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Diffstat (limited to 'doc/faq/FAQ.tex')
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1 files changed, 154 insertions, 88 deletions
diff --git a/doc/faq/FAQ.tex b/doc/faq/FAQ.tex index b03aa64090..48b61827d1 100644 --- a/doc/faq/FAQ.tex +++ b/doc/faq/FAQ.tex @@ -228,7 +228,7 @@ 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 +(no object, no label ...), so that it is highly improbable 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. @@ -509,22 +509,14 @@ library. (Statements in boldface are the most ``interesting'' ones for Coq.) The justification of their validity relies on the interpretability in set theory. -% fig2dev -m 2 -L png axioms.fig axioms.png -% fig2dev -L pdftex axioms.fig axioms.pdf -% fig2dev -L pdftex_t -p axioms.pdf axioms.fig axioms.pdf_t -% fig2dev -L pstex axioms.fig axioms.eps -% fig2dev -L pstex_t -p axioms.eps axioms.fig axioms.eps_t - \begin{figure}[htbp] %HEVEA\imgsrc{axioms.png} %BEGIN LATEX \begin{center} \ifpdf % si on est en pdflatex \scalebox{0.65}{\input{axioms.pdf_t}} -%\includegraphics[width=1.0\textwidth]{axioms.png} \else \scalebox{0.65}{\input{axioms.eps_t}} -%\includegraphics[width=1.0\textwidth]{axioms.eps} \fi \end{center} %END LATEX @@ -718,7 +710,7 @@ 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 +\ref{le-uniqueness} an axiom-free proof of the uniqueness 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 @@ -849,26 +841,41 @@ mapped to {\Prop}. Use some theorem or assumption or use the {\split} tactic. \begin{coq_example} -Goal forall A B:Prop, A->B-> A/\B. +Goal forall A B:Prop, A -> B -> A/\B. intros. split. assumption. assumption. -Qed. \end{coq_example} +\begin{coq_example*} +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: +If you want to decompose a hypothesis into several hypotheses, you can +use the {\destruct} tactic: \begin{coq_example} -Goal forall A B:Prop, A/\B-> B. +Goal forall A B:Prop, A/\B -> B. intros. -decompose [and] H. +destruct H as [H1 H2]. assumption. -Qed. \end{coq_example} +\begin{coq_example*} +Qed. +\end{coq_example*} +You can also perform the destruction at the time of introduction: + +\begin{coq_example} +Goal forall A B:Prop, A/\B -> B. +intros A B [H1 H2]. +assumption. +\end{coq_example} +\begin{coq_example*} +Qed. +\end{coq_example*} \Question{My goal is a disjunction, how can I prove it?} @@ -878,26 +885,28 @@ reasoning step, use the {\tt classic} axiom to prove the right part with the ass that the left part of the disjunction is false. \begin{coq_example} -Goal forall A B:Prop, A-> A\/B. +Goal forall A B:Prop, A -> A\/B. intros. left. assumption. -Qed. \end{coq_example} +\begin{coq_example*} +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]) +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]) +Ltac classical_left := +match goal with +| _:_ |- _ \/ ?X1 => (elim (classic X1);intro;[right;trivial|left]) end. @@ -905,8 +914,10 @@ Goal forall A B:Prop, (~A -> B) -> A\/B. intros. classical_right. auto. -Qed. \end{coq_example} +\begin{coq_example*} +Qed. +\end{coq_example*} \Question{My goal is an universally quantified statement, how can I prove it?} @@ -935,8 +946,10 @@ Goal exists x:nat, forall y, x+y=y. exists 0. intros. auto. -Qed. \end{coq_example} +\begin{coq_example*} +Qed. +\end{coq_example*} \Question{My goal is solvable by some lemma, how can I prove it?} @@ -954,8 +967,10 @@ Qed. Goal 3+0 = 3. apply mylemma. -Qed. \end{coq_example} +\begin{coq_example*} +Qed. +\end{coq_example*} @@ -972,8 +987,10 @@ Just use the {\reflexivity} tactic. Goal forall x, 0+x = x. intros. reflexivity. -Qed. \end{coq_example} +\begin{coq_example*} +Qed. +\end{coq_example*} \Question{My goal is a {\tt let x := a in ...}, how can I prove it?} @@ -987,13 +1004,21 @@ 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. +As with conjunctive hypotheses, you can use the {\destruct} tactic or +the {\intros} tactic to decompose them into several hypotheses. -\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} +\begin{coq_example*} +Require Import Arith. +\end{coq_example*} +\begin{coq_example} +Goal forall x, (exists y, x * y = 1) -> x = 1. +intros x [y H]. +apply mult_is_one in H. +easy. +\end{coq_example} +\begin{coq_example*} +Qed. +\end{coq_example*} \Question{My goal is an equality, how can I swap the left and right hand terms?} @@ -1004,8 +1029,10 @@ Goal forall x y : nat, x=y -> y=x. intros. symmetry. assumption. -Qed. \end{coq_example} +\begin{coq_example*} +Qed. +\end{coq_example*} \Question{My hypothesis is an equality, how can I swap the left and right hand terms?} @@ -1016,8 +1043,10 @@ Goal forall x y : nat, x=y -> y=x. intros. symmetry in H. assumption. -Qed. \end{coq_example} +\begin{coq_example*} +Qed. +\end{coq_example*} \Question{My goal is an equality, how can I prove it by transitivity?} @@ -1029,8 +1058,10 @@ intros. transitivity y. assumption. assumption. -Qed. \end{coq_example} +\begin{coq_example*} +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?} @@ -1066,7 +1097,6 @@ 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?} @@ -1103,8 +1133,10 @@ Use the {\assumption} tactic. Goal 1=1 -> 1=1. intro. assumption. -Qed. \end{coq_example} +\begin{coq_example*} +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?} @@ -1114,8 +1146,10 @@ Use the {\exact} tactic. Goal 1=1 -> 1=1 -> 1=1. intros. exact H0. -Qed. \end{coq_example} +\begin{coq_example*} +Qed. +\end{coq_example*} \Question{What can be the difference between applying one hypothesis or another in the context of the last question?} @@ -1131,8 +1165,10 @@ Just use the {\tauto} tactic. Goal forall A B:Prop, A-> (A\/B) /\ A. intros. tauto. -Qed. \end{coq_example} +\begin{coq_example*} +Qed. +\end{coq_example*} \Question{My goal is a first order formula, how can I prove it?} @@ -1149,8 +1185,10 @@ Just use the {\congruence} tactic. Goal forall a b c d e, a=d -> b=e -> c+b=d -> c+e=a. intros. congruence. -Qed. \end{coq_example} +\begin{coq_example*} +Qed. +\end{coq_example*} \Question{My goal is a disequality solvable by a sequence of rewrites, how can I prove it?} @@ -1161,8 +1199,10 @@ Just use the {\congruence} tactic. Goal forall a b c d, a<>d -> b=a -> d=c+b -> b<>c+b. intros. congruence. -Qed. \end{coq_example} +\begin{coq_example*} +Qed. +\end{coq_example*} \Question{My goal is an equality on some ring (e.g. natural numbers), how can I prove it?} @@ -1173,11 +1213,13 @@ Just use the {\ring} tactic. Require Import ZArith. Require Ring. Local Open Scope Z_scope. -Goal forall a b : Z, (a+b)*(a+b) = a*a + 2*a*b + b*b. +Goal forall a b : Z, (a+b)*(a+b) = a*a + 2*a*b + b*b. intros. ring. -Qed. \end{coq_example} +\begin{coq_example*} +Qed. +\end{coq_example*} \Question{My goal is an equality on some field (e.g. real numbers), how can I prove it?} @@ -1187,30 +1229,30 @@ Just use the {\field} tactic. Require Import Reals. Require Ring. Local Open Scope R_scope. -Goal forall a b : R, b*a<>0 -> (a/b) * (b/a) = 1. +Goal forall a b : R, b*a<>0 -> (a/b) * (b/a) = 1. intros. field. -cut (b*a <>0 -> a<>0). -cut (b*a <>0 -> b<>0). -auto. -auto with real. -auto with real. -Qed. +split ; auto with real. \end{coq_example} +\begin{coq_example*} +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?} +\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. Local Open Scope Z_scope. -Goal forall a : Z, a>0 -> a+a > a. +Goal forall a : Z, a>0 -> a+a > a. intros. omega. -Qed. \end{coq_example} +\begin{coq_example*} +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?} @@ -1233,16 +1275,22 @@ assert (A->C). intro;apply H0;apply H;assumption. apply H2. assumption. +\end{coq_example} +\begin{coq_example*} Qed. +\end{coq_example*} +\begin{coq_example} 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} +\begin{coq_example*} +Qed. +\end{coq_example*} @@ -1333,8 +1381,10 @@ H1 *) intros A B C H H0 H1. repeat split;assumption. -Qed. \end{coq_example} +\begin{coq_example*} +Qed. +\end{coq_example*} \Question{I want to automatize the use of some tactic, how can I do it?} @@ -1347,8 +1397,10 @@ Goal forall A B C : Prop, A -> B/\C -> A/\B/\C. Proof with assumption. intros. split... -Qed. \end{coq_example} +\begin{coq_example*} +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?} @@ -1360,8 +1412,10 @@ Local Open Scope Z_scope. Goal forall a b c : Z, a+b=b+a. Proof with try solve [ring]. intros... -Qed. \end{coq_example} +\begin{coq_example*} +Qed. +\end{coq_example*} \Question{How can I do the opposite of the {\intro} tactic?} @@ -1373,8 +1427,10 @@ intros. generalize H. intro. auto. -Qed. \end{coq_example} +\begin{coq_example*} +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?} @@ -1441,7 +1497,7 @@ 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}. +You can state explicitly what this implicit argument is. See \ref{implicit}. \Question{How can I explicit some implicit argument ?}\label{implicit} @@ -1512,8 +1568,10 @@ You can use the {\discriminate} tactic. Inductive toto : Set := | C1 : toto | C2 : toto. Goal C1 <> C2. discriminate. -Qed. \end{coq_example} +\begin{coq_example*} +Qed. +\end{coq_example*} \Question{During an inductive proof, how to get rid of impossible cases of an inductive definition?} @@ -1561,20 +1619,20 @@ If you type for instance the following ``definition'': Reset Initial. \end{coq_eval} \begin{coq_example} -Definition max (n p : nat) := if n <= p then p else n. +Fail 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, +proven theorems. 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 $p < n$. +$n$ and $p$, determines whether $n\leq p$ 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. @@ -1648,7 +1706,7 @@ immediate, whereas one can't wait for the result of 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 $n\leq p$ +function}, i.e. a complex object, able not only to tell whether $n\leq p$ 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. @@ -1729,7 +1787,7 @@ mergesort} as an example). the arguments of the loop. \begin{coq_eval} -Open Scope R_scope. +Reset Initial. Require Import List. \end{coq_eval} @@ -1742,21 +1800,25 @@ Definition R (a b:list nat) := length a < length b. \begin{coq_example*} Lemma Rwf : well_founded R. \end{coq_example*} +\begin{coq_eval} +Admitted. +\end{coq_eval} \item Define the step function (which needs proofs that recursive calls are on smaller arguments). -\begin{verbatim} -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 := (* ... *) . +\begin{coq_example*} +Definition split (l : list nat) + : {l1: list nat | R l1 l} * {l2 : list nat | R l2 l}. +Admitted. +Definition concat (l1 l2 : list nat) : list nat. +Admitted. 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{verbatim} +\end{coq_example*} \item Define the recursive function by fixpoint on the step function. @@ -1811,9 +1873,9 @@ induction 1. inversion 1. inversion 1. apply IHeven; trivial. \end{coq_example} -\begin{coq_eval} +\begin{coq_example*} Qed. -\end{coq_eval} +\end{coq_example*} In case the type of the second induction hypothesis is not dependent, {\tt inversion} can just be replaced by {\tt destruct}. @@ -1848,10 +1910,10 @@ 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) -> +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 *) @@ -1859,9 +1921,9 @@ induction n; [assumption | apply HmS; apply IHn]. (* case Sm *) intro n; apply HSn; apply IHm. \end{coq_example} -\begin{coq_eval} +\begin{coq_example*} Qed. -\end{coq_eval} +\end{coq_example*} \Question{How to define a function by nested recursion?} @@ -1896,7 +1958,7 @@ Set Implicit Arguments. 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 : +Lemma Stream_unfold : forall n:nat, nats n = Cons n (nats (S n)). Proof. intro; @@ -1904,8 +1966,10 @@ Proof. | Cons x s => Cons x s end). case (nats n); reflexivity. -Qed. \end{coq_example} +\begin{coq_example*} +Qed. +\end{coq_example*} @@ -2340,8 +2404,8 @@ You can use {\tt coqdoc}. \Question{How can I generate some dependency graph from my development?} -You can use the tool \verb|coqgraph| developped by Philippe Audebaud in 2002. -This tool transforms dependancies generated by \verb|coqdep| into 'dot' files which can be visualized using the Graphviz software (http://www.graphviz.org/). +You can use the tool \verb|coqgraph| developed by Philippe Audebaud in 2002. +This tool transforms dependencies generated by \verb|coqdep| into 'dot' files which can be visualized using the Graphviz software (http://www.graphviz.org/). \Question{How can I cite some {\Coq} in my latex document?} @@ -2475,7 +2539,7 @@ modifiers keys available through GTK. The straightest way to get rid of the problem is to edit by hand your coqiderc (either \verb|/home/<user>/.config/coq/coqiderc| under Linux, or \\ \verb|C:\Documents and Settings\<user>\.config\coq\coqiderc| under Windows) - and replace any occurence of \texttt{MOD4} by \texttt{MOD1}. +and replace any occurrence of \texttt{MOD4} by \texttt{MOD1}. @@ -2574,7 +2638,7 @@ existential variable which eventually got erased by some computation. You may backtrack to the faulty occurrence of {\eauto} or {\eapply} and give the missing argument an explicit value. Alternatively, you can use the commands \texttt{Show Existentials.} and -\texttt{Existential.} to display and instantiate the remainig +\texttt{Existential.} to display and instantiate the remaining existential variables. @@ -2586,8 +2650,10 @@ eapply eq_trans. Show Existentials. eassumption. assumption. -Qed. \end{coq_example} +\begin{coq_example*} +Qed. +\end{coq_example*} \Question{What can I do if I get ``Cannot solve a second-order unification problem''?} |