(* Relación 1: Programación funcional en Coq *)
Definition admit {T: Type} : T. Admitted.
(* ---------------------------------------------------------------------
Ejercicio 1. Definir la función
nandb :: bool -> bool -> bool
tal que (nanb x y) se verifica si x e y no son verdaderos.
Demostrar las siguientes propiedades de nand
(nandb true false) = true.
(nandb false false) = true.
(nandb false true) = true.
(nandb true true) = false.
------------------------------------------------------------------ *)
Definition nandb (b1:bool) (b2:bool) : bool :=
admit.
Example prop_nandb1: (nandb true false) = true.
Admitted.
Example prop_nandb2: (nandb false false) = true.
Admitted.
Example prop_nandb3: (nandb false true) = true.
Admitted.
Example prop_nandb4: (nandb true true) = false.
Admitted.
(* ---------------------------------------------------------------------
Ejercicio 2.1. Definir la función
andb3 :: bool -> bool -> bool -> bool
tal que (andb3 x y z) se verifica si x, y y z son verdaderos.
Demostrar las siguientes propiedades de andb3
(andb3 true true true) = true.
(andb3 false true true) = false.
(andb3 true false true) = false.
(andb3 true true false) = false.
------------------------------------------------------------------ *)
Definition andb3 (x:bool) (y:bool) (z:bool) : bool :=
admit.
Example prop_andb31: (andb3 true true true) = true.
Admitted.
Example prop_andb32: (andb3 false true true) = false.
Admitted.
Example prop_andb33: (andb3 true false true) = false.
Admitted.
Example prop_andb34: (andb3 true true false) = false.
Admitted.
(* ---------------------------------------------------------------------
Ejercicio 3. Definir la función
factorial :: nat -> nat1
tal que (factorial n) es el factorial de n.
(factorial 3) = 6.
(factorial 5) = (mult 10 12).
------------------------------------------------------------------ *)
Fixpoint factorial (n:nat) : nat :=
admit.
Example prop_factorial1: (factorial 3) = 6.
Admitted.
Example prop_factorial2: (factorial 5) = (mult 10 12).
Admitted.
(* ---------------------------------------------------------------------
Ejercicio 4. Definir la función
blt_nat :: nat -> nat -> bool
tal que (blt n m) se verifica si n es menor que m.
Demostrar las siguientes propiedades
(blt_nat 2 2) = false.
(blt_nat 2 4) = true.
(blt_nat 4 2) = false.
------------------------------------------------------------------ *)
(* Nota: Se puede usar las funciones beq_nat y leb del texto del tema 1 *)
Fixpoint beq_nat (n m : nat) : bool :=
match n with
| O => match m with
| O => true
| S m' => false
end
| S n' => match m with
| O => false
| S m' => beq_nat n' m'
end
end.
Fixpoint leb (n m : nat) : bool :=
match n with
| O => true
| S n' =>
match m with
| O => false
| S m' => leb n' m'
end
end.
Definition blt_nat (n m : nat) : bool :=
admit.
Example prop_blt_nat1: (blt_nat 2 2) = false.
Admitted.
Example prop_blt_nat2: (blt_nat 2 4) = true.
Admitted.
Example prop_blt_nat3: (blt_nat 4 2) = false.
Admitted.
(* ---------------------------------------------------------------------
Ejercicio 5. Demostrar que
forall n m o : nat,
n = m -> m = o -> n + m = m + o.
------------------------------------------------------------------ *)
Theorem plus_id_exercise: forall n m o : nat,
n = m -> m = o -> n + m = m + o.
Admitted.
(* ---------------------------------------------------------------------
Ejercicio 6. Demostrar que
forall n m : nat,
m = S n ->
m * (1 + n) = m * m.
------------------------------------------------------------------ *)
(* Nota: Se puede usar el lema plus_1_l del texto del tema 1 *)
Theorem plus_1_l : forall n:nat, 1 + n = S n.
Proof.
intros n. reflexivity. Qed.
Theorem mult_S_1 : forall n m : nat,
m = S n ->
m * (1 + n) = m * m.
Admitted.
(* ---------------------------------------------------------------------
Ejercicio 7. Demostrar que
forall b c : bool,
andb b c = true -> c = true.
------------------------------------------------------------------ *)
Theorem andb_true_elim2 : forall b c : bool,
andb b c = true -> c = true.
Admitted.
(* ---------------------------------------------------------------------
Ejercicio 8. Dmostrar que
forall n : nat,
beq_nat 0 (n + 1) = false.
------------------------------------------------------------------ *)
Theorem zero_nbeq_plus_1: forall n : nat,
beq_nat 0 (n + 1) = false.
Admitted.
(* ---------------------------------------------------------------------
Ejercicio 9. Demostrar que
forall (f : bool -> bool),
(forall (x : bool), f x = x) ->
forall (b : bool), f (f b) = b.
------------------------------------------------------------------ *)
Theorem identity_fn_applied_twice :
forall (f : bool -> bool),
(forall (x : bool), f x = x) ->
forall (b : bool), f (f b) = b.
Admitted.
(* ---------------------------------------------------------------------
Ejercicio 10. Demostrar que
forall (b c : bool),
(andb b c = orb b c) -> b = c.
------------------------------------------------------------------ *)
Theorem andb_eq_orb :
forall (b c : bool),
(andb b c = orb b c) -> b = c.
Admitted.
(* ---------------------------------------------------------------------
Ejercicio 11. En este ejercicio se considera la siguiente
representación de los números naturales
Inductive nat2 : Type :=
| C : nat2
| D : nat2 -> nat2
| SD : nat2 -> nat2.
donde C representa el cero, D el doble y SD el siguiente del doble.
Definir la función
nat2Anat :: nat2 -> nat
tal que (nat2Anat x) es el número natural representado por x.
Demostrar que
nat2Anat (SD (SD C)) = 3
nat2Anat (D (SD (SD C))) = 6.
------------------------------------------------------------------ *)
Inductive nat2 : Type :=
| C : nat2
| D : nat2 -> nat2
| SD : nat2 -> nat2.
Fixpoint nat2Anat (x:nat2) : nat :=
admit.
Example prop_nat2Anat1: (nat2Anat (SD (SD C))) = 3.
Admitted.
Example prop_nat2Anat2: (nat2Anat (D (SD (SD C)))) = 6.
Admitted.
