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diff --git a/doc/faq/FAQ.tex b/doc/faq/FAQ.tex new file mode 100644 index 00000000..2b5d898f --- /dev/null +++ b/doc/faq/FAQ.tex @@ -0,0 +1,2481 @@ +\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} |