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Jalonso en 11:00 15 jul 2018

2018-07-15T11:00:11Z

Angruicam1 en 14:58 10 may 2018

2018-05-10T14:58:31Z

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2018-05-09T15:55:26Z

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2018-05-09T15:50:52Z

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2018-05-06T09:00:29Z

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<source lang="ocaml">
(* T5: Tácticas básicas *)

Require Export T4_PolimorfismoyOS.

(* =====================================================================
§ La táctica apply
================================================================== *)

(* ---------------------------------------------------------------------
Ejemplo de demostración donde el objetivo coincide con alguna
hipótesis.
------------------------------------------------------------------ *)

(* Demostración sin apply *)
Theorem silly1 : forall (n m o p : nat),
n = m ->
[n;o] = [n;p] ->
[n;o] = [m;p].
Proof.
intros n m o p eq1 eq2.
(* [n; o] = [m; p] *)
rewrite <- eq1.
(* [n; o] = [n; p] *)
rewrite eq2.
(* [n; p] = [n; p] *)
reflexivity.
Qed.

(* Demostración con apply *)
Theorem silly1' : forall (n m o p : nat),
n = m ->
[n;o] = [n;p] ->
[n;o] = [m;p].
Proof.
intros n m o p eq1 eq2.
(* [n; o] = [m; p] *)
rewrite <- eq1.
(* [n; o] = [n; p] *)
apply eq2.
Qed.

(* ---------------------------------------------------------------------
Ejemplo de aplicación de apply con hipótesis condicionales.
------------------------------------------------------------------ *)

Theorem silly2 : forall (n m o p : nat),
n = m ->
(forall (q r : nat), q = r -> [q;o] = [r;p]) ->
[n;o] = [m;p].
Proof.
intros n m o p eq1 eq2.
(* [n; o] = [m; p] *)
apply eq2.
(* n = m *)
apply eq1.
Qed.

(* ---------------------------------------------------------------------
Ejemplo de aplicación de apply con hipótesis condicionales y
cuantificadores.
------------------------------------------------------------------ *)

Theorem silly2a : forall (n m : nat),
(n,n) = (m,m) ->
(forall (q r : nat), (q,q) = (r,r) -> [q] = [r]) ->
[n] = [m].
Proof.
intros n m eq1 eq2.
(* [n] = [m] *)
apply eq2.
(* (n, n) = (m, m) *)
apply eq1.
Qed.

(* ---------------------------------------------------------------------
Ejercicio 1. Demostrar, sin usar simpl, que
(forall n, evenb n = true -> oddb (S n) = true) ->
evenb 3 = true ->
oddb 4 = true.
------------------------------------------------------------------ *)

Theorem silly_ex :
(forall n, evenb n = true -> oddb (S n) = true) ->
evenb 3 = true ->
oddb 4 = true.
Proof.
(* FILL IN HERE *) Admitted.

(* ---------------------------------------------------------------------
Ejemplo e necesidad de usar symmetry antes de apply.
------------------------------------------------------------------ *)

Theorem silly3_firsttry : forall (n : nat),
true = beq_nat n 5 ->
beq_nat (S (S n)) 7 = true.
Proof.
intros n H.
(* beq_nat (S (S n)) 7 = true *)
symmetry.
(* true = beq_nat (S (S n)) 7 *)
simpl.
(* true = beq_nat n 5 *)
apply H.
Qed.

(* ---------------------------------------------------------------------
Ejercicio 2. Demostrar
forall (l l' : list nat), l = rev l' -> l' = rev l.
------------------------------------------------------------------ *)

Theorem rev_exercise1:
forall (l l' : list nat), l = rev l' -> l' = rev l.
Proof.
(* FILL IN HERE *) Admitted.

(* =====================================================================
§ La táctica apply ... with ...
================================================================== *)

(* ---------------------------------------------------------------------
Ejemplo. Con dos reescrituras.
------------------------------------------------------------------ *)

Example trans_eq_example : forall (a b c d e f : nat),
[a;b] = [c;d] ->
[c;d] = [e;f] ->
[a;b] = [e;f].
Proof.
intros a b c d e f eq1 eq2.
(* [a; b] = [e; f] *)
rewrite -> eq1.
(* [c; d] = [e; f] *)
rewrite -> eq2.
(* [e; f] = [e; f] *)
reflexivity.
Qed.

(* ---------------------------------------------------------------------
Ejemplo de lema para simplificar la demostración anterior.
------------------------------------------------------------------ *)

Theorem trans_eq : forall (X:Type) (n m o : X),
n = m -> m = o -> n = o.
Proof.
intros X n m o eq1 eq2.
(* n = o *)
rewrite -> eq1.
(* m = o *)
rewrite -> eq2.
(* o = o *)
reflexivity.
Qed.

(* ---------------------------------------------------------------------
Ejemplo. De simplificación de la prueba usando el lema.
------------------------------------------------------------------ *)

Example trans_eq_example' : forall (a b c d e f : nat),
[a;b] = [c;d] ->
[c;d] = [e;f] ->
[a;b] = [e;f].
Proof.
intros a b c d e f eq1 eq2.
(* [a; b] = [e; f] *)
apply trans_eq with (m:=[c;d]).
(* [a; b] = [c; d]
[c; d] = [e; f] *)
apply eq1.
(* [c; d] = [e; f] *)
apply eq2.
Qed.

(* ---------------------------------------------------------------------
Ejemplo. Simplificación de la prueba anterior.
------------------------------------------------------------------ *)

Example trans_eq_example'' : forall (a b c d e f : nat),
[a;b] = [c;d] ->
[c;d] = [e;f] ->
[a;b] = [e;f].
Proof.
intros a b c d e f eq1 eq2.
(* [a; b] = [e; f] *)
apply trans_eq with [c;d].
(* [a; b] = [c; d]
[c; d] = [e; f] *)
apply eq1.
(* [c; d] = [e; f] *)
apply eq2.
Qed.

(* ---------------------------------------------------------------------
Ejercicio 3. Demostrar que
forall (n m o p : nat),
m = (minustwo o) ->
(n + p) = m ->
(n + p) = (minustwo o).
------------------------------------------------------------------ *)

Example trans_eq_exercise : forall (n m o p : nat),
m = (minustwo o) ->
(n + p) = m ->
(n + p) = (minustwo o).
Proof.
(* FILL IN HERE *) Admitted.

(* =====================================================================
§ La táctica inversión
================================================================== *)

(* ---------------------------------------------------------------------
Ejemplo de demostración con inversión sobre los naturales.
------------------------------------------------------------------ *)

Theorem S_injective : forall (n m : nat),
S n = S m ->
n = m.
Proof.
intros n m H.
(* n : nat
m : nat
H : S n = S m *)
inversion H.
(* H1 : n = m *)
reflexivity.
Qed.

(* ---------------------------------------------------------------------
Ejemplo de inversión generando varias hipótesis.
------------------------------------------------------------------ *)

Theorem inversion_ex1 : forall (n m o : nat),
[n; m] = [o; o] ->
[n] = [m].
Proof.
intros n m o H.
(* n : nat
m : nat
o : nat
H : [n; m] = [o; o] *)
inversion H.
(* H1 : n = o
H2 : m = o *)
reflexivity.
Qed.

(* ---------------------------------------------------------------------
Ejemplo de nombramiento de las hipótesis generadas por inversión.
------------------------------------------------------------------ *)

Theorem inversion_ex2 : forall (n m : nat),
[n] = [m] ->
n = m.
Proof.
intros n m H.
(* n : nat
m : nat
H : [n] = [m] *)
inversion H as [Hnm].
(* Hnm : n = m *)
reflexivity.
Qed.

(* ---------------------------------------------------------------------
Ejercicio 4. Demostrar que
forall (X : Type) (x y z : X) (l j : list X),
x :: y :: l = z :: j ->
y :: l = x :: j ->
x = y.
------------------------------------------------------------------ *)

Example inversion_ex3 : forall (X : Type) (x y z : X) (l j : list X),
x :: y :: l = z :: j ->
y :: l = x :: j ->
x = y.
Proof.
(* FILL IN HERE *) Admitted.

(* ---------------------------------------------------------------------
Ejemplo de demostración por inversión basada en que los constructores
son disjuntos.
------------------------------------------------------------------ *)

Theorem beq_nat_0_l : forall n,
beq_nat 0 n = true -> n = 0.
Proof.
intros n.
(* beq_nat 0 n = true -> n = 0 *)
destruct n as [| n'].
- (* beq_nat 0 0 = true -> 0 = 0 *)
intros H.
(* 0 = 0 *)
reflexivity.
- (* beq_nat 0 (S n') = true -> S n' = 0 *)
simpl.
(* false = true -> S n' = 0 *)
intros H.
(* S n' = 0 *)
inversion H.
Qed.

(* ---------------------------------------------------------------------
Ejemplos de demostración por inversión sobre los booleanos.
------------------------------------------------------------------ *)

Theorem inversion_ex4 : forall (n : nat),
S n = O ->
2 + 2 = 5.
Proof.
intros n contra.
inversion contra.
Qed.

Theorem inversion_ex5 : forall (n m : nat),
false = true ->
[n] = [m].
Proof.
intros n m contra.
inversion contra.
Qed.

(* ---------------------------------------------------------------------
Ejercicio 5. Demostrar que
forall (X : Type) (x y z : X) (l j : list X),
x :: y :: l = [] ->
y :: l = z :: j ->
x = z.
------------------------------------------------------------------ *)

Example inversion_ex6 :
forall (X : Type) (x y z : X) (l j : list X),
x :: y :: l = [] ->
y :: l = z :: j ->
x = z.
Proof.
(* FILL IN HERE *) Admitted.

(* ---------------------------------------------------------------------
Ejemplo de lema que usaremos más tarde.
------------------------------------------------------------------ *)

Theorem f_equal : forall (A B : Type) (f: A -> B) (x y: A),
x = y -> f x = f y.
Proof.
intros A B f x y eq.
rewrite eq.
reflexivity.
Qed.

(* =====================================================================
§ Uso de tácticas sobre las hipótesis
================================================================== *)

(* ---------------------------------------------------------------------
Ejemplo de demostración con "simpl in ..."
------------------------------------------------------------------ *)

Theorem S_inj : forall (n m : nat) (b : bool),
beq_nat (S n) (S m) = b ->
beq_nat n m = b.
Proof.
intros n m b H.
(* H : beq_nat (S n) (S m) = b *)
simpl in H.
(* H : beq_nat n m = b *)
apply H.
Qed.

(* ---------------------------------------------------------------------
Ejemplo de "razonamiento hacia adelante" con "apply L in H".
------------------------------------------------------------------ *)

Theorem silly3' : forall (n : nat),
(beq_nat n 5 = true -> beq_nat (S (S n)) 7 = true) ->
true = beq_nat n 5 ->
true = beq_nat (S (S n)) 7.
Proof.
intros n eq H.
(* eq : beq_nat n 5 = true -> beq_nat (S (S n)) 7 = true
H : true = beq_nat n 5 *)
symmetry in H.
(* H : beq_nat n 5 = true *)
apply eq in H.
(* H : beq_nat (S (S n)) 7 = true *)
symmetry in H.
(* H : true = beq_nat (S (S n)) 7 *)
apply H.
Qed.

(* ---------------------------------------------------------------------
Ejercicio 6. Demostrar
forall n m,
n + n = m + m ->
n = m.

Nota: Usar plus_n_Sm
------------------------------------------------------------------ *)

Theorem plus_n_n_injective :
forall n m,
n + n = m + m ->
n = m.
Proof.
intros n.
induction n as [| n'].
(* FILL IN HERE *) Admitted.

(* =====================================================================
§ Control de la hipótesis de inducción
================================================================== *)

(* ---------------------------------------------------------------------
Ejemplo de necesidad de controlar la hipótesis de inducción.
------------------------------------------------------------------ *)

Theorem double_injective_FAILED : forall n m,
double n = double m ->
n = m.
Proof.
intros n m.
induction n as [| n'].
- (* double 0 = double m -> 0 = m *)
simpl.
(* 0 = double m -> 0 = m *)
intros eq.
(* 0 = m *)
destruct m as [| m'].
+ (* 0 = O *)
reflexivity.
+ (* 0 = S m' *)
inversion eq.
- (* double (S n') = double m -> S n' = m *)
intros eq.
(* S n' = m *)
destruct m as [| m'].
+ (* S n' = 0 *)
inversion eq.
+ (* S n' = S m' *)
apply f_equal.
(* n' = m' *)
Abort.

Theorem double_injective : forall n m,
double n = double m ->
n = m.
Proof.
intros n.
induction n as [| n'].
- (* forall m : nat, double 0 = double m -> 0 = m *)
simpl.
(* forall m : nat, 0 = double m -> 0 = m *)
intros m eq.
destruct m as [| m'].
+ (* 0 = O *)
reflexivity.
+ (* 0 = S m' *)
inversion eq.
- (* IHn' : forall m : nat, double n' = double m -> n' = m
forall m : nat, double (S n') = double m -> S n' = m *)
simpl.
(* forall m : nat, S (S (double n')) = double m -> S n' = m *)
intros m eq.
destruct m as [| m'].
+ (* S n' = O *)
simpl.
inversion eq.
+ (* S n' = S m' *)
apply f_equal.
(* n' = m' *)
apply IHn'.
(* double n' = double m' *)
inversion eq.
(* double n' = double n' *)
reflexivity.
Qed.

(* ---------------------------------------------------------------------
Comentario sobre la estrategia de generalización.
------------------------------------------------------------------ *)

(* ---------------------------------------------------------------------
Ejercicio 7. Demostrar que
forall n m,
beq_nat n m = true -> n = m.
------------------------------------------------------------------ *)

Theorem beq_nat_true : forall n m,
beq_nat n m = true -> n = m.
Proof.
(* FILL IN HERE *) Admitted.

(* ---------------------------------------------------------------------
Ejemplo de problema por usar intros antes que induction.
------------------------------------------------------------------ *)
Theorem double_injective_take2_FAILED : forall n m,
double n = double m ->
n = m.
Proof.
intros n m.
induction m as [| m'].
- (* m = O *) simpl. intros eq. destruct n as [| n'].
+ (* n = O *) reflexivity.
+ (* n = S n' *) inversion eq.
- (* m = S m' *) intros eq. destruct n as [| n'].
+ (* n = O *) inversion eq.
+ (* n = S n' *) apply f_equal.
Abort.

(* ---------------------------------------------------------------------
Ejemplo con la táctica "generalize dependent"
------------------------------------------------------------------ *)

Theorem double_injective_take2 : forall n m,
double n = double m ->
n = m.
Proof.
intros n m.
generalize dependent n.
induction m as [| m'].
- (* m = O *) simpl. intros n eq. destruct n as [| n'].
+ (* n = O *) reflexivity.
+ (* n = S n' *) inversion eq.
- (* m = S m' *) intros n eq. destruct n as [| n'].
+ (* n = O *) inversion eq.
+ (* n = S n' *) apply f_equal. apply IHm'. inversion eq. reflexivity.
Qed.

(* ---------------------------------------------------------------------
Lema para iso posterior.
------------------------------------------------------------------ *)

Theorem beq_id_true : forall x y,
beq_id x y = true -> x = y.
Proof.
intros [m] [n]. simpl. intros H.
assert (H' : m = n). { apply beq_nat_true. apply H. }
rewrite H'. reflexivity.
Qed.

(* ---------------------------------------------------------------------
Ejercicio 8. Demostra, por inducción sobre l,
forall (n : nat) (X : Type) (l : list X),
length l = n ->
nth_error l n = None.
------------------------------------------------------------------ *)

Theorem nth_error_after_last:
forall (n : nat) (X : Type) (l : list X),
length l = n ->
nth_error l n = None.
Proof.
(* FILL IN HERE *) Admitted.

(* =====================================================================
§ Expnasión de definiciones
================================================================== *)

(* ---------------------------------------------------------------------
Ejemplo de expansión de una definición con unfold.
------------------------------------------------------------------ *)

Definition square n := n * n.

Lemma square_mult : forall n m, square (n * m) = square n * square m.
Proof.
intros n m.
simpl. (* no hace nada *)
unfold square.
rewrite mult_assoc.
assert (H : n * m * n = n * n * m).
{ rewrite mult_comm.
apply mult_assoc. }
rewrite H. rewrite mult_assoc. reflexivity.
Qed.

(* ---------------------------------------------------------------------
Ejemplo de expansión automática de definiciones.
------------------------------------------------------------------ *)

Definition foo (x: nat) := 5.

Fact silly_fact_1 : forall m, foo m + 1 = foo (m + 1) + 1.
Proof.
intros m.
simpl.
reflexivity.
Qed.

(* ---------------------------------------------------------------------
Ejemplo de no expansión automática de definiciones.
------------------------------------------------------------------ *)

Definition bar x :=
match x with
| O => 5
| S _ => 5
end.

Fact silly_fact_2_FAILED : forall m, bar m + 1 = bar (m + 1) + 1.
Proof.
intros m.
simpl. (* No hace nada *)
Abort.

(* Demostración con destruct *)
Fact silly_fact_2 : forall m, bar m + 1 = bar (m + 1) + 1.
Proof.
intros m.
destruct m.
- simpl. reflexivity.
- simpl. reflexivity.
Qed.

(* Demostración con unfold *)
Fact silly_fact_2' : forall m, bar m + 1 = bar (m + 1) + 1.
Proof.
intros m.
unfold bar.
destruct m.
- reflexivity.
- reflexivity.
Qed.

(* =====================================================================
§ Uso de destruct sobre expresiones compuestas
================================================================== *)

(* ---------------------------------------------------------------------
Ejemplos de uso de destruct sobre expresiones compuestas.
------------------------------------------------------------------ *)

Definition sillyfun (n : nat) : bool :=
if beq_nat n 3 then false
else if beq_nat n 5 then false
else false.

Theorem sillyfun_false : forall (n : nat),
sillyfun n = false.
Proof.
intros n. unfold sillyfun.
destruct (beq_nat n 3).
- (* beq_nat n 3 = true *) reflexivity.
- (* beq_nat n 3 = false *) destruct (beq_nat n 5).
+ (* beq_nat n 5 = true *) reflexivity.
+ (* beq_nat n 5 = false *) reflexivity.
Qed.

(* ---------------------------------------------------------------------
Ejercicio 9. Se define la función split por
Fixpoint split {X Y : Type} (l : list (X*Y))
: (list X) * (list Y) :=
match l with
| [] => ([], [])
| (x, y) :: t =>
match split t with
| (lx, ly) => (x :: lx, y :: ly)
end
end.

Demostrar que split y combine son inversas; es decir,
forall X Y (l : list (X * Y)) l1 l2,
split l = (l1, l2) ->
combine l1 l2 = l.
------------------------------------------------------------------ *)

Fixpoint split {X Y : Type} (l : list (X*Y))
: (list X) * (list Y) :=
match l with
| [] => ([], [])
| (x, y) :: t =>
match split t with
| (lx, ly) => (x :: lx, y :: ly)
end
end.

Theorem combine_split :
forall X Y (l : list (X * Y)) l1 l2,
split l = (l1, l2) ->
combine l1 l2 = l.
Proof.
(* FILL IN HERE *) Admitted.

(* ---------------------------------------------------------------------
Ejemplo de precauciones al usar destruct para no perder información.
------------------------------------------------------------------ *)

Definition sillyfun1 (n : nat) : bool :=
if beq_nat n 3 then true
else if beq_nat n 5 then true
else false.

Theorem sillyfun1_odd_FAILED : forall (n : nat),
sillyfun1 n = true ->
oddb n = true.
Proof.
intros n eq. unfold sillyfun1 in eq.
destruct (beq_nat n 3).
(* Falso por falta de información *)
Abort.

(* Solución usando destruct con eqn *)
Theorem sillyfun1_odd : forall (n : nat),
sillyfun1 n = true ->
oddb n = true.
Proof.
intros n eq. unfold sillyfun1 in eq.
destruct (beq_nat n 3) eqn:Heqe3.
- (* e3 = true *) apply beq_nat_true in Heqe3.
rewrite -> Heqe3. reflexivity.
- (* e3 = false *)
destruct (beq_nat n 5) eqn:Heqe5.
+ (* e5 = true *)
apply beq_nat_true in Heqe5.
rewrite -> Heqe5. reflexivity.
+ (* e5 = false *) inversion eq.
Qed.

(* ---------------------------------------------------------------------
Ejercicio 10. Demostrar que
forall (f : bool -> bool) (b : bool),
f (f (f b)) = f b.
------------------------------------------------------------------ *)

Theorem bool_fn_applied_thrice :
forall (f : bool -> bool) (b : bool),
f (f (f b)) = f b.
Proof.
(* FILL IN HERE *) Admitted.

(* =====================================================================
§ Resumen de tácticas básicas
================================================================== *)

(* Tácticas básicas:
- [intros]: move hypotheses/variables from goal to context

- [reflexivity]: finish the proof (when the goal looks like [e = e])

- [apply]: prove goal using a hypothesis, lemma, or constructor

- [apply... in H]: apply a hypothesis, lemma, or constructor to
a hypothesis in the context (forward reasoning)

- [apply... with...]: explicitly specify values for variables
that cannot be determined by pattern matching

- [simpl]: simplify computations in the goal

- [simpl in H]: ... or a hypothesis

- [rewrite]: use an equality hypothesis (or lemma) to rewrite
the goal

- [rewrite ... in H]: ... or a hypothesis

- [symmetry]: changes a goal of the form [t=u] into [u=t]

- [symmetry in H]: changes a hypothesis of the form [t=u] into [u=t]

- [unfold]: replace a defined constant by its right-hand side in
the goal

- [unfold... in H]: ... or a hypothesis

- [destruct... as...]: case analysis on values of inductively
defined types

- [destruct... eqn:...]: specify the name of an equation to be
added to the context, recording the result of the case analysis

- [induction... as...]: induction on values of inductively
defined types

- [inversion]: reason by injectivity and distinctness of constructors

- [assert (H: e)] (or [assert (e) as H]): introduce a "local
lemma" [e] and call it [H]

- [generalize dependent x]: move the variable [x] (and anything
else that depends on it) from the context back to an explicit
hypothesis in the goal formula *)

(* =====================================================================
§ Ejercicios
================================================================== *)

(* ---------------------------------------------------------------------
Ejercicio 11. Demostrar que
forall (n m : nat),
beq_nat n m = beq_nat m n.
------------------------------------------------------------------ *)

Theorem beq_nat_sym :
forall (n m : nat),
beq_nat n m = beq_nat m n.
Proof.
(* FILL IN HERE *) Admitted.

(* ---------------------------------------------------------------------
Ejercicio 12. Demostrar que
forall n m p,
beq_nat n m = true ->
beq_nat m p = true ->
beq_nat n p = true.
------------------------------------------------------------------ *)

Theorem beq_nat_trans :
forall n m p,
beq_nat n m = true ->
beq_nat m p = true ->
beq_nat n p = true.
Proof.
(* FILL IN HERE *) Admitted.

(* ---------------------------------------------------------------------
Ejercicio 13. We proved, in an exercise above, that for all lists of
pairs, [combine] is the inverse of [split]. How would you formalize
the statement that [split] is the inverse of [combine]? When is this
property true?

Complete the definition of [split_combine_statement] below with a
property that states that [split] is the inverse of
[combine]. Then, prove that the property holds. (Be sure to leave
your induction hypothesis general by not doing [intros] on more
things than necessary. Hint: what property do you need of [l1]
and [l2] for [split] [combine l1 l2 = (l1,l2)] to be true?)
------------------------------------------------------------------ *)

Definition split_combine_statement : Prop
(* ("[: Prop]" means that we are giving a name to a
logical proposition here.) *)
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.

Theorem split_combine : split_combine_statement.
Proof.
(* FILL IN HERE *) Admitted.

(* ---------------------------------------------------------------------
Ejercicio 14. Demostrar que
forall (X : Type) (test : X -> bool) (x : X) (l lf : list X),
filter test l = x :: lf ->
test x = true.
------------------------------------------------------------------ *)

Theorem filter_exercise :
forall (X : Type) (test : X -> bool) (x : X) (l lf : list X),
filter test l = x :: lf ->
test x = true.
Proof.
(* FILL IN HERE *) Admitted.

(* ---------------------------------------------------------------------
Ejercicio 15. Definir, por recursión, las funciones forallb y existsb
tales que
+ (forallb p xs) se verifica si todos los elementos de xs cumplen
p. Por ejemplo,
forallb oddb [1;3;5;7;9] = true
forallb negb [false;false] = true
forallb evenb [0;2;4;5] = false
forallb (beq_nat 5) [] = true
+ (existsb p xs) se verifica si algún elemento de xs cumple p. Por
ejemplo,
existsb (beq_nat 5) [0;2;3;6] = false
existsb (andb true) [true;true;false] = true
existsb oddb [1;0;0;0;0;3] = true
existsb evenb [] = false

Redefinir, usando forallb y negb, la función existsb' y demostrar su
equivalencia con existsb.
------------------------------------------------------------------ *)
</source>

← Revisión anterior		Revisión del 11:00 15 jul 2018
Línea 1:		Línea 1:
−	<source lang="~~ocaml~~">	+	<source lang="coq">
	(* T5: Tácticas básicas *)		(* T5: Tácticas básicas *)

@@ Línea 875: / Línea 875: @@
    (* ("[: Prop]" means that we are giving a name to a
       logical proposition here.) *)
-   (* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.
+   := forall (X : Type) (l1 l2 : list X),
 Theorem split_combine : split_combine_statement.
 Proof.
-(* FILL IN HERE *) Admitted.
+  intro. induction l1 as [| h1 l1' IHl1'].
-(* ---------------------------------------------------------------------
+(** **** Exercise: 3 stars, advanced (filter_exercise)  *)
-   Ejercicio 14. Demostrar que
+(** This one is a bit challenging.  Pay attention to the form of your
-      forall (X : Type) (test : X -> bool) (x : X) (l lf : list X),
+    induction hypothesis. *)
-Theorem filter_exercise :
+Theorem filter_exercise : forall (X : Type) (test : X -> bool)
-  forall (X : Type) (test : X -> bool) (x : X) (l lf : list X),
+                             (x : X) (l lf : list X),
-    filter test l = x :: lf ->
+     filter test l = x :: lf ->
-    test x = true.
+     test x = true.
 Proof.
-   (* FILL IN HERE *) Admitted.
+   intros. generalize dependent lf. induction l as [| n l' IHl'].
-(* ---------------------------------------------------------------------
+      forallb (beq_nat 5) [] = true
-   Redefinir, usando forallb y negb, la función existsb' y demostrar su
+    The second checks whether there exists an element in the list that
-   equivalencia con existsb.
+    satisfies a given predicate:
-   ------------------------------------------------------------------ *)
 </source>

@@ Línea 505: / Línea 505: @@
     ------------------------------------------------------------------ *)
 Theorem beq_nat_true : forall n m,
      beq_nat n m = true -> n = m.
 Proof.
-   (* FILL IN HERE *) Admitted.
+ induction n.
+  + induction m.
 (* ---------------------------------------------------------------------
     Ejemplo de problema por usar intros antes que induction.
@@ Línea 694: / Línea 704: @@
    end.
 Theorem combine_split :
    forall X Y (l : list (X * Y)) l1 l2,
@@ Línea 699: / Línea 710: @@
      combine l1 l2 = l.
 Proof.
-   (* FILL IN HERE *) Admitted.
+   induction l as [|(x, y) l'].
+  + intros l1 l2 h1.
 (* ---------------------------------------------------------------------
     Ejemplo de precauciones al usar destruct para no perder información.
@@ Línea 810: / Línea 831: @@
          beq_nat n m = beq_nat m n.
     ------------------------------------------------------------------ *)
+(* alerodrod5 *)
 Theorem beq_nat_sym :
    forall (n m : nat),
      beq_nat n m = beq_nat m n.
 Proof.
-   (* FILL IN HERE *) Admitted.
+   induction n, m.
+  + reflexivity.
 (* ---------------------------------------------------------------------
     Ejercicio 12. Demostrar que