coq

问题 Assume I have two hypotheses in the context, a_b : A -> B and a : A . I should be able to apply a_b to a to gain a further hypothesis, b : B . That is, given the following state: 1 subgoal A : Prop B : Prop C : Prop a_b : A -> B a : A ______________________________________(1/1) C There should be some tactic, foo (a_b a) , to transform this into the following state: 1 subgoal A : Prop B : Prop C : Prop a_b : A -> B a : A b : B ______________________________________(1/1) C But I don't know what

问题 How can I in coq, prove that a function f that accepts a bool true|false and returns a bool true|false (shown below), when applied twice to a single bool true|false would always return that same value true|false : (f:bool -> bool) For example the function f can only do 4 things, lets call the input of the function b : Always return true Always return false Return b (i.e. returns true if b is true vice versa) Return not b (i.e. returns false if b is true and vice vera) So if the function

问题 I've two Fibonacci implementations, seen below, that I want to prove are functionally equivalent. I've already proved properties about natural numbers, but this exercise requires another approach that I cannot figure out. The textbook I'm using have introduced the following syntax of Coq, so it should be possible to prove equality using this notation: <definition> ::= <keyword> <identifier> : <statement> <proof> <keyword> ::= Proposition | Lemma | Theorem | Corollary <statement> ::= {

问题 I’d like to define the following function using Program Fixpoint or Function in Coq: Require Import Coq.Lists.List. Import ListNotations. Require Import Coq.Program.Wf. Require Import Recdef. Inductive Tree := Node : nat -> list Tree -> Tree. Fixpoint height (t : Tree) : nat := match t with | Node x ts => S (fold_right Nat.max 0 (map height ts)) end. Program Fixpoint mapTree (f : nat -> nat) (t : Tree) {measure (height t)} : Tree := match t with Node x ts => Node (f x) (map (fun t => mapTree

问题 How to call proof assistant Coq from external software? Does Coq have some API? Is Coq command line interface rich enough to pass arguments in file and receive response in file? I am interested in Java or C++ bridges. This is legitimate question. Coq is not the usual business software from which one can expect the developer friendly API. I had similary question about Isabelle/HOL and it was legitimate question with non-trivial answer. 回答1: As of today, there are three ways to interact with

问题 I want to prove or falsify forall (P Q : Prop), (P -> Q) -> (Q -> P) -> P = Q. in Coq. Here is my approach. Inductive True2 : Prop := | One : True2 | Two : True2. Lemma True_has_one : forall (t0 t1 : True), t0 = t1. Proof. intros. destruct t0. destruct t1. reflexivity. Qed. Lemma not_True2_has_one : (forall (t0 t1 : True2), t0 = t1) -> False. Proof. intros. specialize (H One Two). inversion H. But, inversion H does nothing. I think maybe it's because the coq's proof independence (I'm not a

问题 I come from a JavaScript/Ruby programming background and am used to this being how true/false works (in JS): !true // false !false // true Then you can use those true/false values with && like var a = true, b = false; a && !b; So and and not (and other logical/boolean operators) are part of a single system; it seems like the "logical" system and the "boolean" system are one and the same. However, in Coq, logics and booleans are two separate things. Why is this? The quote/link below

问题 I was trying out the examples from the Coq documentation Software Foundations (http://www.cis.upenn.edu/~bcpierce/sf/current/Induction.html#lab40) when I noticed that to solve the example give in the link: Theorem andb_true_elim1 : ∀b c : bool, andb b c = true → b = true. Proof. intros b c H. destruct b. Case "b = true". (* <----- here *) reflexivity. Case "b = false". (* <---- and here *) rewrite ← H. reflexivity. Qed. we are destructing b which appears on the left of c for "andb". However

问题 For the inductive type nat , the generated induction principle uses the constructors O and S in its statement: Inductive nat : Set := O : nat | S : nat -> nat nat_ind : forall P : nat -> Prop, P 0 -> (forall n : nat, P n -> P (S n)) -> forall n : nat, P n But for le , the generated statement does not uses the constructors le_n and le_S : Inductive le (n : nat) : nat -> Prop := le_n : n <= n | le_S : forall m : nat, n <= m -> n <= S m le_ind : forall (n : nat) (P : nat -> Prop), P n -> (forall

问题 Is there a way, to apply an hypotesis to our goal in Coq ? For example: Hypothesis: 1 subgoal a : nat l1 : list nat l2 : list nat H : Prefix (a :: l1) l2 IHl1 : Prefix l1 l2 -> sum l1 <= sum l2 Goal ______________________________________(1/1) sum (a :: l1) <= sum l2 I know that if i could do : apply IHl1 , i could have a result like Prefix (a::l1) l2 and after i will be able to do an assumption ! But i can't do the apply because it's giving me this error : Error: Impossible to unify "sum l1 <