Temas

2018-03-24T10:51:20Z

Jalonso: /* Temas de Seminario de lógica computacional (2018) */

== Temas de ''Seminario de lógica computacional (2018)'' ==

En esta página se irán publicando los temas conforme se vayan estudiando.

* Tema 0: Introducción al Seminario
** [http://www.seas.upenn.edu/~cis500/current/lectures/lec01.pdf Introducción de B. Pierce al curso "Software foundations"].
** [https://softwarefoundations.cis.upenn.edu/lf-current/Preface.html Introducción del libro "Software foundations"].
* Tema 1: Programación funcional en Coq
** [[Tema 1 |Exposición]], [https://softwarefoundations.cis.upenn.edu/lf-current/Basics.html Texto] y [http://www.seas.upenn.edu/~cis500/current/sf/lf-current/Basics.v Código].
* Tema 2: Demostraciones por inducción en Coq
** [[Tema 2 |Exposición]], [https://softwarefoundations.cis.upenn.edu/lf-current/Induction.html Texto] y [http://www.seas.upenn.edu/~cis500/current/sf/lf-current/Induction.v Código].
* Tema 3: Datos estructurados en Coq
** [[Tema 3 |Exposición]], [https://softwarefoundations.cis.upenn.edu/lf-current/Lists.html Texto], [http://www.seas.upenn.edu/~cis500/current/sf/lf-current/Lists.v Código].

=== Ponentes ===

* Ponentes de las próximas sesiones:
** Miriam Medrán Navarro
** Samuel Ortiz Morales
** Borja Sierra Miranda
* Ponentes de las sesiones anteriores:
** Ángel Ruiz Campos (Tema 1)
** Alejandro Rodríguez Rodríguez (Tema 2)
** Jorge Catarecha Otero-Saavedra (Tema 3)

=== Participantes ===

Los participantes en el Seminario son
* Alejandro Rodriguez Rodríguez
* Ángel Ruiz Campos
* Borja Sierra Miranda
* Elías Guisado Villalgordo
* Jorge Catarecha Otero-Saavedra
* Marina Jiménez Comez
* Mencía Veas Lerdo de Tejada
* Samuel Ortiz Morales

Tema 2

2018-03-24T10:50:31Z

Jalonso:

Temas

2018-03-24T10:31:48Z

Jalonso: /* Temas de Seminario de lógica computacional (2018) */

== Temas de ''Seminario de lógica computacional (2018)'' ==

En esta página se irán publicando los temas conforme se vayan estudiando.

* Tema 0: Introducción al Seminario
** [http://www.seas.upenn.edu/~cis500/current/lectures/lec01.pdf Introducción de B. Pierce al curso "Software foundations"].
** [https://softwarefoundations.cis.upenn.edu/lf-current/Preface.html Introducción del libro "Software foundations"].
* Tema 1: Programación funcional en Coq
** [[Tema 1 |Exposición]], [https://softwarefoundations.cis.upenn.edu/lf-current/Basics.html Texto] y [http://www.seas.upenn.edu/~cis500/current/sf/lf-current/Basics.v Código].
* Tema 2: Demostraciones por inducción en Coq
** [[Tema 2 |Exposición]], [https://softwarefoundations.cis.upenn.edu/lf-current/Induction.html Texto], [http://www.seas.upenn.edu/~cis500/current/sf/lf-current/Induction.v Código], [[R2 |Enunciado de ejercicios]] y [[Relación 2 | Solución colaborativa]].
* Tema 3: Datos estructurados en Coq
** [[Tema 3 |Exposición]], [https://softwarefoundations.cis.upenn.edu/lf-current/Lists.html Texto], [http://www.seas.upenn.edu/~cis500/current/sf/lf-current/Lists.v Código].

=== Ponentes ===

* Ponentes de las próximas sesiones:
** Miriam Medrán Navarro
** Samuel Ortiz Morales
** Borja Sierra Miranda
* Ponentes de las sesiones anteriores:
** Ángel Ruiz Campos (Tema 1)
** Alejandro Rodríguez Rodríguez (Tema 2)
** Jorge Catarecha Otero-Saavedra (Tema 3)

=== Participantes ===

Los participantes en el Seminario son
* Alejandro Rodriguez Rodríguez
* Ángel Ruiz Campos
* Borja Sierra Miranda
* Elías Guisado Villalgordo
* Jorge Catarecha Otero-Saavedra
* Marina Jiménez Comez
* Mencía Veas Lerdo de Tejada
* Samuel Ortiz Morales

2018-03-22T06:54:35Z

Jalonso:

<source lang="ocaml">
(* Datos estructurados en Coq *)

Require Export Induction.

(* ---------------------------------------------------------------------
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.
Admitted.

(* ---------------------------------------------------------------------
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.
Admitted.

(* =====================================================================
§ 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.

(* ---------------------------------------------------------------------
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.
------------------------------------------------------------------ *)

(* =====================================================================
§ 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?
------------------------------------------------------------------ *)
</source>