SLC2018: Datos estructurados en Coq
En la sesión de hoy del Seminario de Lógica Computacional Jorge Catarecha Otero-Saavedra ha explicado cómo definir datos estructurados (pares, listas, multiconjuntos y diccionarios) en Coq y cómo demostrar sus propiedades.
La teoría, junto con los ejercicios, utilizados en la exposición son
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(* T3: Datos estructurados en Coq *) Require Export T2_Induccion. (* La teoría T2_Induccion se encuentra en http://bit.ly/2pDlxlF *) (* --------------------------------------------------------------------- Nota. Iniciar el módulo NatList. ------------------------------------------------------------------ *) Module NatList. (* ===================================================================== § Pares de números ================================================================== *) (* --------------------------------------------------------------------- Ejemplo. El tipo de los números naturales es natprod y su constructor es pair. ------------------------------------------------------------------ *) Inductive natprod : Type := pair : nat -> nat -> natprod. Check (pair 3 5). (* --------------------------------------------------------------------- Ejemplo. Definir la función fst : natprod -> nat tal que (fst p) es la primera componente de p. ------------------------------------------------------------------ *) Definition fst (p : natprod) : nat := match p with | pair x y => x end. (* --------------------------------------------------------------------- Ejemplo. Evaluar la expresión fst (pair 3 5) ------------------------------------------------------------------ *) Eval compute in (fst (pair 3 5)). (* ===> 3 *) (* --------------------------------------------------------------------- Ejemplo. Definir la función snd : natprod -> nat tal que (snd p) es la segunda componente de p. ------------------------------------------------------------------ *) Definition snd (p : natprod) : nat := match p with | pair x y => y end. (* --------------------------------------------------------------------- Ejemplo. Definir la notación (x,y) como una abreviaura de (pair x y). ------------------------------------------------------------------ *) Notation "( x , y )" := (pair x y). (* --------------------------------------------------------------------- Ejemplo. Evaluar la expresión fst (3,5) ------------------------------------------------------------------ *) Eval compute in (fst (3,5)). (* ===> 3 *) (* --------------------------------------------------------------------- Ejemplo. Redefinir la función fst usando la abreviatura de pares. ------------------------------------------------------------------ *) Definition fst' (p : natprod) : nat := match p with | (x,y) => x end. (* --------------------------------------------------------------------- Ejemplo. Redefinir la función snd usando la abreviatura de pares. ------------------------------------------------------------------ *) Definition snd' (p : natprod) : nat := match p with | (x,y) => y end. (* --------------------------------------------------------------------- Ejemplo. Definir la función swap_pair : natprod -> natprod tal que (swap_pair p) es el par obtenido intercambiando las componentes de p. ------------------------------------------------------------------ *) Definition swap_pair (p : natprod) : natprod := match p with | (x,y) => (y,x) end. (* --------------------------------------------------------------------- Ejemplo. Demostrar que para todos los naturales (n,m) = (fst (n,m), snd (n,m)). ------------------------------------------------------------------ *) Theorem surjective_pairing' : forall (n m : nat), (n,m) = (fst (n,m), snd (n,m)). Proof. reflexivity. Qed. (* --------------------------------------------------------------------- Ejemplo. Demostrar que para todo par de naturales p = (fst p, snd p). ------------------------------------------------------------------ *) Theorem surjective_pairing_stuck : forall (p : natprod), p = (fst p, snd p). Proof. simpl. (* No reduce nada. *) Abort. Theorem surjective_pairing : forall (p : natprod), p = (fst p, snd p). Proof. intros p. destruct p as [n m]. simpl. reflexivity. Qed. (* --------------------------------------------------------------------- Ejercicio 1. Demostrar que para todo par de naturales p, (snd p, fst p) = swap_pair p. ------------------------------------------------------------------ *) Theorem snd_fst_is_swap : forall (p : natprod), (snd p, fst p) = swap_pair p. Proof. intro p. destruct p as [n m]. simpl. reflexivity. Qed. (* --------------------------------------------------------------------- Ejercicio 2. Demostrar que para todo par de naturales p, fst (swap_pair p) = snd p. ------------------------------------------------------------------ *) Theorem fst_swap_is_snd : forall (p : natprod), fst (swap_pair p) = snd p. Proof. intro p. destruct p as [n m]. simpl. reflexivity. Qed. (* ===================================================================== § Listas de números ================================================================== *) (* --------------------------------------------------------------------- Ejemplo. natlist es la lista de los números naturales y sus constructores son + nil (la lista vacía) y + cons (tal que (cons x ys) es la lista obtenida añadiéndole x a ys. ------------------------------------------------------------------ *) Inductive natlist : Type := | nil : natlist | cons : nat -> natlist -> natlist. (* --------------------------------------------------------------------- Ejemplo. Definir la constante mylist : natlist que es la lista cuyos elementos son 1, 2 y 3. ------------------------------------------------------------------ *) Definition mylist := cons 1 (cons 2 (cons 3 nil)). (* --------------------------------------------------------------------- Ejemplo. Definir la notación (x :: ys) como una abreviatura de (cons x ys). ------------------------------------------------------------------ *) Notation "x :: l" := (cons x l) (at level 60, right associativity). (* --------------------------------------------------------------------- Ejemplo. Definir la notación de las listas finitas escribiendo sus elementos entre corchetes y separados por puntos y comas. ------------------------------------------------------------------ *) Notation "[ ]" := nil. Notation "[ x ; .. ; y ]" := (cons x .. (cons y nil) ..). (* --------------------------------------------------------------------- Ejemplo. Distintas representaciones de mylist. ------------------------------------------------------------------ *) Definition mylist1 := 1 :: (2 :: (3 :: nil)). Definition mylist2 := 1 :: 2 :: 3 :: nil. Definition mylist3 := [1;2;3]. (* ===================================================================== §§ Repeat ================================================================== *) (* --------------------------------------------------------------------- Ejemplo. Definir la función repeat : nat -> nat -> natlist tal que (repeat n k) es la lista formada por k veces el número n. ------------------------------------------------------------------ *) Fixpoint repeat (n count : nat) : natlist := match count with | O => nil | S count' => n :: (repeat n count') end. (* ===================================================================== §§ Length ================================================================== *) (* --------------------------------------------------------------------- Ejemplo. Definir la función length : natlist -> nat tal que (length xs) es el número de elementos de xs. ------------------------------------------------------------------ *) Fixpoint length (l:natlist) : nat := match l with | nil => O | h :: t => S (length t) end. (* ===================================================================== §§ Append ================================================================== *) (* --------------------------------------------------------------------- Ejemplo. Definir la función append : natlist -> natlist -> natlist tal que (append xs ys) es la concatenación de xs e ys. ------------------------------------------------------------------ *) Fixpoint app (l1 l2 : natlist) : natlist := match l1 with | nil => l2 | h :: t => h :: (app t l2) end. (* --------------------------------------------------------------------- Ejemplo. Definir la notación (xs ++ ys) como una abreviaura de (append xs ys). ------------------------------------------------------------------ *) Notation "x ++ y" := (app x y) (right associativity, at level 60). (* --------------------------------------------------------------------- Ejemplo. Demostrar que [1;2;3] ++ [4;5] = [1;2;3;4;5]. nil ++ [4;5] = [4;5]. [1;2;3] ++ nil = [1;2;3]. ------------------------------------------------------------------ *) Example test_app1: [1;2;3] ++ [4;5] = [1;2;3;4;5]. Proof. reflexivity. Qed. Example test_app2: nil ++ [4;5] = [4;5]. Proof. reflexivity. Qed. Example test_app3: [1;2;3] ++ nil = [1;2;3]. Proof. reflexivity. Qed. (* ===================================================================== §§ Head y tail ================================================================== *) (* --------------------------------------------------------------------- Ejemplo. Definir la función hd : nat -> natlist -> natlist tal que (hd d xs) es el primer elemento de xs o d, si xs es la lista vacía. ------------------------------------------------------------------ *) Definition hd (default:nat) (l:natlist) : nat := match l with | nil => default | h :: t => h end. (* --------------------------------------------------------------------- Ejemplo. Definir la función tl : natlist -> natlist tal que (tl xs) es el resto de xs. ------------------------------------------------------------------ *) Definition tl (l:natlist) : natlist := match l with | nil => nil | h :: t => t end. (* --------------------------------------------------------------------- Ejemplo. Demostrar que hd 0 [1;2;3] = 1. hd 0 [] = 0. tl [1;2;3] = [2;3]. ------------------------------------------------------------------ *) Example test_hd1: hd 0 [1;2;3] = 1. Proof. reflexivity. Qed. Example test_hd2: hd 0 [] = 0. Proof. reflexivity. Qed. Example test_tl: tl [1;2;3] = [2;3]. Proof. reflexivity. Qed. (* --------------------------------------------------------------------- Ejercicio 3. Definir la función nonzeros : natlist -> natlist tal que (nonzeros xs) es la lista de los elementos de xs distintos de cero. Por ejemplo, nonzeros [0;1;0;2;3;0;0] = [1;2;3]. ------------------------------------------------------------------ *) Fixpoint nonzeros (l:natlist) : natlist := match l with | nil => nil | a::bs => match a with | 0 => nonzeros bs | _ => a:: nonzeros bs end end. Example test_nonzeros: nonzeros [0;1;0;2;3;0;0] = [1;2;3]. Proof. simpl. reflexivity. Qed. Fixpoint nonzeros2 (l:natlist) : natlist := match l with | nil => nil | h :: t => if(beq_nat h 0) then nonzeros2 t else h :: nonzeros2 t end. Example test_nonzeros2: nonzeros2 [0;1;0;2;3;0;0] = [1;2;3]. Proof. reflexivity. Qed. (* --------------------------------------------------------------------- Ejercicio 4. Definir la función oddmembers : natlist -> natlist tal que (oddmembers xs) es la lista de los elementos impares de xs. Por ejemplo, oddmembers [0;1;0;2;3;0;0] = [1;3]. ------------------------------------------------------------------ *) Fixpoint oddmembers (l:natlist) : natlist := match l with | nil => nil | t::xs => if oddb t then t :: oddmembers xs else oddmembers xs end. Example test_oddmembers: oddmembers [0;1;0;2;3;0;0] = [1;3]. Proof. reflexivity. Qed. (* --------------------------------------------------------------------- Ejercicio 5. Definir la función countoddmembers : natlist -> nat tal que (countoddmembers xs) es el número de elementos impares de xs. Por ejemplo, countoddmembers [1;0;3;1;4;5] = 4. countoddmembers [0;2;4] = 0. countoddmembers nil = 0. ------------------------------------------------------------------ *) Definition countoddmembers (l:natlist) : nat := length (oddmembers l). Example test_countoddmembers1: countoddmembers [1;0;3;1;4;5] = 4. Proof. reflexivity. Qed. Example test_countoddmembers2: countoddmembers [0;2;4] = 0. Proof. reflexivity. Qed. Example test_countoddmembers3: countoddmembers nil = 0. Proof. reflexivity. Qed. (* --------------------------------------------------------------------- Ejercicio 6. Definir la función alternate : natlist -> natlist -> natlist tal que (alternate xs ys) es la lista obtenida intercalando los elementos de xs e ys. Por ejemplo, alternate [1;2;3] [4;5;6] = [1;4;2;5;3;6]. alternate [1] [4;5;6] = [1;4;5;6]. alternate [1;2;3] [4] = [1;4;2;3]. alternate [] [20;30] = [20;30]. ------------------------------------------------------------------ *) Fixpoint alternate (l1 l2 : natlist) : natlist := match l1 with | nil => l2 | t::xs => match l2 with | nil => t::xs | p::ys => t::p::alternate xs ys end end. Example test_alternate1: alternate [1;2;3] [4;5;6] = [1;4;2;5;3;6]. Proof. reflexivity. Qed. Example test_alternate2: alternate [1] [4;5;6] = [1;4;5;6]. Proof. reflexivity. Qed. Example test_alternate3: alternate [1;2;3] [4] = [1;4;2;3]. Proof. reflexivity. Qed. Example test_alternate4: alternate [] [20;30] = [20;30]. Proof. reflexivity. Qed. (* ===================================================================== §§ Multiconjuntos como listas ================================================================== *) (* Un multiconjunto es como un conjunto donde los elementos pueden repetirse más de una vez. Podemos implementarlos como listas. *) (* --------------------------------------------------------------------- Ejemplo. Definir el tipo baf de los multiconjuntos de números naturales. ------------------------------------------------------------------ *) Definition bag := natlist. (* --------------------------------------------------------------------- Ejercicio 7. Definir la función count : nat -> bag -> nat tal que (count v s) es el número des veces que aparece el elemento v en el multiconjunto s. Por ejemplo, count 1 [1;2;3;1;4;1] = 3. count 6 [1;2;3;1;4;1] = 0. ------------------------------------------------------------------ *) Fixpoint count (v:nat) (s:bag) : nat := match s with | nil => 0 | t::xs => if beq_nat t v then 1 + count v xs else count v xs end. Example test_count1: count 1 [1;2;3;1;4;1] = 3. Proof. reflexivity. Qed. Example test_count2: count 6 [1;2;3;1;4;1] = 0. Proof. reflexivity. Qed. (* --------------------------------------------------------------------- Ejercicio 8. Definir la función sum : bag -> bag -> bag tal que (sum xs ys) es la suma de los multiconjuntos xs e ys. Por ejemplo, count 1 (sum [1;2;3] [1;4;1]) = 3. ------------------------------------------------------------------ *) Definition sum : bag -> bag -> bag := app. Example test_sum1: count 1 (sum [1;2;3] [1;4;1]) = 3. Proof. reflexivity. Qed. (* --------------------------------------------------------------------- Ejercicio 9. Definir la función add : nat -> bag -> bag tal que (add x ys) es el multiconjunto obtenido añadiendo el elemento x al multiconjunto ys. Por ejemplo, count 1 (add 1 [1;4;1]) = 3. count 5 (add 1 [1;4;1]) = 0. ------------------------------------------------------------------ *) Definition add (v:nat) (s:bag) : bag := v :: s. Example test_add1: count 1 (add 1 [1;4;1]) = 3. Proof. reflexivity. Qed. Example test_add2: count 5 (add 1 [1;4;1]) = 0. Proof. reflexivity. Qed. (* --------------------------------------------------------------------- Ejercicio 10. Definir la función member : nat -> bag -> bool tal que (member x ys) se verfica si x pertenece al multiconjunto ys. Por ejemplo, member 1 [1;4;1] = true. member 2 [1;4;1] = false. ------------------------------------------------------------------ *) Definition member (v:nat) (s:bag) : bool := if beq_nat 0 (count v s) then false else true. Example test_member1: member 1 [1;4;1] = true. Proof. reflexivity. Qed. Example test_member2: member 2 [1;4;1] = false. Proof. reflexivity. Qed. Definition member2 (v:nat) (s:bag) : bool := negb (beq_nat O (count v s)). Example test_member2_1: member 1 [1;4;1] = true. Proof. reflexivity. Qed. Example test_member2_2: member 2 [1;4;1] = false. Proof. reflexivity. Qed. (* --------------------------------------------------------------------- Ejercicio 11. Definir la función remove_one : nat -> bag -> bag tal que (remove_one x ys) es el multiconjunto obtenido eliminando una ocurrencia de x en el multiconjunto ys. Por ejemplo, count 5 (remove_one 5 [2;1;5;4;1]) = 0. count 4 (remove_one 5 [2;1;4;5;1;4]) = 2. count 5 (remove_one 5 [2;1;5;4;5;1;4]) = 1. ------------------------------------------------------------------ *) Fixpoint remove_one (v:nat) (s:bag) : bag := match s with | nil => nil | t :: xs => if beq_nat t v then xs else t :: remove_one v xs end. Example test_remove_one1: count 5 (remove_one 5 [2;1;5;4;1]) = 0. Proof. reflexivity. Qed. Example test_remove_one2: count 5 (remove_one 5 [2;1;4;1]) = 0. Proof. reflexivity. Qed. Example test_remove_one3: count 4 (remove_one 5 [2;1;4;5;1;4]) = 2. Proof. reflexivity. Qed. Example test_remove_one4: count 5 (remove_one 5 [2;1;5;4;5;1;4]) = 1. Proof. reflexivity. Qed. (* --------------------------------------------------------------------- Ejercicio 12. Definir la función remove_all : nat -> bag -> bag tal que (remove_all x ys) es el multiconjunto obtenido eliminando todas las ocurrencias de x en el multiconjunto ys. Por ejemplo, count 5 (remove_all 5 [2;1;5;4;1]) = 0. count 5 (remove_all 5 [2;1;4;1]) = 0. count 4 (remove_all 5 [2;1;4;5;1;4]) = 2. count 5 (remove_all 5 [2;1;5;4;5;1;4;5;1;4]) = 0. ------------------------------------------------------------------ *) Fixpoint remove_all (v:nat) (s:bag) : bag := match s with | nil => nil | t :: xs => if beq_nat t v then remove_all v xs else t :: remove_all v xs end. Example test_remove_all1: count 5 (remove_all 5 [2;1;5;4;1]) = 0. Proof. reflexivity. Qed. Example test_remove_all2: count 5 (remove_all 5 [2;1;4;1]) = 0. Proof. reflexivity. Qed. Example test_remove_all3: count 4 (remove_all 5 [2;1;4;5;1;4]) = 2. Proof. reflexivity. Qed. Example test_remove_all4: count 5 (remove_all 5 [2;1;5;4;5;1;4;5;1;4]) = 0. Proof. reflexivity. Qed. (* --------------------------------------------------------------------- Ejercicio 13. Definir la función subset : bag -> bag -> bool tal que (subset xs ys) se verifica si xs es un sub,ulticonjunto de ys. Por ejemplo, subset [1;2] [2;1;4;1] = true. subset [1;2;2] [2;1;4;1] = false. ------------------------------------------------------------------ *) Fixpoint subset (s1:bag) (s2:bag) : bool := match s1 with | nil => true | x::xs => member x s2 && subset xs (remove_one x s2) end. Example test_subset1: subset [1;2] [2;1;4;1] = true. Proof. reflexivity. Qed. Example test_subset2: subset [1;2;2] [2;1;4;1] = false. Proof. reflexivity. Qed. (* --------------------------------------------------------------------- Ejercicio 14. Escribir un teorema sobre multiconjuntos con las funciones count y add y probarlo. ------------------------------------------------------------------ *) Theorem bag_theorem : forall s1 s2 : bag, forall n : nat, count n s1 + count n s2 = count n (app s1 s2). Proof. intros s1 s2 n. induction s1 as [|s s']. - simpl. reflexivity. - simpl. destruct (beq_nat s n). + simpl. rewrite IHs'. reflexivity. + rewrite IHs'. reflexivity. Qed. (* ===================================================================== § Razonamiento sobre listas ================================================================== *) (* --------------------------------------------------------------------- Ejemplo. Demostrar que, para toda lista de naturales l, [] ++ l = l ------------------------------------------------------------------ *) Theorem nil_app : forall l:natlist, [] ++ l = l. Proof. reflexivity. Qed. (* --------------------------------------------------------------------- Ejemplo. Demostrar que, para toda lista de naturales l, pred (length l) = length (tl l) ------------------------------------------------------------------ *) Theorem tl_length_pred : forall l:natlist, pred (length l) = length (tl l). Proof. intros l. destruct l as [| n l']. - (* l = nil *) reflexivity. - (* l = cons n l' *) reflexivity. Qed. (* ===================================================================== §§ Inducción sobre listas ================================================================== *) (* --------------------------------------------------------------------- Ejemplo. Demostrar que la concatenación de listas de naturales es asociativa. ------------------------------------------------------------------ *) Theorem app_assoc : forall l1 l2 l3 : natlist, (l1 ++ l2) ++ l3 = l1 ++ (l2 ++ l3). Proof. intros l1 l2 l3. induction l1 as [| n l1' IHl1']. - (* l1 = nil *) reflexivity. - (* l1 = cons n l1' *) simpl. rewrite -> IHl1'. reflexivity. Qed. (* Comentar los nombres dados en la hipótesis de inducción. *) (* ===================================================================== §§§ Inversa de una lista ================================================================== *) (* --------------------------------------------------------------------- Ejemplo. Definir la función rev : natlist -> natlist tal que (rev xs) es la inversa de xs. Por ejemplo, rev [1;2;3] = [3;2;1]. rev nil = nil. ------------------------------------------------------------------ *) Fixpoint rev (l:natlist) : natlist := match l with | nil => nil | h :: t => rev t ++ [h] end. Example test_rev1: rev [1;2;3] = [3;2;1]. Proof. reflexivity. Qed. Example test_rev2: rev nil = nil. Proof. reflexivity. Qed. (* ===================================================================== §§§ Propiedaes de la función rev ================================================================== *) (* --------------------------------------------------------------------- Ejemplo. Demostrar que length (rev l) = length l ------------------------------------------------------------------ *) Theorem rev_length_firsttry : forall l : natlist, length (rev l) = length l. Proof. intros l. induction l as [| n l' IHl']. - (* l = [] *) reflexivity. - (* l = n :: l' *) (* Probamos simplificando *) simpl. rewrite <- IHl'. (* Nos encontramos sin más que hacer, así que buscamos un lema que nos ayude. *) Abort. Theorem app_length : forall l1 l2 : natlist, length (l1 ++ l2) = (length l1) + (length l2). Proof. intros l1 l2. induction l1 as [| n l1' IHl1']. - (* l1 = nil *) reflexivity. - (* l1 = cons *) simpl. rewrite -> IHl1'. reflexivity. Qed. (* Ahora completamos la prueba original. *) Theorem rev_length : forall l : natlist, length (rev l) = length l. Proof. intros l. induction l as [| n l' IHl']. - (* l = nil *) reflexivity. - (* l = cons *) simpl. rewrite -> app_length, plus_comm. simpl. rewrite -> IHl'. reflexivity. Qed. (* ===================================================================== § Ejercicios ================================================================== *) (* ===================================================================== §§ Ejercicios: 1ª parte ================================================================== *) (* --------------------------------------------------------------------- Ejercicio 15. Demostrar que la lista vacía es el elemento neutro por la derecha de la concatenación de listas. ------------------------------------------------------------------ *) Theorem app_nil_r : forall l : natlist, l ++ [] = l. Proof. intros l. induction l as [| x xs HI]. - reflexivity. - simpl. rewrite HI. reflexivity. Qed. (* --------------------------------------------------------------------- Ejercicio 16. Demostrar que rev es un endomorfismo en (natlist,++) ------------------------------------------------------------------ *) Theorem rev_app_distr: forall l1 l2 : natlist, rev (l1 ++ l2) = rev l2 ++ rev l1. Proof. intros l1 l2. induction l1 as [|x xs HI]. - simpl. rewrite app_nil_r. reflexivity. - simpl. rewrite HI, app_assoc. reflexivity. Qed. (* --------------------------------------------------------------------- Ejercicio 17. Demostrar que rev es involutiva. ------------------------------------------------------------------ *) Theorem rev_involutive : forall l : natlist, rev (rev l) = l. Proof. induction l as [|x xs HI]. - reflexivity. - simpl. rewrite rev_app_distr. rewrite HI. reflexivity. Qed. (* --------------------------------------------------------------------- Ejercicio 18. Demostrar que l1 ++ (l2 ++ (l3 ++ l4)) = ((l1 ++ l2) ++ l3) ++ l4. ------------------------------------------------------------------ *) Theorem app_assoc4 : forall l1 l2 l3 l4 : natlist, l1 ++ (l2 ++ (l3 ++ l4)) = ((l1 ++ l2) ++ l3) ++ l4. Proof. intros l1 l2 l3 l4. rewrite app_assoc. rewrite app_assoc. reflexivity. Qed. (* --------------------------------------------------------------------- Ejercicio 19. Demostrar que al concatenar dos listas no aparecen ni desaparecen ceros. ------------------------------------------------------------------ *) Lemma nonzeros_app : forall l1 l2 : natlist, nonzeros (l1 ++ l2) = (nonzeros l1) ++ (nonzeros l2). Proof. intros l1 l2. induction l1 as [|x xs HI]. - reflexivity. - simpl. destruct x. + rewrite HI. reflexivity. + simpl. rewrite HI. reflexivity. Qed. (* --------------------------------------------------------------------- Ejercicio 20. Definir la función beq_natlist : natlist -> natlist -> bool tal que (beq_natlist xs ys) se verifica si las listas xs e ys son iguales. Por ejemplo, beq_natlist nil nil = true. beq_natlist [1;2;3] [1;2;3] = true. beq_natlist [1;2;3] [1;2;4] = false. ------------------------------------------------------------------ *) Fixpoint beq_natlist (l1 l2 : natlist) : bool:= match l1, l2 with | nil, nil => true | x::xs, y::ys => beq_nat x y && beq_natlist xs ys | _, _ => false end. Example test_beq_natlist1: (beq_natlist nil nil = true). Proof. reflexivity. Qed. Example test_beq_natlist2: beq_natlist [1;2;3] [1;2;3] = true. Proof. reflexivity. Qed. Example test_beq_natlist3: beq_natlist [1;2;3] [1;2;4] = false. Proof. reflexivity. Qed. (* --------------------------------------------------------------------- Ejercicio 21. Demostrar que la igualdad de listas cumple la propiedad reflexiva. ------------------------------------------------------------------ *) Theorem beq_natlist_refl : forall l:natlist, true = beq_natlist l l. Proof. induction l as [|n xs HI]. - reflexivity. - simpl. rewrite <- HI. replace (beq_nat n n) with true. reflexivity. + rewrite <- beq_nat_refl. reflexivity. Qed. (* ===================================================================== §§ Ejercicios: 1ª parte ================================================================== *) (* --------------------------------------------------------------------- Ejercicio 22. Demostrar que al incluir un elemento en un multiconjunto, ese elemento aparece al menos una vez en el resultado. ------------------------------------------------------------------ *) Theorem count_member_nonzero : forall (s : bag), leb 1 (count 1 (1 :: s)) = true. Proof. intro s. simpl. reflexivity. Qed. (* --------------------------------------------------------------------- Ejercicio 23. Demostrar que cada número natural es menor o igual que su siguiente. ------------------------------------------------------------------ *) Theorem ble_n_Sn : forall n, leb n (S n) = true. Proof. intros n. induction n as [| n' IHn']. - (* 0 *) simpl. reflexivity. - (* S n' *) simpl. rewrite IHn'. reflexivity. Qed. (* --------------------------------------------------------------------- Ejercicio 24. Demostrar que al borrar una ocurrencia de 0 de un multiconjunto el número de ocurrencias de 0 en el resultado es menor o igual que en el original. ------------------------------------------------------------------ *) Theorem remove_decreases_count: forall (s : bag), leb (count 0 (remove_one 0 s)) (count 0 s) = true. Proof. induction s as [|x xs HI]. - reflexivity. - simpl. destruct x. + rewrite ble_n_Sn. reflexivity. + simpl. rewrite HI. reflexivity. Qed. (* --------------------------------------------------------------------- Ejercicio 25. Escribir un teorema con las funciones count y sum de los multiconjuntos. ------------------------------------------------------------------ *) Theorem bag_count_sum: forall n : nat, forall b1 b2 : bag, count n b1 + count n b2 = count n (sum b1 b2). Proof. intros n b1 b2. induction b1 as [|b bs HI]. - reflexivity. - simpl. destruct (beq_nat b n). + simpl. rewrite HI. reflexivity. + rewrite HI. reflexivity. Qed. (* --------------------------------------------------------------------- Ejercicio 26. Demostrar que la función rev es inyectiva; es decir, forall (l1 l2 : natlist), rev l1 = rev l2 -> l1 = l2. ------------------------------------------------------------------ *) Theorem rev_injective : forall (l1 l2 : natlist), rev l1 = rev l2 -> l1 = l2. Proof. intros. rewrite <- rev_involutive, <- H, rev_involutive. reflexivity. Qed. (* ===================================================================== § Opcionales ================================================================== *) (* --------------------------------------------------------------------- Ejemplo. Definir la función nth_bad : natlist -> n -> nat tal que (nth_bad xs n) es el n-ésimo elemento de la lista xs y 42 si la lista tiene menos de n elementos. ------------------------------------------------------------------ *) Fixpoint nth_bad (l:natlist) (n:nat) : nat := match l with | nil => 42 (* un valor arbitrario *) | a :: l' => match beq_nat n O with | true => a | false => nth_bad l' (pred n) end end. (* --------------------------------------------------------------------- Ejemplo. Definir el tipo natoption con los contructores Some : nat -> natoption None : natoption. ------------------------------------------------------------------ *) Inductive natoption : Type := | Some : nat -> natoption | None : natoption. (* --------------------------------------------------------------------- Ejemplo. Definir la función nth_error : natlist -> nat -> natoption tal que (nth_error xs n) es el n-ésimo elemento de la lista xs o None si la lista tiene menos de n elementos. Por ejemplo, nth_error [4;5;6;7] 0 = Some 4. nth_error [4;5;6;7] 3 = Some 7. nth_error [4;5;6;7] 9 = None. ------------------------------------------------------------------ *) Fixpoint nth_error (l:natlist) (n:nat) : natoption := match l with | nil => None | a :: l' => match beq_nat n O with | true => Some a | false => nth_error l' (pred n) end end. Example test_nth_error1 : nth_error [4;5;6;7] 0 = Some 4. Proof. reflexivity. Qed. Example test_nth_error2 : nth_error [4;5;6;7] 3 = Some 7. Proof. reflexivity. Qed. Example test_nth_error3 : nth_error [4;5;6;7] 9 = None. Proof. reflexivity. Qed. (* Introduciendo condicionales nos queda: *) Fixpoint nth_error' (l:natlist) (n:nat) : natoption := match l with | nil => None | a :: l' => if beq_nat n O then Some a else nth_error' l' (pred n) end. (* Nota: Los condicionales funcionan sobre todo tipo inductivo con dos constructores en Coq, sin booleanos. *) (* --------------------------------------------------------------------- Ejemplo. Definir la función option_elim nat -> natoption -> nat tal que (option_elim d o) es el valor de o, si o tienve valor o es d en caso contrario. ------------------------------------------------------------------ *) Definition option_elim (d : nat) (o : natoption) : nat := match o with | Some n' => n' | None => d end. (* --------------------------------------------------------------------- Ejercicio 27. Definir la función hd_error : natlist -> natoption tal que (hd_error xs) es el primer elemento de xs, si xs es no vacía; o es None, en caso contrario. Por ejemplo, hd_error [] = None. hd_error [1] = Some 1. hd_error [5;6] = Some 5. ------------------------------------------------------------------ *) Definition hd_error (l : natlist) : natoption := match l with | nil => None | x::xs => Some x end. Example test_hd_error1 : hd_error [] = None. Proof. reflexivity. Qed. Example test_hd_error2 : hd_error [1] = Some 1. Proof. reflexivity. Qed. Example test_hd_error3 : hd_error [5;6] = Some 5. Proof. reflexivity. Qed. (* --------------------------------------------------------------------- Ejercicio 28. Demostrar que hd default l = option_elim default (hd_error l). ------------------------------------------------------------------ *) Theorem option_elim_hd : forall (l:natlist) (default:nat), hd default l = option_elim default (hd_error l). Proof. intros l default. destruct l as [|x xs]. - reflexivity. - simpl. reflexivity. Qed. (* --------------------------------------------------------------------- Nota. Finalizar el módulo NatList. ------------------------------------------------------------------ *) End NatList. (* ===================================================================== § Funciones parciales (o diccionarios) ================================================================== *) (* --------------------------------------------------------------------- Ejemplo. Definir el tipo id con el constructor Id : nat -> id. La idea es usarlo como clave de los dicccionarios. ------------------------------------------------------------------ *) Inductive id : Type := | Id : nat -> id. (* --------------------------------------------------------------------- Ejemplo. Definir la función beq_id : id -> id -> bool tal que (beq_id x1 x2) se verifcia si tienen la misma clave. ------------------------------------------------------------------ *) Definition beq_id (x1 x2 : id) := match x1, x2 with | Id n1, Id n2 => beq_nat n1 n2 end. (* --------------------------------------------------------------------- Ejercicio 29. Demostrar que beq_id es reflexiva. ------------------------------------------------------------------ *) Theorem beq_id_refl : forall x, true = beq_id x x. Proof. intro x. destruct x. simpl. rewrite <- beq_nat_refl. reflexivity. Qed. (* --------------------------------------------------------------------- Nota. Iniciar el módulo PartialMap que importa a NatList. ------------------------------------------------------------------ *) Module PartialMap. Export NatList. (* --------------------------------------------------------------------- Ejemplo. Definir el tipo partial_map (para representar los diccionarios) con los contructores empty : partial_map record : id -> nat -> partial_map -> partial_map. ------------------------------------------------------------------ *) Inductive partial_map : Type := | empty : partial_map | record : id -> nat -> partial_map -> partial_map. (* --------------------------------------------------------------------- Ejemplo. Definir la función update : partial_map -> id -> nat -> partial_map tal que (update d i v) es el diccionario obtenido a partir del d + si d tiene un elemento con clave i, le cambia su valor a v + en caso contrario, le añade el elemento v con clave i ------------------------------------------------------------------ *) Definition update (d : partial_map) (x : id) (value : nat) : partial_map := record x value d. (* --------------------------------------------------------------------- Ejemplo. Definir la función find : id -> partial_map -> natoption tal que (find i d) es el valor de la entrada de d con clave i, o None si d no tiene ninguna entrada con clave i. ------------------------------------------------------------------ *) Fixpoint find (x : id) (d : partial_map) : natoption := match d with | empty => None | record y v d' => if beq_id x y then Some v else find x d' end. (* --------------------------------------------------------------------- Ejercicio 30. Demostrar que forall (d : partial_map) (x : id) (v: nat), find x (update d x v) = Some v. ------------------------------------------------------------------ *) Theorem update_eq : forall (d : partial_map) (x : id) (v: nat), find x (update d x v) = Some v. Proof. intros d x v. destruct d as [|d' x' v']. - simpl. destruct x. simpl. rewrite <- beq_nat_refl. reflexivity. - simpl. destruct x. simpl. rewrite <- beq_nat_refl. reflexivity. Qed. (* --------------------------------------------------------------------- Ejercicio 31. Demostrar que forall (d : partial_map) (x y : id) (o: nat), beq_id x y = false -> find x (update d y o) = find x d. ------------------------------------------------------------------ *) Theorem update_neq : forall (d : partial_map) (x y : id) (o: nat), beq_id x y = false -> find x (update d y o) = find x d. Proof. intros d x y o p. simpl. rewrite p. reflexivity. Qed. (* --------------------------------------------------------------------- Nota. Finalizr el módulo PartialMap ------------------------------------------------------------------ *) End PartialMap. (* --------------------------------------------------------------------- Ejercicio 32. Se define el tipo baz por Inductive baz : Type := | Baz1 : baz -> baz | Baz2 : baz -> bool -> baz. ¿Cuántos elementos tiene el tipo baz? |