end{coq_example} - -\Question{I wrote my own decision procedure for $\leq$, which -is much faster than yours, but proving such theorems as - \texttt{max\_equiv} seems to be quite difficult} - -Your code is probably the following one: - -\begin{coq_example} -Fixpoint my_le_lt_dec (n p :nat) {struct n}: bool := - match n, p with 0, _ => true - | S n', S p' => my_le_lt_dec n' p' - | _ , _ => false - end. - -Definition my_max (n p:nat) := if my_le_lt_dec n p then p else n. - -Definition my_min (n p:nat) := if my_le_lt_dec n p then n else p. -\end{coq_example} - - -For instance, the computation of \texttt{my\_max 567 321} is almost -immediate, whereas one can't wait for the result of -\texttt{max 56 32}, using \emph{Coq's} \texttt{le\_lt\_dec}. - -This is normal. Your definition is a simple recursive function which -returns a boolean value. Coq's \texttt{le\_lt\_dec} is a \emph{certified -function}, i.e. a complex object, able not only to tell wether $n\leq p$ -or $p<n$, but also of building a complete proof of the correct inequality. -What make \texttt{le\_lt\_dec} inefficient for computing \texttt{min} -and \texttt{max} is the building of a huge proof term. - -Nevertheless, \texttt{le\_lt\_dec} is very useful. Its type -is a strong specification, using the -\texttt{sumbool} type (look at the reference manual or chapter 9 of -\cite{coqart}). Eliminations of the form -``~\texttt{case (le\_lt\_dec n p)}~'' provide proofs of -either $n \leq p$ or $p < n$, allowing to prove easily theorems as in -question~\ref{minmax}. Unfortunately, this not the case of your -\texttt{my\_le\_lt\_dec}, which returns a quite non-informative boolean -value. - - -\begin{coq_example} -Check le_lt_dec. -\end{coq_example} - -You should keep in mind that \texttt{le\_lt\_dec} is useful to build -certified programs which need to compare natural numbers, and is not -designed to compare quickly two numbers. - -Nevertheless, the \emph{extraction} of \texttt{le\_lt\_dec} towards -\emph{Ocaml} or \emph{Haskell}, is a reasonable program for comparing two -natural numbers in Peano form in linear time. - -It is also possible to keep your boolean function as a decision procedure, -but you have to establish yourself the relationship between \texttt{my\_le\_lt\_dec} and the propositions $n\leq p$ and $p<n$: - -\begin{coq_example*} -Theorem my_le_lt_dec_true : - forall n p, my_le_lt_dec n p = true <-> n <= p. - -Theorem my_le_lt_dec_false : - forall n p, my_le_lt_dec n p = false <-> p < n. -\end{coq_example*} - - -\subsection{Recursion} - -\Question{Why can't I define a non terminating program?} - - Because otherwise the decidability of the type-checking -algorithm (which involves evaluation of programs) is not ensured. On -another side, if non terminating proofs were allowed, we could get a -proof of {\tt False}: - -\begin{coq_example*} -(* This is fortunately not allowed! *) -Fixpoint InfiniteProof (n:nat) : False := InfiniteProof n. -Theorem Paradox : False. -Proof (InfiniteProof O). -\end{coq_example*} - - -\Question{Why only structurally well-founded loops are allowed?} - - The structural order on inductive types is a simple and -powerful notion of termination. The consistency of the Calculus of -Inductive Constructions relies on it and another consistency proof -would have to be made for stronger termination arguments (such -as the termination of the evaluation of CIC programs themselves!). - -In spite of this, all non-pathological termination orders can be mapped -to a structural order. Tools to do this are provided in the file -\vfile{\InitWf}{Wf} of the standard library of {\Coq}. - -\Question{How to define loops based on non structurally smaller -recursive calls?} - - The procedure is as follows (we consider the definition of {\tt -mergesort} as an example). - -\begin{itemize} - -\item Define the termination order, say {\tt R} on the type {\tt A} of -the arguments of the loop. - -\begin{coq_eval} -Open Scope R_scope. -Require Import List. -\end{coq_eval} - -\begin{coq_example*} -Definition R (a b:list nat) := length a < length b. -\end{coq_example*} - -\item Prove that this order is well-founded (in fact that all elements in {\tt A} are accessible along {\tt R}). - -\begin{coq_example*} -Lemma Rwf : well_founded R. -\end{coq_example*} - -\item Define the step function (which needs proofs that recursive -calls are on smaller arguments). - -\begin{coq_example*} -Definition split (l : list nat) - : {l1: list nat | R l1 l} * {l2 : list nat | R l2 l} - := (* ... *) . -Definition concat (l1 l2 : list nat) : list nat := (* ... *) . -Definition merge_step (l : list nat) (f: forall l':list nat, R l' l -> list nat) := - let (lH1,lH2) := (split l) in - let (l1,H1) := lH1 in - let (l2,H2) := lH2 in - concat (f l1 H1) (f l2 H2). -\end{coq_example*} - -\item Define the recursive function by fixpoint on the step function. - -\begin{coq_example*} -Definition merge := Fix Rwf (fun _ => list nat) merge_step. -\end{coq_example*} - -\end{itemize} - -\Question{What is behind the accessibility and well-foundedness proofs?} - - Well-foundedness of some relation {\tt R} on some type {\tt A} -is defined as the accessibility of all elements of {\tt A} along {\tt R}. - -\begin{coq_example} -Print well_founded. -Print Acc. -\end{coq_example} - -The structure of the accessibility predicate is a well-founded tree -branching at each node {\tt x} in {\tt A} along all the nodes {\tt x'} -less than {\tt x} along {\tt R}. Any sequence of elements of {\tt A} -decreasing along the order {\tt R} are branches in the accessibility -tree. Hence any decreasing along {\tt R} is mapped into a structural -decreasing in the accessibility tree of {\tt R}. This is emphasised in -the definition of {\tt fix} which recurs not on its argument {\tt x:A} -but on the accessibility of this argument along {\tt R}. - -See file \vfile{\InitWf}{Wf}. - -\Question{How to perform simultaneous double induction?} - - In general a (simultaneous) double induction is simply solved by an -induction on the first hypothesis followed by an inversion over the -second hypothesis. Here is an example - -\begin{coq_eval} -Reset Initial. -\end{coq_eval} - -\begin{coq_example} -Inductive even : nat -> Prop := - | even_O : even 0 - | even_S : forall n:nat, even n -> even (S (S n)). - -Inductive odd : nat -> Prop := - | odd_SO : odd 1 - | odd_S : forall n:nat, odd n -> odd (S (S n)). - -Lemma not_even_and_odd : forall n:nat, even n -> odd n -> False. -induction 1. - inversion 1. - inversion 1. apply IHeven; trivial. -\end{coq_example} -\begin{coq_eval} -Qed. -\end{coq_eval} - -In case the type of the second induction hypothesis is not -dependent, {\tt inversion} can just be replaced by {\tt destruct}. - -\Question{How to define a function by simultaneous double recursion?} - - The same trick applies, you can even use the pattern-matching -compilation algorithm to do the work for you. Here is an example: - -\begin{coq_example} -Fixpoint minus (n m:nat) {struct n} : nat := - match n, m with - | O, _ => 0 - | S k, O => S k - | S k, S l => minus k l - end. -Print minus. -\end{coq_example} - -In case of dependencies in the type of the induction objects -$t_1$ and $t_2$, an extra argument stating $t_1=t_2$ must be given to -the fixpoint definition - -\Question{How to perform nested and double induction?} - - To reason by nested (i.e. lexicographic) induction, just reason by -induction on the successive components. - -\smallskip - -Double induction (or induction on pairs) is a restriction of the -lexicographic induction. Here is an example of double induction. - -\begin{coq_example} -Lemma nat_double_ind : -forall P : nat -> nat -> Prop, P 0 0 -> - (forall m n, P m n -> P m (S n)) -> - (forall m n, P m n -> P (S m) n) -> - forall m n, P m n. -intros P H00 HmS HSn; induction m. -(* case 0 *) -induction n; [assumption | apply HmS; apply IHn]. -(* case Sm *) -intro n; apply HSn; apply IHm. -\end{coq_example} -\begin{coq_eval} -Qed. -\end{coq_eval} - -\Question{How to define a function by nested recursion?} - - The same trick applies. Here is the example of Ackermann -function. - -\begin{coq_example} -Fixpoint ack (n:nat) : nat -> nat := - match n with - | O => S - | S n' => - (fix ack' (m:nat) : nat := - match m with - | O => ack n' 1 - | S m' => ack n' (ack' m') - end) - end. -\end{coq_example} - - -\subsection{Co-inductive types} - -\Question{I have a cofixpoint $t:=F(t)$ and I want to prove $t=F(t)$. How to do it?} - -Just case-expand $F({\tt t})$ then complete by a trivial case analysis. -Here is what it gives on e.g. the type of streams on naturals - -\begin{coq_eval} -Set Implicit Arguments. -\end{coq_eval} -\begin{coq_example} -CoInductive Stream (A:Set) : Set := - Cons : A -> Stream A -> Stream A. -CoFixpoint nats (n:nat) : Stream nat := Cons n (nats (S n)). -Lemma Stream_unfold : - forall n:nat, nats n = Cons n (nats (S n)). -Proof. - intro; - change (nats n = match nats n with - | Cons x s => Cons x s - end). - case (nats n); reflexivity. -Qed. -\end{coq_example} - - - -\section{Syntax and notations} - -\Question{I do not want to type ``forall'' because it is too long, what can I do?} - -You can define your own notation for forall: -\begin{verbatim} -Notation "fa x : t, P" := (forall x:t, P) (at level 200, x ident). -\end{verbatim} -or if your are using {\CoqIde} you can define a pretty symbol for for all and an input method (see \ref{forallcoqide}). - - - -\Question{How can I define a notation for square?} - -You can use for instance: -\begin{verbatim} -Notation "x ^2" := (Rmult x x) (at level 20). -\end{verbatim} -Note that you can not use: -\begin{texttt} -Notation "x $^²$" := (Rmult x x) (at level 20). -\end{texttt} -because ``$^2$'' is an iso-latin character. If you really want this kind of notation you should use UTF-8. - - -\Question{Why ``no associativity'' and ``left associativity'' at the same level does not work?} - -Because we relie on camlp4 for syntactical analysis and camlp4 does not really implement no associativity. By default, non associative operators are defined as right associative. - - - -\Question{How can I know the associativity associated with a level?} - -You can do ``Print Grammar constr'', and decode the output from camlp4, good luck ! - -\section{Modules} - - - - -%%%%%%% -\section{\Ltac} - -\Question{What is {\Ltac}?} - -{\Ltac} is the tactic language for \Coq. It provides the user with a -high-level ``toolbox'' for tactic creation. - -\Question{Why do I always get the same error message?} - - -\Question{Is there any printing command in {\Ltac}?} - -You can use the {\idtac} tactic with a string argument. This string -will be printed out. The same applies to the {\fail} tactic - -\Question{What is the syntax for let in {\Ltac}?} - -If $x_i$ are identifiers and $e_i$ and $expr$ are tactic expressions, then let reads: -\begin{center} -{\tt let $x_1$:=$e_1$ with $x_2$:=$e_2$\ldots with $x_n$:=$e_n$ in -$expr$}. -\end{center} -Beware that if $expr$ is complex (i.e. features at least a sequence) parenthesis -should be added around it. For example: -\begin{coq_example} -Ltac twoIntro := let x:=intro in (x;x). -\end{coq_example} - -\Question{What is the syntax for pattern matching in {\Ltac}?} - -Pattern matching on a term $expr$ (non-linear first order unification) -with patterns $p_i$ and tactic expressions $e_i$ reads: -\begin{center} -\hspace{10ex} -{\tt match $expr$ with -\hspace*{2ex}$p_1$ => $e_1$ -\hspace*{1ex}\textbar$p_2$ => $e_2$ -\hspace*{1ex}\ldots -\hspace*{1ex}\textbar$p_n$ => $e_n$ -\hspace*{1ex}\textbar\ \textunderscore\ => $e_{n+1}$ -end. -} -\end{center} -Underscore matches all terms. - -\Question{What is the semantics for ``match goal''?} - -The semantics of {\tt match goal} depends on whether it returns -tactics or not. The {\tt match goal} expression matches the current -goal against a series of patterns: {$hyp_1 {\ldots} hyp_n$ \textbar- -$ccl$}. It uses a first-order unification algorithm and in case of -success, if the right-hand-side is an expression, it tries to type it -while if the right-hand-side is a tactic, it tries to apply it. If the -typing or the tactic application fails, the {\tt match goal} tries all -the possible combinations of $hyp_i$ before dropping the branch and -moving to the next one. Underscore matches all terms. - -\Question{Why can't I use a ``match goal'' returning a tactic in a non -tail-recursive position?} - -This is precisely because the semantics of {\tt match goal} is to -apply the tactic on the right as soon as a pattern unifies what is -meaningful only in tail-recursive uses. - -The semantics in non tail-recursive call could have been the one used -for terms (i.e. fail if the tactic expression is not typable, but -don't try to apply it). For uniformity of semantics though, this has -been rejected. - -\Question{How can I generate a new name?} - -You can use the following syntax: -{\tt let id:=fresh in \ldots}\\ -For example: -\begin{coq_example} -Ltac introIdGen := let id:=fresh in intro id. -\end{coq_example} - - -\iffalse -\Question{How can I access the type of a term?} - -You can use typeof. -todo -\fi - -\Question{How can I define static and dynamic code?} - -\section{Tactics written in Ocaml} - -\Question{Can you show me an example of a tactic written in OCaml?} - -You have some examples of tactics written in Ocaml in the ``contrib'' directory of {\Coq} sources. - - - - -\section{Case studies} - - -\Question{How can I define vectors or lists of size n?} - -\Question{How to prove that 2 sets are different?} - - You need to find a property true on one set and false on the -other one. As an example we show how to prove that {\tt bool} and {\tt -nat} are discriminable. As discrimination property we take the -property to have no more than 2 elements. - -\begin{coq_example*} -Theorem nat_bool_discr : bool <> nat. -Proof. - pose (discr := - fun X:Set => - ~ (forall a b:X, ~ (forall x:X, x <> a -> x <> b -> False))). - intro Heq; assert (H: discr bool). - intro H; apply (H true false); destruct x; auto. - rewrite Heq in H; apply H; clear H. - destruct a; destruct b as [|n]; intro H0; eauto. - destruct n; [ apply (H0 2); discriminate | eauto ]. -Qed. -\end{coq_example*} - -\Question{Is there an axiom-free proof of Streicher's axiom $K$ for -the equality on {\tt nat}?} -\label{K-nat} - -Yes, because equality is decidable on {\tt nat}. Here is the proof. - -\begin{coq_example*} -Require Import Eqdep_dec. -Require Import Peano_dec. -Theorem K_nat : - forall (x:nat) (P:x = x -> Prop), P (refl_equal x) -> forall p:x = x, P p. -Proof. -intros; apply K_dec_set with (p := p). -apply eq_nat_dec. -assumption. -Qed. -\end{coq_example*} - -Similarly, we have - -\begin{coq_example*} -Theorem eq_rect_eq_nat : - forall (p:nat) (Q:nat->Type) (x:Q p) (h:p=p), x = eq_rect p Q x p h. -Proof. -intros; apply K_nat with (p := h); reflexivity. -Qed. -\end{coq_example*} - -\Question{How to prove that two proofs of {\tt n<=m} on {\tt nat} are equal?} -\label{le-uniqueness} - -This is provable without requiring any axiom because axiom $K$ -directly holds on {\tt nat}. Here is a proof using question \ref{K-nat}. - -\begin{coq_example*} -Require Import Arith. -Scheme le_ind' := Induction for le Sort Prop. -Theorem le_uniqueness_proof : forall (n m : nat) (p q : n <= m), p = q. -Proof. -induction p using le_ind'; intro q. - replace (le_n n) with - (eq_rect _ (fun n0 => n <= n0) (le_n n) _ (refl_equal n)). - 2:reflexivity. - generalize (refl_equal n). - pattern n at 2 4 6 10, q; case q; [intro | intros m l e]. - rewrite <- eq_rect_eq_nat; trivial. - contradiction (le_Sn_n m); rewrite <- e; assumption. - replace (le_S n m p) with - (eq_rect _ (fun n0 => n <= n0) (le_S n m p) _ (refl_equal (S m))). - 2:reflexivity. - generalize (refl_equal (S m)). - pattern (S m) at 1 3 4 6, q; case q; [intro Heq | intros m0 l HeqS]. - contradiction (le_Sn_n m); rewrite Heq; assumption. - injection HeqS; intro Heq; generalize l HeqS. - rewrite <- Heq; intros; rewrite <- eq_rect_eq_nat. - rewrite (IHp l0); reflexivity. -Qed. -\end{coq_example*} - -\Question{How to exploit equalities on sets} - -To extract information from an equality on sets, you need to -find a predicate of sets satisfied by the elements of the sets. As an -example, let's consider the following theorem. - -\begin{coq_example*} -Theorem interval_discr : - forall m n:nat, - {x : nat | x <= m} = {x : nat | x <= n} -> m = n. -\end{coq_example*} - -We have a proof requiring the axiom of proof-irrelevance. We -conjecture that proof-irrelevance can be circumvented by introducing a -primitive definition of discrimination of the proofs of -\verb!{x : nat | x <= m}!. - -\begin{latexonly}% -The proof can be found in file {\tt interval$\_$discr.v} in this directory. -%Here is the proof -%\begin{small} -%\begin{flushleft} -%\begin{texttt} -%\def_{\ifmmode\sb\else\subscr\fi} -%\include{interval_discr.v} -%%% WARNING semantics of \_ has changed ! -%\end{texttt} -%$a\_b\_c$ -%\end{flushleft} -%\end{small} -\end{latexonly}% -\begin{htmlonly}% -\ahref{./interval_discr.v}{Here} is the proof. -\end{htmlonly} - -\Question{I have a problem of dependent elimination on -proofs, how to solve it?} - -\begin{coq_eval} -Reset Initial. -\end{coq_eval} - -\begin{coq_example*} -Inductive Def1 : Set := c1 : Def1. -Inductive DefProp : Def1 -> Prop := - c2 : forall d:Def1, DefProp d. -Inductive Comb : Set := - c3 : forall d:Def1, DefProp d -> Comb. -Lemma eq_comb : - forall (d1 d1':Def1) (d2:DefProp d1) (d2':DefProp d1'), - d1 = d1' -> c3 d1 d2 = c3 d1' d2'. -\end{coq_example*} - - You need to derive the dependent elimination -scheme for DefProp by hand using {\coqtt Scheme}. - -\begin{coq_eval} -Abort. -\end{coq_eval} - -\begin{coq_example*} -Scheme DefProp_elim := Induction for DefProp Sort Prop. -Lemma eq_comb : - forall d1 d1':Def1, - d1 = d1' -> - forall (d2:DefProp d1) (d2':DefProp d1'), c3 d1 d2 = c3 d1' d2'. -intros. -destruct H. -destruct d2 using DefProp_elim. -destruct d2' using DefProp_elim. -reflexivity. -Qed. -\end{coq_example*} - - -\Question{And what if I want to prove the following?} - -\begin{coq_example*} -Inductive natProp : nat -> Prop := - | p0 : natProp 0 - | pS : forall n:nat, natProp n -> natProp (S n). -Inductive package : Set := - pack : forall n:nat, natProp n -> package. -Lemma eq_pack : - forall n n':nat, - n = n' -> - forall (np:natProp n) (np':natProp n'), pack n np = pack n' np'. -\end{coq_example*} - - - -\begin{coq_eval} -Abort. -\end{coq_eval} -\begin{coq_example*} -Scheme natProp_elim := Induction for natProp Sort Prop. -Definition pack_S : package -> package. -destruct 1. -apply (pack (S n)). -apply pS; assumption. -Defined. -Lemma eq_pack : - forall n n':nat, - n = n' -> - forall (np:natProp n) (np':natProp n'), pack n np = pack n' np'. -intros n n' Heq np np'. -generalize dependent n'. -induction np using natProp_elim. -induction np' using natProp_elim; intros; auto. - discriminate Heq. -induction np' using natProp_elim; intros; auto. - discriminate Heq. -change (pack_S (pack n np) = pack_S (pack n0 np')). -apply (f_equal (A:=package)). -apply IHnp. -auto. -Qed. -\end{coq_example*} - - - - - - - -\section{Publishing tools} - -\Question{How can I generate some latex from my development?} - -You can use {\tt coqdoc}. - -\Question{How can I generate some HTML from my development?} - -You can use {\tt coqdoc}. - -\Question{How can I generate some dependency graph from my development?} - -\Question{How can I cite some {\Coq} in my latex document?} - -You can use {\tt coq\_tex}. - -\Question{How can I cite the {\Coq} reference manual?} - -You can use this bibtex entry: -\begin{verbatim} -@Manual{Coq:manual, - title = {The Coq proof assistant reference manual}, - author = {\mbox{The Coq development team}}, - organization = {LogiCal Project}, - note = {Version 8.0}, - year = {2004}, - url = "http://coq.inria.fr" -} -\end{verbatim} - -\Question{Where can I publish my developments in {\Coq}?} - -You can submit your developments as a user contribution to the {\Coq} -development team. This ensures its liveness along the evolution and -possible changes of {\Coq}. - -You can also submit your developments to the HELM/MoWGLI repository at -the University of Bologna (see -\ahref{http://mowgli.cs.unibo.it}{\url{http://mowgli.cs.unibo.it}}). For -developments submitted in this database, it is possible to visualize -the developments in natural language and execute various retrieving -requests. - -\Question{How can I read my proof in natural language?} - -You can submit your proof to the HELM/MoWGLI repository and use the -rendering tool provided by the server (see -\ahref{http://mowgli.cs.unibo.it}{\url{http://mowgli.cs.unibo.it}}). - -\section{\CoqIde} - -\Question{What is {\CoqIde}?} - -{\CoqIde} is a gtk based GUI for \Coq. - -\Question{How to enable Emacs keybindings?} - Insert \texttt{gtk-key-theme-name = "Emacs"} - in your \texttt{.coqide-gtk2rc} file. It may be in the current dir - or in \verb#$HOME# dir. This is done by default. - -%$ juste pour que la coloration emacs marche - -\Question{How to enable antialiased fonts?} - - Set the \verb#GDK_USE_XFT# variable to \verb#1#. This is by default with \verb#Gtk >= 2.2#. - If some of your fonts are not available, set \verb#GDK_USE_XFT# to \verb#0#. - -\Question{How to use those Forall and Exists pretty symbols?}\label{forallcoqide} - Thanks to the notation features in \Coq, you just need to insert these -lines in your {\Coq} buffer:\\ -\begin{texttt} -Notation "$\forall$ x : t, P" := (forall x:t, P) (at level 200, x ident). -\end{texttt}\\ -\begin{texttt} -Notation "$\exists$ x : t, P" := (exists x:t, P) (at level 200, x ident). -\end{texttt} - -Copy/Paste of these lines from this file will not work outside of \CoqIde. -You need to load a file containing these lines or to enter the $\forall$ -using an input method (see \ref{inputmeth}). To try it just use \verb#Require Import utf8# from inside -\CoqIde. -To enable these notations automatically start coqide with -\begin{verbatim} - coqide -l utf8 -\end{verbatim} -In the ide subdir of {\Coq} library, you will find a sample utf8.v with some -pretty simple notations. - -\Question{How to define an input method for non ASCII symbols?}\label{inputmeth} - -\begin{itemize} -\item First solution: type \verb#<CONTROL><SHIFT>2200# to enter a forall in the script widow. - 2200 is the hexadecimal code for forall in unicode charts and is encoded as - in UTF-8. - 2203 is for exists. See \ahref{http://www.unicode.org}{\url{http://www.unicode.org}} for more codes. -\item Second solution: rebind \verb#<AltGr>a# to forall and \verb#<AltGr>e# to exists. - Under X11, you need to use something like -\begin{verbatim} - xmodmap -e "keycode 24 = a A F13 F13" - xmodmap -e "keycode 26 = e E F14 F14" -\end{verbatim} - and then to add -\begin{verbatim} - bind "F13" {"insert-at-cursor" ("")} - bind "F14" {"insert-at-cursor" ("")} -\end{verbatim} - to your "binding "text"" section in \verb#.coqiderc-gtk2rc.# - The strange ("") argument is the UTF-8 encoding for - 0x2200. - You can compute these encodings using the lablgtk2 toplevel with -\begin{verbatim} -Glib.Utf8.from_unichar 0x2200;; -\end{verbatim} - Further symbols can be bound on higher Fxx keys or on even on other keys you - do not need . -\end{itemize} - -\Question{How to build a custom {\CoqIde} with user ml code?} - Use - coqmktop -ide -byte m1.cmo...mi.cmo - or - coqmktop -ide -opt m1.cmx...mi.cmx - -\Question{How to customize the shortcuts for menus?} - Two solutions are offered: -\begin{itemize} -\item Edit \$HOME/.coqide.keys by hand or -\item Add "gtk-can-change-accels = 1" in your .coqide-gtk2rc file. Then - from \CoqIde, you may select a menu entry and press the desired - shortcut. -\end{itemize} - -\Question{What encoding should I use? What is this $\backslash$x\{iiii\} in my file?} - The encoding option is related to the way files are saved. - Keep it as UTF-8 until it becomes important for you to exchange files - with non UTF-8 aware applications. - If you choose something else than UTF-8, then missing characters will - be encoded by $\backslash$x\{....\} or $\backslash$x\{........\} - where each dot is an hex. digit. - The number between braces is the hexadecimal UNICODE index for the - missing character. - - - - -\section{Extraction} - -\Question{What is program extraction?} - -Program extraction consist in generating a program from a constructive proof. - -\Question{Which language can I extract to?} - -You can extract your programs to Objective Caml and Haskell. - -\Question{How can I extract an incomplete proof?} - -You can provide programs for your axioms. - - - -%%%%%%% -\section{Glossary} - -\Question{Can you explain me what an evaluable constant is?} - -An evaluable constant is a constant which is unfoldable. - -\Question{What is a goal?} - -The goal is the statement to be proved. - -\Question{What is a meta variable?} - -A meta variable in {\Coq} represents a ``hole'', i.e. a part of a proof -that is still unknown. - -\Question{What is Gallina?} - -Gallina is the specification language of \Coq. Complete documentation -of this language can be found in the Reference Manual. - -\Question{What is The Vernacular?} - -It is the language of commands of Gallina i.e. definitions, lemmas, {\ldots} - - -\Question{What is a dependent type?} - -A dependant type is a type which depends on some term. For instance -``vector of size n'' is a dependant type representing all the vectors -of size $n$. Its type depends on $n$ - -\Question{What is a proof by reflection?} - -This is a proof generated by some computation which is done using the -internal reduction of {\Coq} (not using the tactic language of {\Coq} -(\Ltac) nor the implementation language for \Coq). An example of -tactic using the reflection mechanism is the {\ring} tactic. The -reflection method consist in reflecting a subset of {\Coq} language (for -example the arithmetical expressions) into an object of the \Coq -language itself (in this case an inductive type denoting arithmetical -expressions). For more information see~\cite{howe,harrison,boutin} -and the last chapter of the Coq'Art. - -\Question{What is intuitionistic logic?} - -This is any logic which does not assume that ``A or not A''. - - -\Question{What is proof-irrelevance?} - -See question \ref{proof-irrelevance} - - -\Question{What is the difference between opaque and transparent?}{\label{opaque}} - -Opaque definitions can not be unfolded but transparent ones can. - - -\section{Troubleshooting} - -\Question{What can I do when {\tt Qed.} is slow?} - -Sometime you can use the {\abstracttac} tactic, which makes as if you had -stated some local lemma, this speeds up the typing process. - -\Question{Why \texttt{Reset Initial.} does not work when using \texttt{coqc}?} - -The initial state corresponds to the state of coqtop when the interactive -session began. It does not make sense in files to compile. - - -\Question{What can I do if I get ``No more subgoals but non-instantiated existential variables''?} - -This means that {\eauto} or {\eapply} didn't instantiate an -existential variable which eventually got erased by some computation. -You have to backtrack to the faulty occurrence of {\eauto} or -{\eapply} and give the missing argument an explicit value. - -\Question{What can I do if I get ``Cannot solve a second-order unification problem''?} - -You can help {\Coq} using the {\pattern} tactic. - -\Question{Why does {\Coq} tell me that \texttt{\{x:A|(P x)\}} is not convertible with \texttt{(sig A P)}?} - - This is because \texttt{\{x:A|P x\}} is a notation for -\texttt{sig (fun x:A => P x)}. Since {\Coq} does not reason up to -$\eta$-conversion, this is different from \texttt{sig P}. - - -\Question{I copy-paste a term and {\Coq} says it is not convertible - to the original term. Sometimes it even says the copied term is not -well-typed.} - - This is probably due to invisible implicit information (implicit -arguments, coercions and Cases annotations) in the printed term, which -is not re-synthesised from the copied-pasted term in the same way as -it is in the original term. - - Consider for instance {\tt (@eq Type True True)}. This term is -printed as {\tt True=True} and re-parsed as {\tt (@eq Prop True -True)}. The two terms are not convertible (hence they fool tactics -like {\tt pattern}). - - There is currently no satisfactory answer to the problem. However, -the command {\tt Set Printing All} is useful for diagnosing the -problem. - - Due to coercions, one may even face type-checking errors. In some -rare cases, the criterion to hide coercions is a bit too loose, which -may result in a typing error message if the parser is not able to find -again the missing coercion. - - - -\section{Conclusion and Farewell.} -\label{ccl} - -\Question{What if my question isn't answered here?} -\label{lastquestion} - -Don't panic \verb+:-)+. You can try the {\Coq} manual~\cite{Coq:manual} for a technical -description of the prover. The Coq'Art~\cite{Coq:coqart} is the first -book written on {\Coq} and provides a comprehensive review of the -theorem prover as well as a number of example and exercises. Finally, -the tutorial~\cite{Coq:Tutorial} provides a smooth introduction to -theorem proving in \Coq. - - -%%%%%%% -\newpage -\nocite{LaTeX:intro} -\nocite{LaTeX:symb} -\bibliography{fk} - -%%%%%%% -\typeout{*********************************************} -\typeout{********* That makes \thequestion{\space} questions **********} -\typeout{*********************************************} - -\end{document} |