Razonamiento automático (2016-17) - Contribuciones del usuario [es]

Relación 10

2017-01-29T14:14:41Z

Manmorjim1:

<source lang="isar">
chapter {* R10: Formalización y argumentación con Isabelle/HOL *}

theory R10_Formalizacion_y_argmentacion
imports Main
begin

text {*
---------------------------------------------------------------------
El objetivo de esta es relación formalizar y demostrar la corrección
de los argumentos automáticamente y detalladamente usando sólo las reglas
básicas de deducción natural.

· conjI: ⟦P; Q⟧ ⟹ P ∧ Q
· conjunct1: P ∧ Q ⟹ P
· conjunct2: P ∧ Q ⟹ Q
· notnotD: ¬¬ P ⟹ P
· mp: ⟦P ⟶ Q; P⟧ ⟹ Q
· impI: (P ⟹ Q) ⟹ P ⟶ Q
· disjI1: P ⟹ P ∨ Q
· disjI2: Q ⟹ P ∨ Q
· disjE: ⟦P ∨ Q; P ⟹ R; Q ⟹ R⟧ ⟹ R
· FalseE: False ⟹ P
· notE: ⟦¬P; P⟧ ⟹ R
· notI: (P ⟹ False) ⟹ ¬P
· iffI: ⟦P ⟹ Q; Q ⟹ P⟧ ⟹ P = Q
· iffD1: ⟦Q = P; Q⟧ ⟹ P
· iffD2: ⟦P = Q; Q⟧ ⟹ P
· ccontr: (¬P ⟹ False) ⟹ P

· allI: ⟦∀x. P x; P x ⟹ R⟧ ⟹ R
· allE: (⋀x. P x) ⟹ ∀x. P x
· exI: P x ⟹ ∃x. P x
· exE: ⟦∃x. P x; ⋀x. P x ⟹ Q⟧ ⟹ Q

· refl: t = t
· subst: ⟦s = t; P s⟧ ⟹ P t
· trans: ⟦r = s; s = t⟧ ⟹ r = t
· sym: s = t ⟹ t = s
· not_sym: t ≠ s ⟹ s ≠ t
· ssubst: ⟦t = s; P s⟧ ⟹ P t
· box_equals: ⟦a = b; a = c; b = d⟧ ⟹ a: = d
· arg_cong: x = y ⟹ f x = f y
· fun_cong: f = g ⟹ f x = g x
· cong: ⟦f = g; x = y⟧ ⟹ f x = g y
---------------------------------------------------------------------
*}

text {*
Se usarán las reglas notnotI, mt, no_ex y no_para_todo que demostramos
a continuación.
*}

lemma notnotI: "P ⟹ ¬¬ P"
by auto

lemma mt: "⟦F ⟶ G; ¬G⟧ ⟹ ¬F"
by auto

lemma no_ex: "¬(∃x. P(x)) ⟹ ∀x. ¬P(x)"
by auto

lemma no_para_todo: "¬(∀x. P(x)) ⟹ ∃x. ¬P(x)"
by auto

text {* ---------------------------------------------------------------
Ejercicio 1. Formalizar, y demostrar la corrección, del siguiente
argumento
Si la válvula está abierta o la monitorización está preparada,
entonces se envía una señal de reconocimiento y un mensaje de
funcionamiento al controlador del ordenador. Si se envía un mensaje
de funcionamiento al controlador del ordenador o el sistema está en
estado normal, entonces se aceptan las órdenes del operador. Por lo
tanto, si la válvula está abierta, entonces se aceptan las órdenes
del operador.
Usar A : La válvula está abierta.
P : La monitorización está preparada.
R : Envía una señal de reconocimiento.
F : Envía un mensaje de funcionamiento.
N : El sistema está en estado normal.
O, AO, OK : Se aceptan órdenes del operador.
------------------------------------------------------------------ *}

(* ivamenjim manmorjim1 *)
(* Buscando, he detectado que 'O' es un carácter especial en Isabelle y que forma parte de su
sintaxis pre-definida, por lo que da problemas a la hora de formalizar y demostrar
el argumento planteado. Por lo tanto, en su lugar he usado "AO: Se aceptan órdenes del operador" *)

lemma ejercicio_1:
assumes 1: "(A ∨ P) ⟶ (R ∧ F)"
assumes 2: "(F ∨ N) ⟶ AO"
shows "A ⟶ AO"
proof -
{assume 3: "A"
have 4: "A ∨ P" using 3 by (rule disjI1)
have 5: "R ∧ F" using 1 4 by (rule mp)
have 6: "F" using 5 by (rule conjunct2)
have 7: "F ∨ N" using 6 by (rule disjI1)
have 8: "AO" using 2 7 by (rule mp)}
then show "A ⟶ AO" by (rule impI)
qed

(*danrodcha ferrenseg anaprarod *)

lemma ej_1:
assumes "A ∨ P ⟶ R ∧ F" and
"F ∨ N ⟶ OK"
shows "A ⟶ OK"
proof (rule impI)
{assume "A"
hence "A ∨ P" by (rule disjI1)
with assms(1) have "R ∧ F" by (rule mp)
hence "F" by (rule conjE)
hence "F ∨ N" by (rule disjI1)
with assms(2) show "OK" by (rule mp)}
qed

text {* ---------------------------------------------------------------
Ejercicio 2. Formalizar, y decidir la corrección, del siguiente
argumento
Hay estudiantes inteligentes y hay estudiantes trabajadores. Por
tanto, hay estudiantes inteligentes y trabajadores.
Usar I(x) para x es inteligente
T(x) para x es trabajador
------------------------------------------------------------------ *}

(* ivamenjim ferrenseg danrodcha anaprarod *)

lemma ejercicio_2:
assumes "(∃x. I(x)) ∧ (∃x. T(x))"
shows "∃x. (I(x) ∧ T(x))"
quickcheck
oops

(* Encontrando el contraejemplo:
I = {a1}
x = a1
T = {a2}
xa = a2
*)

text {* ---------------------------------------------------------------
Ejercicio 3. Formalizar, y decidir la corrección, del siguiente
argumento
Los hermanos tienen el mismo padre. Juan es hermano de Luis. Carlos
es padre de Luis. Por tanto, Carlos es padre de Juan.
Usar H(x,y) para x es hermano de y
P(x,y) para x es padre de y
j para Juan
l para Luis
c para Carlos
------------------------------------------------------------------ *}

(* ivamenjim *)

lemma ejercicio_3:
assumes 1: "∀x y. P(x,y) ⟶ (∀z. (H(z,y) ⟶ P(x,z)))"
assumes 2: "H(j,l)"
assumes 3: "P(c,l)"
shows "P(c,j)"
proof -
have 4 : "∀y. P(c,y) ⟶ (∀z. (H(z,y) ⟶ P(c,z)))" using 1 by (rule allE)
have 5 : "P(c,l) ⟶ (∀z. (H(z,l) ⟶ P(c,z)))" using 4 by (rule allE)
then have 6 : "(∀z. (H(z,l) ⟶ P(c,z)))" using 3 by (rule mp)
have 7 : "H(j,l) ⟶ P(c,j)" using 6 by (rule allE)
then show "P(c,j)" using 2 by (rule mp)
qed

(* danrodcha anaprarod *)
(* es casi igual que la anterior *)
lemma ej_3:
assumes "∀x y. P(x,y) ⟶ (∀z. (H(z,y) ⟶ P(x,z)))"
assumes "H(j,l)"
assumes "P(c,l)"
shows "P(c,j)"
proof (rule mp)
have 4 : "∀y. P(c,y) ⟶ (∀z. (H(z,y) ⟶ P(c,z)))" using assms(1) by (rule allE)
hence "P(c,l) ⟶ (∀z. (H(z,l) ⟶ P(c,z)))" by (rule allE)
hence "(∀z. (H(z,l) ⟶ P(c,z)))" using assms(3) by (rule mp)
thus "H(j,l) ⟶ P(c,j)" by (rule allE)
next
show "H(j,l)" using assms(2) by this
qed

text {* ---------------------------------------------------------------
Ejercicio 4. Formalizar, y decidir la corrección, del siguiente
argumento
Los aficionados al fútbol aplauden a cualquier futbolista
extranjero. Juanito no aplaude a futbolistas extranjeros. Por
tanto, si hay algún futbolista extranjero nacionalizado español,
Juanito no es aficionado al fútbol.
Usar Af(x) para x es aficicionado al fútbol
Ap(x,y) para x aplaude a y
E(x) para x es un futbolista extranjero
N(x) para x es un futbolista nacionalizado español
j para Juanito
------------------------------------------------------------------ *}
(* danrodcha *)

lemma ej_4:
assumes "∀x y. Af(x) ∧ E(y) ⟶ Ap(x,y)"
"∀x. Ap(j,x) ⟶ ¬ E(x)"
shows "(∃x. E(x) ∧ N(x)) ⟶ ¬Af(j)"
proof (rule impI)
assume "∃x. E(x) ∧ N(x)"
then obtain a where "E(a) ∧ N(a)" by (rule exE)
hence "E(a)" by (rule conjE)
show "¬ Af(j)"
proof (rule notI)
assume "Af(j)"
hence "Af(j) ∧ E(a)" using `E(a)` by (rule conjI)
have "∀y. Af(j) ∧ E(y) ⟶ Ap(j,y)" using assms(1) by (rule allE)
hence "Af(j) ∧ E(a) ⟶ Ap(j,a)" by (rule allE)
hence "Ap(j,a)" using `Af(j) ∧ E(a)` by (rule mp)
have "Ap(j,a) ⟶ ¬ E(a)" using assms(2) by (rule allE)
hence "¬ E(a)" using `Ap(j,a)` by (rule mp)
thus False using `E(a)` by (rule notE)
qed
qed

(* ivamenjim *)

lemma ejercicio_4:
assumes 1: "∀x y. Af(x) ∧ E(y) ⟶ Ap(x,y)"
assumes 2: "¬(∃x. E(x) ∧ Ap(j,x))"
shows "(∃x. E(x) ∧ N(x)) ⟶ ¬Af(j)"
proof (rule impI)
assume 3: "∃x. E(x) ∧ N(x)"
then obtain a where 4: "E(a) ∧ N(a)" by (rule exE)
then have 5: "E(a)" by (rule conjunct1)
show 6: "¬Af(j)"
proof (rule notI)
assume 7: "Af(j)"
then have 8: "Af(j) ∧ E(a)" using 5 by (rule conjI)
have 9: "∀y. Af(j) ∧ E(y) ⟶ Ap(j,y)" using 1 by (rule allE)
have 10: "Af(j) ∧ E(a) ⟶ Ap(j,a)" using 9 by (rule allE)
have 11: "Ap(j,a)" using 10 8 by (rule mp)
have 12: "E(a) ∧ Ap(j,a)" using 5 11 by (rule conjI)
have 13: "∃x. E(x) ∧ Ap(j,x)" using 12 by (rule exI)
show "False" using 2 13 by (rule notE)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 5. Formalizar, y decidir la corrección, del siguiente
argumento
El esposo de la hermana de Toni es Roberto. La hermana de Toni es
María. Por tanto, el esposo de María es Roberto.
Usar e(x) para el esposo de x
h para la hermana de Toni
m para María
r para Roberto
------------------------------------------------------------------ *}

(* ivamenjim *)

lemma ejercicio_5:
assumes 1: "e(h) = r"
assumes 2: "h = m"
shows "e(m) = r"
proof -
have 3: "e(h) = e(m)" using 2 by (rule arg_cong)
have 4: "e(m) = e(h)" using 3 by (rule sym)
then show "e(m) = r" using 1 by (rule trans)
qed

(* danrodcha *)
lemma ej_5:
assumes "e(h) = r" and
"h = m"
shows "e(m) = r"
proof -
show "e(m) = r" using assms(2) assms(1) by (rule subst)
qed

end
</source>

Relación 10

2017-01-29T14:14:13Z

Manmorjim1:

<source lang="isar">
chapter {* R10: Formalización y argumentación con Isabelle/HOL *}

theory R10_Formalizacion_y_argmentacion
imports Main
begin

text {*
---------------------------------------------------------------------
El objetivo de esta es relación formalizar y demostrar la corrección
de los argumentos automáticamente y detalladamente usando sólo las reglas
básicas de deducción natural.

· conjI: ⟦P; Q⟧ ⟹ P ∧ Q
· conjunct1: P ∧ Q ⟹ P
· conjunct2: P ∧ Q ⟹ Q
· notnotD: ¬¬ P ⟹ P
· mp: ⟦P ⟶ Q; P⟧ ⟹ Q
· impI: (P ⟹ Q) ⟹ P ⟶ Q
· disjI1: P ⟹ P ∨ Q
· disjI2: Q ⟹ P ∨ Q
· disjE: ⟦P ∨ Q; P ⟹ R; Q ⟹ R⟧ ⟹ R
· FalseE: False ⟹ P
· notE: ⟦¬P; P⟧ ⟹ R
· notI: (P ⟹ False) ⟹ ¬P
· iffI: ⟦P ⟹ Q; Q ⟹ P⟧ ⟹ P = Q
· iffD1: ⟦Q = P; Q⟧ ⟹ P
· iffD2: ⟦P = Q; Q⟧ ⟹ P
· ccontr: (¬P ⟹ False) ⟹ P

· allI: ⟦∀x. P x; P x ⟹ R⟧ ⟹ R
· allE: (⋀x. P x) ⟹ ∀x. P x
· exI: P x ⟹ ∃x. P x
· exE: ⟦∃x. P x; ⋀x. P x ⟹ Q⟧ ⟹ Q

· refl: t = t
· subst: ⟦s = t; P s⟧ ⟹ P t
· trans: ⟦r = s; s = t⟧ ⟹ r = t
· sym: s = t ⟹ t = s
· not_sym: t ≠ s ⟹ s ≠ t
· ssubst: ⟦t = s; P s⟧ ⟹ P t
· box_equals: ⟦a = b; a = c; b = d⟧ ⟹ a: = d
· arg_cong: x = y ⟹ f x = f y
· fun_cong: f = g ⟹ f x = g x
· cong: ⟦f = g; x = y⟧ ⟹ f x = g y
---------------------------------------------------------------------
*}

text {*
Se usarán las reglas notnotI, mt, no_ex y no_para_todo que demostramos
a continuación.
*}

lemma notnotI: "P ⟹ ¬¬ P"
by auto

lemma mt: "⟦F ⟶ G; ¬G⟧ ⟹ ¬F"
by auto

lemma no_ex: "¬(∃x. P(x)) ⟹ ∀x. ¬P(x)"
by auto

lemma no_para_todo: "¬(∀x. P(x)) ⟹ ∃x. ¬P(x)"
by auto

text {* ---------------------------------------------------------------
Ejercicio 1. Formalizar, y demostrar la corrección, del siguiente
argumento
Si la válvula está abierta o la monitorización está preparada,
entonces se envía una señal de reconocimiento y un mensaje de
funcionamiento al controlador del ordenador. Si se envía un mensaje
de funcionamiento al controlador del ordenador o el sistema está en
estado normal, entonces se aceptan las órdenes del operador. Por lo
tanto, si la válvula está abierta, entonces se aceptan las órdenes
del operador.
Usar A : La válvula está abierta.
P : La monitorización está preparada.
R : Envía una señal de reconocimiento.
F : Envía un mensaje de funcionamiento.
N : El sistema está en estado normal.
O, AO, OK : Se aceptan órdenes del operador.
------------------------------------------------------------------ *}

(* ivamenjim *)
(* Buscando, he detectado que 'O' es un carácter especial en Isabelle y que forma parte de su
sintaxis pre-definida, por lo que da problemas a la hora de formalizar y demostrar
el argumento planteado. Por lo tanto, en su lugar he usado "AO: Se aceptan órdenes del operador" *)

lemma ejercicio_1:
assumes 1: "(A ∨ P) ⟶ (R ∧ F)"
assumes 2: "(F ∨ N) ⟶ AO"
shows "A ⟶ AO"
proof -
{assume 3: "A"
have 4: "A ∨ P" using 3 by (rule disjI1)
have 5: "R ∧ F" using 1 4 by (rule mp)
have 6: "F" using 5 by (rule conjunct2)
have 7: "F ∨ N" using 6 by (rule disjI1)
have 8: "AO" using 2 7 by (rule mp)}
then show "A ⟶ AO" by (rule impI)
qed

(*danrodcha ferrenseg anaprarod manmorjim1 *)

lemma ej_1:
assumes "A ∨ P ⟶ R ∧ F" and
"F ∨ N ⟶ OK"
shows "A ⟶ OK"
proof (rule impI)
{assume "A"
hence "A ∨ P" by (rule disjI1)
with assms(1) have "R ∧ F" by (rule mp)
hence "F" by (rule conjE)
hence "F ∨ N" by (rule disjI1)
with assms(2) show "OK" by (rule mp)}
qed

text {* ---------------------------------------------------------------
Ejercicio 2. Formalizar, y decidir la corrección, del siguiente
argumento
Hay estudiantes inteligentes y hay estudiantes trabajadores. Por
tanto, hay estudiantes inteligentes y trabajadores.
Usar I(x) para x es inteligente
T(x) para x es trabajador
------------------------------------------------------------------ *}

(* ivamenjim ferrenseg danrodcha anaprarod *)

lemma ejercicio_2:
assumes "(∃x. I(x)) ∧ (∃x. T(x))"
shows "∃x. (I(x) ∧ T(x))"
quickcheck
oops

(* Encontrando el contraejemplo:
I = {a1}
x = a1
T = {a2}
xa = a2
*)

text {* ---------------------------------------------------------------
Ejercicio 3. Formalizar, y decidir la corrección, del siguiente
argumento
Los hermanos tienen el mismo padre. Juan es hermano de Luis. Carlos
es padre de Luis. Por tanto, Carlos es padre de Juan.
Usar H(x,y) para x es hermano de y
P(x,y) para x es padre de y
j para Juan
l para Luis
c para Carlos
------------------------------------------------------------------ *}

(* ivamenjim *)

lemma ejercicio_3:
assumes 1: "∀x y. P(x,y) ⟶ (∀z. (H(z,y) ⟶ P(x,z)))"
assumes 2: "H(j,l)"
assumes 3: "P(c,l)"
shows "P(c,j)"
proof -
have 4 : "∀y. P(c,y) ⟶ (∀z. (H(z,y) ⟶ P(c,z)))" using 1 by (rule allE)
have 5 : "P(c,l) ⟶ (∀z. (H(z,l) ⟶ P(c,z)))" using 4 by (rule allE)
then have 6 : "(∀z. (H(z,l) ⟶ P(c,z)))" using 3 by (rule mp)
have 7 : "H(j,l) ⟶ P(c,j)" using 6 by (rule allE)
then show "P(c,j)" using 2 by (rule mp)
qed

(* danrodcha anaprarod *)
(* es casi igual que la anterior *)
lemma ej_3:
assumes "∀x y. P(x,y) ⟶ (∀z. (H(z,y) ⟶ P(x,z)))"
assumes "H(j,l)"
assumes "P(c,l)"
shows "P(c,j)"
proof (rule mp)
have 4 : "∀y. P(c,y) ⟶ (∀z. (H(z,y) ⟶ P(c,z)))" using assms(1) by (rule allE)
hence "P(c,l) ⟶ (∀z. (H(z,l) ⟶ P(c,z)))" by (rule allE)
hence "(∀z. (H(z,l) ⟶ P(c,z)))" using assms(3) by (rule mp)
thus "H(j,l) ⟶ P(c,j)" by (rule allE)
next
show "H(j,l)" using assms(2) by this
qed

text {* ---------------------------------------------------------------
Ejercicio 4. Formalizar, y decidir la corrección, del siguiente
argumento
Los aficionados al fútbol aplauden a cualquier futbolista
extranjero. Juanito no aplaude a futbolistas extranjeros. Por
tanto, si hay algún futbolista extranjero nacionalizado español,
Juanito no es aficionado al fútbol.
Usar Af(x) para x es aficicionado al fútbol
Ap(x,y) para x aplaude a y
E(x) para x es un futbolista extranjero
N(x) para x es un futbolista nacionalizado español
j para Juanito
------------------------------------------------------------------ *}
(* danrodcha *)

lemma ej_4:
assumes "∀x y. Af(x) ∧ E(y) ⟶ Ap(x,y)"
"∀x. Ap(j,x) ⟶ ¬ E(x)"
shows "(∃x. E(x) ∧ N(x)) ⟶ ¬Af(j)"
proof (rule impI)
assume "∃x. E(x) ∧ N(x)"
then obtain a where "E(a) ∧ N(a)" by (rule exE)
hence "E(a)" by (rule conjE)
show "¬ Af(j)"
proof (rule notI)
assume "Af(j)"
hence "Af(j) ∧ E(a)" using `E(a)` by (rule conjI)
have "∀y. Af(j) ∧ E(y) ⟶ Ap(j,y)" using assms(1) by (rule allE)
hence "Af(j) ∧ E(a) ⟶ Ap(j,a)" by (rule allE)
hence "Ap(j,a)" using `Af(j) ∧ E(a)` by (rule mp)
have "Ap(j,a) ⟶ ¬ E(a)" using assms(2) by (rule allE)
hence "¬ E(a)" using `Ap(j,a)` by (rule mp)
thus False using `E(a)` by (rule notE)
qed
qed

(* ivamenjim *)

lemma ejercicio_4:
assumes 1: "∀x y. Af(x) ∧ E(y) ⟶ Ap(x,y)"
assumes 2: "¬(∃x. E(x) ∧ Ap(j,x))"
shows "(∃x. E(x) ∧ N(x)) ⟶ ¬Af(j)"
proof (rule impI)
assume 3: "∃x. E(x) ∧ N(x)"
then obtain a where 4: "E(a) ∧ N(a)" by (rule exE)
then have 5: "E(a)" by (rule conjunct1)
show 6: "¬Af(j)"
proof (rule notI)
assume 7: "Af(j)"
then have 8: "Af(j) ∧ E(a)" using 5 by (rule conjI)
have 9: "∀y. Af(j) ∧ E(y) ⟶ Ap(j,y)" using 1 by (rule allE)
have 10: "Af(j) ∧ E(a) ⟶ Ap(j,a)" using 9 by (rule allE)
have 11: "Ap(j,a)" using 10 8 by (rule mp)
have 12: "E(a) ∧ Ap(j,a)" using 5 11 by (rule conjI)
have 13: "∃x. E(x) ∧ Ap(j,x)" using 12 by (rule exI)
show "False" using 2 13 by (rule notE)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 5. Formalizar, y decidir la corrección, del siguiente
argumento
El esposo de la hermana de Toni es Roberto. La hermana de Toni es
María. Por tanto, el esposo de María es Roberto.
Usar e(x) para el esposo de x
h para la hermana de Toni
m para María
r para Roberto
------------------------------------------------------------------ *}

(* ivamenjim *)

lemma ejercicio_5:
assumes 1: "e(h) = r"
assumes 2: "h = m"
shows "e(m) = r"
proof -
have 3: "e(h) = e(m)" using 2 by (rule arg_cong)
have 4: "e(m) = e(h)" using 3 by (rule sym)
then show "e(m) = r" using 1 by (rule trans)
qed

(* danrodcha *)
lemma ej_5:
assumes "e(h) = r" and
"h = m"
shows "e(m) = r"
proof -
show "e(m) = r" using assms(2) assms(1) by (rule subst)
qed

end
</source>

Relación 8

2017-01-16T22:37:47Z

Manmorjim1:

<source lang="isar">
chapter {* R8: Deducción natural proposicional en Isabelle/HOL *}

theory R8_Deduccion_natural_proposicional
imports Main
begin

text {*
Demostrar o refutar los siguientes lemas usando sólo las reglas
básicas de deducción natural de la lógica proposicional, de los
cuantificadores y de la igualdad:
· conjI: ⟦P; Q⟧ ⟹ P ∧ Q
· conjunct1: P ∧ Q ⟹ P
· conjunct2: P ∧ Q ⟹ Q
· notnotD: ¬¬ P ⟹ P
· mp: ⟦P ⟶ Q; P⟧ ⟹ Q
· impI: (P ⟹ Q) ⟹ P ⟶ Q
· disjI1: P ⟹ P ∨ Q
· disjI2: Q ⟹ P ∨ Q
· disjE: ⟦P ∨ Q; P ⟹ R; Q ⟹ R⟧ ⟹ R
· FalseE: False ⟹ P
· notE: ⟦¬P; P⟧ ⟹ R
· notI: (P ⟹ False) ⟹ ¬P
· iffI: ⟦P ⟹ Q; Q ⟹ P⟧ ⟹ P = Q
· iffD1: ⟦Q = P; Q⟧ ⟹ P
· iffD2: ⟦P = Q; Q⟧ ⟹ P
· ccontr: (¬P ⟹ False) ⟹ P

· allI: ⟦∀x. P x; P x ⟹ R⟧ ⟹ R
· allE: (⋀x. P x) ⟹ ∀x. P x
· exI: P x ⟹ ∃x. P x
· exE: ⟦∃x. P x; ⋀x. P x ⟹ Q⟧ ⟹ Q

· refl: t = t
· subst: ⟦s = t; P s⟧ ⟹ P t
· trans: ⟦r = s; s = t⟧ ⟹ r = t
· sym: s = t ⟹ t = s
· not_sym: t ≠ s ⟹ s ≠ t
· ssubst: ⟦t = s; P s⟧ ⟹ P t
· box_equals: ⟦a = b; a = c; b = d⟧ ⟹ a: = d
· arg_cong: x = y ⟹ f x = f y
· fun_cong: f = g ⟹ f x = g x
· cong: ⟦f = g; x = y⟧ ⟹ f x = g y
*}

text {*
Se usarán las reglas notnotI, mt y not_ex que demostramos a continuación.
*}

lemma notnotI: "P ⟹ ¬¬ P"
by auto

lemma mt: "⟦F ⟶ G; ¬G⟧ ⟹ ¬F"
by auto

lemma no_ex: "¬(∃x. P(x)) ⟹ ∀x. ¬P(x)"
by auto

text {* ---------------------------------------------------------------
Ejercicio 1. Demostrar
¬q ⟶ ¬p ⊢ p ⟶ q
------------------------------------------------------------------ *}
(* marcarmor13 *)
--"usando un supuesto ¬¬p"
lemma ejercicio_1:
assumes 1: "¬q ⟶ ¬p" and
2: "¬¬p"
shows "p ⟶ q"
proof -
have 3: "¬¬q" using 1 2 by (rule mt)
have 4: "q" using 3 by (rule notnotD)
show "p ⟶ q" using 4 by (rule impI)
qed

(*pablucoto jeamacpov *)
lemma ejercicio_1_2:
assumes "¬q ⟶ ¬p"
shows "p ⟶ q"
proof -
{assume "p"
hence "¬¬p" by (rule notnotI)
with `¬q ⟶ ¬p` have "¬¬q" by (rule mt)
hence "q" by (rule notnotD)}
then show "p ⟶ q" by (rule impI)
qed

(* ivamenjim serrodcal anaprarod marpoldia1 manmorjim1 *)
lemma ejercicio_1_3:
assumes 1: "¬q ⟶ ¬p"
shows "p ⟶ q"
proof -
{assume 2:"p"
then have 3: "¬¬p" by (rule notnotI)
have 4: "¬¬q" using 1 3 by (rule mt)
then have 5: "q" by (rule notnotD)}
thus "p ⟶ q" by (rule impI)
qed

(* bowma *)
lemma ejercicio_1_4:
assumes "¬q ⟶ ¬p"
shows "p ⟶ q"
proof -
{assume "p"
hence "¬¬p" by (rule notnotI)
with assms have "¬¬q" by (rule mt)
then have "q" by (rule notnotD)}
thus "p ⟶ q" by (rule impI)
qed

（* bowma danrodcha *)
"quita la limitación de -"
lemma ejercicio_1_5:
assumes "¬q ⟶ ¬p"
shows "p ⟶ q"
proof
assume "p"
hence "¬¬p" by (rule notnotI)
with assms have "¬¬q" by (rule mt)
thus "q" by (rule notnotD)
qed

text {* ---------------------------------------------------------------
Ejercicio 2. Demostrar
¬(¬p ∧ ¬q) ⊢ p ∨ q
------------------------------------------------------------------ *}
(* marcarmor13 *)
--"usando un supuesto ¬p ∧ ¬q"
lemma ejercicio_2:
assumes 1: "¬(¬p ∧ ¬q)" and
2:"¬p ∧ ¬q"
shows "p ∨ q"
proof-
have 3: "p"using 1 2 by (rule notE)
show "p ∨ q" using 3 by (rule disjI1)
qed

(* ivamenjim serrodcal marpoldia1 *)
lemma ejercicio_2_2:
assumes 1: "¬(¬p ∧ ¬q)"
shows "p ∨ q"
proof -
{assume 2:"(¬p ∧ ¬q)"
have "p" using 1 2 by (rule notE)
then have "p ∨ q" by (rule disjI1)}
thus "p ∨ q" by auto
qed

(* pablucoto jeamacpov *)

lemma aux_ejercicio2:
assumes "¬(p ∨ q)"
shows "¬p ∧ ¬q"
proof
{assume "p"
hence "p ∨ q" by (rule disjI1)
with `¬(p ∨ q)` have "False" by (rule notE)}
then show "¬p" ..
next
{assume "q"
hence "p ∨ q" by (rule disjI2)
with `¬(p ∨ q)` have "False" by (rule notE)}
then show "¬q" ..
qed

lemma ejercicio_2_3:
assumes "¬(¬p ∧ ¬q)"
shows "p ∨ q"
proof -
{assume 2:"¬(p ∨ q)"
hence "¬p ∧ ¬q" by (rule aux_ejercicio2)
with `¬(¬p ∧ ¬q)` have "False" ..}
then show "p ∨ q" by (rule ccontr)
qed

(* danrodcha *)
lemma ej_2:
assumes "¬(¬p ∧ ¬q)"
shows "p ∨ q"
proof -
have "¬p ∨ p" by (rule excluded_middle)
thus "p ∨ q"
proof (rule disjE)
{ assume "¬p"
have "¬q ∨ q" by (rule excluded_middle)
thus "p ∨ q"
proof (rule disjE)
{ assume "¬q"
with `¬p` have "¬p ∧ ¬q" by (rule conjI)
with assms show "p ∨ q" by (rule notE)}
next
{ assume "q"
then show "p ∨ q" by (rule disjI2)}
qed}
next
{ assume "p"
thus "p ∨ q" by (rule disjI1)}
qed
qed

(* anaprarod *)
(* Igual que el anterior pero con etiquetas *)
lemma ejercicio_2:
assumes 0: "¬(¬p ∧ ¬q)"
shows "p ∨ q"
proof-
have "¬p ∨ p" by (rule excluded_middle)
thus "p ∨ q"
proof (rule disjE)
{assume 1: "¬p"
have "¬q ∨ q" by (rule excluded_middle)
thus "p ∨ q"
proof (rule disjE)
{assume 2: "¬q"
have 3: "(¬p ∧ ¬q)" using 1 2 by (rule conjI)
have "p ∨ q" using 0 3 by (rule notE)
thus "p ∨ q" by this}
next
{assume 4: "q"
have "p ∨ q" using 4 by (rule disjI2)
thus "p ∨ q" by this}
qed}
next
{assume 5: "p"
have "p ∨ q" using 5 by (rule disjI1)
thus "p ∨ q" by this}
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 3. Demostrar
¬(¬p ∨ ¬q) ⊢ p ∧ q
------------------------------------------------------------------ *}
(* marcarmor13 serrodcal marpoldia1 *)
--"usando un supuesto ¬p ∨ ¬q"
lemma ejercicio_3:
assumes 1: "¬(¬p ∨ ¬q)" and
2:"¬p ∨ ¬q"
shows "p ∧ q"
proof-
have 3: "p"using 1 2 by (rule notE)
have 4: "q"using 1 2 by (rule notE)
show "p ∧ q" using 3 4 by (rule conjI)
qed

(* pablucoto jeamacpov *)

lemma ejercicio_3_2:
assumes "¬(¬p ∨ ¬q)"
shows "p ∧ q"
proof
have "¬¬p ∧ ¬¬q" using assms(1) by (rule aux_ejercicio2)
hence "¬¬p" by (rule conjunct1)
show "p" using `¬¬p` by (rule notnotD)
next
have "¬¬p ∧ ¬¬q" using assms(1) by (rule aux_ejercicio2)
have "¬¬q" using `¬¬p ∧ ¬¬q` by (rule conjunct2)
show "q" using `¬¬q` by (rule notnotD)
qed

(* ivamenjim *)
lemma aux: "¬(p ∨ q) ⟹ ¬p ∧ ¬q"
by auto

lemma ejercicio_3_3:
assumes 1: "¬(¬p ∨ ¬q)"
shows "p ∧ q"
proof
have 2: "¬¬p ∧ ¬¬q" using 1 by (rule aux)
have 3: "¬¬p" using 2 ..
have 4: "¬¬q" using 2 ..
show "p" using 3 by (rule notnotD)
show "q" using 4 by (rule notnotD)
qed

(* danrodcha *)
lemma ej_3:
assumes "¬(¬p ∨ ¬q)"
shows "p ∧ q"
proof (rule conjI)
{ assume "¬p"
hence "¬p ∨ ¬q" by (rule disjI1)
with assms have False by (rule notE)}
then show "p" by (rule ccontr)
{ assume "¬q"
hence "¬p ∨ ¬q" by (rule disjI2)
with assms have False by (rule notE)}
then show "q" by (rule ccontr)
qed

(* anaprarod *)
(* Igual que el anterior pero con etiquetas *)
lemma ejercicio_3:
assumes "¬(¬p ∨ ¬q)"
shows "p ∧ q"
proof (rule conjI)
{assume 1: "¬p"
hence 2: "¬p ∨ ¬q" by (rule disjI1)
have "False" using assms 2 by (rule notE)}
thus 3: "p" by (rule ccontr)
{assume 4: "¬q"
hence 5: "¬p ∨ ¬q" by (rule disjI2)
have "False" using assms 5 by (rule notE)}
thus 6: "q" by (rule ccontr)
qed

text {* ---------------------------------------------------------------
Ejercicio 4. Demostrar
¬(p ∧ q) ⊢ ¬p ∨ ¬q
------------------------------------------------------------------ *}
(* marcarmor13 serrodcal *)
--"usando un supuesto p ∧ q"
lemma ejercicio_4:
assumes 1: " ¬(p ∧ q)" and
2:"p ∧ q"
shows "¬p ∨ ¬q"
proof-
have 3: "¬p"using 1 2 by (rule notE)
show "¬p ∨ ¬q" using 3 by (rule disjI1)
qed

(* ivamenjim marpoldia1 *)
lemma ejercicio_4:
assumes 1: "¬(p ∧ q)"
shows "¬p ∨ ¬q"
proof -
{assume 2:"(p ∧ q)"
have "¬p" using 1 2 by (rule notE)
then have "¬p ∨ ¬q" by (rule disjI1)}
thus "¬p ∨ ¬q" by auto
qed

( * pablucoto jeamacpov *)
lemma ejercicio_4:
assumes " ¬(p ∧ q)"
shows " ¬p ∨ ¬q"
proof -
{ assume 2: "¬(¬p ∨ ¬q)"
hence "p ∧ q" by (rule ejercicio_3_2)
with assms(1) have "False" .. }
then show " ¬p ∨ ¬q" by (rule ccontr)
qed

(*danrodcha anaprarod*)
lemma ej_4:
assumes "¬(p ∧ q)"
shows "¬p ∨ ¬q"
proof (rule ccontr)
assume "¬ (¬ p ∨ ¬ q)"
hence "p ∧ q" by (rule ej_3)
with assms show False by (rule notE)
qed

(* anaprarod *)
(* sin usar el ejercicio anterior *)
lemma ejercicio_4:
assumes "¬(p ∧ q)"
shows "¬p ∨ ¬q"
proof -
have "¬p ∨ p" by (rule excluded_middle)
thus "¬p ∨ ¬q"
proof (rule disjE)
{assume "¬p"
thus "¬p ∨ ¬q" by (rule disjI1)}
next
{assume 1:"p"
have "¬q ∨ q" by (rule excluded_middle)
thus "¬p ∨ ¬q"
proof (rule disjE)
{assume "¬q"
thus "¬p ∨ ¬q" by (rule disjI2)}
next
{assume 2:"q"
have 3:"p ∧ q" using 1 2 by (rule conjI)
have "¬p ∨ ¬q" using assms 3 by (rule notE)
thus "¬p ∨ ¬q" by this}
qed}
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 5. Demostrar
⊢ (p ⟶ q) ∨ (q ⟶ p)
------------------------------------------------------------------ *}
(* marcarmor13 jeamacpov serrodcal *)
--"usando un supuesto q"
lemma ejercicio_5:
assumes 1: "q"

shows "(p ⟶ q) ∨ (q ⟶ p)"
proof-
have 2: "p ⟶ q" using 1 by (rule impI)
show "(p ⟶ q) ∨ (q ⟶ p)" using 2 by (rule disjI1)
qed

(* ivamenjim marpoldia1 *)
lemma ejercicio_5:
shows "(p ⟶ q) ∨ (q ⟶ p)"
proof -
{assume 1:"q"
have "(p ⟶ q)" using 1 by (rule impI)
then have "(p ⟶ q) ∨ (q ⟶ p)" by (rule disjI1)}
thus "(p ⟶ q) ∨ (q ⟶ p)" by auto
qed

(* danrodcha pablucoto *)
lemma ej_5:
shows "(p ⟶ q) ∨ (q ⟶ p)"
proof -
have "¬p ∨ p" by (rule excluded_middle)
thus "(p ⟶ q) ∨ (q ⟶ p)"
proof (rule disjE)
{ assume "¬p"
have "¬q ∨ q" by (rule excluded_middle)
thus "(p ⟶ q) ∨ (q ⟶ p)"
proof (rule disjE)
{assume "¬q"
hence "¬p ⟶ ¬q" by (rule impI)
{ assume "q"
hence "¬¬q" by (rule notnotI)
with `¬p ⟶ ¬q` have "¬¬p" by (rule mt)
hence "p" by (rule notnotD)}
hence "q ⟶ p" by (rule impI)
thus "(p ⟶ q) ∨ (q ⟶ p)" by (rule disjI2)}
next
{assume "q"
hence "(p ⟶ q)" by (rule impI)
thus "(p ⟶ q) ∨ (q ⟶ p)" by (rule disjI1)}
qed}
next
{assume "p"
hence "q ⟶ p" by (rule impI)
thus "(p ⟶ q) ∨ (q ⟶ p)" by (rule disjI2)}
qed
qed

end
</source>

Relación 7

2016-12-20T21:40:19Z

Manmorjim1:

<source lang="isar">
chapter {* R7: Árboles binarios completos *}

theory R7_Arboles_binarios_completos
imports Main
begin

text {*
En esta relación se piden demostraciones automáticas (lo más cortas
posibles). Para ello, en algunos casos es necesario incluir lemas
auxiliares (que se demuestran automáticamente) y usar ejercicios
anteriores.

---------------------------------------------------------------------
Ejercicio 1. Definir el tipo de datos arbol para representar los
árboles binarios que no tienen información ni en los nodos y ni en las
hojas. Por ejemplo, el árbol
·
/ \
/ \
· ·
/ \ / \
· · · ·
se representa por "N (N H H) (N H H)".
---------------------------------------------------------------------
*}

datatype arbol = H | N arbol arbol

value "N (N H H) (N H H) = (N (N H H) (N H H) :: arbol)"

text {*
---------------------------------------------------------------------
Ejercicio 2. Definir la función
hojas :: "arbol => nat"
tal que (hojas a) es el número de hojas del árbol a. Por ejemplo,
hojas (N (N H H) (N H H)) = 4
---------------------------------------------------------------------
*}

(* fraortmoy marcarmor13 josgarsan*)
fun hojas :: "arbol => nat" where
"hojas H = Suc 0"
| "hojas (N a b) = hojas a + hojas b"

value "hojas (N (N H H) (N H H)) = 4"

(* marpoldia1 anaprarod paupeddeg migtermor wilmorort pablucoto
ivamenjim serrodcal crigomgom rubgonmar danrodcha ferrenseg manmorjim1 juacabsou *)
(* Es muy parecida a la definición anterior *)
fun hojas2 :: "arbol => nat" where
"hojas2 H = 1" |
"hojas2 (N i d) = hojas2 i + hojas2 d"

value "hojas2 (N (N H H) (N H H)) = 4"

(* anaprarod *)
(* Equivalencia de las definiciones *)
lemma "hojas a = hojas2 a"
by (induct a) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 4. Definir la función
profundidad :: "arbol => nat"
tal que (profundidad a) es la profundidad del árbol a. Por ejemplo,
profundidad (N (N H H) (N H H)) = 2
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 anaprarod migtermor wilmorort marcarmor13*)
fun profundidad :: "arbol => nat" where
"profundidad H = 0"
| "profundidad (N a b) = (if profundidad a > profundidad b
then 1 + profundidad a
else 1 + profundidad b)"

value "profundidad (N (N H H) (N H H)) = 2"

(* anaprarod wilmorort pablucoto ivamenjim serrodcal crigomgom rubgonmar
danrodcha ferrenseg josgarsan juacabsou *)
fun profundidad2 :: "arbol => nat" where
"profundidad2 H = 0"
|"profundidad2 (N i d) = 1 + (max (profundidad2 i) (profundidad2 d))"

value "profundidad2 (N (N H H) (N H H)) = 2"

(* anaprarod *)
(* Equivalencia de las definiciones *)
lemma "profundidad a= profundidad2 a"
by (induct a) auto

(* paupeddeg *)
fun maximo :: "nat × nat => nat" where
"maximo (a,b) = (if a > b
then a else b)"

fun profundidad :: "arbol => nat" where
"profundidad H = 0"
| "profundidad (N i d) = 1 + maximo(profundidad i, profundidad d)"

(* ivamenjim: llamando a la función anterior profundidad3 *)

lemma "profundidad a = profundidad3 a"
by (induct a) auto

(* ivamenjim manmorjim1 *)
fun profundidad4 :: "arbol => nat" where
"profundidad4 H = 0"
| "profundidad4 (N i d) = Suc (max (profundidad4 i)(profundidad4 d))"

lemma "profundidad a = profundidad4 a"
by (induct a) auto

text {*
---------------------------------------------------------------------
Ejercicio 5. Definir la función
abc :: "nat ⇒ arbol"
tal que (abc n) es el árbol binario completo de profundidad n. Por
ejemplo,
abc 3 = N (N (N H H) (N H H)) (N (N H H) (N H H))
---------------------------------------------------------------------
*}
(* fraortmoy marpoldia1 anaprarod paupeddeg migtermor wilmorort
serrodcal crigomgom rubgonmar danrodcha ferrenseg josgarsan manmorjim1 juacabsou *)
fun abc :: "nat ⇒ arbol" where
"abc 0 = H"
| "abc (Suc n) = (N (abc n) (abc n))"

value "abc 3 = N (N (N H H) (N H H)) (N (N H H) (N H H))"

(* ivamenjim pablucoto marcarmor13*)
fun abc2 :: "nat ⇒ arbol" where
"abc2 0 = H"
| "abc2 t = N (abc2 (t-1)) (abc2 (t-1))"

value "abc2 3 = N (N (N H H) (N H H)) (N (N H H) (N H H))"

(* ivamenjim: Metaejercicio de demostración *)
lemma "abc t = abc2 t"
by (induct t) auto

text {*
---------------------------------------------------------------------
Ejercicio 6. Un árbol binario a es completo respecto de la medida f si
a es una hoja o bien a es de la forma (N i d) y se cumple que tanto i
como d son árboles binarios completos respecto de f y, además,
f(i) = f(r).

Definir la función
es_abc :: "(arbol => 'a) => arbol => bool
tal que (es_abc f a) se verifica si a es un árbol binario completo
respecto de f.
---------------------------------------------------------------------
*}

(* fraortmoy anaprarod migtermor serrodcal crigomgom rubgonmar
danrodcha ferrenseg juacabsou *)
fun es_abc :: "(arbol => 'a) => arbol => bool" where
"es_abc _ H = True"
| "es_abc f (N a b) = (es_abc f a ∧ es_abc f b ∧ (f a = f b))"

(* marpoldia1 paupeddeg ivamenjim pablucoto marcarmor13 manmorjim1 *)
fun es_abc2 :: "(arbol => 'a) => arbol => bool" where
"es_abc2 f H = True" |
"es_abc2 f (N i d) = ((f i = f d) ∧ (es_abc2 f i) ∧ (es_abc2 f d))"

(* anaprarod *)
(* Equivalencia de las definiciones *)
lemma "es_abc f a = es_abc2 f a"
by (induct a) auto

text {*
---------------------------------------------------------------------
Nota. (size a) es el número de nodos del árbol a. Por ejemplo,
size (N (N H H) (N H H)) = 3
---------------------------------------------------------------------
*}

value "size (N (N H H) (N H H)) = 3"
value "size (N (N (N H H) (N H H)) (N (N H H) (N H H))) = 7"

text {*
---------------------------------------------------------------------
Nota. Tenemos 3 funciones de medida sobre los árboles: número de
hojas, número de nodos y profundidad. A cada una le corresponde un
concepto de completitud. En los siguientes ejercicios demostraremos
que los tres conceptos de completitud son iguales.
---------------------------------------------------------------------
*}

text {*
---------------------------------------------------------------------
Ejercicio 7. Demostrar que un árbol binario a es completo respecto de
la profundidad syss es completo respecto del número de hojas.
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 migtermor wilmorort pablucoto serrodcal *)

lemma abc_prof_num_hojas:
assumes "es_abc profundidad a"
shows "hojas a = 2^(profundidad a)"
using assms
by (induct a) auto

(* Otra forma de escribir lo mismo *)
(* anaprarod crigomgom ivamenjim paupeddeg juacabsou *)
lemma AUX7: "es_abc profundidad a ⟶ (hojas a = 2^(profundidad a))"
by (induct a) simp_all

(* Otra forma de escribir lo mismo *)
(* danrodcha *)
lemma aux1: "es_abc profundidad (a::arbol) ⟹ (hojas a = 2^ (profundidad a))"
by (induct a) simp_all

(* fraortmoy marpoldia1 migtermor anaprarod wilmorort serrodcal crigomgom
rubgonmar ivamenjim danrodcha marcarmor13 paupeddeg juacabsou*)
(* También funciona con AUX7 *)

lemma lej7: "es_abc profundidad a = es_abc hojas a"
by (induct a) (auto simp add: abc_prof_num_hojas)

(* danrodcha *)
lemma 7: "es_abc profundidad a = es_abc hojas a"
apply (induct a)
apply simp
apply (auto simp add: aux1)
done

(* ferrenseg *)
lemma [simp]: "es_abc profundidad a ⟶ hojas a = 2 ^ (profundidad a)"
by (induct a) auto

theorem es_abc_profundidad_hojas: "es_abc profundidad a = es_abc hojas a"
by (induct a) auto

text {*
---------------------------------------------------------------------
Ejercicio 8. Demostrar que un árbol binario a es completo respecto del
número de hojas syss es completo respecto del número de nodos.
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 migtermor wilmorort pablucoto serrodcal marcarmor13*)

lemma abc_hojas_num_nodos:
assumes "es_abc hojas a"
shows "Suc(size a) = hojas a"
using assms
by (induct a) auto

(* Otra forma de escribir lo mismo *)
(* anaprarod crigomgom paupeddeg juacabsou*)
lemma AUX8: "es_abc hojas a ⟶ (hojas a = (Suc (size a)))"
by (induct a) auto

(* fraortmoy marpoldia1 migtermor anaprarod wilmorort pablucoto serrodcal *)

lemma lej8: "es_abc hojas a = es_abc size a"
by (induct a) (auto simp add:abc_hojas_num_nodos [symmetric])

(* anaprarod crigomgom paupeddeg juacabsou*)
(* Usando AUX8 *)
lemma L8: "es_abc hojas a= es_abc size a"
by (induct a) (auto simp add: AUX8)

(* ivamenjim *)
(* Teorema auxiliar *)
lemma auxEj8: "hojas a = size a + 1"
by (induct a) auto

(* ivamenjim *)
lemma lej8: "es_abc hojas a = es_abc size a"
by (induct a) (auto simp add: auxEj8)

(* danrodcha *)
lemma aux3: "es_abc hojas a ⟹ (hojas a = 1 + size a)"
by (induct a) simp_all

(* danrodcha *)
lemma 8: "es_abc hojas a = es_abc size a"
apply (induct a)
apply simp
apply (auto simp add: aux3)
done

(* ferrenseg *)
lemma [simp]: "hojas a = size a + 1"
by (induct a) auto

theorem es_abc_hojas_size: "es_abc hojas a = es_abc size a"
by (induct a) auto

text {*
---------------------------------------------------------------------
Ejercicio 9. Demostrar que un árbol binario a es completo respecto de
la profundidad syss es completo respecto del número de nodos.
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 migtermor anaprarod wilmorort pablucoto
serrodcal crigomgom rubgonmar danrodcha ivamenjim marcarmor13 paupeddeg juacabsou
*)

lemma lej9: "es_abc profundidad a = es_abc size a"
by (simp add: lej7 lej8)

(* ferrenseg *)

corollary es_abc_size_profundidad: "es_abc size a = es_abc profundidad a"
by (simp add: es_abc_profundidad_hojas es_abc_hojas_size)

text {*
---------------------------------------------------------------------
Ejercicio 10. Demostrar que (abc n) es un árbol binario completo.
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 migtermor wilmorort pablucoto serrodcal crigomgom marcarmor13 paupeddeg juacabsou*)

lemma lej10: "es_abc profundidad (abc n)"
by (induct n) auto

(* anaprarod rubgonmar danrodcha ferrenseg ivamenjim paupeddeg juacabsou*)
(* con un demostrador más débil *)
(* y en general para cualquier medida *)
lemma L10: "es_abc f (abc a)"
by (induct a) simp_all

(* ivamenjim *)
(* Igual que el anterior pero usando auto *)
lemma lej10: "es_abc f (abc n)"
by (induct n) auto

text {*
---------------------------------------------------------------------
Ejercicio 11. Demostrar que si a es un árbolo binario completo
respecto de la profundidad, entonces a es igual a
(abc (profundidad a)).
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 migtermor wilmorort pablucoto serrodcal marcarmor13 *)

lemma lej11:
assumes " es_abc profundidad a"
shows "a = (abc (profundidad a))"
using assms
by (induct a) auto

(* Otra forma de escribir lo mismo *)
(* anaprarod crigomgom rubgonmar ferrenseg ivamenjim paupeddeg juacabsou *)
lemma "es_abc profundidad a ⟶ (a = (abc (profundidad a)))"
by (induct a) auto

(* danrodcha*)
lemma 11:"es_abc profundidad a ⟹ (a = (abc (profundidad a)))"
by (induct a) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 12. Encontrar una medida f tal que (es_abc f) es distinto de
(es_abc size).
---------------------------------------------------------------------
*}

(* migtermor *)
fun medida_nula :: "arbol => nat" where
"medida_nula H = 0"
| "medida_nula (N i d) = 0"

lemma "es_abc medida_nula a = es_abc size a"
quickcheck
oops
(* Quickcheck encuentra el siguiente contraejemplo:
a= N H (N H H)
Tras evaluar:
es_abc medida_nula a = True
es_abc size a = False*)

(* anaprarod wilmorort pablucoto serrodcal danrodcha marcarmor13 ferrenseg ivamenjim paupeddeg juacabsou *)
lemma "es_abc f a = es_abc size a"
quickcheck
(* Quickcheck found a counterexample:
f = λx. a⇩1
a = N H (N H H)
Evaluated terms:
es_abc f a = True
es_abc size a = False *)
oops
(* el contraejemplo que encuentra es la medida constante a1 *)

(*crigomgom *)
(* Como en la primera de las soluciones he usado la función constante 0 pero he usado una expresión lambda*)
lemma "es_abc (λx. 0::nat) a = es_abc size a"
quickcheck
oops

end
</source>

Relación 7

2016-12-20T20:56:21Z

Manmorjim1:

<source lang="isar">
chapter {* R7: Árboles binarios completos *}

theory R7_Arboles_binarios_completos
imports Main
begin

text {*
En esta relación se piden demostraciones automáticas (lo más cortas
posibles). Para ello, en algunos casos es necesario incluir lemas
auxiliares (que se demuestran automáticamente) y usar ejercicios
anteriores.

---------------------------------------------------------------------
Ejercicio 1. Definir el tipo de datos arbol para representar los
árboles binarios que no tienen información ni en los nodos y ni en las
hojas. Por ejemplo, el árbol
·
/ \
/ \
· ·
/ \ / \
· · · ·
se representa por "N (N H H) (N H H)".
---------------------------------------------------------------------
*}

datatype arbol = H | N arbol arbol

value "N (N H H) (N H H) = (N (N H H) (N H H) :: arbol)"

text {*
---------------------------------------------------------------------
Ejercicio 2. Definir la función
hojas :: "arbol => nat"
tal que (hojas a) es el número de hojas del árbol a. Por ejemplo,
hojas (N (N H H) (N H H)) = 4
---------------------------------------------------------------------
*}

(* fraortmoy marcarmor13 josgarsan*)
fun hojas :: "arbol => nat" where
"hojas H = Suc 0"
| "hojas (N a b) = hojas a + hojas b"

value "hojas (N (N H H) (N H H)) = 4"

(* marpoldia1 anaprarod paupeddeg migtermor wilmorort pablucoto
ivamenjim serrodcal crigomgom rubgonmar danrodcha ferrenseg manmorjim1 *)
(* Es muy parecida a la definición anterior *)
fun hojas2 :: "arbol => nat" where
"hojas2 H = 1" |
"hojas2 (N i d) = hojas2 i + hojas2 d"

value "hojas2 (N (N H H) (N H H)) = 4"

(* anaprarod *)
(* Equivalencia de las definiciones *)
lemma "hojas a = hojas2 a"
by (induct a) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 4. Definir la función
profundidad :: "arbol => nat"
tal que (profundidad a) es la profundidad del árbol a. Por ejemplo,
profundidad (N (N H H) (N H H)) = 2
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 anaprarod migtermor wilmorort marcarmor13*)
fun profundidad :: "arbol => nat" where
"profundidad H = 0"
| "profundidad (N a b) = (if profundidad a > profundidad b
then 1 + profundidad a
else 1 + profundidad b)"

value "profundidad (N (N H H) (N H H)) = 2"

(* anaprarod wilmorort pablucoto ivamenjim serrodcal crigomgom rubgonmar
danrodcha ferrenseg josgarsan *)
fun profundidad2 :: "arbol => nat" where
"profundidad2 H = 0"
|"profundidad2 (N i d) = 1 + (max (profundidad2 i) (profundidad2 d))"

value "profundidad2 (N (N H H) (N H H)) = 2"

(* anaprarod *)
(* Equivalencia de las definiciones *)
lemma "profundidad a= profundidad2 a"
by (induct a) auto

(* paupeddeg *)
fun maximo :: "nat × nat => nat" where
"maximo (a,b) = (if a > b
then a else b)"

fun profundidad :: "arbol => nat" where
"profundidad H = 0"
| "profundidad (N i d) = 1 + maximo(profundidad i, profundidad d)"

(* ivamenjim: llamando a la función anterior profundidad3 *)

lemma "profundidad a = profundidad3 a"
by (induct a) auto

(* ivamenjim manmorjim1 *)
fun profundidad4 :: "arbol => nat" where
"profundidad4 H = 0"
| "profundidad4 (N i d) = Suc (max (profundidad4 i)(profundidad4 d))"

lemma "profundidad a = profundidad4 a"
by (induct a) auto

text {*
---------------------------------------------------------------------
Ejercicio 5. Definir la función
abc :: "nat ⇒ arbol"
tal que (abc n) es el árbol binario completo de profundidad n. Por
ejemplo,
abc 3 = N (N (N H H) (N H H)) (N (N H H) (N H H))
---------------------------------------------------------------------
*}
(* fraortmoy marpoldia1 anaprarod paupeddeg migtermor wilmorort
serrodcal crigomgom rubgonmar danrodcha ferrenseg josgarsan manmorjim1 *)
fun abc :: "nat ⇒ arbol" where
"abc 0 = H"
| "abc (Suc n) = (N (abc n) (abc n))"

value "abc 3 = N (N (N H H) (N H H)) (N (N H H) (N H H))"

(* ivamenjim pablucoto marcarmor13*)
fun abc2 :: "nat ⇒ arbol" where
"abc2 0 = H"
| "abc2 t = N (abc2 (t-1)) (abc2 (t-1))"

value "abc2 3 = N (N (N H H) (N H H)) (N (N H H) (N H H))"

(* ivamenjim: Metaejercicio de demostración *)
lemma "abc t = abc2 t"
by (induct t) auto

text {*
---------------------------------------------------------------------
Ejercicio 6. Un árbol binario a es completo respecto de la medida f si
a es una hoja o bien a es de la forma (N i d) y se cumple que tanto i
como d son árboles binarios completos respecto de f y, además,
f(i) = f(r).

Definir la función
es_abc :: "(arbol => 'a) => arbol => bool
tal que (es_abc f a) se verifica si a es un árbol binario completo
respecto de f.
---------------------------------------------------------------------
*}

(* fraortmoy anaprarod migtermor serrodcal crigomgom rubgonmar
danrodcha ferrenseg *)
fun es_abc :: "(arbol => 'a) => arbol => bool" where
"es_abc _ H = True"
| "es_abc f (N a b) = (es_abc f a ∧ es_abc f b ∧ (f a = f b))"

(* marpoldia1 paupeddeg ivamenjim pablucoto marcarmor13*)
fun es_abc2 :: "(arbol => 'a) => arbol => bool" where
"es_abc2 f H = True" |
"es_abc2 f (N i d) = ((f i = f d) ∧ (es_abc2 f i) ∧ (es_abc2 f d))"

(* anaprarod *)
(* Equivalencia de las definiciones *)
lemma "es_abc f a = es_abc2 f a"
by (induct a) auto

text {*
---------------------------------------------------------------------
Nota. (size a) es el número de nodos del árbol a. Por ejemplo,
size (N (N H H) (N H H)) = 3
---------------------------------------------------------------------
*}

value "size (N (N H H) (N H H)) = 3"
value "size (N (N (N H H) (N H H)) (N (N H H) (N H H))) = 7"

text {*
---------------------------------------------------------------------
Nota. Tenemos 3 funciones de medida sobre los árboles: número de
hojas, número de nodos y profundidad. A cada una le corresponde un
concepto de completitud. En los siguientes ejercicios demostraremos
que los tres conceptos de completitud son iguales.
---------------------------------------------------------------------
*}

text {*
---------------------------------------------------------------------
Ejercicio 7. Demostrar que un árbol binario a es completo respecto de
la profundidad syss es completo respecto del número de hojas.
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 migtermor wilmorort pablucoto serrodcal *)

lemma abc_prof_num_hojas:
assumes "es_abc profundidad a"
shows "hojas a = 2^(profundidad a)"
using assms
by (induct a) auto

(* Otra forma de escribir lo mismo *)
(* anaprarod crigomgom ivamenjim *)
lemma AUX7: "es_abc profundidad a ⟶ (hojas a = 2^(profundidad a))"
by (induct a) simp_all

(* Otra forma de escribir lo mismo *)
(* danrodcha *)
lemma aux1: "es_abc profundidad (a::arbol) ⟹ (hojas a = 2^ (profundidad a))"
by (induct a) simp_all

(* fraortmoy marpoldia1 migtermor anaprarod wilmorort serrodcal crigomgom
rubgonmar ivamenjim danrodcha marcarmor13 *)
(* También funciona con AUX7 *)

lemma lej7: "es_abc profundidad a = es_abc hojas a"
by (induct a) (auto simp add: abc_prof_num_hojas)

(* danrodcha *)
lemma 7: "es_abc profundidad a = es_abc hojas a"
apply (induct a)
apply simp
apply (auto simp add: aux1)
done

(* ferrenseg *)
lemma [simp]: "es_abc profundidad a ⟶ hojas a = 2 ^ (profundidad a)"
by (induct a) auto

theorem es_abc_profundidad_hojas: "es_abc profundidad a = es_abc hojas a"
by (induct a) auto

text {*
---------------------------------------------------------------------
Ejercicio 8. Demostrar que un árbol binario a es completo respecto del
número de hojas syss es completo respecto del número de nodos.
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 migtermor wilmorort pablucoto serrodcal marcarmor13*)

lemma abc_hojas_num_nodos:
assumes "es_abc hojas a"
shows "Suc(size a) = hojas a"
using assms
by (induct a) auto

(* Otra forma de escribir lo mismo *)
(* anaprarod crigomgom*)
lemma AUX8: "es_abc hojas a ⟶ (hojas a = (Suc (size a)))"
by (induct a) auto

(* fraortmoy marpoldia1 migtermor anaprarod wilmorort pablucoto serrodcal *)

lemma lej8: "es_abc hojas a = es_abc size a"
by (induct a) (auto simp add:abc_hojas_num_nodos [symmetric])

(* anaprarod crigomgom*)
(* Usando AUX8 *)
lemma L8: "es_abc hojas a= es_abc size a"
by (induct a) (auto simp add: AUX8)

(* ivamenjim *)
(* Teorema auxiliar *)
lemma auxEj8: "hojas a = size a + 1"
by (induct a) auto

(* ivamenjim *)
lemma lej8: "es_abc hojas a = es_abc size a"
by (induct a) (auto simp add: auxEj8)

(* danrodcha *)
lemma aux3: "es_abc hojas a ⟹ (hojas a = 1 + size a)"
by (induct a) simp_all

(* danrodcha *)
lemma 8: "es_abc hojas a = es_abc size a"
apply (induct a)
apply simp
apply (auto simp add: aux3)
done

(* ferrenseg *)
lemma [simp]: "hojas a = size a + 1"
by (induct a) auto

theorem es_abc_hojas_size: "es_abc hojas a = es_abc size a"
by (induct a) auto

text {*
---------------------------------------------------------------------
Ejercicio 9. Demostrar que un árbol binario a es completo respecto de
la profundidad syss es completo respecto del número de nodos.
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 migtermor anaprarod wilmorort pablucoto
serrodcal crigomgom rubgonmar danrodcha ivamenjim marcarmor13
*)

lemma lej9: "es_abc profundidad a = es_abc size a"
by (simp add: lej7 lej8)

(* ferrenseg *)

corollary es_abc_size_profundidad: "es_abc size a = es_abc profundidad a"
by (simp add: es_abc_profundidad_hojas es_abc_hojas_size)

text {*
---------------------------------------------------------------------
Ejercicio 10. Demostrar que (abc n) es un árbol binario completo.
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 migtermor wilmorort pablucoto serrodcal crigomgom marcarmor13*)

lemma lej10: "es_abc profundidad (abc n)"
by (induct n) auto

(* anaprarod rubgonmar danrodcha ferrenseg ivamenjim *)
(* con un demostrador más débil *)
(* y en general para cualquier medida *)
lemma L10: "es_abc f (abc a)"
by (induct a) simp_all

(* ivamenjim *)
(* Igual que el anterior pero usando auto *)
lemma lej10: "es_abc f (abc n)"
by (induct n) auto

text {*
---------------------------------------------------------------------
Ejercicio 11. Demostrar que si a es un árbolo binario completo
respecto de la profundidad, entonces a es igual a
(abc (profundidad a)).
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 migtermor wilmorort pablucoto serrodcal marcarmor13 *)

lemma lej11:
assumes " es_abc profundidad a"
shows "a = (abc (profundidad a))"
using assms
by (induct a) auto

(* Otra forma de escribir lo mismo *)
(* anaprarod crigomgom rubgonmar ferrenseg ivamenjim *)
lemma "es_abc profundidad a ⟶ (a = (abc (profundidad a)))"
by (induct a) auto

(* danrodcha*)
lemma 11:"es_abc profundidad a ⟹ (a = (abc (profundidad a)))"
by (induct a) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 12. Encontrar una medida f tal que (es_abc f) es distinto de
(es_abc size).
---------------------------------------------------------------------
*}

(* migtermor *)
fun medida_nula :: "arbol => nat" where
"medida_nula H = 0"
| "medida_nula (N i d) = 0"

lemma "es_abc medida_nula a = es_abc size a"
quickcheck
oops
(* Quickcheck encuentra el siguiente contraejemplo:
a= N H (N H H)
Tras evaluar:
es_abc medida_nula a = True
es_abc size a = False*)

(* anaprarod wilmorort pablucoto serrodcal danrodcha marcarmor13 ferrenseg ivamenjim *)
lemma "es_abc f a = es_abc size a"
quickcheck
(* Quickcheck found a counterexample:
f = λx. a⇩1
a = N H (N H H)
Evaluated terms:
es_abc f a = True
es_abc size a = False *)
oops
(* el contraejemplo que encuentra es la medida constante a1 *)

(*crigomgom *)
(* Como en la primera de las soluciones he usado la función constante 0 pero he usado una expresión lambda*)
lemma "es_abc (λx. 0::nat) a = es_abc size a"
quickcheck
oops

end
</source>

Relación 7

2016-12-20T20:43:35Z

Manmorjim1:

<source lang="isar">
chapter {* R7: Árboles binarios completos *}

theory R7_Arboles_binarios_completos
imports Main
begin

text {*
En esta relación se piden demostraciones automáticas (lo más cortas
posibles). Para ello, en algunos casos es necesario incluir lemas
auxiliares (que se demuestran automáticamente) y usar ejercicios
anteriores.

---------------------------------------------------------------------
Ejercicio 1. Definir el tipo de datos arbol para representar los
árboles binarios que no tienen información ni en los nodos y ni en las
hojas. Por ejemplo, el árbol
·
/ \
/ \
· ·
/ \ / \
· · · ·
se representa por "N (N H H) (N H H)".
---------------------------------------------------------------------
*}

datatype arbol = H | N arbol arbol

value "N (N H H) (N H H) = (N (N H H) (N H H) :: arbol)"

text {*
---------------------------------------------------------------------
Ejercicio 2. Definir la función
hojas :: "arbol => nat"
tal que (hojas a) es el número de hojas del árbol a. Por ejemplo,
hojas (N (N H H) (N H H)) = 4
---------------------------------------------------------------------
*}

(* fraortmoy marcarmor13 josgarsan*)
fun hojas :: "arbol => nat" where
"hojas H = Suc 0"
| "hojas (N a b) = hojas a + hojas b"

value "hojas (N (N H H) (N H H)) = 4"

(* marpoldia1 anaprarod paupeddeg migtermor wilmorort pablucoto
ivamenjim serrodcal crigomgom rubgonmar danrodcha ferrenseg manmorjim1 *)
(* Es muy parecida a la definición anterior *)
fun hojas2 :: "arbol => nat" where
"hojas2 H = 1" |
"hojas2 (N i d) = hojas2 i + hojas2 d"

value "hojas2 (N (N H H) (N H H)) = 4"

(* anaprarod *)
(* Equivalencia de las definiciones *)
lemma "hojas a = hojas2 a"
by (induct a) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 4. Definir la función
profundidad :: "arbol => nat"
tal que (profundidad a) es la profundidad del árbol a. Por ejemplo,
profundidad (N (N H H) (N H H)) = 2
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 anaprarod migtermor wilmorort marcarmor13*)
fun profundidad :: "arbol => nat" where
"profundidad H = 0"
| "profundidad (N a b) = (if profundidad a > profundidad b
then 1 + profundidad a
else 1 + profundidad b)"

value "profundidad (N (N H H) (N H H)) = 2"

(* anaprarod wilmorort pablucoto ivamenjim serrodcal crigomgom rubgonmar
danrodcha ferrenseg josgarsan *)
fun profundidad2 :: "arbol => nat" where
"profundidad2 H = 0"
|"profundidad2 (N i d) = 1 + (max (profundidad2 i) (profundidad2 d))"

value "profundidad2 (N (N H H) (N H H)) = 2"

(* anaprarod *)
(* Equivalencia de las definiciones *)
lemma "profundidad a= profundidad2 a"
by (induct a) auto

(* paupeddeg *)
fun maximo :: "nat × nat => nat" where
"maximo (a,b) = (if a > b
then a else b)"

fun profundidad :: "arbol => nat" where
"profundidad H = 0"
| "profundidad (N i d) = 1 + maximo(profundidad i, profundidad d)"

(* ivamenjim: llamando a la función anterior profundidad3 *)

lemma "profundidad a = profundidad3 a"
by (induct a) auto

(* ivamenjim manmorjim1 *)
fun profundidad4 :: "arbol => nat" where
"profundidad4 H = 0"
| "profundidad4 (N i d) = Suc (max (profundidad4 i)(profundidad4 d))"

lemma "profundidad a = profundidad4 a"
by (induct a) auto

text {*
---------------------------------------------------------------------
Ejercicio 5. Definir la función
abc :: "nat ⇒ arbol"
tal que (abc n) es el árbol binario completo de profundidad n. Por
ejemplo,
abc 3 = N (N (N H H) (N H H)) (N (N H H) (N H H))
---------------------------------------------------------------------
*}
(* fraortmoy marpoldia1 anaprarod paupeddeg migtermor wilmorort
serrodcal crigomgom rubgonmar danrodcha ferrenseg josgarsan*)
fun abc :: "nat ⇒ arbol" where
"abc 0 = H"
| "abc (Suc n) = (N (abc n) (abc n))"

value "abc 3 = N (N (N H H) (N H H)) (N (N H H) (N H H))"

(* ivamenjim pablucoto marcarmor13*)
fun abc2 :: "nat ⇒ arbol" where
"abc2 0 = H"
| "abc2 t = N (abc2 (t-1)) (abc2 (t-1))"

value "abc2 3 = N (N (N H H) (N H H)) (N (N H H) (N H H))"

(* ivamenjim: Metaejercicio de demostración *)
lemma "abc t = abc2 t"
by (induct t) auto

text {*
---------------------------------------------------------------------
Ejercicio 6. Un árbol binario a es completo respecto de la medida f si
a es una hoja o bien a es de la forma (N i d) y se cumple que tanto i
como d son árboles binarios completos respecto de f y, además,
f(i) = f(r).

Definir la función
es_abc :: "(arbol => 'a) => arbol => bool
tal que (es_abc f a) se verifica si a es un árbol binario completo
respecto de f.
---------------------------------------------------------------------
*}

(* fraortmoy anaprarod migtermor serrodcal crigomgom rubgonmar
danrodcha ferrenseg *)
fun es_abc :: "(arbol => 'a) => arbol => bool" where
"es_abc _ H = True"
| "es_abc f (N a b) = (es_abc f a ∧ es_abc f b ∧ (f a = f b))"

(* marpoldia1 paupeddeg ivamenjim pablucoto marcarmor13*)
fun es_abc2 :: "(arbol => 'a) => arbol => bool" where
"es_abc2 f H = True" |
"es_abc2 f (N i d) = ((f i = f d) ∧ (es_abc2 f i) ∧ (es_abc2 f d))"

(* anaprarod *)
(* Equivalencia de las definiciones *)
lemma "es_abc f a = es_abc2 f a"
by (induct a) auto

text {*
---------------------------------------------------------------------
Nota. (size a) es el número de nodos del árbol a. Por ejemplo,
size (N (N H H) (N H H)) = 3
---------------------------------------------------------------------
*}

value "size (N (N H H) (N H H)) = 3"
value "size (N (N (N H H) (N H H)) (N (N H H) (N H H))) = 7"

text {*
---------------------------------------------------------------------
Nota. Tenemos 3 funciones de medida sobre los árboles: número de
hojas, número de nodos y profundidad. A cada una le corresponde un
concepto de completitud. En los siguientes ejercicios demostraremos
que los tres conceptos de completitud son iguales.
---------------------------------------------------------------------
*}

text {*
---------------------------------------------------------------------
Ejercicio 7. Demostrar que un árbol binario a es completo respecto de
la profundidad syss es completo respecto del número de hojas.
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 migtermor wilmorort pablucoto serrodcal *)

lemma abc_prof_num_hojas:
assumes "es_abc profundidad a"
shows "hojas a = 2^(profundidad a)"
using assms
by (induct a) auto

(* Otra forma de escribir lo mismo *)
(* anaprarod crigomgom ivamenjim *)
lemma AUX7: "es_abc profundidad a ⟶ (hojas a = 2^(profundidad a))"
by (induct a) simp_all

(* Otra forma de escribir lo mismo *)
(* danrodcha *)
lemma aux1: "es_abc profundidad (a::arbol) ⟹ (hojas a = 2^ (profundidad a))"
by (induct a) simp_all

(* fraortmoy marpoldia1 migtermor anaprarod wilmorort serrodcal crigomgom
rubgonmar ivamenjim danrodcha marcarmor13 *)
(* También funciona con AUX7 *)

lemma lej7: "es_abc profundidad a = es_abc hojas a"
by (induct a) (auto simp add: abc_prof_num_hojas)

(* danrodcha *)
lemma 7: "es_abc profundidad a = es_abc hojas a"
apply (induct a)
apply simp
apply (auto simp add: aux1)
done

(* ferrenseg *)
lemma [simp]: "es_abc profundidad a ⟶ hojas a = 2 ^ (profundidad a)"
by (induct a) auto

theorem es_abc_profundidad_hojas: "es_abc profundidad a = es_abc hojas a"
by (induct a) auto

text {*
---------------------------------------------------------------------
Ejercicio 8. Demostrar que un árbol binario a es completo respecto del
número de hojas syss es completo respecto del número de nodos.
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 migtermor wilmorort pablucoto serrodcal marcarmor13*)

lemma abc_hojas_num_nodos:
assumes "es_abc hojas a"
shows "Suc(size a) = hojas a"
using assms
by (induct a) auto

(* Otra forma de escribir lo mismo *)
(* anaprarod crigomgom*)
lemma AUX8: "es_abc hojas a ⟶ (hojas a = (Suc (size a)))"
by (induct a) auto

(* fraortmoy marpoldia1 migtermor anaprarod wilmorort pablucoto serrodcal *)

lemma lej8: "es_abc hojas a = es_abc size a"
by (induct a) (auto simp add:abc_hojas_num_nodos [symmetric])

(* anaprarod crigomgom*)
(* Usando AUX8 *)
lemma L8: "es_abc hojas a= es_abc size a"
by (induct a) (auto simp add: AUX8)

(* ivamenjim *)
(* Teorema auxiliar *)
lemma auxEj8: "hojas a = size a + 1"
by (induct a) auto

(* ivamenjim *)
lemma lej8: "es_abc hojas a = es_abc size a"
by (induct a) (auto simp add: auxEj8)

(* danrodcha *)
lemma aux3: "es_abc hojas a ⟹ (hojas a = 1 + size a)"
by (induct a) simp_all

(* danrodcha *)
lemma 8: "es_abc hojas a = es_abc size a"
apply (induct a)
apply simp
apply (auto simp add: aux3)
done

(* ferrenseg *)
lemma [simp]: "hojas a = size a + 1"
by (induct a) auto

theorem es_abc_hojas_size: "es_abc hojas a = es_abc size a"
by (induct a) auto

text {*
---------------------------------------------------------------------
Ejercicio 9. Demostrar que un árbol binario a es completo respecto de
la profundidad syss es completo respecto del número de nodos.
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 migtermor anaprarod wilmorort pablucoto
serrodcal crigomgom rubgonmar danrodcha ivamenjim marcarmor13
*)

lemma lej9: "es_abc profundidad a = es_abc size a"
by (simp add: lej7 lej8)

(* ferrenseg *)

corollary es_abc_size_profundidad: "es_abc size a = es_abc profundidad a"
by (simp add: es_abc_profundidad_hojas es_abc_hojas_size)

text {*
---------------------------------------------------------------------
Ejercicio 10. Demostrar que (abc n) es un árbol binario completo.
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 migtermor wilmorort pablucoto serrodcal crigomgom marcarmor13*)

lemma lej10: "es_abc profundidad (abc n)"
by (induct n) auto

(* anaprarod rubgonmar danrodcha ferrenseg ivamenjim *)
(* con un demostrador más débil *)
(* y en general para cualquier medida *)
lemma L10: "es_abc f (abc a)"
by (induct a) simp_all

(* ivamenjim *)
(* Igual que el anterior pero usando auto *)
lemma lej10: "es_abc f (abc n)"
by (induct n) auto

text {*
---------------------------------------------------------------------
Ejercicio 11. Demostrar que si a es un árbolo binario completo
respecto de la profundidad, entonces a es igual a
(abc (profundidad a)).
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 migtermor wilmorort pablucoto serrodcal marcarmor13 *)

lemma lej11:
assumes " es_abc profundidad a"
shows "a = (abc (profundidad a))"
using assms
by (induct a) auto

(* Otra forma de escribir lo mismo *)
(* anaprarod crigomgom rubgonmar ferrenseg ivamenjim *)
lemma "es_abc profundidad a ⟶ (a = (abc (profundidad a)))"
by (induct a) auto

(* danrodcha*)
lemma 11:"es_abc profundidad a ⟹ (a = (abc (profundidad a)))"
by (induct a) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 12. Encontrar una medida f tal que (es_abc f) es distinto de
(es_abc size).
---------------------------------------------------------------------
*}

(* migtermor *)
fun medida_nula :: "arbol => nat" where
"medida_nula H = 0"
| "medida_nula (N i d) = 0"

lemma "es_abc medida_nula a = es_abc size a"
quickcheck
oops
(* Quickcheck encuentra el siguiente contraejemplo:
a= N H (N H H)
Tras evaluar:
es_abc medida_nula a = True
es_abc size a = False*)

(* anaprarod wilmorort pablucoto serrodcal danrodcha marcarmor13 ferrenseg ivamenjim *)
lemma "es_abc f a = es_abc size a"
quickcheck
(* Quickcheck found a counterexample:
f = λx. a⇩1
a = N H (N H H)
Evaluated terms:
es_abc f a = True
es_abc size a = False *)
oops
(* el contraejemplo que encuentra es la medida constante a1 *)

(*crigomgom *)
(* Como en la primera de las soluciones he usado la función constante 0 pero he usado una expresión lambda*)
lemma "es_abc (λx. 0::nat) a = es_abc size a"
quickcheck
oops

end
</source>

Relación 7

2016-12-20T20:41:59Z

Manmorjim1:

<source lang="isar">
chapter {* R7: Árboles binarios completos *}

theory R7_Arboles_binarios_completos
imports Main
begin

text {*
En esta relación se piden demostraciones automáticas (lo más cortas
posibles). Para ello, en algunos casos es necesario incluir lemas
auxiliares (que se demuestran automáticamente) y usar ejercicios
anteriores.

---------------------------------------------------------------------
Ejercicio 1. Definir el tipo de datos arbol para representar los
árboles binarios que no tienen información ni en los nodos y ni en las
hojas. Por ejemplo, el árbol
·
/ \
/ \
· ·
/ \ / \
· · · ·
se representa por "N (N H H) (N H H)".
---------------------------------------------------------------------
*}

datatype arbol = H | N arbol arbol

value "N (N H H) (N H H) = (N (N H H) (N H H) :: arbol)"

text {*
---------------------------------------------------------------------
Ejercicio 2. Definir la función
hojas :: "arbol => nat"
tal que (hojas a) es el número de hojas del árbol a. Por ejemplo,
hojas (N (N H H) (N H H)) = 4
---------------------------------------------------------------------
*}

(* fraortmoy marcarmor13 josgarsan*)
fun hojas :: "arbol => nat" where
"hojas H = Suc 0"
| "hojas (N a b) = hojas a + hojas b"

value "hojas (N (N H H) (N H H)) = 4"

(* marpoldia1 anaprarod paupeddeg migtermor wilmorort pablucoto
ivamenjim serrodcal crigomgom rubgonmar danrodcha ferrenseg *)
(* Es muy parecida a la definición anterior manmorjim1 *)
fun hojas2 :: "arbol => nat" where
"hojas2 H = 1" |
"hojas2 (N i d) = hojas2 i + hojas2 d"

value "hojas2 (N (N H H) (N H H)) = 4"

(* anaprarod *)
(* Equivalencia de las definiciones *)
lemma "hojas a = hojas2 a"
by (induct a) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 4. Definir la función
profundidad :: "arbol => nat"
tal que (profundidad a) es la profundidad del árbol a. Por ejemplo,
profundidad (N (N H H) (N H H)) = 2
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 anaprarod migtermor wilmorort marcarmor13*)
fun profundidad :: "arbol => nat" where
"profundidad H = 0"
| "profundidad (N a b) = (if profundidad a > profundidad b
then 1 + profundidad a
else 1 + profundidad b)"

value "profundidad (N (N H H) (N H H)) = 2"

(* anaprarod wilmorort pablucoto ivamenjim serrodcal crigomgom rubgonmar
danrodcha ferrenseg josgarsan manmorjim1 *)
fun profundidad2 :: "arbol => nat" where
"profundidad2 H = 0"
|"profundidad2 (N i d) = 1 + (max (profundidad2 i) (profundidad2 d))"

value "profundidad2 (N (N H H) (N H H)) = 2"

(* anaprarod *)
(* Equivalencia de las definiciones *)
lemma "profundidad a= profundidad2 a"
by (induct a) auto

(* paupeddeg *)
fun maximo :: "nat × nat => nat" where
"maximo (a,b) = (if a > b
then a else b)"

fun profundidad :: "arbol => nat" where
"profundidad H = 0"
| "profundidad (N i d) = 1 + maximo(profundidad i, profundidad d)"

(* ivamenjim: llamando a la función anterior profundidad3 *)

lemma "profundidad a = profundidad3 a"
by (induct a) auto

(* ivamenjim *)
fun profundidad4 :: "arbol => nat" where
"profundidad4 H = 0"
| "profundidad4 (N i d) = Suc (max (profundidad4 i)(profundidad4 d))"

lemma "profundidad a = profundidad4 a"
by (induct a) auto

text {*
---------------------------------------------------------------------
Ejercicio 5. Definir la función
abc :: "nat ⇒ arbol"
tal que (abc n) es el árbol binario completo de profundidad n. Por
ejemplo,
abc 3 = N (N (N H H) (N H H)) (N (N H H) (N H H))
---------------------------------------------------------------------
*}
(* fraortmoy marpoldia1 anaprarod paupeddeg migtermor wilmorort
serrodcal crigomgom rubgonmar danrodcha ferrenseg josgarsan*)
fun abc :: "nat ⇒ arbol" where
"abc 0 = H"
| "abc (Suc n) = (N (abc n) (abc n))"

value "abc 3 = N (N (N H H) (N H H)) (N (N H H) (N H H))"

(* ivamenjim pablucoto marcarmor13*)
fun abc2 :: "nat ⇒ arbol" where
"abc2 0 = H"
| "abc2 t = N (abc2 (t-1)) (abc2 (t-1))"

value "abc2 3 = N (N (N H H) (N H H)) (N (N H H) (N H H))"

(* ivamenjim: Metaejercicio de demostración *)
lemma "abc t = abc2 t"
by (induct t) auto

text {*
---------------------------------------------------------------------
Ejercicio 6. Un árbol binario a es completo respecto de la medida f si
a es una hoja o bien a es de la forma (N i d) y se cumple que tanto i
como d son árboles binarios completos respecto de f y, además,
f(i) = f(r).

Definir la función
es_abc :: "(arbol => 'a) => arbol => bool
tal que (es_abc f a) se verifica si a es un árbol binario completo
respecto de f.
---------------------------------------------------------------------
*}

(* fraortmoy anaprarod migtermor serrodcal crigomgom rubgonmar
danrodcha ferrenseg *)
fun es_abc :: "(arbol => 'a) => arbol => bool" where
"es_abc _ H = True"
| "es_abc f (N a b) = (es_abc f a ∧ es_abc f b ∧ (f a = f b))"

(* marpoldia1 paupeddeg ivamenjim pablucoto marcarmor13*)
fun es_abc2 :: "(arbol => 'a) => arbol => bool" where
"es_abc2 f H = True" |
"es_abc2 f (N i d) = ((f i = f d) ∧ (es_abc2 f i) ∧ (es_abc2 f d))"

(* anaprarod *)
(* Equivalencia de las definiciones *)
lemma "es_abc f a = es_abc2 f a"
by (induct a) auto

text {*
---------------------------------------------------------------------
Nota. (size a) es el número de nodos del árbol a. Por ejemplo,
size (N (N H H) (N H H)) = 3
---------------------------------------------------------------------
*}

value "size (N (N H H) (N H H)) = 3"
value "size (N (N (N H H) (N H H)) (N (N H H) (N H H))) = 7"

text {*
---------------------------------------------------------------------
Nota. Tenemos 3 funciones de medida sobre los árboles: número de
hojas, número de nodos y profundidad. A cada una le corresponde un
concepto de completitud. En los siguientes ejercicios demostraremos
que los tres conceptos de completitud son iguales.
---------------------------------------------------------------------
*}

text {*
---------------------------------------------------------------------
Ejercicio 7. Demostrar que un árbol binario a es completo respecto de
la profundidad syss es completo respecto del número de hojas.
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 migtermor wilmorort pablucoto serrodcal *)

lemma abc_prof_num_hojas:
assumes "es_abc profundidad a"
shows "hojas a = 2^(profundidad a)"
using assms
by (induct a) auto

(* Otra forma de escribir lo mismo *)
(* anaprarod crigomgom ivamenjim *)
lemma AUX7: "es_abc profundidad a ⟶ (hojas a = 2^(profundidad a))"
by (induct a) simp_all

(* Otra forma de escribir lo mismo *)
(* danrodcha *)
lemma aux1: "es_abc profundidad (a::arbol) ⟹ (hojas a = 2^ (profundidad a))"
by (induct a) simp_all

(* fraortmoy marpoldia1 migtermor anaprarod wilmorort serrodcal crigomgom
rubgonmar ivamenjim danrodcha marcarmor13 *)
(* También funciona con AUX7 *)

lemma lej7: "es_abc profundidad a = es_abc hojas a"
by (induct a) (auto simp add: abc_prof_num_hojas)

(* danrodcha *)
lemma 7: "es_abc profundidad a = es_abc hojas a"
apply (induct a)
apply simp
apply (auto simp add: aux1)
done

(* ferrenseg *)
lemma [simp]: "es_abc profundidad a ⟶ hojas a = 2 ^ (profundidad a)"
by (induct a) auto

theorem es_abc_profundidad_hojas: "es_abc profundidad a = es_abc hojas a"
by (induct a) auto

text {*
---------------------------------------------------------------------
Ejercicio 8. Demostrar que un árbol binario a es completo respecto del
número de hojas syss es completo respecto del número de nodos.
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 migtermor wilmorort pablucoto serrodcal marcarmor13*)

lemma abc_hojas_num_nodos:
assumes "es_abc hojas a"
shows "Suc(size a) = hojas a"
using assms
by (induct a) auto

(* Otra forma de escribir lo mismo *)
(* anaprarod crigomgom*)
lemma AUX8: "es_abc hojas a ⟶ (hojas a = (Suc (size a)))"
by (induct a) auto

(* fraortmoy marpoldia1 migtermor anaprarod wilmorort pablucoto serrodcal *)

lemma lej8: "es_abc hojas a = es_abc size a"
by (induct a) (auto simp add:abc_hojas_num_nodos [symmetric])

(* anaprarod crigomgom*)
(* Usando AUX8 *)
lemma L8: "es_abc hojas a= es_abc size a"
by (induct a) (auto simp add: AUX8)

(* ivamenjim *)
(* Teorema auxiliar *)
lemma auxEj8: "hojas a = size a + 1"
by (induct a) auto

(* ivamenjim *)
lemma lej8: "es_abc hojas a = es_abc size a"
by (induct a) (auto simp add: auxEj8)

(* danrodcha *)
lemma aux3: "es_abc hojas a ⟹ (hojas a = 1 + size a)"
by (induct a) simp_all

(* danrodcha *)
lemma 8: "es_abc hojas a = es_abc size a"
apply (induct a)
apply simp
apply (auto simp add: aux3)
done

(* ferrenseg *)
lemma [simp]: "hojas a = size a + 1"
by (induct a) auto

theorem es_abc_hojas_size: "es_abc hojas a = es_abc size a"
by (induct a) auto

text {*
---------------------------------------------------------------------
Ejercicio 9. Demostrar que un árbol binario a es completo respecto de
la profundidad syss es completo respecto del número de nodos.
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 migtermor anaprarod wilmorort pablucoto
serrodcal crigomgom rubgonmar danrodcha ivamenjim marcarmor13
*)

lemma lej9: "es_abc profundidad a = es_abc size a"
by (simp add: lej7 lej8)

(* ferrenseg *)

corollary es_abc_size_profundidad: "es_abc size a = es_abc profundidad a"
by (simp add: es_abc_profundidad_hojas es_abc_hojas_size)

text {*
---------------------------------------------------------------------
Ejercicio 10. Demostrar que (abc n) es un árbol binario completo.
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 migtermor wilmorort pablucoto serrodcal crigomgom marcarmor13*)

lemma lej10: "es_abc profundidad (abc n)"
by (induct n) auto

(* anaprarod rubgonmar danrodcha ferrenseg ivamenjim *)
(* con un demostrador más débil *)
(* y en general para cualquier medida *)
lemma L10: "es_abc f (abc a)"
by (induct a) simp_all

(* ivamenjim *)
(* Igual que el anterior pero usando auto *)
lemma lej10: "es_abc f (abc n)"
by (induct n) auto

text {*
---------------------------------------------------------------------
Ejercicio 11. Demostrar que si a es un árbolo binario completo
respecto de la profundidad, entonces a es igual a
(abc (profundidad a)).
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 migtermor wilmorort pablucoto serrodcal marcarmor13 *)

lemma lej11:
assumes " es_abc profundidad a"
shows "a = (abc (profundidad a))"
using assms
by (induct a) auto

(* Otra forma de escribir lo mismo *)
(* anaprarod crigomgom rubgonmar ferrenseg ivamenjim *)
lemma "es_abc profundidad a ⟶ (a = (abc (profundidad a)))"
by (induct a) auto

(* danrodcha*)
lemma 11:"es_abc profundidad a ⟹ (a = (abc (profundidad a)))"
by (induct a) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 12. Encontrar una medida f tal que (es_abc f) es distinto de
(es_abc size).
---------------------------------------------------------------------
*}

(* migtermor *)
fun medida_nula :: "arbol => nat" where
"medida_nula H = 0"
| "medida_nula (N i d) = 0"

lemma "es_abc medida_nula a = es_abc size a"
quickcheck
oops
(* Quickcheck encuentra el siguiente contraejemplo:
a= N H (N H H)
Tras evaluar:
es_abc medida_nula a = True
es_abc size a = False*)

(* anaprarod wilmorort pablucoto serrodcal danrodcha marcarmor13 ferrenseg ivamenjim *)
lemma "es_abc f a = es_abc size a"
quickcheck
(* Quickcheck found a counterexample:
f = λx. a⇩1
a = N H (N H H)
Evaluated terms:
es_abc f a = True
es_abc size a = False *)
oops
(* el contraejemplo que encuentra es la medida constante a1 *)

(*crigomgom *)
(* Como en la primera de las soluciones he usado la función constante 0 pero he usado una expresión lambda*)
lemma "es_abc (λx. 0::nat) a = es_abc size a"
quickcheck
oops

end
</source>

Relación 7

2016-12-20T20:37:00Z

Manmorjim1:

<source lang="isar">
chapter {* R7: Árboles binarios completos *}

theory R7_Arboles_binarios_completos
imports Main
begin

text {*
En esta relación se piden demostraciones automáticas (lo más cortas
posibles). Para ello, en algunos casos es necesario incluir lemas
auxiliares (que se demuestran automáticamente) y usar ejercicios
anteriores.

---------------------------------------------------------------------
Ejercicio 1. Definir el tipo de datos arbol para representar los
árboles binarios que no tienen información ni en los nodos y ni en las
hojas. Por ejemplo, el árbol
·
/ \
/ \
· ·
/ \ / \
· · · ·
se representa por "N (N H H) (N H H)".
---------------------------------------------------------------------
*}

datatype arbol = H | N arbol arbol

value "N (N H H) (N H H) = (N (N H H) (N H H) :: arbol)"

text {*
---------------------------------------------------------------------
Ejercicio 2. Definir la función
hojas :: "arbol => nat"
tal que (hojas a) es el número de hojas del árbol a. Por ejemplo,
hojas (N (N H H) (N H H)) = 4
---------------------------------------------------------------------
*}

(* fraortmoy marcarmor13 josgarsan*)
fun hojas :: "arbol => nat" where
"hojas H = Suc 0"
| "hojas (N a b) = hojas a + hojas b"

value "hojas (N (N H H) (N H H)) = 4"

(* marpoldia1 anaprarod paupeddeg migtermor wilmorort pablucoto
ivamenjim serrodcal crigomgom rubgonmar danrodcha ferrenseg *)
(* Es muy parecida a la definición anterior manmorjim1 *)
fun hojas2 :: "arbol => nat" where
"hojas2 H = 1" |
"hojas2 (N i d) = hojas2 i + hojas2 d"

value "hojas2 (N (N H H) (N H H)) = 4"

(* anaprarod *)
(* Equivalencia de las definiciones *)
lemma "hojas a = hojas2 a"
by (induct a) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 4. Definir la función
profundidad :: "arbol => nat"
tal que (profundidad a) es la profundidad del árbol a. Por ejemplo,
profundidad (N (N H H) (N H H)) = 2
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 anaprarod migtermor wilmorort marcarmor13*)
fun profundidad :: "arbol => nat" where
"profundidad H = 0"
| "profundidad (N a b) = (if profundidad a > profundidad b
then 1 + profundidad a
else 1 + profundidad b)"

value "profundidad (N (N H H) (N H H)) = 2"

(* anaprarod wilmorort pablucoto ivamenjim serrodcal crigomgom rubgonmar
danrodcha ferrenseg josgarsan *)
fun profundidad2 :: "arbol => nat" where
"profundidad2 H = 0"
|"profundidad2 (N i d) = 1 + (max (profundidad2 i) (profundidad2 d))"

value "profundidad2 (N (N H H) (N H H)) = 2"

(* anaprarod *)
(* Equivalencia de las definiciones *)
lemma "profundidad a= profundidad2 a"
by (induct a) auto

(* paupeddeg *)
fun maximo :: "nat × nat => nat" where
"maximo (a,b) = (if a > b
then a else b)"

fun profundidad :: "arbol => nat" where
"profundidad H = 0"
| "profundidad (N i d) = 1 + maximo(profundidad i, profundidad d)"

(* ivamenjim: llamando a la función anterior profundidad3 *)

lemma "profundidad a = profundidad3 a"
by (induct a) auto

(* ivamenjim *)
fun profundidad4 :: "arbol => nat" where
"profundidad4 H = 0"
| "profundidad4 (N i d) = Suc (max (profundidad4 i)(profundidad4 d))"

lemma "profundidad a = profundidad4 a"
by (induct a) auto

text {*
---------------------------------------------------------------------
Ejercicio 5. Definir la función
abc :: "nat ⇒ arbol"
tal que (abc n) es el árbol binario completo de profundidad n. Por
ejemplo,
abc 3 = N (N (N H H) (N H H)) (N (N H H) (N H H))
---------------------------------------------------------------------
*}
(* fraortmoy marpoldia1 anaprarod paupeddeg migtermor wilmorort
serrodcal crigomgom rubgonmar danrodcha ferrenseg josgarsan*)
fun abc :: "nat ⇒ arbol" where
"abc 0 = H"
| "abc (Suc n) = (N (abc n) (abc n))"

value "abc 3 = N (N (N H H) (N H H)) (N (N H H) (N H H))"

(* ivamenjim pablucoto marcarmor13*)
fun abc2 :: "nat ⇒ arbol" where
"abc2 0 = H"
| "abc2 t = N (abc2 (t-1)) (abc2 (t-1))"

value "abc2 3 = N (N (N H H) (N H H)) (N (N H H) (N H H))"

(* ivamenjim: Metaejercicio de demostración *)
lemma "abc t = abc2 t"
by (induct t) auto

text {*
---------------------------------------------------------------------
Ejercicio 6. Un árbol binario a es completo respecto de la medida f si
a es una hoja o bien a es de la forma (N i d) y se cumple que tanto i
como d son árboles binarios completos respecto de f y, además,
f(i) = f(r).

Definir la función
es_abc :: "(arbol => 'a) => arbol => bool
tal que (es_abc f a) se verifica si a es un árbol binario completo
respecto de f.
---------------------------------------------------------------------
*}

(* fraortmoy anaprarod migtermor serrodcal crigomgom rubgonmar
danrodcha ferrenseg *)
fun es_abc :: "(arbol => 'a) => arbol => bool" where
"es_abc _ H = True"
| "es_abc f (N a b) = (es_abc f a ∧ es_abc f b ∧ (f a = f b))"

(* marpoldia1 paupeddeg ivamenjim pablucoto marcarmor13*)
fun es_abc2 :: "(arbol => 'a) => arbol => bool" where
"es_abc2 f H = True" |
"es_abc2 f (N i d) = ((f i = f d) ∧ (es_abc2 f i) ∧ (es_abc2 f d))"

(* anaprarod *)
(* Equivalencia de las definiciones *)
lemma "es_abc f a = es_abc2 f a"
by (induct a) auto

text {*
---------------------------------------------------------------------
Nota. (size a) es el número de nodos del árbol a. Por ejemplo,
size (N (N H H) (N H H)) = 3
---------------------------------------------------------------------
*}

value "size (N (N H H) (N H H)) = 3"
value "size (N (N (N H H) (N H H)) (N (N H H) (N H H))) = 7"

text {*
---------------------------------------------------------------------
Nota. Tenemos 3 funciones de medida sobre los árboles: número de
hojas, número de nodos y profundidad. A cada una le corresponde un
concepto de completitud. En los siguientes ejercicios demostraremos
que los tres conceptos de completitud son iguales.
---------------------------------------------------------------------
*}

text {*
---------------------------------------------------------------------
Ejercicio 7. Demostrar que un árbol binario a es completo respecto de
la profundidad syss es completo respecto del número de hojas.
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 migtermor wilmorort pablucoto serrodcal *)

lemma abc_prof_num_hojas:
assumes "es_abc profundidad a"
shows "hojas a = 2^(profundidad a)"
using assms
by (induct a) auto

(* Otra forma de escribir lo mismo *)
(* anaprarod crigomgom ivamenjim *)
lemma AUX7: "es_abc profundidad a ⟶ (hojas a = 2^(profundidad a))"
by (induct a) simp_all

(* Otra forma de escribir lo mismo *)
(* danrodcha *)
lemma aux1: "es_abc profundidad (a::arbol) ⟹ (hojas a = 2^ (profundidad a))"
by (induct a) simp_all

(* fraortmoy marpoldia1 migtermor anaprarod wilmorort serrodcal crigomgom
rubgonmar ivamenjim danrodcha marcarmor13 *)
(* También funciona con AUX7 *)

lemma lej7: "es_abc profundidad a = es_abc hojas a"
by (induct a) (auto simp add: abc_prof_num_hojas)

(* danrodcha *)
lemma 7: "es_abc profundidad a = es_abc hojas a"
apply (induct a)
apply simp
apply (auto simp add: aux1)
done

(* ferrenseg *)
lemma [simp]: "es_abc profundidad a ⟶ hojas a = 2 ^ (profundidad a)"
by (induct a) auto

theorem es_abc_profundidad_hojas: "es_abc profundidad a = es_abc hojas a"
by (induct a) auto

text {*
---------------------------------------------------------------------
Ejercicio 8. Demostrar que un árbol binario a es completo respecto del
número de hojas syss es completo respecto del número de nodos.
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 migtermor wilmorort pablucoto serrodcal marcarmor13*)

lemma abc_hojas_num_nodos:
assumes "es_abc hojas a"
shows "Suc(size a) = hojas a"
using assms
by (induct a) auto

(* Otra forma de escribir lo mismo *)
(* anaprarod crigomgom*)
lemma AUX8: "es_abc hojas a ⟶ (hojas a = (Suc (size a)))"
by (induct a) auto

(* fraortmoy marpoldia1 migtermor anaprarod wilmorort pablucoto serrodcal *)

lemma lej8: "es_abc hojas a = es_abc size a"
by (induct a) (auto simp add:abc_hojas_num_nodos [symmetric])

(* anaprarod crigomgom*)
(* Usando AUX8 *)
lemma L8: "es_abc hojas a= es_abc size a"
by (induct a) (auto simp add: AUX8)

(* ivamenjim *)
(* Teorema auxiliar *)
lemma auxEj8: "hojas a = size a + 1"
by (induct a) auto

(* ivamenjim *)
lemma lej8: "es_abc hojas a = es_abc size a"
by (induct a) (auto simp add: auxEj8)

(* danrodcha *)
lemma aux3: "es_abc hojas a ⟹ (hojas a = 1 + size a)"
by (induct a) simp_all

(* danrodcha *)
lemma 8: "es_abc hojas a = es_abc size a"
apply (induct a)
apply simp
apply (auto simp add: aux3)
done

(* ferrenseg *)
lemma [simp]: "hojas a = size a + 1"
by (induct a) auto

theorem es_abc_hojas_size: "es_abc hojas a = es_abc size a"
by (induct a) auto

text {*
---------------------------------------------------------------------
Ejercicio 9. Demostrar que un árbol binario a es completo respecto de
la profundidad syss es completo respecto del número de nodos.
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 migtermor anaprarod wilmorort pablucoto
serrodcal crigomgom rubgonmar danrodcha ivamenjim marcarmor13
*)

lemma lej9: "es_abc profundidad a = es_abc size a"
by (simp add: lej7 lej8)

(* ferrenseg *)

corollary es_abc_size_profundidad: "es_abc size a = es_abc profundidad a"
by (simp add: es_abc_profundidad_hojas es_abc_hojas_size)

text {*
---------------------------------------------------------------------
Ejercicio 10. Demostrar que (abc n) es un árbol binario completo.
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 migtermor wilmorort pablucoto serrodcal crigomgom marcarmor13*)

lemma lej10: "es_abc profundidad (abc n)"
by (induct n) auto

(* anaprarod rubgonmar danrodcha ferrenseg ivamenjim *)
(* con un demostrador más débil *)
(* y en general para cualquier medida *)
lemma L10: "es_abc f (abc a)"
by (induct a) simp_all

(* ivamenjim *)
(* Igual que el anterior pero usando auto *)
lemma lej10: "es_abc f (abc n)"
by (induct n) auto

text {*
---------------------------------------------------------------------
Ejercicio 11. Demostrar que si a es un árbolo binario completo
respecto de la profundidad, entonces a es igual a
(abc (profundidad a)).
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 migtermor wilmorort pablucoto serrodcal marcarmor13 *)

lemma lej11:
assumes " es_abc profundidad a"
shows "a = (abc (profundidad a))"
using assms
by (induct a) auto

(* Otra forma de escribir lo mismo *)
(* anaprarod crigomgom rubgonmar ferrenseg ivamenjim *)
lemma "es_abc profundidad a ⟶ (a = (abc (profundidad a)))"
by (induct a) auto

(* danrodcha*)
lemma 11:"es_abc profundidad a ⟹ (a = (abc (profundidad a)))"
by (induct a) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 12. Encontrar una medida f tal que (es_abc f) es distinto de
(es_abc size).
---------------------------------------------------------------------
*}

(* migtermor *)
fun medida_nula :: "arbol => nat" where
"medida_nula H = 0"
| "medida_nula (N i d) = 0"

lemma "es_abc medida_nula a = es_abc size a"
quickcheck
oops
(* Quickcheck encuentra el siguiente contraejemplo:
a= N H (N H H)
Tras evaluar:
es_abc medida_nula a = True
es_abc size a = False*)

(* anaprarod wilmorort pablucoto serrodcal danrodcha marcarmor13 ferrenseg ivamenjim *)
lemma "es_abc f a = es_abc size a"
quickcheck
(* Quickcheck found a counterexample:
f = λx. a⇩1
a = N H (N H H)
Evaluated terms:
es_abc f a = True
es_abc size a = False *)
oops
(* el contraejemplo que encuentra es la medida constante a1 *)

(*crigomgom *)
(* Como en la primera de las soluciones he usado la función constante 0 pero he usado una expresión lambda*)
lemma "es_abc (λx. 0::nat) a = es_abc size a"
quickcheck
oops

end
</source>

Relación 6

2016-12-03T17:02:32Z

Manmorjim1:

<source lang="isar">
chapter {* R6: Recorridos de árboles *}

theory R6_Recorridos_de_arboles
imports Main
begin

text {*
---------------------------------------------------------------------
Ejercicio 1. Definir el tipo de datos arbol para representar los
árboles binarios que tiene información en los nodos y en las hojas.
Por ejemplo, el árbol
e
/ \
/ \
c g
/ \ / \
a d f h
se representa por "N e (N c (H a) (H d)) (N g (H f) (H h))".
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 manmorjim1 *)

datatype 'a arbol = H "'a" | N "'a" "'a arbol" "'a arbol"

value "N e (N c (H a) (H d)) (N g (H f) (H h))"

text {*
---------------------------------------------------------------------
Ejercicio 2. Definir la función
preOrden :: "'a arbol ⇒ 'a list"
tal que (preOrden a) es el recorrido pre orden del árbol a. Por
ejemplo,
preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1*)

fun preOrden :: "'a arbol ⇒ 'a list" where
"preOrden (H t) = [t]"
| "preOrden (N t i d) = [t] @ (preOrden i) @ (preOrden d)"

(* danrodcha crigomgom manmorjim1 *)
fun preOrden1 :: "'a arbol ⇒ 'a list" where
"preOrden1 (H x) = [x]"
| "preOrden1 (N x i d) = x#preOrden1 i @ preOrden1 d"

value "preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"
value "preOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"

(* danrodcha *)
lemma "preOrden a = preOrden1 a"
by (induct a) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 3. Definir la función
postOrden :: "'a arbol ⇒ 'a list"
tal que (postOrden a) es el recorrido post orden del árbol a. Por
ejemplo,
postOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 *)

fun postOrden :: "'a arbol ⇒ 'a list" where
"postOrden (H t) = [t]"
| "postOrden (N t i d) = (postOrden i) @ (postOrden d) @ [t]"

value "postOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,d,c,f,h,g,e]"

text {*
---------------------------------------------------------------------
Ejercicio 4. Definir la función
inOrden :: "'a arbol ⇒ 'a list"
tal que (inOrden a) es el recorrido in orden del árbol a. Por
ejemplo,
inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom marpoldia1*)

fun inOrden :: "'a arbol ⇒ 'a list" where
"inOrden (H t) = [t]"
| "inOrden (N t i d) = (inOrden i) @ [t] @ (inOrden d)"

(* danrodcha manmorjim1 *)
fun inOrden1 :: "'a arbol ⇒ 'a list" where
"inOrden1 (H t) = [t]"
| "inOrden1 (N t i d) = inOrden1 i @ t#inOrden1 d"

value "inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"
value "inOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"

(* manmorjim1 *)
lemma "inOrden t = inOrden1 t"
apply (induct t)
apply auto
done

text {*
---------------------------------------------------------------------
Ejercicio 5. Definir la función
espejo :: "'a arbol ⇒ 'a arbol"
tal que (espejo a) es la imagen especular del árbol a. Por ejemplo,
espejo (N e (N c (H a) (H d)) (N g (H f) (H h)))
= N e (N g (H h) (H f)) (N c (H d) (H a))
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 *)

fun espejo :: "'a arbol ⇒ 'a arbol" where
"espejo (H t) = H t"
| "espejo (N t i d) = N t (espejo d) (espejo i)"

value "espejo (N e (N c (H a) (H d)) (N g (H f) (H h)))
= N e (N g (H h) (H f)) (N c (H d) (H a))"

text {*
---------------------------------------------------------------------
Ejercicio 6. Demostrar que
preOrden (espejo a) = rev (postOrden a)
---------------------------------------------------------------------
*}

(* ivamenjim *)

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))" by simp
also have "... = [x] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp
also have "... = [x] @ rev (postOrden d) @ rev (postOrden i)" using h1 h2 by simp
finally show "preOrden (espejo (N x i d)) = rev (postOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)"
by (induct a) simp_all

(* danrodcha crigomgom*)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))"
by (simp only: espejo.simps(2))
also have "… = x#preOrden (espejo d) @ preOrden (espejo i)"
by (simp only: preOrden.simps(2))
also have"… = x#rev (postOrden d) @ rev (postOrden i)"
using HIi HId by simp
also have "… = rev (postOrden (N x i d))" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 7. Demostrar que
postOrden (espejo a) = rev (preOrden a)
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom*)

lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "postOrden (espejo (N x i d)) = postOrden (N x (espejo d) (espejo i))" by simp
also have "... = (postOrden (espejo d)) @ (postOrden (espejo i)) @ [x]" by simp
also have "... = rev (preOrden d) @ rev (preOrden i) @ [x]" using h1 h2 by simp
finally show "postOrden (espejo (N x i d)) = rev (preOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 8. Demostrar que
inOrden (espejo a) = rev (inOrden a)
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom*)

theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "inOrden (espejo (N x i d)) = inOrden (N x (espejo d) (espejo i))" by simp
also have "... = (inOrden (espejo d)) @ [x] @ (inOrden (espejo i))" by simp
also have "... = rev (inOrden d) @ [x] @ rev (inOrden i)" using h1 h2 by simp
finally show "inOrden (espejo (N x i d)) = rev (inOrden (N x i d))" by simp
qed

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)"
by (induct a) simp_all

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 9. Definir la función
raiz :: "'a arbol ⇒ 'a"
tal que (raiz a) es la raiz del árbol a. Por ejemplo,
raiz (N e (N c (H a) (H d)) (N g (H f) (H h))) = e
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 *)
fun raiz :: "'a arbol ⇒ 'a" where
"raiz (H x) = x"
| "raiz (N x i d) = x"

value "raiz (N e (N c (H a) (H d)) (N g (H f) (H h))) = e"

text {*
---------------------------------------------------------------------
Ejercicio 10. Definir la función
extremo_izquierda :: "'a arbol ⇒ 'a"
tal que (extremo_izquierda a) es el nodo más a la izquierda del árbol
a. Por ejemplo,
extremo_izquierda (N e (N c (H a) (H d)) (N g (H f) (H h))) = a
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 *)
fun extremo_izquierda :: "'a arbol ⇒ 'a" where
"extremo_izquierda (H x) = x"
| "extremo_izquierda (N x i d) = extremo_izquierda i"

value "extremo_izquierda (N e (N c (H a) (H d)) (N g (H f) (H h))) = a"

text {*
---------------------------------------------------------------------
Ejercicio 11. Definir la función
extremo_derecha :: "'a arbol ⇒ 'a"
tal que (extremo_derecha a) es el nodo más a la derecha del árbol
a. Por ejemplo,
extremo_derecha (N e (N c (H a) (H d)) (N g (H f) (H h))) = h
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 *)
fun extremo_derecha :: "'a arbol ⇒ 'a" where
"extremo_derecha (H x) = x"
| "extremo_derecha (N x i d) = extremo_derecha d"

value "extremo_derecha (N e (N c (H a) (H d)) (N g (H f) (H h))) = h"

text {*
---------------------------------------------------------------------
Ejercicio 12. Demostrar o refutar
last (inOrden a) = extremo_derecha a
---------------------------------------------------------------------
*}

(* danrodcha *)
lemma aux_ej12: "inOrden a ≠ []"
apply (induct a)
apply simp
apply simp
done

(* danrodcha *)
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (inOrden (N x i d)) = last (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = last (inOrden d)" by (simp add: aux_ej12)
also have "… = extremo_derecha d" using HId by simp
also have "… = extremo_derecha (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 13. Demostrar o refutar
hd (inOrden a) = extremo_izquierda a
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (inOrden (N x i d)) = hd (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = hd (inOrden i)" by (simp add: aux_ej12)
also have "… = extremo_izquierda i" using HIi by simp
also have "… = extremo_izquierda (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 14. Demostrar o refutar
hd (preOrden a) = last (postOrden a)
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = last (postOrden i @ postOrden d @ [x])" by simp
also have "… = last (postOrden (N x i d))"
by (simp only: postOrden.simps(2))
finally show "?P (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 15. Demostrar o refutar
hd (preOrden a) = raiz a
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 16. Demostrar o refutar
hd (inOrden a) = raiz a
---------------------------------------------------------------------
*}

(*crigomgom*)
theorem "hd (inOrden a) = raiz a"
quickcheck
oops

(* danrodcha:
Auto Quickcheck found a counterexample:
a = N a⇩1 (H a⇩2) (H a⇩1)
Evaluated terms:
hd (inOrden a) = a⇩2
raiz a = a⇩1 *)

text {*
---------------------------------------------------------------------
Ejercicio 17. Demostrar o refutar
last (postOrden a) = raiz a
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (postOrden (N x i d)) = last (postOrden i @ postOrden d @ [x])"
by (simp only: postOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by (simp only: raiz.simps(2))
finally show "?P (N x i d)" by simp
qed

end
</source>

Relación 6

2016-12-03T17:01:18Z

Manmorjim1:

<source lang="isar">
chapter {* R6: Recorridos de árboles *}

theory R6_Recorridos_de_arboles
imports Main
begin

text {*
---------------------------------------------------------------------
Ejercicio 1. Definir el tipo de datos arbol para representar los
árboles binarios que tiene información en los nodos y en las hojas.
Por ejemplo, el árbol
e
/ \
/ \
c g
/ \ / \
a d f h
se representa por "N e (N c (H a) (H d)) (N g (H f) (H h))".
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 manmorjim1 *)

datatype 'a arbol = H "'a" | N "'a" "'a arbol" "'a arbol"

value "N e (N c (H a) (H d)) (N g (H f) (H h))"

text {*
---------------------------------------------------------------------
Ejercicio 2. Definir la función
preOrden :: "'a arbol ⇒ 'a list"
tal que (preOrden a) es el recorrido pre orden del árbol a. Por
ejemplo,
preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1*)

fun preOrden :: "'a arbol ⇒ 'a list" where
"preOrden (H t) = [t]"
| "preOrden (N t i d) = [t] @ (preOrden i) @ (preOrden d)"

(* danrodcha crigomgom manmorjim1 *)
fun preOrden1 :: "'a arbol ⇒ 'a list" where
"preOrden1 (H x) = [x]"
| "preOrden1 (N x i d) = x#preOrden1 i @ preOrden1 d"

value "preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"
value "preOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"

(* danrodcha *)
lemma "preOrden a = preOrden1 a"
by (induct a) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 3. Definir la función
postOrden :: "'a arbol ⇒ 'a list"
tal que (postOrden a) es el recorrido post orden del árbol a. Por
ejemplo,
postOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 *)

fun postOrden :: "'a arbol ⇒ 'a list" where
"postOrden (H t) = [t]"
| "postOrden (N t i d) = (postOrden i) @ (postOrden d) @ [t]"

value "postOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,d,c,f,h,g,e]"

text {*
---------------------------------------------------------------------
Ejercicio 4. Definir la función
inOrden :: "'a arbol ⇒ 'a list"
tal que (inOrden a) es el recorrido in orden del árbol a. Por
ejemplo,
inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom marpoldia1*)

fun inOrden :: "'a arbol ⇒ 'a list" where
"inOrden (H t) = [t]"
| "inOrden (N t i d) = (inOrden i) @ [t] @ (inOrden d)"

(* danrodcha manmorjim1 *)
fun inOrden1 :: "'a arbol ⇒ 'a list" where
"inOrden1 (H t) = [t]"
| "inOrden1 (N t i d) = inOrden1 i @ t#inOrden1 d"

value "inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"
value "inOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"

(* manmorjim1 *)
lemma "inOrden t = inOrden1 t"
apply (induct t)
apply auto
done

text {*
---------------------------------------------------------------------
Ejercicio 5. Definir la función
espejo :: "'a arbol ⇒ 'a arbol"
tal que (espejo a) es la imagen especular del árbol a. Por ejemplo,
espejo (N e (N c (H a) (H d)) (N g (H f) (H h)))
= N e (N g (H h) (H f)) (N c (H d) (H a))
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 *)

fun espejo :: "'a arbol ⇒ 'a arbol" where
"espejo (H t) = H t"
| "espejo (N t i d) = N t (espejo d) (espejo i)"

value "espejo (N e (N c (H a) (H d)) (N g (H f) (H h)))
= N e (N g (H h) (H f)) (N c (H d) (H a))"

text {*
---------------------------------------------------------------------
Ejercicio 6. Demostrar que
preOrden (espejo a) = rev (postOrden a)
---------------------------------------------------------------------
*}

(* ivamenjim *)

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))" by simp
also have "... = [x] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp
also have "... = [x] @ rev (postOrden d) @ rev (postOrden i)" using h1 h2 by simp
finally show "preOrden (espejo (N x i d)) = rev (postOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)"
by (induct a) simp_all

(* danrodcha crigomgom*)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))"
by (simp only: espejo.simps(2))
also have "… = x#preOrden (espejo d) @ preOrden (espejo i)"
by (simp only: preOrden.simps(2))
also have"… = x#rev (postOrden d) @ rev (postOrden i)"
using HIi HId by simp
also have "… = rev (postOrden (N x i d))" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 7. Demostrar que
postOrden (espejo a) = rev (preOrden a)
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom*)

lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "postOrden (espejo (N x i d)) = postOrden (N x (espejo d) (espejo i))" by simp
also have "... = (postOrden (espejo d)) @ (postOrden (espejo i)) @ [x]" by simp
also have "... = rev (preOrden d) @ rev (preOrden i) @ [x]" using h1 h2 by simp
finally show "postOrden (espejo (N x i d)) = rev (preOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 8. Demostrar que
inOrden (espejo a) = rev (inOrden a)
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom*)

theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "inOrden (espejo (N x i d)) = inOrden (N x (espejo d) (espejo i))" by simp
also have "... = (inOrden (espejo d)) @ [x] @ (inOrden (espejo i))" by simp
also have "... = rev (inOrden d) @ [x] @ rev (inOrden i)" using h1 h2 by simp
finally show "inOrden (espejo (N x i d)) = rev (inOrden (N x i d))" by simp
qed

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)"
by (induct a) simp_all

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 9. Definir la función
raiz :: "'a arbol ⇒ 'a"
tal que (raiz a) es la raiz del árbol a. Por ejemplo,
raiz (N e (N c (H a) (H d)) (N g (H f) (H h))) = e
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 *)
fun raiz :: "'a arbol ⇒ 'a" where
"raiz (H x) = x"
| "raiz (N x i d) = x"

value "raiz (N e (N c (H a) (H d)) (N g (H f) (H h))) = e"

text {*
---------------------------------------------------------------------
Ejercicio 10. Definir la función
extremo_izquierda :: "'a arbol ⇒ 'a"
tal que (extremo_izquierda a) es el nodo más a la izquierda del árbol
a. Por ejemplo,
extremo_izquierda (N e (N c (H a) (H d)) (N g (H f) (H h))) = a
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 *)
fun extremo_izquierda :: "'a arbol ⇒ 'a" where
"extremo_izquierda (H x) = x"
| "extremo_izquierda (N x i d) = extremo_izquierda i"

value "extremo_izquierda (N e (N c (H a) (H d)) (N g (H f) (H h))) = a"

text {*
---------------------------------------------------------------------
Ejercicio 11. Definir la función
extremo_derecha :: "'a arbol ⇒ 'a"
tal que (extremo_derecha a) es el nodo más a la derecha del árbol
a. Por ejemplo,
extremo_derecha (N e (N c (H a) (H d)) (N g (H f) (H h))) = h
---------------------------------------------------------------------
*}

(* danrodcha crigomgom*)
fun extremo_derecha :: "'a arbol ⇒ 'a" where
"extremo_derecha (H x) = x"
| "extremo_derecha (N x i d) = extremo_derecha d"

value "extremo_derecha (N e (N c (H a) (H d)) (N g (H f) (H h))) = h"

text {*
---------------------------------------------------------------------
Ejercicio 12. Demostrar o refutar
last (inOrden a) = extremo_derecha a
---------------------------------------------------------------------
*}

(* danrodcha *)
lemma aux_ej12: "inOrden a ≠ []"
apply (induct a)
apply simp
apply simp
done

(* danrodcha *)
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (inOrden (N x i d)) = last (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = last (inOrden d)" by (simp add: aux_ej12)
also have "… = extremo_derecha d" using HId by simp
also have "… = extremo_derecha (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 13. Demostrar o refutar
hd (inOrden a) = extremo_izquierda a
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (inOrden (N x i d)) = hd (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = hd (inOrden i)" by (simp add: aux_ej12)
also have "… = extremo_izquierda i" using HIi by simp
also have "… = extremo_izquierda (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 14. Demostrar o refutar
hd (preOrden a) = last (postOrden a)
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = last (postOrden i @ postOrden d @ [x])" by simp
also have "… = last (postOrden (N x i d))"
by (simp only: postOrden.simps(2))
finally show "?P (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 15. Demostrar o refutar
hd (preOrden a) = raiz a
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 16. Demostrar o refutar
hd (inOrden a) = raiz a
---------------------------------------------------------------------
*}

(*crigomgom*)
theorem "hd (inOrden a) = raiz a"
quickcheck
oops

(* danrodcha:
Auto Quickcheck found a counterexample:
a = N a⇩1 (H a⇩2) (H a⇩1)
Evaluated terms:
hd (inOrden a) = a⇩2
raiz a = a⇩1 *)

text {*
---------------------------------------------------------------------
Ejercicio 17. Demostrar o refutar
last (postOrden a) = raiz a
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (postOrden (N x i d)) = last (postOrden i @ postOrden d @ [x])"
by (simp only: postOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by (simp only: raiz.simps(2))
finally show "?P (N x i d)" by simp
qed

end
</source>

Relación 6

2016-12-03T16:59:15Z

Manmorjim1:

<source lang="isar">
chapter {* R6: Recorridos de árboles *}

theory R6_Recorridos_de_arboles
imports Main
begin

text {*
---------------------------------------------------------------------
Ejercicio 1. Definir el tipo de datos arbol para representar los
árboles binarios que tiene información en los nodos y en las hojas.
Por ejemplo, el árbol
e
/ \
/ \
c g
/ \ / \
a d f h
se representa por "N e (N c (H a) (H d)) (N g (H f) (H h))".
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 manmorjim1 *)

datatype 'a arbol = H "'a" | N "'a" "'a arbol" "'a arbol"

value "N e (N c (H a) (H d)) (N g (H f) (H h))"

text {*
---------------------------------------------------------------------
Ejercicio 2. Definir la función
preOrden :: "'a arbol ⇒ 'a list"
tal que (preOrden a) es el recorrido pre orden del árbol a. Por
ejemplo,
preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1*)

fun preOrden :: "'a arbol ⇒ 'a list" where
"preOrden (H t) = [t]"
| "preOrden (N t i d) = [t] @ (preOrden i) @ (preOrden d)"

(* danrodcha crigomgom manmorjim1 *)
fun preOrden1 :: "'a arbol ⇒ 'a list" where
"preOrden1 (H x) = [x]"
| "preOrden1 (N x i d) = x#preOrden1 i @ preOrden1 d"

value "preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"
value "preOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"

(* danrodcha *)
lemma "preOrden a = preOrden1 a"
by (induct a) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 3. Definir la función
postOrden :: "'a arbol ⇒ 'a list"
tal que (postOrden a) es el recorrido post orden del árbol a. Por
ejemplo,
postOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 *)

fun postOrden :: "'a arbol ⇒ 'a list" where
"postOrden (H t) = [t]"
| "postOrden (N t i d) = (postOrden i) @ (postOrden d) @ [t]"

value "postOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,d,c,f,h,g,e]"

text {*
---------------------------------------------------------------------
Ejercicio 4. Definir la función
inOrden :: "'a arbol ⇒ 'a list"
tal que (inOrden a) es el recorrido in orden del árbol a. Por
ejemplo,
inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom marpoldia1*)

fun inOrden :: "'a arbol ⇒ 'a list" where
"inOrden (H t) = [t]"
| "inOrden (N t i d) = (inOrden i) @ [t] @ (inOrden d)"

(* danrodcha manmorjim1 *)
fun inOrden1 :: "'a arbol ⇒ 'a list" where
"inOrden1 (H t) = [t]"
| "inOrden1 (N t i d) = inOrden1 i @ t#inOrden1 d"

value "inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"
value "inOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"

(* manmorjim1 *)
lemma "inOrden t = inOrden1 t"
apply (induct t)
apply auto
done

text {*
---------------------------------------------------------------------
Ejercicio 5. Definir la función
espejo :: "'a arbol ⇒ 'a arbol"
tal que (espejo a) es la imagen especular del árbol a. Por ejemplo,
espejo (N e (N c (H a) (H d)) (N g (H f) (H h)))
= N e (N g (H h) (H f)) (N c (H d) (H a))
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 *)

fun espejo :: "'a arbol ⇒ 'a arbol" where
"espejo (H t) = H t"
| "espejo (N t i d) = N t (espejo d) (espejo i)"

value "espejo (N e (N c (H a) (H d)) (N g (H f) (H h)))
= N e (N g (H h) (H f)) (N c (H d) (H a))"

text {*
---------------------------------------------------------------------
Ejercicio 6. Demostrar que
preOrden (espejo a) = rev (postOrden a)
---------------------------------------------------------------------
*}

(* ivamenjim *)

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))" by simp
also have "... = [x] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp
also have "... = [x] @ rev (postOrden d) @ rev (postOrden i)" using h1 h2 by simp
finally show "preOrden (espejo (N x i d)) = rev (postOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)"
by (induct a) simp_all

(* danrodcha crigomgom*)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))"
by (simp only: espejo.simps(2))
also have "… = x#preOrden (espejo d) @ preOrden (espejo i)"
by (simp only: preOrden.simps(2))
also have"… = x#rev (postOrden d) @ rev (postOrden i)"
using HIi HId by simp
also have "… = rev (postOrden (N x i d))" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 7. Demostrar que
postOrden (espejo a) = rev (preOrden a)
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom*)

lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "postOrden (espejo (N x i d)) = postOrden (N x (espejo d) (espejo i))" by simp
also have "... = (postOrden (espejo d)) @ (postOrden (espejo i)) @ [x]" by simp
also have "... = rev (preOrden d) @ rev (preOrden i) @ [x]" using h1 h2 by simp
finally show "postOrden (espejo (N x i d)) = rev (preOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 8. Demostrar que
inOrden (espejo a) = rev (inOrden a)
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom*)

theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "inOrden (espejo (N x i d)) = inOrden (N x (espejo d) (espejo i))" by simp
also have "... = (inOrden (espejo d)) @ [x] @ (inOrden (espejo i))" by simp
also have "... = rev (inOrden d) @ [x] @ rev (inOrden i)" using h1 h2 by simp
finally show "inOrden (espejo (N x i d)) = rev (inOrden (N x i d))" by simp
qed

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)"
by (induct a) simp_all

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 9. Definir la función
raiz :: "'a arbol ⇒ 'a"
tal que (raiz a) es la raiz del árbol a. Por ejemplo,
raiz (N e (N c (H a) (H d)) (N g (H f) (H h))) = e
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 *)
fun raiz :: "'a arbol ⇒ 'a" where
"raiz (H x) = x"
| "raiz (N x i d) = x"

value "raiz (N e (N c (H a) (H d)) (N g (H f) (H h))) = e"

text {*
---------------------------------------------------------------------
Ejercicio 10. Definir la función
extremo_izquierda :: "'a arbol ⇒ 'a"
tal que (extremo_izquierda a) es el nodo más a la izquierda del árbol
a. Por ejemplo,
extremo_izquierda (N e (N c (H a) (H d)) (N g (H f) (H h))) = a
---------------------------------------------------------------------
*}

(* danrodcha crigomgom*)
fun extremo_izquierda :: "'a arbol ⇒ 'a" where
"extremo_izquierda (H x) = x"
| "extremo_izquierda (N x i d) = extremo_izquierda i"

value "extremo_izquierda (N e (N c (H a) (H d)) (N g (H f) (H h))) = a"

text {*
---------------------------------------------------------------------
Ejercicio 11. Definir la función
extremo_derecha :: "'a arbol ⇒ 'a"
tal que (extremo_derecha a) es el nodo más a la derecha del árbol
a. Por ejemplo,
extremo_derecha (N e (N c (H a) (H d)) (N g (H f) (H h))) = h
---------------------------------------------------------------------
*}

(* danrodcha crigomgom*)
fun extremo_derecha :: "'a arbol ⇒ 'a" where
"extremo_derecha (H x) = x"
| "extremo_derecha (N x i d) = extremo_derecha d"

value "extremo_derecha (N e (N c (H a) (H d)) (N g (H f) (H h))) = h"

text {*
---------------------------------------------------------------------
Ejercicio 12. Demostrar o refutar
last (inOrden a) = extremo_derecha a
---------------------------------------------------------------------
*}

(* danrodcha *)
lemma aux_ej12: "inOrden a ≠ []"
apply (induct a)
apply simp
apply simp
done

(* danrodcha *)
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (inOrden (N x i d)) = last (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = last (inOrden d)" by (simp add: aux_ej12)
also have "… = extremo_derecha d" using HId by simp
also have "… = extremo_derecha (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 13. Demostrar o refutar
hd (inOrden a) = extremo_izquierda a
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (inOrden (N x i d)) = hd (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = hd (inOrden i)" by (simp add: aux_ej12)
also have "… = extremo_izquierda i" using HIi by simp
also have "… = extremo_izquierda (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 14. Demostrar o refutar
hd (preOrden a) = last (postOrden a)
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = last (postOrden i @ postOrden d @ [x])" by simp
also have "… = last (postOrden (N x i d))"
by (simp only: postOrden.simps(2))
finally show "?P (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 15. Demostrar o refutar
hd (preOrden a) = raiz a
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 16. Demostrar o refutar
hd (inOrden a) = raiz a
---------------------------------------------------------------------
*}

(*crigomgom*)
theorem "hd (inOrden a) = raiz a"
quickcheck
oops

(* danrodcha:
Auto Quickcheck found a counterexample:
a = N a⇩1 (H a⇩2) (H a⇩1)
Evaluated terms:
hd (inOrden a) = a⇩2
raiz a = a⇩1 *)

text {*
---------------------------------------------------------------------
Ejercicio 17. Demostrar o refutar
last (postOrden a) = raiz a
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (postOrden (N x i d)) = last (postOrden i @ postOrden d @ [x])"
by (simp only: postOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by (simp only: raiz.simps(2))
finally show "?P (N x i d)" by simp
qed

end
</source>

Relación 6

2016-12-03T16:38:19Z

Manmorjim1:

<source lang="isar">
chapter {* R6: Recorridos de árboles *}

theory R6_Recorridos_de_arboles
imports Main
begin

text {*
---------------------------------------------------------------------
Ejercicio 1. Definir el tipo de datos arbol para representar los
árboles binarios que tiene información en los nodos y en las hojas.
Por ejemplo, el árbol
e
/ \
/ \
c g
/ \ / \
a d f h
se representa por "N e (N c (H a) (H d)) (N g (H f) (H h))".
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 manmorjim1 *)

datatype 'a arbol = H "'a" | N "'a" "'a arbol" "'a arbol"

value "N e (N c (H a) (H d)) (N g (H f) (H h))"

text {*
---------------------------------------------------------------------
Ejercicio 2. Definir la función
preOrden :: "'a arbol ⇒ 'a list"
tal que (preOrden a) es el recorrido pre orden del árbol a. Por
ejemplo,
preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1*)

fun preOrden :: "'a arbol ⇒ 'a list" where
"preOrden (H t) = [t]"
| "preOrden (N t i d) = [t] @ (preOrden i) @ (preOrden d)"

(* danrodcha crigomgom manmorjim1 *)
fun preOrden1 :: "'a arbol ⇒ 'a list" where
"preOrden1 (H x) = [x]"
| "preOrden1 (N x i d) = x#preOrden1 i @ preOrden1 d"

value "preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"
value "preOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"

(* danrodcha *)
lemma "preOrden a = preOrden1 a"
by (induct a) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 3. Definir la función
postOrden :: "'a arbol ⇒ 'a list"
tal que (postOrden a) es el recorrido post orden del árbol a. Por
ejemplo,
postOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 *)

fun postOrden :: "'a arbol ⇒ 'a list" where
"postOrden (H t) = [t]"
| "postOrden (N t i d) = (postOrden i) @ (postOrden d) @ [t]"

value "postOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,d,c,f,h,g,e]"

text {*
---------------------------------------------------------------------
Ejercicio 4. Definir la función
inOrden :: "'a arbol ⇒ 'a list"
tal que (inOrden a) es el recorrido in orden del árbol a. Por
ejemplo,
inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom marpoldia1*)

fun inOrden :: "'a arbol ⇒ 'a list" where
"inOrden (H t) = [t]"
| "inOrden (N t i d) = (inOrden i) @ [t] @ (inOrden d)"

(* danrodcha manmorjim1 *)
fun inOrden1 :: "'a arbol ⇒ 'a list" where
"inOrden1 (H t) = [t]"
| "inOrden1 (N t i d) = inOrden1 i @ t#inOrden1 d"

value "inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"
value "inOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"

(* manmorjim1 *)
lemma "inOrden t = inOrden1 t"
apply (induct t)
apply auto
done

text {*
---------------------------------------------------------------------
Ejercicio 5. Definir la función
espejo :: "'a arbol ⇒ 'a arbol"
tal que (espejo a) es la imagen especular del árbol a. Por ejemplo,
espejo (N e (N c (H a) (H d)) (N g (H f) (H h)))
= N e (N g (H h) (H f)) (N c (H d) (H a))
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 *)

fun espejo :: "'a arbol ⇒ 'a arbol" where
"espejo (H t) = H t"
| "espejo (N t i d) = N t (espejo d) (espejo i)"

value "espejo (N e (N c (H a) (H d)) (N g (H f) (H h)))
= N e (N g (H h) (H f)) (N c (H d) (H a))"

text {*
---------------------------------------------------------------------
Ejercicio 6. Demostrar que
preOrden (espejo a) = rev (postOrden a)
---------------------------------------------------------------------
*}

(* ivamenjim *)

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))" by simp
also have "... = [x] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp
also have "... = [x] @ rev (postOrden d) @ rev (postOrden i)" using h1 h2 by simp
finally show "preOrden (espejo (N x i d)) = rev (postOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)"
by (induct a) simp_all

(* danrodcha crigomgom*)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))"
by (simp only: espejo.simps(2))
also have "… = x#preOrden (espejo d) @ preOrden (espejo i)"
by (simp only: preOrden.simps(2))
also have"… = x#rev (postOrden d) @ rev (postOrden i)"
using HIi HId by simp
also have "… = rev (postOrden (N x i d))" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 7. Demostrar que
postOrden (espejo a) = rev (preOrden a)
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom*)

lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "postOrden (espejo (N x i d)) = postOrden (N x (espejo d) (espejo i))" by simp
also have "... = (postOrden (espejo d)) @ (postOrden (espejo i)) @ [x]" by simp
also have "... = rev (preOrden d) @ rev (preOrden i) @ [x]" using h1 h2 by simp
finally show "postOrden (espejo (N x i d)) = rev (preOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 8. Demostrar que
inOrden (espejo a) = rev (inOrden a)
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom*)

theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "inOrden (espejo (N x i d)) = inOrden (N x (espejo d) (espejo i))" by simp
also have "... = (inOrden (espejo d)) @ [x] @ (inOrden (espejo i))" by simp
also have "... = rev (inOrden d) @ [x] @ rev (inOrden i)" using h1 h2 by simp
finally show "inOrden (espejo (N x i d)) = rev (inOrden (N x i d))" by simp
qed

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)"
by (induct a) simp_all

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 9. Definir la función
raiz :: "'a arbol ⇒ 'a"
tal que (raiz a) es la raiz del árbol a. Por ejemplo,
raiz (N e (N c (H a) (H d)) (N g (H f) (H h))) = e
---------------------------------------------------------------------
*}

(* danrodcha crigomgom*)
fun raiz :: "'a arbol ⇒ 'a" where
"raiz (H x) = x"
| "raiz (N x i d) = x"

value "raiz (N e (N c (H a) (H d)) (N g (H f) (H h))) = e"

text {*
---------------------------------------------------------------------
Ejercicio 10. Definir la función
extremo_izquierda :: "'a arbol ⇒ 'a"
tal que (extremo_izquierda a) es el nodo más a la izquierda del árbol
a. Por ejemplo,
extremo_izquierda (N e (N c (H a) (H d)) (N g (H f) (H h))) = a
---------------------------------------------------------------------
*}

(* danrodcha crigomgom*)
fun extremo_izquierda :: "'a arbol ⇒ 'a" where
"extremo_izquierda (H x) = x"
| "extremo_izquierda (N x i d) = extremo_izquierda i"

value "extremo_izquierda (N e (N c (H a) (H d)) (N g (H f) (H h))) = a"

text {*
---------------------------------------------------------------------
Ejercicio 11. Definir la función
extremo_derecha :: "'a arbol ⇒ 'a"
tal que (extremo_derecha a) es el nodo más a la derecha del árbol
a. Por ejemplo,
extremo_derecha (N e (N c (H a) (H d)) (N g (H f) (H h))) = h
---------------------------------------------------------------------
*}

(* danrodcha crigomgom*)
fun extremo_derecha :: "'a arbol ⇒ 'a" where
"extremo_derecha (H x) = x"
| "extremo_derecha (N x i d) = extremo_derecha d"

value "extremo_derecha (N e (N c (H a) (H d)) (N g (H f) (H h))) = h"

text {*
---------------------------------------------------------------------
Ejercicio 12. Demostrar o refutar
last (inOrden a) = extremo_derecha a
---------------------------------------------------------------------
*}

(* danrodcha *)
lemma aux_ej12: "inOrden a ≠ []"
apply (induct a)
apply simp
apply simp
done

(* danrodcha *)
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (inOrden (N x i d)) = last (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = last (inOrden d)" by (simp add: aux_ej12)
also have "… = extremo_derecha d" using HId by simp
also have "… = extremo_derecha (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 13. Demostrar o refutar
hd (inOrden a) = extremo_izquierda a
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (inOrden (N x i d)) = hd (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = hd (inOrden i)" by (simp add: aux_ej12)
also have "… = extremo_izquierda i" using HIi by simp
also have "… = extremo_izquierda (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 14. Demostrar o refutar
hd (preOrden a) = last (postOrden a)
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = last (postOrden i @ postOrden d @ [x])" by simp
also have "… = last (postOrden (N x i d))"
by (simp only: postOrden.simps(2))
finally show "?P (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 15. Demostrar o refutar
hd (preOrden a) = raiz a
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 16. Demostrar o refutar
hd (inOrden a) = raiz a
---------------------------------------------------------------------
*}

(*crigomgom*)
theorem "hd (inOrden a) = raiz a"
quickcheck
oops

(* danrodcha:
Auto Quickcheck found a counterexample:
a = N a⇩1 (H a⇩2) (H a⇩1)
Evaluated terms:
hd (inOrden a) = a⇩2
raiz a = a⇩1 *)

text {*
---------------------------------------------------------------------
Ejercicio 17. Demostrar o refutar
last (postOrden a) = raiz a
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (postOrden (N x i d)) = last (postOrden i @ postOrden d @ [x])"
by (simp only: postOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by (simp only: raiz.simps(2))
finally show "?P (N x i d)" by simp
qed

end
</source>

Relación 6

2016-12-03T16:37:30Z

Manmorjim1:

<source lang="isar">
chapter {* R6: Recorridos de árboles *}

theory R6_Recorridos_de_arboles
imports Main
begin

text {*
---------------------------------------------------------------------
Ejercicio 1. Definir el tipo de datos arbol para representar los
árboles binarios que tiene información en los nodos y en las hojas.
Por ejemplo, el árbol
e
/ \
/ \
c g
/ \ / \
a d f h
se representa por "N e (N c (H a) (H d)) (N g (H f) (H h))".
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 manmorjim1 *)

datatype 'a arbol = H "'a" | N "'a" "'a arbol" "'a arbol"

value "N e (N c (H a) (H d)) (N g (H f) (H h))"

text {*
---------------------------------------------------------------------
Ejercicio 2. Definir la función
preOrden :: "'a arbol ⇒ 'a list"
tal que (preOrden a) es el recorrido pre orden del árbol a. Por
ejemplo,
preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1*)

fun preOrden :: "'a arbol ⇒ 'a list" where
"preOrden (H t) = [t]"
| "preOrden (N t i d) = [t] @ (preOrden i) @ (preOrden d)"

(* danrodcha crigomgom manmorjim1 *)
fun preOrden1 :: "'a arbol ⇒ 'a list" where
"preOrden1 (H x) = [x]"
| "preOrden1 (N x i d) = x#preOrden1 i @ preOrden1 d"

value "preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"
value "preOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"

(* danrodcha *)
lemma "preOrden a = preOrden1 a"
by (induct a) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 3. Definir la función
postOrden :: "'a arbol ⇒ 'a list"
tal que (postOrden a) es el recorrido post orden del árbol a. Por
ejemplo,
postOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 *)

fun postOrden :: "'a arbol ⇒ 'a list" where
"postOrden (H t) = [t]"
| "postOrden (N t i d) = (postOrden i) @ (postOrden d) @ [t]"

value "postOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,d,c,f,h,g,e]"

text {*
---------------------------------------------------------------------
Ejercicio 4. Definir la función
inOrden :: "'a arbol ⇒ 'a list"
tal que (inOrden a) es el recorrido in orden del árbol a. Por
ejemplo,
inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom marpoldia1*)

fun inOrden :: "'a arbol ⇒ 'a list" where
"inOrden (H t) = [t]"
| "inOrden (N t i d) = (inOrden i) @ [t] @ (inOrden d)"

(* danrodcha manmorjim1 *)
fun inOrden1 :: "'a arbol ⇒ 'a list" where
"inOrden1 (H t) = [t]"
| "inOrden1 (N t i d) = inOrden1 i @ t#inOrden1 d"

value "inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"
value "inOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"

(* manmorjim1 *)
lemma "inOrden t = inOrden1 t"
apply (induct t)
apply auto
done

text {*
---------------------------------------------------------------------
Ejercicio 5. Definir la función
espejo :: "'a arbol ⇒ 'a arbol"
tal que (espejo a) es la imagen especular del árbol a. Por ejemplo,
espejo (N e (N c (H a) (H d)) (N g (H f) (H h)))
= N e (N g (H h) (H f)) (N c (H d) (H a))
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1*)

fun espejo :: "'a arbol ⇒ 'a arbol" where
"espejo (H t) = H t"
| "espejo (N t i d) = N t (espejo d) (espejo i)"

value "espejo (N e (N c (H a) (H d)) (N g (H f) (H h)))
= N e (N g (H h) (H f)) (N c (H d) (H a))"

text {*
---------------------------------------------------------------------
Ejercicio 6. Demostrar que
preOrden (espejo a) = rev (postOrden a)
---------------------------------------------------------------------
*}

(* ivamenjim *)

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))" by simp
also have "... = [x] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp
also have "... = [x] @ rev (postOrden d) @ rev (postOrden i)" using h1 h2 by simp
finally show "preOrden (espejo (N x i d)) = rev (postOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)"
by (induct a) simp_all

(* danrodcha crigomgom*)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))"
by (simp only: espejo.simps(2))
also have "… = x#preOrden (espejo d) @ preOrden (espejo i)"
by (simp only: preOrden.simps(2))
also have"… = x#rev (postOrden d) @ rev (postOrden i)"
using HIi HId by simp
also have "… = rev (postOrden (N x i d))" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 7. Demostrar que
postOrden (espejo a) = rev (preOrden a)
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom*)

lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "postOrden (espejo (N x i d)) = postOrden (N x (espejo d) (espejo i))" by simp
also have "... = (postOrden (espejo d)) @ (postOrden (espejo i)) @ [x]" by simp
also have "... = rev (preOrden d) @ rev (preOrden i) @ [x]" using h1 h2 by simp
finally show "postOrden (espejo (N x i d)) = rev (preOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 8. Demostrar que
inOrden (espejo a) = rev (inOrden a)
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom*)

theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "inOrden (espejo (N x i d)) = inOrden (N x (espejo d) (espejo i))" by simp
also have "... = (inOrden (espejo d)) @ [x] @ (inOrden (espejo i))" by simp
also have "... = rev (inOrden d) @ [x] @ rev (inOrden i)" using h1 h2 by simp
finally show "inOrden (espejo (N x i d)) = rev (inOrden (N x i d))" by simp
qed

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)"
by (induct a) simp_all

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 9. Definir la función
raiz :: "'a arbol ⇒ 'a"
tal que (raiz a) es la raiz del árbol a. Por ejemplo,
raiz (N e (N c (H a) (H d)) (N g (H f) (H h))) = e
---------------------------------------------------------------------
*}

(* danrodcha crigomgom*)
fun raiz :: "'a arbol ⇒ 'a" where
"raiz (H x) = x"
| "raiz (N x i d) = x"

value "raiz (N e (N c (H a) (H d)) (N g (H f) (H h))) = e"

text {*
---------------------------------------------------------------------
Ejercicio 10. Definir la función
extremo_izquierda :: "'a arbol ⇒ 'a"
tal que (extremo_izquierda a) es el nodo más a la izquierda del árbol
a. Por ejemplo,
extremo_izquierda (N e (N c (H a) (H d)) (N g (H f) (H h))) = a
---------------------------------------------------------------------
*}

(* danrodcha crigomgom*)
fun extremo_izquierda :: "'a arbol ⇒ 'a" where
"extremo_izquierda (H x) = x"
| "extremo_izquierda (N x i d) = extremo_izquierda i"

value "extremo_izquierda (N e (N c (H a) (H d)) (N g (H f) (H h))) = a"

text {*
---------------------------------------------------------------------
Ejercicio 11. Definir la función
extremo_derecha :: "'a arbol ⇒ 'a"
tal que (extremo_derecha a) es el nodo más a la derecha del árbol
a. Por ejemplo,
extremo_derecha (N e (N c (H a) (H d)) (N g (H f) (H h))) = h
---------------------------------------------------------------------
*}

(* danrodcha crigomgom*)
fun extremo_derecha :: "'a arbol ⇒ 'a" where
"extremo_derecha (H x) = x"
| "extremo_derecha (N x i d) = extremo_derecha d"

value "extremo_derecha (N e (N c (H a) (H d)) (N g (H f) (H h))) = h"

text {*
---------------------------------------------------------------------
Ejercicio 12. Demostrar o refutar
last (inOrden a) = extremo_derecha a
---------------------------------------------------------------------
*}

(* danrodcha *)
lemma aux_ej12: "inOrden a ≠ []"
apply (induct a)
apply simp
apply simp
done

(* danrodcha *)
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (inOrden (N x i d)) = last (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = last (inOrden d)" by (simp add: aux_ej12)
also have "… = extremo_derecha d" using HId by simp
also have "… = extremo_derecha (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 13. Demostrar o refutar
hd (inOrden a) = extremo_izquierda a
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (inOrden (N x i d)) = hd (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = hd (inOrden i)" by (simp add: aux_ej12)
also have "… = extremo_izquierda i" using HIi by simp
also have "… = extremo_izquierda (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 14. Demostrar o refutar
hd (preOrden a) = last (postOrden a)
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = last (postOrden i @ postOrden d @ [x])" by simp
also have "… = last (postOrden (N x i d))"
by (simp only: postOrden.simps(2))
finally show "?P (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 15. Demostrar o refutar
hd (preOrden a) = raiz a
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 16. Demostrar o refutar
hd (inOrden a) = raiz a
---------------------------------------------------------------------
*}

(*crigomgom*)
theorem "hd (inOrden a) = raiz a"
quickcheck
oops

(* danrodcha:
Auto Quickcheck found a counterexample:
a = N a⇩1 (H a⇩2) (H a⇩1)
Evaluated terms:
hd (inOrden a) = a⇩2
raiz a = a⇩1 *)

text {*
---------------------------------------------------------------------
Ejercicio 17. Demostrar o refutar
last (postOrden a) = raiz a
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (postOrden (N x i d)) = last (postOrden i @ postOrden d @ [x])"
by (simp only: postOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by (simp only: raiz.simps(2))
finally show "?P (N x i d)" by simp
qed

end
</source>

Relación 6

2016-12-03T16:03:18Z

Manmorjim1:

<source lang="isar">
chapter {* R6: Recorridos de árboles *}

theory R6_Recorridos_de_arboles
imports Main
begin

text {*
---------------------------------------------------------------------
Ejercicio 1. Definir el tipo de datos arbol para representar los
árboles binarios que tiene información en los nodos y en las hojas.
Por ejemplo, el árbol
e
/ \
/ \
c g
/ \ / \
a d f h
se representa por "N e (N c (H a) (H d)) (N g (H f) (H h))".
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 manmorjim1 *)

datatype 'a arbol = H "'a" | N "'a" "'a arbol" "'a arbol"

value "N e (N c (H a) (H d)) (N g (H f) (H h))"

text {*
---------------------------------------------------------------------
Ejercicio 2. Definir la función
preOrden :: "'a arbol ⇒ 'a list"
tal que (preOrden a) es el recorrido pre orden del árbol a. Por
ejemplo,
preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1*)

fun preOrden :: "'a arbol ⇒ 'a list" where
"preOrden (H t) = [t]"
| "preOrden (N t i d) = [t] @ (preOrden i) @ (preOrden d)"

(* danrodcha crigomgom manmorjim1 *)
fun preOrden1 :: "'a arbol ⇒ 'a list" where
"preOrden1 (H x) = [x]"
| "preOrden1 (N x i d) = x#preOrden1 i @ preOrden1 d"

value "preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"
value "preOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"

(* danrodcha *)
lemma "preOrden a = preOrden1 a"
by (induct a) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 3. Definir la función
postOrden :: "'a arbol ⇒ 'a list"
tal que (postOrden a) es el recorrido post orden del árbol a. Por
ejemplo,
postOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 *)

fun postOrden :: "'a arbol ⇒ 'a list" where
"postOrden (H t) = [t]"
| "postOrden (N t i d) = (postOrden i) @ (postOrden d) @ [t]"

value "postOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,d,c,f,h,g,e]"

text {*
---------------------------------------------------------------------
Ejercicio 4. Definir la función
inOrden :: "'a arbol ⇒ 'a list"
tal que (inOrden a) es el recorrido in orden del árbol a. Por
ejemplo,
inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom marpoldia1*)

fun inOrden :: "'a arbol ⇒ 'a list" where
"inOrden (H t) = [t]"
| "inOrden (N t i d) = (inOrden i) @ [t] @ (inOrden d)"

(* danrodcha manmorjim1 *)
fun inOrden1 :: "'a arbol ⇒ 'a list" where
"inOrden1 (H t) = [t]"
| "inOrden1 (N t i d) = inOrden1 i @ t#inOrden1 d"

value "inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"
value "inOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"

(* manmorjim1 *)
lemma "inOrden t = inOrden1 t"
apply (induct t)
apply auto

text {*
---------------------------------------------------------------------
Ejercicio 5. Definir la función
espejo :: "'a arbol ⇒ 'a arbol"
tal que (espejo a) es la imagen especular del árbol a. Por ejemplo,
espejo (N e (N c (H a) (H d)) (N g (H f) (H h)))
= N e (N g (H h) (H f)) (N c (H d) (H a))
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1*)

fun espejo :: "'a arbol ⇒ 'a arbol" where
"espejo (H t) = H t"
| "espejo (N t i d) = N t (espejo d) (espejo i)"

value "espejo (N e (N c (H a) (H d)) (N g (H f) (H h)))
= N e (N g (H h) (H f)) (N c (H d) (H a))"

text {*
---------------------------------------------------------------------
Ejercicio 6. Demostrar que
preOrden (espejo a) = rev (postOrden a)
---------------------------------------------------------------------
*}

(* ivamenjim *)

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))" by simp
also have "... = [x] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp
also have "... = [x] @ rev (postOrden d) @ rev (postOrden i)" using h1 h2 by simp
finally show "preOrden (espejo (N x i d)) = rev (postOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)"
by (induct a) simp_all

(* danrodcha crigomgom*)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))"
by (simp only: espejo.simps(2))
also have "… = x#preOrden (espejo d) @ preOrden (espejo i)"
by (simp only: preOrden.simps(2))
also have"… = x#rev (postOrden d) @ rev (postOrden i)"
using HIi HId by simp
also have "… = rev (postOrden (N x i d))" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 7. Demostrar que
postOrden (espejo a) = rev (preOrden a)
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom*)

lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "postOrden (espejo (N x i d)) = postOrden (N x (espejo d) (espejo i))" by simp
also have "... = (postOrden (espejo d)) @ (postOrden (espejo i)) @ [x]" by simp
also have "... = rev (preOrden d) @ rev (preOrden i) @ [x]" using h1 h2 by simp
finally show "postOrden (espejo (N x i d)) = rev (preOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 8. Demostrar que
inOrden (espejo a) = rev (inOrden a)
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom*)

theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "inOrden (espejo (N x i d)) = inOrden (N x (espejo d) (espejo i))" by simp
also have "... = (inOrden (espejo d)) @ [x] @ (inOrden (espejo i))" by simp
also have "... = rev (inOrden d) @ [x] @ rev (inOrden i)" using h1 h2 by simp
finally show "inOrden (espejo (N x i d)) = rev (inOrden (N x i d))" by simp
qed

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)"
by (induct a) simp_all

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 9. Definir la función
raiz :: "'a arbol ⇒ 'a"
tal que (raiz a) es la raiz del árbol a. Por ejemplo,
raiz (N e (N c (H a) (H d)) (N g (H f) (H h))) = e
---------------------------------------------------------------------
*}

(* danrodcha crigomgom*)
fun raiz :: "'a arbol ⇒ 'a" where
"raiz (H x) = x"
| "raiz (N x i d) = x"

value "raiz (N e (N c (H a) (H d)) (N g (H f) (H h))) = e"

text {*
---------------------------------------------------------------------
Ejercicio 10. Definir la función
extremo_izquierda :: "'a arbol ⇒ 'a"
tal que (extremo_izquierda a) es el nodo más a la izquierda del árbol
a. Por ejemplo,
extremo_izquierda (N e (N c (H a) (H d)) (N g (H f) (H h))) = a
---------------------------------------------------------------------
*}

(* danrodcha crigomgom*)
fun extremo_izquierda :: "'a arbol ⇒ 'a" where
"extremo_izquierda (H x) = x"
| "extremo_izquierda (N x i d) = extremo_izquierda i"

value "extremo_izquierda (N e (N c (H a) (H d)) (N g (H f) (H h))) = a"

text {*
---------------------------------------------------------------------
Ejercicio 11. Definir la función
extremo_derecha :: "'a arbol ⇒ 'a"
tal que (extremo_derecha a) es el nodo más a la derecha del árbol
a. Por ejemplo,
extremo_derecha (N e (N c (H a) (H d)) (N g (H f) (H h))) = h
---------------------------------------------------------------------
*}

(* danrodcha crigomgom*)
fun extremo_derecha :: "'a arbol ⇒ 'a" where
"extremo_derecha (H x) = x"
| "extremo_derecha (N x i d) = extremo_derecha d"

value "extremo_derecha (N e (N c (H a) (H d)) (N g (H f) (H h))) = h"

text {*
---------------------------------------------------------------------
Ejercicio 12. Demostrar o refutar
last (inOrden a) = extremo_derecha a
---------------------------------------------------------------------
*}

(* danrodcha *)
lemma aux_ej12: "inOrden a ≠ []"
apply (induct a)
apply simp
apply simp
done

(* danrodcha *)
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (inOrden (N x i d)) = last (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = last (inOrden d)" by (simp add: aux_ej12)
also have "… = extremo_derecha d" using HId by simp
also have "… = extremo_derecha (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 13. Demostrar o refutar
hd (inOrden a) = extremo_izquierda a
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (inOrden (N x i d)) = hd (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = hd (inOrden i)" by (simp add: aux_ej12)
also have "… = extremo_izquierda i" using HIi by simp
also have "… = extremo_izquierda (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 14. Demostrar o refutar
hd (preOrden a) = last (postOrden a)
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = last (postOrden i @ postOrden d @ [x])" by simp
also have "… = last (postOrden (N x i d))"
by (simp only: postOrden.simps(2))
finally show "?P (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 15. Demostrar o refutar
hd (preOrden a) = raiz a
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 16. Demostrar o refutar
hd (inOrden a) = raiz a
---------------------------------------------------------------------
*}

(*crigomgom*)
theorem "hd (inOrden a) = raiz a"
quickcheck
oops

(* danrodcha:
Auto Quickcheck found a counterexample:
a = N a⇩1 (H a⇩2) (H a⇩1)
Evaluated terms:
hd (inOrden a) = a⇩2
raiz a = a⇩1 *)

text {*
---------------------------------------------------------------------
Ejercicio 17. Demostrar o refutar
last (postOrden a) = raiz a
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (postOrden (N x i d)) = last (postOrden i @ postOrden d @ [x])"
by (simp only: postOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by (simp only: raiz.simps(2))
finally show "?P (N x i d)" by simp
qed

end
</source>

Relación 6

2016-12-03T16:01:00Z

Manmorjim1:

<source lang="isar">
chapter {* R6: Recorridos de árboles *}

theory R6_Recorridos_de_arboles
imports Main
begin

text {*
---------------------------------------------------------------------
Ejercicio 1. Definir el tipo de datos arbol para representar los
árboles binarios que tiene información en los nodos y en las hojas.
Por ejemplo, el árbol
e
/ \
/ \
c g
/ \ / \
a d f h
se representa por "N e (N c (H a) (H d)) (N g (H f) (H h))".
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 manmorjim1 *)

datatype 'a arbol = H "'a" | N "'a" "'a arbol" "'a arbol"

value "N e (N c (H a) (H d)) (N g (H f) (H h))"

text {*
---------------------------------------------------------------------
Ejercicio 2. Definir la función
preOrden :: "'a arbol ⇒ 'a list"
tal que (preOrden a) es el recorrido pre orden del árbol a. Por
ejemplo,
preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1*)

fun preOrden :: "'a arbol ⇒ 'a list" where
"preOrden (H t) = [t]"
| "preOrden (N t i d) = [t] @ (preOrden i) @ (preOrden d)"

(* danrodcha crigomgom manmorjim1 *)
fun preOrden1 :: "'a arbol ⇒ 'a list" where
"preOrden1 (H x) = [x]"
| "preOrden1 (N x i d) = x#preOrden1 i @ preOrden1 d"

value "preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"
value "preOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"

(* danrodcha *)
lemma "preOrden a = preOrden1 a"
by (induct a) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 3. Definir la función
postOrden :: "'a arbol ⇒ 'a list"
tal que (postOrden a) es el recorrido post orden del árbol a. Por
ejemplo,
postOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 *)

fun postOrden :: "'a arbol ⇒ 'a list" where
"postOrden (H t) = [t]"
| "postOrden (N t i d) = (postOrden i) @ (postOrden d) @ [t]"

value "postOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,d,c,f,h,g,e]"

text {*
---------------------------------------------------------------------
Ejercicio 4. Definir la función
inOrden :: "'a arbol ⇒ 'a list"
tal que (inOrden a) es el recorrido in orden del árbol a. Por
ejemplo,
inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom marpoldia1*)

fun inOrden :: "'a arbol ⇒ 'a list" where
"inOrden (H t) = [t]"
| "inOrden (N t i d) = (inOrden i) @ [t] @ (inOrden d)"

(* danrodcha manmorjim1 *)
fun inOrden1 :: "'a arbol ⇒ 'a list" where
"inOrden1 (H t) = [t]"
| "inOrden1 (N t i d) = inOrden1 i @ t#inOrden1 d"

value "inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"
value "inOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"

text {*
---------------------------------------------------------------------
Ejercicio 5. Definir la función
espejo :: "'a arbol ⇒ 'a arbol"
tal que (espejo a) es la imagen especular del árbol a. Por ejemplo,
espejo (N e (N c (H a) (H d)) (N g (H f) (H h)))
= N e (N g (H h) (H f)) (N c (H d) (H a))
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1*)

fun espejo :: "'a arbol ⇒ 'a arbol" where
"espejo (H t) = H t"
| "espejo (N t i d) = N t (espejo d) (espejo i)"

value "espejo (N e (N c (H a) (H d)) (N g (H f) (H h)))
= N e (N g (H h) (H f)) (N c (H d) (H a))"

text {*
---------------------------------------------------------------------
Ejercicio 6. Demostrar que
preOrden (espejo a) = rev (postOrden a)
---------------------------------------------------------------------
*}

(* ivamenjim *)

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))" by simp
also have "... = [x] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp
also have "... = [x] @ rev (postOrden d) @ rev (postOrden i)" using h1 h2 by simp
finally show "preOrden (espejo (N x i d)) = rev (postOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)"
by (induct a) simp_all

(* danrodcha crigomgom*)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))"
by (simp only: espejo.simps(2))
also have "… = x#preOrden (espejo d) @ preOrden (espejo i)"
by (simp only: preOrden.simps(2))
also have"… = x#rev (postOrden d) @ rev (postOrden i)"
using HIi HId by simp
also have "… = rev (postOrden (N x i d))" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 7. Demostrar que
postOrden (espejo a) = rev (preOrden a)
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom*)

lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "postOrden (espejo (N x i d)) = postOrden (N x (espejo d) (espejo i))" by simp
also have "... = (postOrden (espejo d)) @ (postOrden (espejo i)) @ [x]" by simp
also have "... = rev (preOrden d) @ rev (preOrden i) @ [x]" using h1 h2 by simp
finally show "postOrden (espejo (N x i d)) = rev (preOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 8. Demostrar que
inOrden (espejo a) = rev (inOrden a)
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom*)

theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "inOrden (espejo (N x i d)) = inOrden (N x (espejo d) (espejo i))" by simp
also have "... = (inOrden (espejo d)) @ [x] @ (inOrden (espejo i))" by simp
also have "... = rev (inOrden d) @ [x] @ rev (inOrden i)" using h1 h2 by simp
finally show "inOrden (espejo (N x i d)) = rev (inOrden (N x i d))" by simp
qed

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)"
by (induct a) simp_all

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 9. Definir la función
raiz :: "'a arbol ⇒ 'a"
tal que (raiz a) es la raiz del árbol a. Por ejemplo,
raiz (N e (N c (H a) (H d)) (N g (H f) (H h))) = e
---------------------------------------------------------------------
*}

(* danrodcha crigomgom*)
fun raiz :: "'a arbol ⇒ 'a" where
"raiz (H x) = x"
| "raiz (N x i d) = x"

value "raiz (N e (N c (H a) (H d)) (N g (H f) (H h))) = e"

text {*
---------------------------------------------------------------------
Ejercicio 10. Definir la función
extremo_izquierda :: "'a arbol ⇒ 'a"
tal que (extremo_izquierda a) es el nodo más a la izquierda del árbol
a. Por ejemplo,
extremo_izquierda (N e (N c (H a) (H d)) (N g (H f) (H h))) = a
---------------------------------------------------------------------
*}

(* danrodcha crigomgom*)
fun extremo_izquierda :: "'a arbol ⇒ 'a" where
"extremo_izquierda (H x) = x"
| "extremo_izquierda (N x i d) = extremo_izquierda i"

value "extremo_izquierda (N e (N c (H a) (H d)) (N g (H f) (H h))) = a"

text {*
---------------------------------------------------------------------
Ejercicio 11. Definir la función
extremo_derecha :: "'a arbol ⇒ 'a"
tal que (extremo_derecha a) es el nodo más a la derecha del árbol
a. Por ejemplo,
extremo_derecha (N e (N c (H a) (H d)) (N g (H f) (H h))) = h
---------------------------------------------------------------------
*}

(* danrodcha crigomgom*)
fun extremo_derecha :: "'a arbol ⇒ 'a" where
"extremo_derecha (H x) = x"
| "extremo_derecha (N x i d) = extremo_derecha d"

value "extremo_derecha (N e (N c (H a) (H d)) (N g (H f) (H h))) = h"

text {*
---------------------------------------------------------------------
Ejercicio 12. Demostrar o refutar
last (inOrden a) = extremo_derecha a
---------------------------------------------------------------------
*}

(* danrodcha *)
lemma aux_ej12: "inOrden a ≠ []"
apply (induct a)
apply simp
apply simp
done

(* danrodcha *)
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (inOrden (N x i d)) = last (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = last (inOrden d)" by (simp add: aux_ej12)
also have "… = extremo_derecha d" using HId by simp
also have "… = extremo_derecha (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 13. Demostrar o refutar
hd (inOrden a) = extremo_izquierda a
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (inOrden (N x i d)) = hd (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = hd (inOrden i)" by (simp add: aux_ej12)
also have "… = extremo_izquierda i" using HIi by simp
also have "… = extremo_izquierda (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 14. Demostrar o refutar
hd (preOrden a) = last (postOrden a)
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = last (postOrden i @ postOrden d @ [x])" by simp
also have "… = last (postOrden (N x i d))"
by (simp only: postOrden.simps(2))
finally show "?P (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 15. Demostrar o refutar
hd (preOrden a) = raiz a
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 16. Demostrar o refutar
hd (inOrden a) = raiz a
---------------------------------------------------------------------
*}

(*crigomgom*)
theorem "hd (inOrden a) = raiz a"
quickcheck
oops

(* danrodcha:
Auto Quickcheck found a counterexample:
a = N a⇩1 (H a⇩2) (H a⇩1)
Evaluated terms:
hd (inOrden a) = a⇩2
raiz a = a⇩1 *)

text {*
---------------------------------------------------------------------
Ejercicio 17. Demostrar o refutar
last (postOrden a) = raiz a
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (postOrden (N x i d)) = last (postOrden i @ postOrden d @ [x])"
by (simp only: postOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by (simp only: raiz.simps(2))
finally show "?P (N x i d)" by simp
qed

end
</source>

Relación 6

2016-12-03T15:54:19Z

Manmorjim1:

<source lang="isar">
chapter {* R6: Recorridos de árboles *}

theory R6_Recorridos_de_arboles
imports Main
begin

text {*
---------------------------------------------------------------------
Ejercicio 1. Definir el tipo de datos arbol para representar los
árboles binarios que tiene información en los nodos y en las hojas.
Por ejemplo, el árbol
e
/ \
/ \
c g
/ \ / \
a d f h
se representa por "N e (N c (H a) (H d)) (N g (H f) (H h))".
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 manmorjim1 *)

datatype 'a arbol = H "'a" | N "'a" "'a arbol" "'a arbol"

value "N e (N c (H a) (H d)) (N g (H f) (H h))"

text {*
---------------------------------------------------------------------
Ejercicio 2. Definir la función
preOrden :: "'a arbol ⇒ 'a list"
tal que (preOrden a) es el recorrido pre orden del árbol a. Por
ejemplo,
preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1*)

fun preOrden :: "'a arbol ⇒ 'a list" where
"preOrden (H t) = [t]"
| "preOrden (N t i d) = [t] @ (preOrden i) @ (preOrden d)"

(* danrodcha crigomgom manmorjim1 *)
fun preOrden1 :: "'a arbol ⇒ 'a list" where
"preOrden1 (H x) = [x]"
| "preOrden1 (N x i d) = x#preOrden1 i @ preOrden1 d"

value "preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"
value "preOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"

(* danrodcha *)
lemma "preOrden a = preOrden1 a"
by (induct a) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 3. Definir la función
postOrden :: "'a arbol ⇒ 'a list"
tal que (postOrden a) es el recorrido post orden del árbol a. Por
ejemplo,
postOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 *)

fun postOrden :: "'a arbol ⇒ 'a list" where
"postOrden (H t) = [t]"
| "postOrden (N t i d) = (postOrden i) @ (postOrden d) @ [t]"

value "postOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,d,c,f,h,g,e]"

text {*
---------------------------------------------------------------------
Ejercicio 4. Definir la función
inOrden :: "'a arbol ⇒ 'a list"
tal que (inOrden a) es el recorrido in orden del árbol a. Por
ejemplo,
inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom marpoldia1*)

fun inOrden :: "'a arbol ⇒ 'a list" where
"inOrden (H t) = [t]"
| "inOrden (N t i d) = (inOrden i) @ [t] @ (inOrden d)"

(* danrodcha *)
fun inOrden1 :: "'a arbol ⇒ 'a list" where
"inOrden1 (H t) = [t]"
| "inOrden1 (N t i d) = inOrden1 i @ t#inOrden1 d"

value "inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"
value "inOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"

text {*
---------------------------------------------------------------------
Ejercicio 5. Definir la función
espejo :: "'a arbol ⇒ 'a arbol"
tal que (espejo a) es la imagen especular del árbol a. Por ejemplo,
espejo (N e (N c (H a) (H d)) (N g (H f) (H h)))
= N e (N g (H h) (H f)) (N c (H d) (H a))
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1*)

fun espejo :: "'a arbol ⇒ 'a arbol" where
"espejo (H t) = H t"
| "espejo (N t i d) = N t (espejo d) (espejo i)"

value "espejo (N e (N c (H a) (H d)) (N g (H f) (H h)))
= N e (N g (H h) (H f)) (N c (H d) (H a))"

text {*
---------------------------------------------------------------------
Ejercicio 6. Demostrar que
preOrden (espejo a) = rev (postOrden a)
---------------------------------------------------------------------
*}

(* ivamenjim *)

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))" by simp
also have "... = [x] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp
also have "... = [x] @ rev (postOrden d) @ rev (postOrden i)" using h1 h2 by simp
finally show "preOrden (espejo (N x i d)) = rev (postOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)"
by (induct a) simp_all

(* danrodcha crigomgom*)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))"
by (simp only: espejo.simps(2))
also have "… = x#preOrden (espejo d) @ preOrden (espejo i)"
by (simp only: preOrden.simps(2))
also have"… = x#rev (postOrden d) @ rev (postOrden i)"
using HIi HId by simp
also have "… = rev (postOrden (N x i d))" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 7. Demostrar que
postOrden (espejo a) = rev (preOrden a)
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom*)

lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "postOrden (espejo (N x i d)) = postOrden (N x (espejo d) (espejo i))" by simp
also have "... = (postOrden (espejo d)) @ (postOrden (espejo i)) @ [x]" by simp
also have "... = rev (preOrden d) @ rev (preOrden i) @ [x]" using h1 h2 by simp
finally show "postOrden (espejo (N x i d)) = rev (preOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 8. Demostrar que
inOrden (espejo a) = rev (inOrden a)
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom*)

theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "inOrden (espejo (N x i d)) = inOrden (N x (espejo d) (espejo i))" by simp
also have "... = (inOrden (espejo d)) @ [x] @ (inOrden (espejo i))" by simp
also have "... = rev (inOrden d) @ [x] @ rev (inOrden i)" using h1 h2 by simp
finally show "inOrden (espejo (N x i d)) = rev (inOrden (N x i d))" by simp
qed

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)"
by (induct a) simp_all

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 9. Definir la función
raiz :: "'a arbol ⇒ 'a"
tal que (raiz a) es la raiz del árbol a. Por ejemplo,
raiz (N e (N c (H a) (H d)) (N g (H f) (H h))) = e
---------------------------------------------------------------------
*}

(* danrodcha crigomgom*)
fun raiz :: "'a arbol ⇒ 'a" where
"raiz (H x) = x"
| "raiz (N x i d) = x"

value "raiz (N e (N c (H a) (H d)) (N g (H f) (H h))) = e"

text {*
---------------------------------------------------------------------
Ejercicio 10. Definir la función
extremo_izquierda :: "'a arbol ⇒ 'a"
tal que (extremo_izquierda a) es el nodo más a la izquierda del árbol
a. Por ejemplo,
extremo_izquierda (N e (N c (H a) (H d)) (N g (H f) (H h))) = a
---------------------------------------------------------------------
*}

(* danrodcha crigomgom*)
fun extremo_izquierda :: "'a arbol ⇒ 'a" where
"extremo_izquierda (H x) = x"
| "extremo_izquierda (N x i d) = extremo_izquierda i"

value "extremo_izquierda (N e (N c (H a) (H d)) (N g (H f) (H h))) = a"

text {*
---------------------------------------------------------------------
Ejercicio 11. Definir la función
extremo_derecha :: "'a arbol ⇒ 'a"
tal que (extremo_derecha a) es el nodo más a la derecha del árbol
a. Por ejemplo,
extremo_derecha (N e (N c (H a) (H d)) (N g (H f) (H h))) = h
---------------------------------------------------------------------
*}

(* danrodcha crigomgom*)
fun extremo_derecha :: "'a arbol ⇒ 'a" where
"extremo_derecha (H x) = x"
| "extremo_derecha (N x i d) = extremo_derecha d"

value "extremo_derecha (N e (N c (H a) (H d)) (N g (H f) (H h))) = h"

text {*
---------------------------------------------------------------------
Ejercicio 12. Demostrar o refutar
last (inOrden a) = extremo_derecha a
---------------------------------------------------------------------
*}

(* danrodcha *)
lemma aux_ej12: "inOrden a ≠ []"
apply (induct a)
apply simp
apply simp
done

(* danrodcha *)
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (inOrden (N x i d)) = last (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = last (inOrden d)" by (simp add: aux_ej12)
also have "… = extremo_derecha d" using HId by simp
also have "… = extremo_derecha (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 13. Demostrar o refutar
hd (inOrden a) = extremo_izquierda a
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (inOrden (N x i d)) = hd (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = hd (inOrden i)" by (simp add: aux_ej12)
also have "… = extremo_izquierda i" using HIi by simp
also have "… = extremo_izquierda (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 14. Demostrar o refutar
hd (preOrden a) = last (postOrden a)
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = last (postOrden i @ postOrden d @ [x])" by simp
also have "… = last (postOrden (N x i d))"
by (simp only: postOrden.simps(2))
finally show "?P (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 15. Demostrar o refutar
hd (preOrden a) = raiz a
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 16. Demostrar o refutar
hd (inOrden a) = raiz a
---------------------------------------------------------------------
*}

(*crigomgom*)
theorem "hd (inOrden a) = raiz a"
quickcheck
oops

(* danrodcha:
Auto Quickcheck found a counterexample:
a = N a⇩1 (H a⇩2) (H a⇩1)
Evaluated terms:
hd (inOrden a) = a⇩2
raiz a = a⇩1 *)

text {*
---------------------------------------------------------------------
Ejercicio 17. Demostrar o refutar
last (postOrden a) = raiz a
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (postOrden (N x i d)) = last (postOrden i @ postOrden d @ [x])"
by (simp only: postOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by (simp only: raiz.simps(2))
finally show "?P (N x i d)" by simp
qed

end
</source>

Relación 6

2016-12-03T15:49:03Z

Manmorjim1:

<source lang="isar">
chapter {* R6: Recorridos de árboles *}

theory R6_Recorridos_de_arboles
imports Main
begin

text {*
---------------------------------------------------------------------
Ejercicio 1. Definir el tipo de datos arbol para representar los
árboles binarios que tiene información en los nodos y en las hojas.
Por ejemplo, el árbol
e
/ \
/ \
c g
/ \ / \
a d f h
se representa por "N e (N c (H a) (H d)) (N g (H f) (H h))".
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 manmorjim1 *)

datatype 'a arbol = H "'a" | N "'a" "'a arbol" "'a arbol"

value "N e (N c (H a) (H d)) (N g (H f) (H h))"

text {*
---------------------------------------------------------------------
Ejercicio 2. Definir la función
preOrden :: "'a arbol ⇒ 'a list"
tal que (preOrden a) es el recorrido pre orden del árbol a. Por
ejemplo,
preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1*)

fun preOrden :: "'a arbol ⇒ 'a list" where
"preOrden (H t) = [t]"
| "preOrden (N t i d) = [t] @ (preOrden i) @ (preOrden d)"

(* danrodcha crigomgom manmorjim1 *)
fun preOrden1 :: "'a arbol ⇒ 'a list" where
"preOrden1 (H x) = [x]"
| "preOrden1 (N x i d) = x#preOrden1 i @ preOrden1 d"

value "preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"
value "preOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"

(* danrodcha *)
lemma "preOrden a = preOrden1 a"
by (induct a) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 3. Definir la función
postOrden :: "'a arbol ⇒ 'a list"
tal que (postOrden a) es el recorrido post orden del árbol a. Por
ejemplo,
postOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 *)

fun postOrden :: "'a arbol ⇒ 'a list" where
"postOrden (H t) = [t]"
| "postOrden (N t i d) = (postOrden i) @ (postOrden d) @ [t]"

value "postOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,d,c,f,h,g,e]"

text {*
---------------------------------------------------------------------
Ejercicio 4. Definir la función
inOrden :: "'a arbol ⇒ 'a list"
tal que (inOrden a) es el recorrido in orden del árbol a. Por
ejemplo,
inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom marpoldia1*)

fun inOrden :: "'a arbol ⇒ 'a list" where
"inOrden (H t) = [t]"
| "inOrden (N t i d) = (inOrden i) @ [t] @ (inOrden d)"

(* danrodcha *)
fun inOrden1 :: "'a arbol ⇒ 'a list" where
"inOrden1 (H t) = [t]"
| "inOrden1 (N t i d) = inOrden1 i @ t#inOrden1 d"

value "inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"
value "inOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"

text {*
---------------------------------------------------------------------
Ejercicio 5. Definir la función
espejo :: "'a arbol ⇒ 'a arbol"
tal que (espejo a) es la imagen especular del árbol a. Por ejemplo,
espejo (N e (N c (H a) (H d)) (N g (H f) (H h)))
= N e (N g (H h) (H f)) (N c (H d) (H a))
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1*)

fun espejo :: "'a arbol ⇒ 'a arbol" where
"espejo (H t) = H t"
| "espejo (N t i d) = N t (espejo d) (espejo i)"

value "espejo (N e (N c (H a) (H d)) (N g (H f) (H h)))
= N e (N g (H h) (H f)) (N c (H d) (H a))"

text {*
---------------------------------------------------------------------
Ejercicio 6. Demostrar que
preOrden (espejo a) = rev (postOrden a)
---------------------------------------------------------------------
*}

(* ivamenjim *)

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))" by simp
also have "... = [x] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp
also have "... = [x] @ rev (postOrden d) @ rev (postOrden i)" using h1 h2 by simp
finally show "preOrden (espejo (N x i d)) = rev (postOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)"
by (induct a) simp_all

(* danrodcha crigomgom*)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))"
by (simp only: espejo.simps(2))
also have "… = x#preOrden (espejo d) @ preOrden (espejo i)"
by (simp only: preOrden.simps(2))
also have"… = x#rev (postOrden d) @ rev (postOrden i)"
using HIi HId by simp
also have "… = rev (postOrden (N x i d))" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 7. Demostrar que
postOrden (espejo a) = rev (preOrden a)
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom*)

lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "postOrden (espejo (N x i d)) = postOrden (N x (espejo d) (espejo i))" by simp
also have "... = (postOrden (espejo d)) @ (postOrden (espejo i)) @ [x]" by simp
also have "... = rev (preOrden d) @ rev (preOrden i) @ [x]" using h1 h2 by simp
finally show "postOrden (espejo (N x i d)) = rev (preOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 8. Demostrar que
inOrden (espejo a) = rev (inOrden a)
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom*)

theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "inOrden (espejo (N x i d)) = inOrden (N x (espejo d) (espejo i))" by simp
also have "... = (inOrden (espejo d)) @ [x] @ (inOrden (espejo i))" by simp
also have "... = rev (inOrden d) @ [x] @ rev (inOrden i)" using h1 h2 by simp
finally show "inOrden (espejo (N x i d)) = rev (inOrden (N x i d))" by simp
qed

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)"
by (induct a) simp_all

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 9. Definir la función
raiz :: "'a arbol ⇒ 'a"
tal que (raiz a) es la raiz del árbol a. Por ejemplo,
raiz (N e (N c (H a) (H d)) (N g (H f) (H h))) = e
---------------------------------------------------------------------
*}

(* danrodcha crigomgom*)
fun raiz :: "'a arbol ⇒ 'a" where
"raiz (H x) = x"
| "raiz (N x i d) = x"

value "raiz (N e (N c (H a) (H d)) (N g (H f) (H h))) = e"

text {*
---------------------------------------------------------------------
Ejercicio 10. Definir la función
extremo_izquierda :: "'a arbol ⇒ 'a"
tal que (extremo_izquierda a) es el nodo más a la izquierda del árbol
a. Por ejemplo,
extremo_izquierda (N e (N c (H a) (H d)) (N g (H f) (H h))) = a
---------------------------------------------------------------------
*}

(* danrodcha crigomgom*)
fun extremo_izquierda :: "'a arbol ⇒ 'a" where
"extremo_izquierda (H x) = x"
| "extremo_izquierda (N x i d) = extremo_izquierda i"

value "extremo_izquierda (N e (N c (H a) (H d)) (N g (H f) (H h))) = a"

text {*
---------------------------------------------------------------------
Ejercicio 11. Definir la función
extremo_derecha :: "'a arbol ⇒ 'a"
tal que (extremo_derecha a) es el nodo más a la derecha del árbol
a. Por ejemplo,
extremo_derecha (N e (N c (H a) (H d)) (N g (H f) (H h))) = h
---------------------------------------------------------------------
*}

(* danrodcha crigomgom*)
fun extremo_derecha :: "'a arbol ⇒ 'a" where
"extremo_derecha (H x) = x"
| "extremo_derecha (N x i d) = extremo_derecha d"

value "extremo_derecha (N e (N c (H a) (H d)) (N g (H f) (H h))) = h"

text {*
---------------------------------------------------------------------
Ejercicio 12. Demostrar o refutar
last (inOrden a) = extremo_derecha a
---------------------------------------------------------------------
*}

(* danrodcha *)
lemma aux_ej12: "inOrden a ≠ []"
apply (induct a)
apply simp
apply simp
done

(* danrodcha *)
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (inOrden (N x i d)) = last (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = last (inOrden d)" by (simp add: aux_ej12)
also have "… = extremo_derecha d" using HId by simp
also have "… = extremo_derecha (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 13. Demostrar o refutar
hd (inOrden a) = extremo_izquierda a
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (inOrden (N x i d)) = hd (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = hd (inOrden i)" by (simp add: aux_ej12)
also have "… = extremo_izquierda i" using HIi by simp
also have "… = extremo_izquierda (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 14. Demostrar o refutar
hd (preOrden a) = last (postOrden a)
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = last (postOrden i @ postOrden d @ [x])" by simp
also have "… = last (postOrden (N x i d))"
by (simp only: postOrden.simps(2))
finally show "?P (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 15. Demostrar o refutar
hd (preOrden a) = raiz a
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 16. Demostrar o refutar
hd (inOrden a) = raiz a
---------------------------------------------------------------------
*}

(*crigomgom*)
theorem "hd (inOrden a) = raiz a"
quickcheck
oops

(* danrodcha:
Auto Quickcheck found a counterexample:
a = N a⇩1 (H a⇩2) (H a⇩1)
Evaluated terms:
hd (inOrden a) = a⇩2
raiz a = a⇩1 *)

text {*
---------------------------------------------------------------------
Ejercicio 17. Demostrar o refutar
last (postOrden a) = raiz a
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (postOrden (N x i d)) = last (postOrden i @ postOrden d @ [x])"
by (simp only: postOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by (simp only: raiz.simps(2))
finally show "?P (N x i d)" by simp
qed

end
</source>

Relación 6

2016-12-03T15:39:21Z

Manmorjim1:

<source lang="isar">
chapter {* R6: Recorridos de árboles *}

theory R6_Recorridos_de_arboles
imports Main
begin

text {*
---------------------------------------------------------------------
Ejercicio 1. Definir el tipo de datos arbol para representar los
árboles binarios que tiene información en los nodos y en las hojas.
Por ejemplo, el árbol
e
/ \
/ \
c g
/ \ / \
a d f h
se representa por "N e (N c (H a) (H d)) (N g (H f) (H h))".
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 manmorjim1 *)

datatype 'a arbol = H "'a" | N "'a" "'a arbol" "'a arbol"

value "N e (N c (H a) (H d)) (N g (H f) (H h))"

text {*
---------------------------------------------------------------------
Ejercicio 2. Definir la función
preOrden :: "'a arbol ⇒ 'a list"
tal que (preOrden a) es el recorrido pre orden del árbol a. Por
ejemplo,
preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1*)

fun preOrden :: "'a arbol ⇒ 'a list" where
"preOrden (H t) = [t]"
| "preOrden (N t i d) = [t] @ (preOrden i) @ (preOrden d)"

(* danrodcha crigomgom *)
fun preOrden1 :: "'a arbol ⇒ 'a list" where
"preOrden1 (H x) = [x]"
| "preOrden1 (N x i d) = x#preOrden1 i @ preOrden1 d"

value "preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"
value "preOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"

(* danrodcha *)
lemma "preOrden a = preOrden1 a"
by (induct a) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 3. Definir la función
postOrden :: "'a arbol ⇒ 'a list"
tal que (postOrden a) es el recorrido post orden del árbol a. Por
ejemplo,
postOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 *)

fun postOrden :: "'a arbol ⇒ 'a list" where
"postOrden (H t) = [t]"
| "postOrden (N t i d) = (postOrden i) @ (postOrden d) @ [t]"

value "postOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,d,c,f,h,g,e]"

text {*
---------------------------------------------------------------------
Ejercicio 4. Definir la función
inOrden :: "'a arbol ⇒ 'a list"
tal que (inOrden a) es el recorrido in orden del árbol a. Por
ejemplo,
inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom marpoldia1*)

fun inOrden :: "'a arbol ⇒ 'a list" where
"inOrden (H t) = [t]"
| "inOrden (N t i d) = (inOrden i) @ [t] @ (inOrden d)"

(* danrodcha *)
fun inOrden1 :: "'a arbol ⇒ 'a list" where
"inOrden1 (H t) = [t]"
| "inOrden1 (N t i d) = inOrden1 i @ t#inOrden1 d"

value "inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"
value "inOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"

text {*
---------------------------------------------------------------------
Ejercicio 5. Definir la función
espejo :: "'a arbol ⇒ 'a arbol"
tal que (espejo a) es la imagen especular del árbol a. Por ejemplo,
espejo (N e (N c (H a) (H d)) (N g (H f) (H h)))
= N e (N g (H h) (H f)) (N c (H d) (H a))
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1*)

fun espejo :: "'a arbol ⇒ 'a arbol" where
"espejo (H t) = H t"
| "espejo (N t i d) = N t (espejo d) (espejo i)"

value "espejo (N e (N c (H a) (H d)) (N g (H f) (H h)))
= N e (N g (H h) (H f)) (N c (H d) (H a))"

text {*
---------------------------------------------------------------------
Ejercicio 6. Demostrar que
preOrden (espejo a) = rev (postOrden a)
---------------------------------------------------------------------
*}

(* ivamenjim *)

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))" by simp
also have "... = [x] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp
also have "... = [x] @ rev (postOrden d) @ rev (postOrden i)" using h1 h2 by simp
finally show "preOrden (espejo (N x i d)) = rev (postOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)"
by (induct a) simp_all

(* danrodcha crigomgom*)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))"
by (simp only: espejo.simps(2))
also have "… = x#preOrden (espejo d) @ preOrden (espejo i)"
by (simp only: preOrden.simps(2))
also have"… = x#rev (postOrden d) @ rev (postOrden i)"
using HIi HId by simp
also have "… = rev (postOrden (N x i d))" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 7. Demostrar que
postOrden (espejo a) = rev (preOrden a)
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom*)

lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "postOrden (espejo (N x i d)) = postOrden (N x (espejo d) (espejo i))" by simp
also have "... = (postOrden (espejo d)) @ (postOrden (espejo i)) @ [x]" by simp
also have "... = rev (preOrden d) @ rev (preOrden i) @ [x]" using h1 h2 by simp
finally show "postOrden (espejo (N x i d)) = rev (preOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 8. Demostrar que
inOrden (espejo a) = rev (inOrden a)
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom*)

theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "inOrden (espejo (N x i d)) = inOrden (N x (espejo d) (espejo i))" by simp
also have "... = (inOrden (espejo d)) @ [x] @ (inOrden (espejo i))" by simp
also have "... = rev (inOrden d) @ [x] @ rev (inOrden i)" using h1 h2 by simp
finally show "inOrden (espejo (N x i d)) = rev (inOrden (N x i d))" by simp
qed

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)"
by (induct a) simp_all

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 9. Definir la función
raiz :: "'a arbol ⇒ 'a"
tal que (raiz a) es la raiz del árbol a. Por ejemplo,
raiz (N e (N c (H a) (H d)) (N g (H f) (H h))) = e
---------------------------------------------------------------------
*}

(* danrodcha crigomgom*)
fun raiz :: "'a arbol ⇒ 'a" where
"raiz (H x) = x"
| "raiz (N x i d) = x"

value "raiz (N e (N c (H a) (H d)) (N g (H f) (H h))) = e"

text {*
---------------------------------------------------------------------
Ejercicio 10. Definir la función
extremo_izquierda :: "'a arbol ⇒ 'a"
tal que (extremo_izquierda a) es el nodo más a la izquierda del árbol
a. Por ejemplo,
extremo_izquierda (N e (N c (H a) (H d)) (N g (H f) (H h))) = a
---------------------------------------------------------------------
*}

(* danrodcha crigomgom*)
fun extremo_izquierda :: "'a arbol ⇒ 'a" where
"extremo_izquierda (H x) = x"
| "extremo_izquierda (N x i d) = extremo_izquierda i"

value "extremo_izquierda (N e (N c (H a) (H d)) (N g (H f) (H h))) = a"

text {*
---------------------------------------------------------------------
Ejercicio 11. Definir la función
extremo_derecha :: "'a arbol ⇒ 'a"
tal que (extremo_derecha a) es el nodo más a la derecha del árbol
a. Por ejemplo,
extremo_derecha (N e (N c (H a) (H d)) (N g (H f) (H h))) = h
---------------------------------------------------------------------
*}

(* danrodcha crigomgom*)
fun extremo_derecha :: "'a arbol ⇒ 'a" where
"extremo_derecha (H x) = x"
| "extremo_derecha (N x i d) = extremo_derecha d"

value "extremo_derecha (N e (N c (H a) (H d)) (N g (H f) (H h))) = h"

text {*
---------------------------------------------------------------------
Ejercicio 12. Demostrar o refutar
last (inOrden a) = extremo_derecha a
---------------------------------------------------------------------
*}

(* danrodcha *)
lemma aux_ej12: "inOrden a ≠ []"
apply (induct a)
apply simp
apply simp
done

(* danrodcha *)
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (inOrden (N x i d)) = last (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = last (inOrden d)" by (simp add: aux_ej12)
also have "… = extremo_derecha d" using HId by simp
also have "… = extremo_derecha (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 13. Demostrar o refutar
hd (inOrden a) = extremo_izquierda a
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (inOrden (N x i d)) = hd (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = hd (inOrden i)" by (simp add: aux_ej12)
also have "… = extremo_izquierda i" using HIi by simp
also have "… = extremo_izquierda (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 14. Demostrar o refutar
hd (preOrden a) = last (postOrden a)
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = last (postOrden i @ postOrden d @ [x])" by simp
also have "… = last (postOrden (N x i d))"
by (simp only: postOrden.simps(2))
finally show "?P (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 15. Demostrar o refutar
hd (preOrden a) = raiz a
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 16. Demostrar o refutar
hd (inOrden a) = raiz a
---------------------------------------------------------------------
*}

(*crigomgom*)
theorem "hd (inOrden a) = raiz a"
quickcheck
oops

(* danrodcha:
Auto Quickcheck found a counterexample:
a = N a⇩1 (H a⇩2) (H a⇩1)
Evaluated terms:
hd (inOrden a) = a⇩2
raiz a = a⇩1 *)

text {*
---------------------------------------------------------------------
Ejercicio 17. Demostrar o refutar
last (postOrden a) = raiz a
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (postOrden (N x i d)) = last (postOrden i @ postOrden d @ [x])"
by (simp only: postOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by (simp only: raiz.simps(2))
finally show "?P (N x i d)" by simp
qed

end
</source>

Relación 5

2016-11-29T23:38:20Z

Manmorjim1:

<source lang="isar">
chapter {* R5: Eliminación de duplicados *}

theory R5_Eliminacion_de_duplicados
imports Main
begin

text {*
---------------------------------------------------------------------
Ejercicio 1. Definir la funcion primitiva recursiva
estaEn :: 'a ⇒ 'a list ⇒ bool
tal que (estaEn x xs) se verifica si el elemento x está en la lista
xs. Por ejemplo,
estaEn (2::nat) [3,2,4] = True
estaEn (1::nat) [3,2,4] = False
---------------------------------------------------------------------
*}

(* crigomgom rubgonmar bowma wilmorort pablucoto serrodcal
anaprarod migtermor paupeddeg fraortmoy marpoldia1
danrodcha manmorjim1 *)
fun estaEn :: "'a ⇒ 'a list ⇒ bool" where
"estaEn _ [] = False"
| "estaEn x (a#xs) = ((a = x) ∨ (estaEn x xs))"

value "estaEn (2::nat) [3,2,4] = True"
value "estaEn (1::nat) [3,2,4] = False"

(* ivamenjim, ferrenseg josgarsan juacabsou dancorgar *)
(* Igual que la anterior pero con x en lugar de _ en el caso base *)

fun estaEn1 :: "'a ⇒ 'a list ⇒ bool" where
"estaEn1 x [] = False"
| "estaEn1 x (a#xs) = ((x=a) ∨ estaEn1 x xs)"

value "estaEn1 (2::nat) [3,2,4] = True"
value "estaEn1 (1::nat) [3,2,4] = False"

(* wilmorort *)
(* reutilizando la funcion "algunos" de R4.thy*)
fun estaEn2 :: "'a ⇒ 'a list ⇒ bool" where
"estaEn2 a xs = algunos (λx. x = a) xs"

value "estaEn2 (2::nat) [3,2,4] = True"
value "estaEn2 (1::nat) [3,2,4] = False"

text {*
---------------------------------------------------------------------
Ejercicio 2. Definir la función primitiva recursiva
sinDuplicados :: 'a list ⇒ bool
tal que (sinDuplicados xs) se verifica si la lista xs no contiene
duplicados. Por ejemplo,
sinDuplicados [1::nat,4,2] = True
sinDuplicados [1::nat,4,2,4] = False
---------------------------------------------------------------------
*}

(* crigomgom rubgonmar ivamenjim wilmorort bowma pablucoto
serrodcal anaprarod migtermor paupeddeg fraortmoy marpoldia1
ferrenseg josgarsan danrodcha manmorjim1 juacabsou dancorgar *)
fun sinDuplicados :: "'a list ⇒ bool" where
"sinDuplicados [] = True"
| "sinDuplicados (x#xs) = (¬ estaEn x xs ∧ sinDuplicados xs)"

value "sinDuplicados [1::nat,4,2] = True"
value "sinDuplicados [1::nat,4,2,4] = False"

(* paupeddeg *)
(* Utilizando la función ∉ de Isabelle *)
fun sinDuplicados2 :: "'a list ⇒ bool" where
"sinDuplicados2 [] = True"
| "sinDuplicados2 (a#xs) = ((a ∉ set xs) ∧ sinDuplicados2 xs ) "

text {*
---------------------------------------------------------------------
Ejercicio 3. Definir la función primitiva recursiva
borraDuplicados :: 'a list ⇒ bool
tal que (borraDuplicados xs) es la lista obtenida eliminando los
elementos duplicados de la lista xs. Por ejemplo,
borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]

Nota: La función borraDuplicados es equivalente a la predefinida
remdups.
---------------------------------------------------------------------
*}

(* crigomgom rubgonmar wilmorort bowma pablucoto serrodcal
anaprarod migtermor paupeddeg fraortmoy marpoldia1 ferrenseg
josgarsan danrodcha juacabsou dancorgar manmorjim1 *)
fun borraDuplicados :: "'a list ⇒ 'a list" where
"borraDuplicados [] = []"
| "borraDuplicados (x#xs) =( if estaEn x xs then borraDuplicados xs else x#borraDuplicados xs)"

value "borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]"

(* ivamenjim *)
(* Utilizando la negación primero *)

fun borraDuplicados :: "'a list ⇒ 'a list" where
"borraDuplicados [] = []"
| "borraDuplicados (x#xs) = (if ¬(estaEn x xs) then (x#(borraDuplicados xs)) else borraDuplicados xs)"

(* rubgonmar *)
(* Otra forma Sin usar if
Utilizando case aunque se le sacaría más partido con más de 2 casos *)

fun borraDuplicados1 :: "'a list ⇒ 'a list" where
"borraDuplicados1 [] = []" |
"borraDuplicados1 (x#xs) = ( case estaEn x xs of False => x#borraDuplicados1 xs | True => borraDuplicados1 xs )"

(* rubgonmar *)
(*Otra forma utilizando let*)
fun borraDuplicados2 :: "'a list ⇒ 'a list" where
"borraDuplicados2 [] = []" |
"borraDuplicados2 (x#xs) = (let condicion = estaEn x xs::bool in
if condicion then borraDuplicados2 xs else x # borraDuplicados2 xs)"

value "borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]"

text {*
---------------------------------------------------------------------
Ejercicio 4.1. Demostrar o refutar automáticamente
length (borraDuplicados xs) ≤ length xs
---------------------------------------------------------------------
*}

-- "La demostración automática es"
(*crigomgom anaprarod ferrenseg*)
lemma length_borraDuplicados:
"length (borraDuplicados xs) ≤ length xs"
by (induct xs, simp_all)

(* rubgonmar wilmorort pablucoto serrodcal migtermor paupeddeg
fraortmoy marpoldia1 danrodcha juacabsou dancorgar *)
lemma length_borraDuplicados:
"length ( borraDuplicados xs ) ≤ length xs"
by (induct xs) auto

(* manmorjim1 *)
(* soy de ponerlo mejor por pasos *)
lemma length_borraDuplicados:
"length ( borraDuplicados xs ) ≤ length xs"
apply (induct xs)
apply auto
done

(* ivamenjim *)
(* Demostrando objetivo a objetivo *)
lemma length_borraDuplicados:
"length (borraDuplicados xs) ≤ length xs"
apply (induct xs)
apply simp
apply auto
done

(* bowma anaprarod *)
lemma length_borraDuplicados:
"length (borraDuplicados xs) ≤ length xs"
apply (induct xs)
apply (simp, simp) (* creo que es mejor poner aquí simp_all *)
done

(* ferrenseg *)

lemma length_borraDuplicados:
"length (borraDuplicados xs) ≤ length xs"
by (induct xs) simp_all (* Creo que se puede poner simp_all fuera de parentesis *)

text {*
---------------------------------------------------------------------
Ejercicio 4.2. Demostrar o refutar detalladamente
length (borraDuplicados xs) ≤ length xs
---------------------------------------------------------------------
*}

-- "La demostración estructurada es"
(* crigomgom *)
lemma length_borraDuplicados_2:
"length (borraDuplicados xs) ≤ length xs"
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix x xs
assume HI: "length (borraDuplicados xs) ≤ length xs"
show "length (borraDuplicados (x#xs)) ≤ length (x#xs)"
proof (cases)
assume "estaEn x xs"
then have "length (borraDuplicados (x#xs)) = length (borraDuplicados xs)" by simp
also have "... ≤ length xs" using HI by simp
also have "... ≤ length (x#xs)" by simp
finally show "length (borraDuplicados (x#xs)) ≤ length (x#xs)" by simp
next
assume "(¬ estaEn x xs)"
then have "length (borraDuplicados (x#xs)) = length (x#borraDuplicados xs)" by simp
also have "... = 1 + length (borraDuplicados xs)" by simp
also have "... ≤ 1 + length xs" using HI by simp
also have "... = length (x#xs)" by simp
finally show "length (borraDuplicados (x#xs)) ≤ length (x#xs)" by simp
qed
qed

(* ivamenjim wilmorort ferrenseg rubgonmar juacabsou dancorgar *)
lemma length_borraDuplicados_2:
"length (borraDuplicados xs) ≤ length xs"
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix a xs
assume HI: "length (borraDuplicados xs) ≤ length xs"
have "length (borraDuplicados (a # xs)) ≤ 1+length (borraDuplicados xs)" by simp
also have "... ≤ 1+length xs" using HI by simp
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs)" by simp
qed

(* serrodcal anaprarod danrodcha *)
lemma length_borraDuplicados_2: "length (borraDuplicados xs) ≤ length xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "length (borraDuplicados (a # xs)) ≤ 1+length (borraDuplicados xs)" by simp
also have "... ≤ 1+length xs" using HI by simp
finally show "?P (a # xs)" by simp
qed

(* pablucoto *)
lemma length_borraDuplicados_2:
"length (borraDuplicados xs) ≤ length xs"
proof(induct xs)
show "length (borraDuplicados []) ≤ length [] " by simp
next
fix a xs
assume HI: " length (borraDuplicados xs) ≤ length xs "
have "length (borraDuplicados (a # xs)) ≤ 1 + length(borraDuplicados xs)" by simp
also have "... ≤ 1 + length xs" using HI by simp
also have "... ≤ length (a#xs)" by simp
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs) " by simp
qed

(* bowma *)
lemma length_borraDuplicados_3:
"length (borraDuplicados xs) ≤ length xs" (is "?p xs")
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix a xs
assume HI: "?p xs"
have "length (a#xs) = 1 + length xs" by simp
have "length(borraDuplicados (a#xs)) ≤ 1 + length(borraDuplicados xs)" by simp
also have "... ≤ 1+length xs" using HI by simp
also have "... ≤ length (a#xs)" by simp
finally show "?p (a # xs)" by simp
qed

(* migtermor *)

lemma length_borraDuplicados_2:
"length (borraDuplicados xs) ≤ length xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "length (borraDuplicados (a#xs)) ≤ (length (a#xs))"
proof (cases)
assume "(estaEn a xs)"
then have Aux: "length (borraDuplicados (a#xs)) = length (borraDuplicados xs)" by simp
also have "… ≤ length (a#xs)" using HI by simp
then show "length (borraDuplicados (a#xs)) ≤ (length (a#xs))" using Aux by simp
next
assume "¬ (estaEn a xs)"
then have Aux: "length (borraDuplicados (a#xs)) = 1+ length (borraDuplicados xs)" by simp
also have "… ≤ length (a#xs)" using HI by simp
then show "length (borraDuplicados (a#xs)) ≤ (length (a#xs))" using Aux by simp
qed
then show "?P (a#xs)" by simp
qed

(* paupeddeg marpoldia1*)
lemma length_borraDuplicados_4:
"length (borraDuplicados xs) ≤ length xs"
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix a xs
assume HI: "length (borraDuplicados xs) ≤ length xs"
have "length (borraDuplicados (a # xs)) ≤ length [a] + length (borraDuplicados xs) " by simp
also have "... ≤ 1 + length (borraDuplicados xs)" by simp
also have "... ≤ 1 + length xs" using HI by simp
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs)" by simp
qed

(* fraortmoy *)
(* muy parecida a alguna anterior, pero yo dí mas pasos *)
lemma length_borraDuplicados_2:
"length (borraDuplicados xs) ≤ length xs"
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix a xs
assume H1: "length (borraDuplicados xs) ≤ length xs"
have "length (borraDuplicados (a # xs)) ≤ length(borraDuplicados [a])+length (borraDuplicados xs)" by simp
also have "… ≤ 1 + length (borraDuplicados xs)" by simp
also have "… ≤ 1 + length xs" using H1 by simp
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 5.1. Demostrar o refutar automáticamente
estaEn a (borraDuplicados xs) = estaEn a xs
---------------------------------------------------------------------
*}

-- "La demostración automática es"
(* crigomgom rubgonmar wilmorort pablucoto serrodcal bowma
migtermor fraortmoy marpoldia1 ferrenseg danrodcha juacabsou paupeddeg*)
lemma estaEn_borraDuplicados:
"estaEn a (borraDuplicados xs) = estaEn a xs"
by (induct xs) auto

(* ivamenjim manmorjim1 *)
lemma estaEn_borraDuplicados:
"estaEn a (borraDuplicados xs) = estaEn a xs"
apply (induct xs)
apply auto
done

(* anaprarod *)
lemma estaEn_borraDuplicados:
"estaEn a (borraDuplicados xs) = estaEn a xs"
by (induct xs, simp_all, blast)

(* anaprarod *)
lemma estaEn_borraDuplicados':
"estaEn a (borraDuplicados xs) = estaEn a xs"
apply (induct xs)
apply (cases "estaEn x xs")
apply (simp_all)
apply blast
done

(* bowma *)
lemma estaEn_borraDuplicados:
"estaEn a (borraDuplicados xs) = estaEn a xs"
apply (induct xs)
apply simp
apply (simp, blast)
done

text {*
---------------------------------------------------------------------
Ejercicio 5.2. Demostrar o refutar detalladamente
estaEn a (borraDuplicados xs) = estaEn a xs
Nota: Para la demostración de la equivalencia se puede usar
proof (rule iffI)
La regla iffI es
⟦P ⟹ Q ; Q ⟹ P⟧ ⟹ P = Q
---------------------------------------------------------------------
*}

-- "La demostración estructurada es"

(* wilmorort *)

lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs"
proof (induct xs)
show "estaEn a (borraDuplicados []) = estaEn a []" by simp
next
fix b xs
assume HI: "estaEn a (borraDuplicados xs) = estaEn a xs"
show "estaEn a (borraDuplicados (b#xs)) = estaEn a (b#xs)"
proof (rule iffI)
assume H1: "estaEn a (borraDuplicados (b#xs))"
show "estaEn a (b#xs)"
proof (cases)
assume "estaEn b xs"
then have "estaEn a (borraDuplicados xs)" using H1 by simp
then have "estaEn a xs" using HI by simp
then show "estaEn a (b#xs)" by simp
next
assume "¬ estaEn b xs"
then have "estaEn a (b#(borraDuplicados xs))" using H1 by simp
then have "a=b ∨ (estaEn a (borraDuplicados xs))" by simp
then have " a=b ∨ (estaEn a xs)" using HI by simp
then show "estaEn a (b#xs)" by simp
qed
next
assume H2: "estaEn a (b#xs)"
show "estaEn a (borraDuplicados (b#xs))"
proof (cases)
assume "a=b"
then have "estaEn b (borraDuplicados xs) = estaEn b xs" using HI by simp
then have "(estaEn b xs ⟶ estaEn b (borraDuplicados xs)) ∧
(¬ estaEn b xs ⟶ estaEn b (b # borraDuplicados xs))" by simp
then have "estaEn b (borraDuplicados (b#xs))" by simp
then show "estaEn a (borraDuplicados (b#xs))" using `a=b` by simp
next
assume "a≠b"
then have "estaEn a (b#xs)" using H2 by simp
then have "a = b ∨ estaEn a xs" by simp
then have "False ∨ estaEn a xs " using `a≠b` by simp
then have "estaEn a xs" by simp
then have "estaEn a (borraDuplicados xs)" using HI by simp
then show "estaEn a (borraDuplicados (b#xs))" using `a≠b` by simp
qed
qed
qed

(* anaprarod marpoldia1 ferrenseg juacabsou *)
lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix x xs
assume HI: "?P xs"
show "?P (x#xs)"
proof (cases)
assume "estaEn x xs"
then have "estaEn a (borraDuplicados (x#xs)) = estaEn a (borraDuplicados xs)" by simp
also have "...= estaEn a xs" using HI by simp
finally show "?P (x#xs)" by auto
next
assume "¬estaEn x xs"
then have "estaEn a (borraDuplicados (x#xs)) = estaEn a (x#borraDuplicados xs)" by simp
also have "...= (x = a ∨ estaEn a (borraDuplicados xs))" by simp
finally show "?P (x#xs)" using HI by simp
qed
qed

(* migtermor *)

lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix aa xs
assume HI: "?P xs"
have P1: "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)"
proof (cases)
assume C1: "(estaEn aa xs)"
have "estaEn a (borraDuplicados (aa#xs)) = estaEn a (borraDuplicados xs)"
using C1 by simp
also have P3: "… = estaEn a xs" using HI by simp
also have "… = estaEn a (aa#xs)"
proof (cases)
assume "(a=aa)"
then show "estaEn a xs = estaEn a (aa#xs)" using C1 by simp
next
assume "¬(a=aa)"
then show "estaEn a xs = estaEn a (aa#xs)" by simp
qed
then show "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)" using P3 by simp
next
assume C2: "¬(estaEn aa xs)"
then show "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)" using HI by simp
qed
also have Conc: "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)" using P1 by simp
finally show "?P (aa#xs)" using Conc by simp
qed

(* crigomgom paupeddeg *)
-- "La demostración estructurada es"
lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs"
proof (induct xs)
show "estaEn a (borraDuplicados []) = estaEn a []" by simp
next
fix x xs
assume HI: "estaEn a (borraDuplicados xs) = estaEn a xs"
show "estaEn a (borraDuplicados (x#xs)) = estaEn a (x#xs)"
proof (rule iffI)
assume a1: "estaEn a (borraDuplicados (x#xs))"
show "estaEn a (x#xs)"
proof (cases)
assume "estaEn x xs"
then have "estaEn a (borraDuplicados xs)" using a1 by simp
then have "estaEn a xs" using HI by simp
then show "estaEn a (x#xs)" by simp
next
assume "¬ estaEn x xs"
then have "estaEn a (x#(borraDuplicados xs))" using a1 by simp
then have " x=a ∨ (estaEn a (borraDuplicados xs))" by simp
then have " x=a ∨ (estaEn a xs)" using HI by simp
then show "estaEn a (x#xs)" by simp
qed
next
assume a2: "estaEn a (x#xs)"
show "estaEn a (borraDuplicados (x#xs))"
proof (cases)
assume "a=x"
then show "estaEn a (borraDuplicados (x#xs))" using HI by simp
next
assume b1: "a≠x"
then have "estaEn a (x#xs)" using a2 by simp
then have "x = a ∨ estaEn a xs" by simp
then have "estaEn a xs " using b1 by simp
then have "estaEn a (borraDuplicados xs)" using HI by simp
then show "estaEn a (borraDuplicados (x#xs))" using b1 by simp
qed
qed
qed

(* rubgonmar *)
lemma estaEn_borraDuplicados_2:
"estaEn a ( borraDuplicados xs ) = estaEn a xs"
proof (induct xs)
show "estaEn a (borraDuplicados []) = estaEn a []" by simp
next
fix x
fix xs
assume HI: "estaEn a (borraDuplicados xs) = estaEn a xs"
show "estaEn a (borraDuplicados (x#xs)) = estaEn a (x#xs)"
proof (rule iffI) (* usamos proof de la regla dada iffI*)
assume cprim: "estaEn a (borraDuplicados (x#xs))"
show "estaEn a (x#xs)"
proof (cases)
assume "estaEn x xs"
then show "estaEn a (x#xs)" using cprim HI by simp
next
assume "¬ estaEn x xs"
then show "estaEn a (x#xs)" using cprim HI by simp
qed
next
assume cseg: "estaEn a (x#xs)"
show "estaEn a (borraDuplicados (x#xs))"
proof (cases)
assume "a=x"
then show "estaEn a (borraDuplicados (x#xs))" using HI by auto
next
assume "a≠x"
then show "estaEn a (borraDuplicados (x#xs))" using `a≠x` cseg HI by simp
qed
qed
qed

(*
Aplico la regla iffI:
⟦P ⟹ Q ; Q ⟹ P⟧ ⟹ P = Q
Así:
[estaEn a (borraDuplicados (x # xs)) ⟹ estaEn a (x # xs); estaEn a (x # xs) ⟹ estaEn a (borraDuplicados (x # xs))]
⟹ estaEn a (borraDuplicados (x # xs)) = estaEn a (x # xs)
*)

(* bowma ivamenjim *)
lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?p xs")
proof (induct xs)
show "?p []" by simp
next
fix x xs
assume HI: "?p xs"
show "?p (x#xs)"
proof (cases)
assume H1:"estaEn x xs"
then have "estaEn a (borraDuplicados (x#xs)) = estaEn a (borraDuplicados xs)" by simp
also have "... = estaEn a xs" using HI by simp
also have "... = estaEn a (x#xs)"
proof(cases)
assume "x=a"
then show "estaEn a xs = estaEn a (x#xs)" using H1 by simp
next
assume "x≠a"
then show "estaEn a xs = estaEn a (x#xs)" by simp
qed
finally show "?p (x#xs)" by simp
next
assume H2:"¬estaEn x xs"
then have "estaEn a (borraDuplicados (x#xs)) = estaEn a (x#borraDuplicados xs)" by simp
also have "... = ((x=a) ∨ estaEn a (borraDuplicados xs))" by simp
also have "... = ((x=a) ∨ estaEn a xs)" using HI by simp
also have "... = estaEn a (x#xs)" by simp
finally show "?p (x#xs)" by simp
qed

(*danrodcha *)
(* es como la de ruben pero con diferencias de estilo *)
lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix x xs assume HI: "?P xs"
show "?P (x#xs)"
proof (rule iffI)
assume H1: "estaEn a (borraDuplicados (x # xs))"
show "estaEn a (x#xs)"
proof (cases "estaEn x xs")
case True
then show "estaEn a (x#xs)" using H1 HI by simp
next
case False
then show "estaEn a (x#xs)" using H1 HI by simp
qed
next
assume H2: "estaEn a (x#xs)"
show "estaEn a (borraDuplicados (x # xs))"
proof (cases "x=a")
case True
then show "estaEn a (borraDuplicados (x # xs))" using HI by simp
next
case False
then show "estaEn a (borraDuplicados (x # xs))" using H2 HI by simp
qed
qed
qed

text {*
---------------------------------------------------------------------
Ejercicio 6.1. Demostrar o refutar automáticamente
sinDuplicados (borraDuplicados xs)
---------------------------------------------------------------------
*}

-- "La demostración automática"

(* ivamenjim wilmorort serrodcal crigomgom anaprarod fraortmoy
marpoldia1 ferrenseg danrodcha juacabsou paupeddeg*)
lemma sinDuplicados_borraDuplicados:
"sinDuplicados (borraDuplicados xs)"
by (induct xs) (auto simp add: estaEn_borraDuplicados)

(* migtermor bowma *)
lemma sinDuplicados_borraDuplicados:
"sinDuplicados (borraDuplicados xs)"
by (induct xs, simp_all add: estaEn_borraDuplicados_2)

(* manmorjim1 no caí en usar la demostración anterior y he realizado
la demostración de que si un elemento no estaba en una lista seguirá sin estar
después de eliminar los duplicados en esa lista... *)
lemma noEsta_tras_borrarDuplicados:
"(¬estaEn x xs) ⟶ (¬estaEn x (borraDuplicados xs))"
apply (induct xs)
apply auto
done

lemma sinDuplicados_borraDuplicados:
"sinDuplicados (borraDuplicados xs)"
apply (induct xs)
apply simp
apply (induct xs)
apply auto
apply (simp add: noEsta_tras_borrarDuplicados)
done

text {*
---------------------------------------------------------------------
Ejercicio 6.2. Demostrar o refutar detalladamente
sinDuplicados (borraDuplicados xs)
---------------------------------------------------------------------
*}

(*wilmorort*)

-- "La demostración estructurada es"
lemma sinDuplicados_borraDuplicados_2:
"sinDuplicados (borraDuplicados xs)"
proof (induct xs)
show " sinDuplicados (borraDuplicados [])" by simp
next
fix a xs
assume HI: "sinDuplicados (borraDuplicados xs)"
show "sinDuplicados (borraDuplicados (a # xs))"
proof (cases)
assume "estaEn a xs"
then show "sinDuplicados (borraDuplicados (a#xs))" using HI by simp
next
assume"¬ estaEn a xs"
then have "¬ (estaEn a xs) ∧ sinDuplicados (borraDuplicados xs)" using HI by simp
then have "¬ estaEn a (borraDuplicados xs) ∧ sinDuplicados (borraDuplicados xs)"
by (simp add: estaEn_borraDuplicados)
then have " sinDuplicados (a#borraDuplicados xs)" by simp
then show " sinDuplicados (borraDuplicados(a #xs))" by simp
qed
qed

(* ivamenjim migtermor crigomgom rubgonmar fraortmoy marpoldia1 ferrenseg bowma juacabsou *)

lemma sinDuplicados_borraDuplicados_2:
"sinDuplicados (borraDuplicados xs)"
proof (induct xs)
show "sinDuplicados (borraDuplicados [])" by simp
next
fix a xs
assume HI: "sinDuplicados (borraDuplicados xs)"
show "sinDuplicados (borraDuplicados (a # xs))"
proof (cases)
assume "estaEn a xs"
then show "sinDuplicados (borraDuplicados (a # xs))" using HI by simp
next
assume "¬(estaEn a xs)"
then show "sinDuplicados (borraDuplicados (a # xs))" using HI by (simp add: estaEn_borraDuplicados)
qed
qed

(* anaprarod paupeddeg*)
lemma sinDuplicados_borraDuplicados_2:
"sinDuplicados (borraDuplicados xs)" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix x xs
assume HI: "?P xs"
show "?P (x#xs)"
proof (cases)
assume c1: "estaEn x xs"
then show "sinDuplicados (borraDuplicados (x#xs))" using HI by simp
next
assume c2: "¬ estaEn x xs"
then have "sinDuplicados (borraDuplicados (x#xs)) =sinDuplicados (x#borraDuplicados xs)" by simp
also have "…= (¬estaEn x (borraDuplicados xs) ∧ sinDuplicados (borraDuplicados xs))" by simp
also have "… = (¬estaEn x (borraDuplicados xs))" using HI by simp
also have "… = (¬(estaEn x xs))" by (simp add:estaEn_borraDuplicados)
also have "… = True" using c2 by simp
finally show "?P (x#xs)" by simp
qed
qed

(* danrodcha *)
lemma sinDuplicados_borraDuplicados_2:
"sinDuplicados (borraDuplicados xs)" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix x xs assume HI: "?P xs"
show "?P (x#xs)"
proof (cases "estaEn x xs")
case True
then have 1: "sinDuplicados (borraDuplicados (x#xs))
= sinDuplicados (borraDuplicados xs)" by (simp add: estaEn_borraDuplicados_2)
show "?P (x#xs)" using HI 1 by simp
next
case False
then have "sinDuplicados (borraDuplicados (x#xs))
= sinDuplicados (x#borraDuplicados xs)" by simp
also have "… = (¬ (estaEn x (borraDuplicados xs)) ∧
sinDuplicados (borraDuplicados xs))" by simp
also have "… = True" using `¬ estaEn x xs` HI by (simp add:estaEn_borraDuplicados)
finally show "?P (x#xs)" by simp
qed
qed

text {*
---------------------------------------------------------------------
Ejercicio 7. Demostrar o refutar:
borraDuplicados (rev xs) = rev (borraDuplicados xs)
---------------------------------------------------------------------
*}

(*crigomgom rubgonmar ivamenjim wilmorort pablucoto migtermor
anaprarod fraortmoy ferrenseg marpoldia1 bowma danrodcha juacabsou paupeddeg manmorjim1 *)
lemma "borraDuplicados (rev xs) = rev (borraDuplicados xs)"
quickcheck
oops

(* ivamenjim: Quickcheck encuentra el siguiente contraejemplo:
xs = [a1, a2, a1]
Por lo que:
· "borraDuplicados (rev xs) = [a2, a1]"
· "rev (borraDuplicados xs) = [a1, a2]" *)

end
</source>

Relación 5

2016-11-29T23:32:55Z

Manmorjim1:

<source lang="isar">
chapter {* R5: Eliminación de duplicados *}

theory R5_Eliminacion_de_duplicados
imports Main
begin

text {*
---------------------------------------------------------------------
Ejercicio 1. Definir la funcion primitiva recursiva
estaEn :: 'a ⇒ 'a list ⇒ bool
tal que (estaEn x xs) se verifica si el elemento x está en la lista
xs. Por ejemplo,
estaEn (2::nat) [3,2,4] = True
estaEn (1::nat) [3,2,4] = False
---------------------------------------------------------------------
*}

(* crigomgom rubgonmar bowma wilmorort pablucoto serrodcal
anaprarod migtermor paupeddeg fraortmoy marpoldia1
danrodcha manmorjim1 *)
fun estaEn :: "'a ⇒ 'a list ⇒ bool" where
"estaEn _ [] = False"
| "estaEn x (a#xs) = ((a = x) ∨ (estaEn x xs))"

value "estaEn (2::nat) [3,2,4] = True"
value "estaEn (1::nat) [3,2,4] = False"

(* ivamenjim, ferrenseg josgarsan juacabsou dancorgar *)
(* Igual que la anterior pero con x en lugar de _ en el caso base *)

fun estaEn1 :: "'a ⇒ 'a list ⇒ bool" where
"estaEn1 x [] = False"
| "estaEn1 x (a#xs) = ((x=a) ∨ estaEn1 x xs)"

value "estaEn1 (2::nat) [3,2,4] = True"
value "estaEn1 (1::nat) [3,2,4] = False"

(* wilmorort *)
(* reutilizando la funcion "algunos" de R4.thy*)
fun estaEn2 :: "'a ⇒ 'a list ⇒ bool" where
"estaEn2 a xs = algunos (λx. x = a) xs"

value "estaEn2 (2::nat) [3,2,4] = True"
value "estaEn2 (1::nat) [3,2,4] = False"

text {*
---------------------------------------------------------------------
Ejercicio 2. Definir la función primitiva recursiva
sinDuplicados :: 'a list ⇒ bool
tal que (sinDuplicados xs) se verifica si la lista xs no contiene
duplicados. Por ejemplo,
sinDuplicados [1::nat,4,2] = True
sinDuplicados [1::nat,4,2,4] = False
---------------------------------------------------------------------
*}

(* crigomgom rubgonmar ivamenjim wilmorort bowma pablucoto
serrodcal anaprarod migtermor paupeddeg fraortmoy marpoldia1
ferrenseg josgarsan danrodcha manmorjim1 juacabsou dancorgar *)
fun sinDuplicados :: "'a list ⇒ bool" where
"sinDuplicados [] = True"
| "sinDuplicados (x#xs) = (¬ estaEn x xs ∧ sinDuplicados xs)"

value "sinDuplicados [1::nat,4,2] = True"
value "sinDuplicados [1::nat,4,2,4] = False"

(* paupeddeg *)
(* Utilizando la función ∉ de Isabelle *)
fun sinDuplicados2 :: "'a list ⇒ bool" where
"sinDuplicados2 [] = True"
| "sinDuplicados2 (a#xs) = ((a ∉ set xs) ∧ sinDuplicados2 xs ) "

text {*
---------------------------------------------------------------------
Ejercicio 3. Definir la función primitiva recursiva
borraDuplicados :: 'a list ⇒ bool
tal que (borraDuplicados xs) es la lista obtenida eliminando los
elementos duplicados de la lista xs. Por ejemplo,
borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]

Nota: La función borraDuplicados es equivalente a la predefinida
remdups.
---------------------------------------------------------------------
*}

(* crigomgom rubgonmar wilmorort bowma pablucoto serrodcal
anaprarod migtermor paupeddeg fraortmoy marpoldia1 ferrenseg
josgarsan danrodcha juacabsou dancorgar manmorjim1 *)
fun borraDuplicados :: "'a list ⇒ 'a list" where
"borraDuplicados [] = []"
| "borraDuplicados (x#xs) =( if estaEn x xs then borraDuplicados xs else x#borraDuplicados xs)"

value "borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]"

(* ivamenjim *)
(* Utilizando la negación primero *)

fun borraDuplicados :: "'a list ⇒ 'a list" where
"borraDuplicados [] = []"
| "borraDuplicados (x#xs) = (if ¬(estaEn x xs) then (x#(borraDuplicados xs)) else borraDuplicados xs)"

(* rubgonmar *)
(* Otra forma Sin usar if
Utilizando case aunque se le sacaría más partido con más de 2 casos *)

fun borraDuplicados1 :: "'a list ⇒ 'a list" where
"borraDuplicados1 [] = []" |
"borraDuplicados1 (x#xs) = ( case estaEn x xs of False => x#borraDuplicados1 xs | True => borraDuplicados1 xs )"

(* rubgonmar *)
(*Otra forma utilizando let*)
fun borraDuplicados2 :: "'a list ⇒ 'a list" where
"borraDuplicados2 [] = []" |
"borraDuplicados2 (x#xs) = (let condicion = estaEn x xs::bool in
if condicion then borraDuplicados2 xs else x # borraDuplicados2 xs)"

value "borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]"

text {*
---------------------------------------------------------------------
Ejercicio 4.1. Demostrar o refutar automáticamente
length (borraDuplicados xs) ≤ length xs
---------------------------------------------------------------------
*}

-- "La demostración automática es"
(*crigomgom anaprarod ferrenseg*)
lemma length_borraDuplicados:
"length (borraDuplicados xs) ≤ length xs"
by (induct xs, simp_all)

(* rubgonmar wilmorort pablucoto serrodcal migtermor paupeddeg
fraortmoy marpoldia1 danrodcha juacabsou dancorgar *)
lemma length_borraDuplicados:
"length ( borraDuplicados xs ) ≤ length xs"
by (induct xs) auto

(* manmorjim1 *)
(* soy de ponerlo mejor por pasos *)
lemma length_borraDuplicados:
"length ( borraDuplicados xs ) ≤ length xs"
apply (induct xs)
apply auto
done

(* ivamenjim *)
(* Demostrando objetivo a objetivo *)
lemma length_borraDuplicados:
"length (borraDuplicados xs) ≤ length xs"
apply (induct xs)
apply simp
apply auto
done

(* bowma anaprarod *)
lemma length_borraDuplicados:
"length (borraDuplicados xs) ≤ length xs"
apply (induct xs)
apply (simp, simp) (* creo que es mejor poner aquí simp_all *)
done

(* ferrenseg *)

lemma length_borraDuplicados:
"length (borraDuplicados xs) ≤ length xs"
by (induct xs) simp_all (* Creo que se puede poner simp_all fuera de parentesis *)

text {*
---------------------------------------------------------------------
Ejercicio 4.2. Demostrar o refutar detalladamente
length (borraDuplicados xs) ≤ length xs
---------------------------------------------------------------------
*}

-- "La demostración estructurada es"
(* crigomgom *)
lemma length_borraDuplicados_2:
"length (borraDuplicados xs) ≤ length xs"
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix x xs
assume HI: "length (borraDuplicados xs) ≤ length xs"
show "length (borraDuplicados (x#xs)) ≤ length (x#xs)"
proof (cases)
assume "estaEn x xs"
then have "length (borraDuplicados (x#xs)) = length (borraDuplicados xs)" by simp
also have "... ≤ length xs" using HI by simp
also have "... ≤ length (x#xs)" by simp
finally show "length (borraDuplicados (x#xs)) ≤ length (x#xs)" by simp
next
assume "(¬ estaEn x xs)"
then have "length (borraDuplicados (x#xs)) = length (x#borraDuplicados xs)" by simp
also have "... = 1 + length (borraDuplicados xs)" by simp
also have "... ≤ 1 + length xs" using HI by simp
also have "... = length (x#xs)" by simp
finally show "length (borraDuplicados (x#xs)) ≤ length (x#xs)" by simp
qed
qed

(* ivamenjim wilmorort ferrenseg rubgonmar juacabsou dancorgar *)
lemma length_borraDuplicados_2:
"length (borraDuplicados xs) ≤ length xs"
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix a xs
assume HI: "length (borraDuplicados xs) ≤ length xs"
have "length (borraDuplicados (a # xs)) ≤ 1+length (borraDuplicados xs)" by simp
also have "... ≤ 1+length xs" using HI by simp
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs)" by simp
qed

(* serrodcal anaprarod danrodcha *)
lemma length_borraDuplicados_2: "length (borraDuplicados xs) ≤ length xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "length (borraDuplicados (a # xs)) ≤ 1+length (borraDuplicados xs)" by simp
also have "... ≤ 1+length xs" using HI by simp
finally show "?P (a # xs)" by simp
qed

(* pablucoto *)
lemma length_borraDuplicados_2:
"length (borraDuplicados xs) ≤ length xs"
proof(induct xs)
show "length (borraDuplicados []) ≤ length [] " by simp
next
fix a xs
assume HI: " length (borraDuplicados xs) ≤ length xs "
have "length (borraDuplicados (a # xs)) ≤ 1 + length(borraDuplicados xs)" by simp
also have "... ≤ 1 + length xs" using HI by simp
also have "... ≤ length (a#xs)" by simp
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs) " by simp
qed

(* bowma *)
lemma length_borraDuplicados_3:
"length (borraDuplicados xs) ≤ length xs" (is "?p xs")
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix a xs
assume HI: "?p xs"
have "length (a#xs) = 1 + length xs" by simp
have "length(borraDuplicados (a#xs)) ≤ 1 + length(borraDuplicados xs)" by simp
also have "... ≤ 1+length xs" using HI by simp
also have "... ≤ length (a#xs)" by simp
finally show "?p (a # xs)" by simp
qed

(* migtermor *)

lemma length_borraDuplicados_2:
"length (borraDuplicados xs) ≤ length xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "length (borraDuplicados (a#xs)) ≤ (length (a#xs))"
proof (cases)
assume "(estaEn a xs)"
then have Aux: "length (borraDuplicados (a#xs)) = length (borraDuplicados xs)" by simp
also have "… ≤ length (a#xs)" using HI by simp
then show "length (borraDuplicados (a#xs)) ≤ (length (a#xs))" using Aux by simp
next
assume "¬ (estaEn a xs)"
then have Aux: "length (borraDuplicados (a#xs)) = 1+ length (borraDuplicados xs)" by simp
also have "… ≤ length (a#xs)" using HI by simp
then show "length (borraDuplicados (a#xs)) ≤ (length (a#xs))" using Aux by simp
qed
then show "?P (a#xs)" by simp
qed

(* paupeddeg marpoldia1*)
lemma length_borraDuplicados_4:
"length (borraDuplicados xs) ≤ length xs"
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix a xs
assume HI: "length (borraDuplicados xs) ≤ length xs"
have "length (borraDuplicados (a # xs)) ≤ length [a] + length (borraDuplicados xs) " by simp
also have "... ≤ 1 + length (borraDuplicados xs)" by simp
also have "... ≤ 1 + length xs" using HI by simp
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs)" by simp
qed

(* fraortmoy *)
(* muy parecida a alguna anterior, pero yo dí mas pasos *)
lemma length_borraDuplicados_2:
"length (borraDuplicados xs) ≤ length xs"
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix a xs
assume H1: "length (borraDuplicados xs) ≤ length xs"
have "length (borraDuplicados (a # xs)) ≤ length(borraDuplicados [a])+length (borraDuplicados xs)" by simp
also have "… ≤ 1 + length (borraDuplicados xs)" by simp
also have "… ≤ 1 + length xs" using H1 by simp
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 5.1. Demostrar o refutar automáticamente
estaEn a (borraDuplicados xs) = estaEn a xs
---------------------------------------------------------------------
*}

-- "La demostración automática es"
(* crigomgom rubgonmar wilmorort pablucoto serrodcal bowma
migtermor fraortmoy marpoldia1 ferrenseg danrodcha juacabsou paupeddeg*)
lemma estaEn_borraDuplicados:
"estaEn a (borraDuplicados xs) = estaEn a xs"
by (induct xs) auto

(* ivamenjim manmorjim1 *)
lemma estaEn_borraDuplicados:
"estaEn a (borraDuplicados xs) = estaEn a xs"
apply (induct xs)
apply auto
done

(* anaprarod *)
lemma estaEn_borraDuplicados:
"estaEn a (borraDuplicados xs) = estaEn a xs"
by (induct xs, simp_all, blast)

(* anaprarod *)
lemma estaEn_borraDuplicados':
"estaEn a (borraDuplicados xs) = estaEn a xs"
apply (induct xs)
apply (cases "estaEn x xs")
apply (simp_all)
apply blast
done

(* bowma *)
lemma estaEn_borraDuplicados:
"estaEn a (borraDuplicados xs) = estaEn a xs"
apply (induct xs)
apply simp
apply (simp, blast)
done

text {*
---------------------------------------------------------------------
Ejercicio 5.2. Demostrar o refutar detalladamente
estaEn a (borraDuplicados xs) = estaEn a xs
Nota: Para la demostración de la equivalencia se puede usar
proof (rule iffI)
La regla iffI es
⟦P ⟹ Q ; Q ⟹ P⟧ ⟹ P = Q
---------------------------------------------------------------------
*}

-- "La demostración estructurada es"

(* wilmorort *)

lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs"
proof (induct xs)
show "estaEn a (borraDuplicados []) = estaEn a []" by simp
next
fix b xs
assume HI: "estaEn a (borraDuplicados xs) = estaEn a xs"
show "estaEn a (borraDuplicados (b#xs)) = estaEn a (b#xs)"
proof (rule iffI)
assume H1: "estaEn a (borraDuplicados (b#xs))"
show "estaEn a (b#xs)"
proof (cases)
assume "estaEn b xs"
then have "estaEn a (borraDuplicados xs)" using H1 by simp
then have "estaEn a xs" using HI by simp
then show "estaEn a (b#xs)" by simp
next
assume "¬ estaEn b xs"
then have "estaEn a (b#(borraDuplicados xs))" using H1 by simp
then have "a=b ∨ (estaEn a (borraDuplicados xs))" by simp
then have " a=b ∨ (estaEn a xs)" using HI by simp
then show "estaEn a (b#xs)" by simp
qed
next
assume H2: "estaEn a (b#xs)"
show "estaEn a (borraDuplicados (b#xs))"
proof (cases)
assume "a=b"
then have "estaEn b (borraDuplicados xs) = estaEn b xs" using HI by simp
then have "(estaEn b xs ⟶ estaEn b (borraDuplicados xs)) ∧
(¬ estaEn b xs ⟶ estaEn b (b # borraDuplicados xs))" by simp
then have "estaEn b (borraDuplicados (b#xs))" by simp
then show "estaEn a (borraDuplicados (b#xs))" using `a=b` by simp
next
assume "a≠b"
then have "estaEn a (b#xs)" using H2 by simp
then have "a = b ∨ estaEn a xs" by simp
then have "False ∨ estaEn a xs " using `a≠b` by simp
then have "estaEn a xs" by simp
then have "estaEn a (borraDuplicados xs)" using HI by simp
then show "estaEn a (borraDuplicados (b#xs))" using `a≠b` by simp
qed
qed
qed

(* anaprarod marpoldia1 ferrenseg juacabsou *)
lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix x xs
assume HI: "?P xs"
show "?P (x#xs)"
proof (cases)
assume "estaEn x xs"
then have "estaEn a (borraDuplicados (x#xs)) = estaEn a (borraDuplicados xs)" by simp
also have "...= estaEn a xs" using HI by simp
finally show "?P (x#xs)" by auto
next
assume "¬estaEn x xs"
then have "estaEn a (borraDuplicados (x#xs)) = estaEn a (x#borraDuplicados xs)" by simp
also have "...= (x = a ∨ estaEn a (borraDuplicados xs))" by simp
finally show "?P (x#xs)" using HI by simp
qed
qed

(* migtermor *)

lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix aa xs
assume HI: "?P xs"
have P1: "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)"
proof (cases)
assume C1: "(estaEn aa xs)"
have "estaEn a (borraDuplicados (aa#xs)) = estaEn a (borraDuplicados xs)"
using C1 by simp
also have P3: "… = estaEn a xs" using HI by simp
also have "… = estaEn a (aa#xs)"
proof (cases)
assume "(a=aa)"
then show "estaEn a xs = estaEn a (aa#xs)" using C1 by simp
next
assume "¬(a=aa)"
then show "estaEn a xs = estaEn a (aa#xs)" by simp
qed
then show "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)" using P3 by simp
next
assume C2: "¬(estaEn aa xs)"
then show "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)" using HI by simp
qed
also have Conc: "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)" using P1 by simp
finally show "?P (aa#xs)" using Conc by simp
qed

(* crigomgom paupeddeg *)
-- "La demostración estructurada es"
lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs"
proof (induct xs)
show "estaEn a (borraDuplicados []) = estaEn a []" by simp
next
fix x xs
assume HI: "estaEn a (borraDuplicados xs) = estaEn a xs"
show "estaEn a (borraDuplicados (x#xs)) = estaEn a (x#xs)"
proof (rule iffI)
assume a1: "estaEn a (borraDuplicados (x#xs))"
show "estaEn a (x#xs)"
proof (cases)
assume "estaEn x xs"
then have "estaEn a (borraDuplicados xs)" using a1 by simp
then have "estaEn a xs" using HI by simp
then show "estaEn a (x#xs)" by simp
next
assume "¬ estaEn x xs"
then have "estaEn a (x#(borraDuplicados xs))" using a1 by simp
then have " x=a ∨ (estaEn a (borraDuplicados xs))" by simp
then have " x=a ∨ (estaEn a xs)" using HI by simp
then show "estaEn a (x#xs)" by simp
qed
next
assume a2: "estaEn a (x#xs)"
show "estaEn a (borraDuplicados (x#xs))"
proof (cases)
assume "a=x"
then show "estaEn a (borraDuplicados (x#xs))" using HI by simp
next
assume b1: "a≠x"
then have "estaEn a (x#xs)" using a2 by simp
then have "x = a ∨ estaEn a xs" by simp
then have "estaEn a xs " using b1 by simp
then have "estaEn a (borraDuplicados xs)" using HI by simp
then show "estaEn a (borraDuplicados (x#xs))" using b1 by simp
qed
qed
qed

(* rubgonmar *)
lemma estaEn_borraDuplicados_2:
"estaEn a ( borraDuplicados xs ) = estaEn a xs"
proof (induct xs)
show "estaEn a (borraDuplicados []) = estaEn a []" by simp
next
fix x
fix xs
assume HI: "estaEn a (borraDuplicados xs) = estaEn a xs"
show "estaEn a (borraDuplicados (x#xs)) = estaEn a (x#xs)"
proof (rule iffI) (* usamos proof de la regla dada iffI*)
assume cprim: "estaEn a (borraDuplicados (x#xs))"
show "estaEn a (x#xs)"
proof (cases)
assume "estaEn x xs"
then show "estaEn a (x#xs)" using cprim HI by simp
next
assume "¬ estaEn x xs"
then show "estaEn a (x#xs)" using cprim HI by simp
qed
next
assume cseg: "estaEn a (x#xs)"
show "estaEn a (borraDuplicados (x#xs))"
proof (cases)
assume "a=x"
then show "estaEn a (borraDuplicados (x#xs))" using HI by auto
next
assume "a≠x"
then show "estaEn a (borraDuplicados (x#xs))" using `a≠x` cseg HI by simp
qed
qed
qed

(*
Aplico la regla iffI:
⟦P ⟹ Q ; Q ⟹ P⟧ ⟹ P = Q
Así:
[estaEn a (borraDuplicados (x # xs)) ⟹ estaEn a (x # xs); estaEn a (x # xs) ⟹ estaEn a (borraDuplicados (x # xs))]
⟹ estaEn a (borraDuplicados (x # xs)) = estaEn a (x # xs)
*)

(* bowma ivamenjim *)
lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?p xs")
proof (induct xs)
show "?p []" by simp
next
fix x xs
assume HI: "?p xs"
show "?p (x#xs)"
proof (cases)
assume H1:"estaEn x xs"
then have "estaEn a (borraDuplicados (x#xs)) = estaEn a (borraDuplicados xs)" by simp
also have "... = estaEn a xs" using HI by simp
also have "... = estaEn a (x#xs)"
proof(cases)
assume "x=a"
then show "estaEn a xs = estaEn a (x#xs)" using H1 by simp
next
assume "x≠a"
then show "estaEn a xs = estaEn a (x#xs)" by simp
qed
finally show "?p (x#xs)" by simp
next
assume H2:"¬estaEn x xs"
then have "estaEn a (borraDuplicados (x#xs)) = estaEn a (x#borraDuplicados xs)" by simp
also have "... = ((x=a) ∨ estaEn a (borraDuplicados xs))" by simp
also have "... = ((x=a) ∨ estaEn a xs)" using HI by simp
also have "... = estaEn a (x#xs)" by simp
finally show "?p (x#xs)" by simp
qed

(*danrodcha *)
(* es como la de ruben pero con diferencias de estilo *)
lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix x xs assume HI: "?P xs"
show "?P (x#xs)"
proof (rule iffI)
assume H1: "estaEn a (borraDuplicados (x # xs))"
show "estaEn a (x#xs)"
proof (cases "estaEn x xs")
case True
then show "estaEn a (x#xs)" using H1 HI by simp
next
case False
then show "estaEn a (x#xs)" using H1 HI by simp
qed
next
assume H2: "estaEn a (x#xs)"
show "estaEn a (borraDuplicados (x # xs))"
proof (cases "x=a")
case True
then show "estaEn a (borraDuplicados (x # xs))" using HI by simp
next
case False
then show "estaEn a (borraDuplicados (x # xs))" using H2 HI by simp
qed
qed
qed

text {*
---------------------------------------------------------------------
Ejercicio 6.1. Demostrar o refutar automáticamente
sinDuplicados (borraDuplicados xs)
---------------------------------------------------------------------
*}

-- "La demostración automática"

(* ivamenjim wilmorort serrodcal crigomgom anaprarod fraortmoy
marpoldia1 ferrenseg danrodcha juacabsou paupeddeg*)
lemma sinDuplicados_borraDuplicados:
"sinDuplicados (borraDuplicados xs)"
by (induct xs) (auto simp add: estaEn_borraDuplicados)

(* migtermor bowma *)
lemma sinDuplicados_borraDuplicados:
"sinDuplicados (borraDuplicados xs)"
by (induct xs, simp_all add: estaEn_borraDuplicados_2)

(* manmorjim1 no caí en usar la demostración anterior y he realizado
la demostración de que si un elemento no estaba en una lista seguirá sin estar
después de eliminar los duplicados en esa lista... *)
lemma noEsta_tras_borrarDuplicados:
"(¬estaEn x xs) ⟶ (¬estaEn x (borraDuplicados xs))"
apply (induct xs)
apply auto
done

lemma sinDuplicados_borraDuplicados:
"sinDuplicados (borraDuplicados xs)"
apply (induct xs)
apply simp
apply (induct xs)
apply auto
apply (simp add: noEsta_tras_borrarDuplicados)
done

text {*
---------------------------------------------------------------------
Ejercicio 6.2. Demostrar o refutar detalladamente
sinDuplicados (borraDuplicados xs)
---------------------------------------------------------------------
*}

(*wilmorort*)

-- "La demostración estructurada es"
lemma sinDuplicados_borraDuplicados_2:
"sinDuplicados (borraDuplicados xs)"
proof (induct xs)
show " sinDuplicados (borraDuplicados [])" by simp
next
fix a xs
assume HI: "sinDuplicados (borraDuplicados xs)"
show "sinDuplicados (borraDuplicados (a # xs))"
proof (cases)
assume "estaEn a xs"
then show "sinDuplicados (borraDuplicados (a#xs))" using HI by simp
next
assume"¬ estaEn a xs"
then have "¬ (estaEn a xs) ∧ sinDuplicados (borraDuplicados xs)" using HI by simp
then have "¬ estaEn a (borraDuplicados xs) ∧ sinDuplicados (borraDuplicados xs)"
by (simp add: estaEn_borraDuplicados)
then have " sinDuplicados (a#borraDuplicados xs)" by simp
then show " sinDuplicados (borraDuplicados(a #xs))" by simp
qed
qed

(* ivamenjim migtermor crigomgom rubgonmar fraortmoy marpoldia1 ferrenseg bowma juacabsou *)

lemma sinDuplicados_borraDuplicados_2:
"sinDuplicados (borraDuplicados xs)"
proof (induct xs)
show "sinDuplicados (borraDuplicados [])" by simp
next
fix a xs
assume HI: "sinDuplicados (borraDuplicados xs)"
show "sinDuplicados (borraDuplicados (a # xs))"
proof (cases)
assume "estaEn a xs"
then show "sinDuplicados (borraDuplicados (a # xs))" using HI by simp
next
assume "¬(estaEn a xs)"
then show "sinDuplicados (borraDuplicados (a # xs))" using HI by (simp add: estaEn_borraDuplicados)
qed
qed

(* anaprarod paupeddeg*)
lemma sinDuplicados_borraDuplicados_2:
"sinDuplicados (borraDuplicados xs)" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix x xs
assume HI: "?P xs"
show "?P (x#xs)"
proof (cases)
assume c1: "estaEn x xs"
then show "sinDuplicados (borraDuplicados (x#xs))" using HI by simp
next
assume c2: "¬ estaEn x xs"
then have "sinDuplicados (borraDuplicados (x#xs)) =sinDuplicados (x#borraDuplicados xs)" by simp
also have "…= (¬estaEn x (borraDuplicados xs) ∧ sinDuplicados (borraDuplicados xs))" by simp
also have "… = (¬estaEn x (borraDuplicados xs))" using HI by simp
also have "… = (¬(estaEn x xs))" by (simp add:estaEn_borraDuplicados)
also have "… = True" using c2 by simp
finally show "?P (x#xs)" by simp
qed
qed

(* danrodcha *)
lemma sinDuplicados_borraDuplicados_2:
"sinDuplicados (borraDuplicados xs)" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix x xs assume HI: "?P xs"
show "?P (x#xs)"
proof (cases "estaEn x xs")
case True
then have 1: "sinDuplicados (borraDuplicados (x#xs))
= sinDuplicados (borraDuplicados xs)" by (simp add: estaEn_borraDuplicados_2)
show "?P (x#xs)" using HI 1 by simp
next
case False
then have "sinDuplicados (borraDuplicados (x#xs))
= sinDuplicados (x#borraDuplicados xs)" by simp
also have "… = (¬ (estaEn x (borraDuplicados xs)) ∧
sinDuplicados (borraDuplicados xs))" by simp
also have "… = True" using `¬ estaEn x xs` HI by (simp add:estaEn_borraDuplicados)
finally show "?P (x#xs)" by simp
qed
qed

text {*
---------------------------------------------------------------------
Ejercicio 7. Demostrar o refutar:
borraDuplicados (rev xs) = rev (borraDuplicados xs)
---------------------------------------------------------------------
*}

(*crigomgom rubgonmar ivamenjim wilmorort pablucoto migtermor
anaprarod fraortmoy ferrenseg marpoldia1 bowma danrodcha juacabsou paupeddeg *)
lemma "borraDuplicados (rev xs) = rev (borraDuplicados xs)"
quickcheck
oops

(* ivamenjim: Quickcheck encuentra el siguiente contraejemplo:
xs = [a1, a2, a1]
Por lo que:
· "borraDuplicados (rev xs) = [a2, a1]"
· "rev (borraDuplicados xs) = [a1, a2]" *)

end
</source>

Relación 5

2016-11-29T22:25:10Z

Manmorjim1:

<source lang="isar">
chapter {* R5: Eliminación de duplicados *}

theory R5_Eliminacion_de_duplicados
imports Main
begin

text {*
---------------------------------------------------------------------
Ejercicio 1. Definir la funcion primitiva recursiva
estaEn :: 'a ⇒ 'a list ⇒ bool
tal que (estaEn x xs) se verifica si el elemento x está en la lista
xs. Por ejemplo,
estaEn (2::nat) [3,2,4] = True
estaEn (1::nat) [3,2,4] = False
---------------------------------------------------------------------
*}

(* crigomgom rubgonmar bowma wilmorort pablucoto serrodcal
anaprarod migtermor paupeddeg fraortmoy marpoldia1
danrodcha manmorjim1 *)
fun estaEn :: "'a ⇒ 'a list ⇒ bool" where
"estaEn _ [] = False"
| "estaEn x (a#xs) = ((a = x) ∨ (estaEn x xs))"

value "estaEn (2::nat) [3,2,4] = True"
value "estaEn (1::nat) [3,2,4] = False"

(* ivamenjim, ferrenseg josgarsan juacabsou dancorgar *)
(* Igual que la anterior pero con x en lugar de _ en el caso base *)

fun estaEn1 :: "'a ⇒ 'a list ⇒ bool" where
"estaEn1 x [] = False"
| "estaEn1 x (a#xs) = ((x=a) ∨ estaEn1 x xs)"

value "estaEn1 (2::nat) [3,2,4] = True"
value "estaEn1 (1::nat) [3,2,4] = False"

(* wilmorort *)
(* reutilizando la funcion "algunos" de R4.thy*)
fun estaEn2 :: "'a ⇒ 'a list ⇒ bool" where
"estaEn2 a xs = algunos (λx. x = a) xs"

value "estaEn2 (2::nat) [3,2,4] = True"
value "estaEn2 (1::nat) [3,2,4] = False"

text {*
---------------------------------------------------------------------
Ejercicio 2. Definir la función primitiva recursiva
sinDuplicados :: 'a list ⇒ bool
tal que (sinDuplicados xs) se verifica si la lista xs no contiene
duplicados. Por ejemplo,
sinDuplicados [1::nat,4,2] = True
sinDuplicados [1::nat,4,2,4] = False
---------------------------------------------------------------------
*}

(* crigomgom rubgonmar ivamenjim wilmorort bowma pablucoto
serrodcal anaprarod migtermor paupeddeg fraortmoy marpoldia1
ferrenseg josgarsan danrodcha manmorjim1 juacabsou dancorgar *)
fun sinDuplicados :: "'a list ⇒ bool" where
"sinDuplicados [] = True"
| "sinDuplicados (x#xs) = (¬ estaEn x xs ∧ sinDuplicados xs)"

value "sinDuplicados [1::nat,4,2] = True"
value "sinDuplicados [1::nat,4,2,4] = False"

(* paupeddeg *)
(* Utilizando la función ∉ de Isabelle *)
fun sinDuplicados2 :: "'a list ⇒ bool" where
"sinDuplicados2 [] = True"
| "sinDuplicados2 (a#xs) = ((a ∉ set xs) ∧ sinDuplicados2 xs ) "

text {*
---------------------------------------------------------------------
Ejercicio 3. Definir la función primitiva recursiva
borraDuplicados :: 'a list ⇒ bool
tal que (borraDuplicados xs) es la lista obtenida eliminando los
elementos duplicados de la lista xs. Por ejemplo,
borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]

Nota: La función borraDuplicados es equivalente a la predefinida
remdups.
---------------------------------------------------------------------
*}

(* crigomgom rubgonmar wilmorort bowma pablucoto serrodcal
anaprarod migtermor paupeddeg fraortmoy marpoldia1 ferrenseg
josgarsan danrodcha juacabsou dancorgar manmorjim1 *)
fun borraDuplicados :: "'a list ⇒ 'a list" where
"borraDuplicados [] = []"
| "borraDuplicados (x#xs) =( if estaEn x xs then borraDuplicados xs else x#borraDuplicados xs)"

value "borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]"

(* ivamenjim *)
(* Utilizando la negación primero *)

fun borraDuplicados :: "'a list ⇒ 'a list" where
"borraDuplicados [] = []"
| "borraDuplicados (x#xs) = (if ¬(estaEn x xs) then (x#(borraDuplicados xs)) else borraDuplicados xs)"

(* rubgonmar *)
(* Otra forma Sin usar if
Utilizando case aunque se le sacaría más partido con más de 2 casos *)

fun borraDuplicados1 :: "'a list ⇒ 'a list" where
"borraDuplicados1 [] = []" |
"borraDuplicados1 (x#xs) = ( case estaEn x xs of False => x#borraDuplicados1 xs | True => borraDuplicados1 xs )"

(* rubgonmar *)
(*Otra forma utilizando let*)
fun borraDuplicados2 :: "'a list ⇒ 'a list" where
"borraDuplicados2 [] = []" |
"borraDuplicados2 (x#xs) = (let condicion = estaEn x xs::bool in
if condicion then borraDuplicados2 xs else x # borraDuplicados2 xs)"

value "borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]"

text {*
---------------------------------------------------------------------
Ejercicio 4.1. Demostrar o refutar automáticamente
length (borraDuplicados xs) ≤ length xs
---------------------------------------------------------------------
*}

-- "La demostración automática es"
(*crigomgom anaprarod ferrenseg*)
lemma length_borraDuplicados:
"length (borraDuplicados xs) ≤ length xs"
by (induct xs, simp_all)

(* rubgonmar wilmorort pablucoto serrodcal migtermor paupeddeg
fraortmoy marpoldia1 danrodcha juacabsou *)
lemma length_borraDuplicados:
"length ( borraDuplicados xs ) ≤ length xs"
by (induct xs) auto

(* manmorjim1 *)
(* soy de ponerlo mejor por pasos *)
lemma length_borraDuplicados:
"length ( borraDuplicados xs ) ≤ length xs"
apply (induct xs)
apply auto
done

(* ivamenjim *)
(* Demostrando objetivo a objetivo *)
lemma length_borraDuplicados:
"length (borraDuplicados xs) ≤ length xs"
apply (induct xs)
apply simp
apply auto
done

(* bowma anaprarod *)
lemma length_borraDuplicados:
"length (borraDuplicados xs) ≤ length xs"
apply (induct xs)
apply (simp, simp) (* creo que es mejor poner aquí simp_all *)
done

(* ferrenseg *)

lemma length_borraDuplicados:
"length (borraDuplicados xs) ≤ length xs"
by (induct xs) simp_all (* Creo que se puede poner simp_all fuera de parentesis *)

text {*
---------------------------------------------------------------------
Ejercicio 4.2. Demostrar o refutar detalladamente
length (borraDuplicados xs) ≤ length xs
---------------------------------------------------------------------
*}

-- "La demostración estructurada es"
(* crigomgom *)
lemma length_borraDuplicados_2:
"length (borraDuplicados xs) ≤ length xs"
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix x xs
assume HI: "length (borraDuplicados xs) ≤ length xs"
show "length (borraDuplicados (x#xs)) ≤ length (x#xs)"
proof (cases)
assume "estaEn x xs"
then have "length (borraDuplicados (x#xs)) = length (borraDuplicados xs)" by simp
also have "... ≤ length xs" using HI by simp
also have "... ≤ length (x#xs)" by simp
finally show "length (borraDuplicados (x#xs)) ≤ length (x#xs)" by simp
next
assume "(¬ estaEn x xs)"
then have "length (borraDuplicados (x#xs)) = length (x#borraDuplicados xs)" by simp
also have "... = 1 + length (borraDuplicados xs)" by simp
also have "... ≤ 1 + length xs" using HI by simp
also have "... = length (x#xs)" by simp
finally show "length (borraDuplicados (x#xs)) ≤ length (x#xs)" by simp
qed
qed

(* ivamenjim wilmorort ferrenseg rubgonmar juacabsou *)
lemma length_borraDuplicados_2:
"length (borraDuplicados xs) ≤ length xs"
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix a xs
assume HI: "length (borraDuplicados xs) ≤ length xs"
have "length (borraDuplicados (a # xs)) ≤ 1+length (borraDuplicados xs)" by simp
also have "... ≤ 1+length xs" using HI by simp
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs)" by simp
qed

(* serrodcal anaprarod danrodcha *)
lemma length_borraDuplicados_2: "length (borraDuplicados xs) ≤ length xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "length (borraDuplicados (a # xs)) ≤ 1+length (borraDuplicados xs)" by simp
also have "... ≤ 1+length xs" using HI by simp
finally show "?P (a # xs)" by simp
qed

(* pablucoto *)
lemma length_borraDuplicados_2:
"length (borraDuplicados xs) ≤ length xs"
proof(induct xs)
show "length (borraDuplicados []) ≤ length [] " by simp
next
fix a xs
assume HI: " length (borraDuplicados xs) ≤ length xs "
have "length (borraDuplicados (a # xs)) ≤ 1 + length(borraDuplicados xs)" by simp
also have "... ≤ 1 + length xs" using HI by simp
also have "... ≤ length (a#xs)" by simp
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs) " by simp
qed

(* bowma *)
lemma length_borraDuplicados_3:
"length (borraDuplicados xs) ≤ length xs" (is "?p xs")
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix a xs
assume HI: "?p xs"
have "length (a#xs) = 1 + length xs" by simp
have "length(borraDuplicados (a#xs)) ≤ 1 + length(borraDuplicados xs)" by simp
also have "... ≤ 1+length xs" using HI by simp
also have "... ≤ length (a#xs)" by simp
finally show "?p (a # xs)" by simp
qed

(* migtermor *)

lemma length_borraDuplicados_2:
"length (borraDuplicados xs) ≤ length xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "length (borraDuplicados (a#xs)) ≤ (length (a#xs))"
proof (cases)
assume "(estaEn a xs)"
then have Aux: "length (borraDuplicados (a#xs)) = length (borraDuplicados xs)" by simp
also have "… ≤ length (a#xs)" using HI by simp
then show "length (borraDuplicados (a#xs)) ≤ (length (a#xs))" using Aux by simp
next
assume "¬ (estaEn a xs)"
then have Aux: "length (borraDuplicados (a#xs)) = 1+ length (borraDuplicados xs)" by simp
also have "… ≤ length (a#xs)" using HI by simp
then show "length (borraDuplicados (a#xs)) ≤ (length (a#xs))" using Aux by simp
qed
then show "?P (a#xs)" by simp
qed

(* paupeddeg marpoldia1*)
lemma length_borraDuplicados_4:
"length (borraDuplicados xs) ≤ length xs"
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix a xs
assume HI: "length (borraDuplicados xs) ≤ length xs"
have "length (borraDuplicados (a # xs)) ≤ length [a] + length (borraDuplicados xs) " by simp
also have "... ≤ 1 + length (borraDuplicados xs)" by simp
also have "... ≤ 1 + length xs" using HI by simp
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs)" by simp
qed

(* fraortmoy *)
(* muy parecida a alguna anterior, pero yo dí mas pasos *)
lemma length_borraDuplicados_2:
"length (borraDuplicados xs) ≤ length xs"
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix a xs
assume H1: "length (borraDuplicados xs) ≤ length xs"
have "length (borraDuplicados (a # xs)) ≤ length(borraDuplicados [a])+length (borraDuplicados xs)" by simp
also have "… ≤ 1 + length (borraDuplicados xs)" by simp
also have "… ≤ 1 + length xs" using H1 by simp
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 5.1. Demostrar o refutar automáticamente
estaEn a (borraDuplicados xs) = estaEn a xs
---------------------------------------------------------------------
*}

-- "La demostración automática es"
(* crigomgom rubgonmar wilmorort pablucoto serrodcal bowma
migtermor fraortmoy marpoldia1 ferrenseg danrodcha juacabsou paupeddeg*)
lemma estaEn_borraDuplicados:
"estaEn a (borraDuplicados xs) = estaEn a xs"
by (induct xs) auto

(* ivamenjim manmorjim1 *)
lemma estaEn_borraDuplicados:
"estaEn a (borraDuplicados xs) = estaEn a xs"
apply (induct xs)
apply auto
done

(* anaprarod *)
lemma estaEn_borraDuplicados:
"estaEn a (borraDuplicados xs) = estaEn a xs"
by (induct xs, simp_all, blast)

(* anaprarod *)
lemma estaEn_borraDuplicados':
"estaEn a (borraDuplicados xs) = estaEn a xs"
apply (induct xs)
apply (cases "estaEn x xs")
apply (simp_all)
apply blast
done

(* bowma *)
lemma estaEn_borraDuplicados:
"estaEn a (borraDuplicados xs) = estaEn a xs"
apply (induct xs)
apply simp
apply (simp, blast)
done

text {*
---------------------------------------------------------------------
Ejercicio 5.2. Demostrar o refutar detalladamente
estaEn a (borraDuplicados xs) = estaEn a xs
Nota: Para la demostración de la equivalencia se puede usar
proof (rule iffI)
La regla iffI es
⟦P ⟹ Q ; Q ⟹ P⟧ ⟹ P = Q
---------------------------------------------------------------------
*}

-- "La demostración estructurada es"

(* wilmorort *)

lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs"
proof (induct xs)
show "estaEn a (borraDuplicados []) = estaEn a []" by simp
next
fix b xs
assume HI: "estaEn a (borraDuplicados xs) = estaEn a xs"
show "estaEn a (borraDuplicados (b#xs)) = estaEn a (b#xs)"
proof (rule iffI)
assume H1: "estaEn a (borraDuplicados (b#xs))"
show "estaEn a (b#xs)"
proof (cases)
assume "estaEn b xs"
then have "estaEn a (borraDuplicados xs)" using H1 by simp
then have "estaEn a xs" using HI by simp
then show "estaEn a (b#xs)" by simp
next
assume "¬ estaEn b xs"
then have "estaEn a (b#(borraDuplicados xs))" using H1 by simp
then have "a=b ∨ (estaEn a (borraDuplicados xs))" by simp
then have " a=b ∨ (estaEn a xs)" using HI by simp
then show "estaEn a (b#xs)" by simp
qed
next
assume H2: "estaEn a (b#xs)"
show "estaEn a (borraDuplicados (b#xs))"
proof (cases)
assume "a=b"
then have "estaEn b (borraDuplicados xs) = estaEn b xs" using HI by simp
then have "(estaEn b xs ⟶ estaEn b (borraDuplicados xs)) ∧
(¬ estaEn b xs ⟶ estaEn b (b # borraDuplicados xs))" by simp
then have "estaEn b (borraDuplicados (b#xs))" by simp
then show "estaEn a (borraDuplicados (b#xs))" using `a=b` by simp
next
assume "a≠b"
then have "estaEn a (b#xs)" using H2 by simp
then have "a = b ∨ estaEn a xs" by simp
then have "False ∨ estaEn a xs " using `a≠b` by simp
then have "estaEn a xs" by simp
then have "estaEn a (borraDuplicados xs)" using HI by simp
then show "estaEn a (borraDuplicados (b#xs))" using `a≠b` by simp
qed
qed
qed

(* anaprarod marpoldia1 ferrenseg juacabsou *)
lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix x xs
assume HI: "?P xs"
show "?P (x#xs)"
proof (cases)
assume "estaEn x xs"
then have "estaEn a (borraDuplicados (x#xs)) = estaEn a (borraDuplicados xs)" by simp
also have "...= estaEn a xs" using HI by simp
finally show "?P (x#xs)" by auto
next
assume "¬estaEn x xs"
then have "estaEn a (borraDuplicados (x#xs)) = estaEn a (x#borraDuplicados xs)" by simp
also have "...= (x = a ∨ estaEn a (borraDuplicados xs))" by simp
finally show "?P (x#xs)" using HI by simp
qed
qed

(* migtermor *)

lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix aa xs
assume HI: "?P xs"
have P1: "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)"
proof (cases)
assume C1: "(estaEn aa xs)"
have "estaEn a (borraDuplicados (aa#xs)) = estaEn a (borraDuplicados xs)"
using C1 by simp
also have P3: "… = estaEn a xs" using HI by simp
also have "… = estaEn a (aa#xs)"
proof (cases)
assume "(a=aa)"
then show "estaEn a xs = estaEn a (aa#xs)" using C1 by simp
next
assume "¬(a=aa)"
then show "estaEn a xs = estaEn a (aa#xs)" by simp
qed
then show "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)" using P3 by simp
next
assume C2: "¬(estaEn aa xs)"
then show "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)" using HI by simp
qed
also have Conc: "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)" using P1 by simp
finally show "?P (aa#xs)" using Conc by simp
qed

(* crigomgom *)
-- "La demostración estructurada es"
lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs"
proof (induct xs)
show "estaEn a (borraDuplicados []) = estaEn a []" by simp
next
fix x xs
assume HI: "estaEn a (borraDuplicados xs) = estaEn a xs"
show "estaEn a (borraDuplicados (x#xs)) = estaEn a (x#xs)"
proof (rule iffI)
assume a1: "estaEn a (borraDuplicados (x#xs))"
show "estaEn a (x#xs)"
proof (cases)
assume "estaEn x xs"
then have "estaEn a (borraDuplicados xs)" using a1 by simp
then have "estaEn a xs" using HI by simp
then show "estaEn a (x#xs)" by simp
next
assume "¬ estaEn x xs"
then have "estaEn a (x#(borraDuplicados xs))" using a1 by simp
then have " x=a ∨ (estaEn a (borraDuplicados xs))" by simp
then have " x=a ∨ (estaEn a xs)" using HI by simp
then show "estaEn a (x#xs)" by simp
qed
next
assume a2: "estaEn a (x#xs)"
show "estaEn a (borraDuplicados (x#xs))"
proof (cases)
assume "a=x"
then show "estaEn a (borraDuplicados (x#xs))" using HI by simp
next
assume b1: "a≠x"
then have "estaEn a (x#xs)" using a2 by simp
then have "x = a ∨ estaEn a xs" by simp
then have "estaEn a xs " using b1 by simp
then have "estaEn a (borraDuplicados xs)" using HI by simp
then show "estaEn a (borraDuplicados (x#xs))" using b1 by simp
qed
qed
qed

(* rubgonmar *)
lemma estaEn_borraDuplicados_2:
"estaEn a ( borraDuplicados xs ) = estaEn a xs"
proof (induct xs)
show "estaEn a (borraDuplicados []) = estaEn a []" by simp
next
fix x
fix xs
assume HI: "estaEn a (borraDuplicados xs) = estaEn a xs"
show "estaEn a (borraDuplicados (x#xs)) = estaEn a (x#xs)"
proof (rule iffI) (* usamos proof de la regla dada iffI*)
assume cprim: "estaEn a (borraDuplicados (x#xs))"
show "estaEn a (x#xs)"
proof (cases)
assume "estaEn x xs"
then show "estaEn a (x#xs)" using cprim HI by simp
next
assume "¬ estaEn x xs"
then show "estaEn a (x#xs)" using cprim HI by simp
qed
next
assume cseg: "estaEn a (x#xs)"
show "estaEn a (borraDuplicados (x#xs))"
proof (cases)
assume "a=x"
then show "estaEn a (borraDuplicados (x#xs))" using HI by auto
next
assume "a≠x"
then show "estaEn a (borraDuplicados (x#xs))" using `a≠x` cseg HI by simp
qed
qed
qed

(*
Aplico la regla iffI:
⟦P ⟹ Q ; Q ⟹ P⟧ ⟹ P = Q
Así:
[estaEn a (borraDuplicados (x # xs)) ⟹ estaEn a (x # xs); estaEn a (x # xs) ⟹ estaEn a (borraDuplicados (x # xs))]
⟹ estaEn a (borraDuplicados (x # xs)) = estaEn a (x # xs)
*)

(* bowma ivamenjim *)
lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?p xs")
proof (induct xs)
show "?p []" by simp
next
fix x xs
assume HI: "?p xs"
show "?p (x#xs)"
proof (cases)
assume H1:"estaEn x xs"
then have "estaEn a (borraDuplicados (x#xs)) = estaEn a (borraDuplicados xs)" by simp
also have "... = estaEn a xs" using HI by simp
also have "... = estaEn a (x#xs)"
proof(cases)
assume "x=a"
then show "estaEn a xs = estaEn a (x#xs)" using H1 by simp
next
assume "x≠a"
then show "estaEn a xs = estaEn a (x#xs)" by simp
qed
finally show "?p (x#xs)" by simp
next
assume H2:"¬estaEn x xs"
then have "estaEn a (borraDuplicados (x#xs)) = estaEn a (x#borraDuplicados xs)" by simp
also have "... = ((x=a) ∨ estaEn a (borraDuplicados xs))" by simp
also have "... = ((x=a) ∨ estaEn a xs)" using HI by simp
also have "... = estaEn a (x#xs)" by simp
finally show "?p (x#xs)" by simp
qed

(*danrodcha *)
(* es como la de ruben pero con diferencias de estilo *)
lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix x xs assume HI: "?P xs"
show "?P (x#xs)"
proof (rule iffI)
assume H1: "estaEn a (borraDuplicados (x # xs))"
show "estaEn a (x#xs)"
proof (cases "estaEn x xs")
case True
then show "estaEn a (x#xs)" using H1 HI by simp
next
case False
then show "estaEn a (x#xs)" using H1 HI by simp
qed
next
assume H2: "estaEn a (x#xs)"
show "estaEn a (borraDuplicados (x # xs))"
proof (cases "x=a")
case True
then show "estaEn a (borraDuplicados (x # xs))" using HI by simp
next
case False
then show "estaEn a (borraDuplicados (x # xs))" using H2 HI by simp
qed
qed
qed

text {*
---------------------------------------------------------------------
Ejercicio 6.1. Demostrar o refutar automáticamente
sinDuplicados (borraDuplicados xs)
---------------------------------------------------------------------
*}

-- "La demostración automática"

(* ivamenjim wilmorort serrodcal crigomgom anaprarod fraortmoy
marpoldia1 ferrenseg danrodcha juacabsou *)
lemma sinDuplicados_borraDuplicados:
"sinDuplicados (borraDuplicados xs)"
by (induct xs) (auto simp add: estaEn_borraDuplicados)

(* migtermor bowma *)
lemma sinDuplicados_borraDuplicados:
"sinDuplicados (borraDuplicados xs)"
by (induct xs, simp_all add: estaEn_borraDuplicados_2)

text {*
---------------------------------------------------------------------
Ejercicio 6.2. Demostrar o refutar detalladamente
sinDuplicados (borraDuplicados xs)
---------------------------------------------------------------------
*}

(*wilmorort*)

-- "La demostración estructurada es"
lemma sinDuplicados_borraDuplicados_2:
"sinDuplicados (borraDuplicados xs)"
proof (induct xs)
show " sinDuplicados (borraDuplicados [])" by simp
next
fix a xs
assume HI: "sinDuplicados (borraDuplicados xs)"
show "sinDuplicados (borraDuplicados (a # xs))"
proof (cases)
assume "estaEn a xs"
then show "sinDuplicados (borraDuplicados (a#xs))" using HI by simp
next
assume"¬ estaEn a xs"
then have "¬ (estaEn a xs) ∧ sinDuplicados (borraDuplicados xs)" using HI by simp
then have "¬ estaEn a (borraDuplicados xs) ∧ sinDuplicados (borraDuplicados xs)"
by (simp add: estaEn_borraDuplicados)
then have " sinDuplicados (a#borraDuplicados xs)" by simp
then show " sinDuplicados (borraDuplicados(a #xs))" by simp
qed
qed

(* ivamenjim migtermor crigomgom rubgonmar fraortmoy marpoldia1 ferrenseg bowma juacabsou *)

lemma sinDuplicados_borraDuplicados_2:
"sinDuplicados (borraDuplicados xs)"
proof (induct xs)
show "sinDuplicados (borraDuplicados [])" by simp
next
fix a xs
assume HI: "sinDuplicados (borraDuplicados xs)"
show "sinDuplicados (borraDuplicados (a # xs))"
proof (cases)
assume "estaEn a xs"
then show "sinDuplicados (borraDuplicados (a # xs))" using HI by simp
next
assume "¬(estaEn a xs)"
then show "sinDuplicados (borraDuplicados (a # xs))" using HI by (simp add: estaEn_borraDuplicados)
qed
qed

(* anaprarod *)
lemma sinDuplicados_borraDuplicados_2:
"sinDuplicados (borraDuplicados xs)" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix x xs
assume HI: "?P xs"
show "?P (x#xs)"
proof (cases)
assume c1: "estaEn x xs"
then show "sinDuplicados (borraDuplicados (x#xs))" using HI by simp
next
assume c2: "¬ estaEn x xs"
then have "sinDuplicados (borraDuplicados (x#xs)) =sinDuplicados (x#borraDuplicados xs)" by simp
also have "…= (¬estaEn x (borraDuplicados xs) ∧ sinDuplicados (borraDuplicados xs))" by simp
also have "… = (¬estaEn x (borraDuplicados xs))" using HI by simp
also have "… = (¬(estaEn x xs))" by (simp add:estaEn_borraDuplicados)
also have "… = True" using c2 by simp
finally show "?P (x#xs)" by simp
qed
qed

(* danrodcha *)
lemma sinDuplicados_borraDuplicados_2:
"sinDuplicados (borraDuplicados xs)" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix x xs assume HI: "?P xs"
show "?P (x#xs)"
proof (cases "estaEn x xs")
case True
then have 1: "sinDuplicados (borraDuplicados (x#xs))
= sinDuplicados (borraDuplicados xs)" by (simp add: estaEn_borraDuplicados_2)
show "?P (x#xs)" using HI 1 by simp
next
case False
then have "sinDuplicados (borraDuplicados (x#xs))
= sinDuplicados (x#borraDuplicados xs)" by simp
also have "… = (¬ (estaEn x (borraDuplicados xs)) ∧
sinDuplicados (borraDuplicados xs))" by simp
also have "… = True" using `¬ estaEn x xs` HI by (simp add:estaEn_borraDuplicados)
finally show "?P (x#xs)" by simp
qed
qed

text {*
---------------------------------------------------------------------
Ejercicio 7. Demostrar o refutar:
borraDuplicados (rev xs) = rev (borraDuplicados xs)
---------------------------------------------------------------------
*}

(*crigomgom rubgonmar ivamenjim wilmorort pablucoto migtermor
anaprarod fraortmoy ferrenseg marpoldia1 bowma danrodcha juacabsou paupeddeg *)
lemma "borraDuplicados (rev xs) = rev (borraDuplicados xs)"
quickcheck
oops

(* ivamenjim: Quickcheck encuentra el siguiente contraejemplo:
xs = [a1, a2, a1]
Por lo que:
· "borraDuplicados (rev xs) = [a2, a1]"
· "rev (borraDuplicados xs) = [a1, a2]" *)

end
</source>

Relación 5

2016-11-29T22:23:06Z

Manmorjim1:

<source lang="isar">
chapter {* R5: Eliminación de duplicados *}

theory R5_Eliminacion_de_duplicados
imports Main
begin

text {*
---------------------------------------------------------------------
Ejercicio 1. Definir la funcion primitiva recursiva
estaEn :: 'a ⇒ 'a list ⇒ bool
tal que (estaEn x xs) se verifica si el elemento x está en la lista
xs. Por ejemplo,
estaEn (2::nat) [3,2,4] = True
estaEn (1::nat) [3,2,4] = False
---------------------------------------------------------------------
*}

(* crigomgom rubgonmar bowma wilmorort pablucoto serrodcal
anaprarod migtermor paupeddeg fraortmoy marpoldia1
danrodcha manmorjim1 *)
fun estaEn :: "'a ⇒ 'a list ⇒ bool" where
"estaEn _ [] = False"
| "estaEn x (a#xs) = ((a = x) ∨ (estaEn x xs))"

value "estaEn (2::nat) [3,2,4] = True"
value "estaEn (1::nat) [3,2,4] = False"

(* ivamenjim, ferrenseg josgarsan juacabsou dancorgar *)
(* Igual que la anterior pero con x en lugar de _ en el caso base *)

fun estaEn1 :: "'a ⇒ 'a list ⇒ bool" where
"estaEn1 x [] = False"
| "estaEn1 x (a#xs) = ((x=a) ∨ estaEn1 x xs)"

value "estaEn1 (2::nat) [3,2,4] = True"
value "estaEn1 (1::nat) [3,2,4] = False"

(* wilmorort *)
(* reutilizando la funcion "algunos" de R4.thy*)
fun estaEn2 :: "'a ⇒ 'a list ⇒ bool" where
"estaEn2 a xs = algunos (λx. x = a) xs"

value "estaEn2 (2::nat) [3,2,4] = True"
value "estaEn2 (1::nat) [3,2,4] = False"

text {*
---------------------------------------------------------------------
Ejercicio 2. Definir la función primitiva recursiva
sinDuplicados :: 'a list ⇒ bool
tal que (sinDuplicados xs) se verifica si la lista xs no contiene
duplicados. Por ejemplo,
sinDuplicados [1::nat,4,2] = True
sinDuplicados [1::nat,4,2,4] = False
---------------------------------------------------------------------
*}

(* crigomgom rubgonmar ivamenjim wilmorort bowma pablucoto
serrodcal anaprarod migtermor paupeddeg fraortmoy marpoldia1
ferrenseg josgarsan danrodcha manmorjim1 juacabsou dancorgar *)
fun sinDuplicados :: "'a list ⇒ bool" where
"sinDuplicados [] = True"
| "sinDuplicados (x#xs) = (¬ estaEn x xs ∧ sinDuplicados xs)"

value "sinDuplicados [1::nat,4,2] = True"
value "sinDuplicados [1::nat,4,2,4] = False"

(* paupeddeg *)
(* Utilizando la función ∉ de Isabelle *)
fun sinDuplicados2 :: "'a list ⇒ bool" where
"sinDuplicados2 [] = True"
| "sinDuplicados2 (a#xs) = ((a ∉ set xs) ∧ sinDuplicados2 xs ) "

text {*
---------------------------------------------------------------------
Ejercicio 3. Definir la función primitiva recursiva
borraDuplicados :: 'a list ⇒ bool
tal que (borraDuplicados xs) es la lista obtenida eliminando los
elementos duplicados de la lista xs. Por ejemplo,
borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]

Nota: La función borraDuplicados es equivalente a la predefinida
remdups.
---------------------------------------------------------------------
*}

(* crigomgom rubgonmar wilmorort bowma pablucoto serrodcal
anaprarod migtermor paupeddeg fraortmoy marpoldia1 ferrenseg
josgarsan danrodcha juacabsou dancorgar manmorjim1 *)
fun borraDuplicados :: "'a list ⇒ 'a list" where
"borraDuplicados [] = []"
| "borraDuplicados (x#xs) =( if estaEn x xs then borraDuplicados xs else x#borraDuplicados xs)"

value "borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]"

(* ivamenjim *)
(* Utilizando la negación primero *)

fun borraDuplicados :: "'a list ⇒ 'a list" where
"borraDuplicados [] = []"
| "borraDuplicados (x#xs) = (if ¬(estaEn x xs) then (x#(borraDuplicados xs)) else borraDuplicados xs)"

(* rubgonmar *)
(* Otra forma Sin usar if
Utilizando case aunque se le sacaría más partido con más de 2 casos *)

fun borraDuplicados1 :: "'a list ⇒ 'a list" where
"borraDuplicados1 [] = []" |
"borraDuplicados1 (x#xs) = ( case estaEn x xs of False => x#borraDuplicados1 xs | True => borraDuplicados1 xs )"

(* rubgonmar *)
(*Otra forma utilizando let*)
fun borraDuplicados2 :: "'a list ⇒ 'a list" where
"borraDuplicados2 [] = []" |
"borraDuplicados2 (x#xs) = (let condicion = estaEn x xs::bool in
if condicion then borraDuplicados2 xs else x # borraDuplicados2 xs)"

value "borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]"

text {*
---------------------------------------------------------------------
Ejercicio 4.1. Demostrar o refutar automáticamente
length (borraDuplicados xs) ≤ length xs
---------------------------------------------------------------------
*}

-- "La demostración automática es"
(*crigomgom anaprarod ferrenseg*)
lemma length_borraDuplicados:
"length (borraDuplicados xs) ≤ length xs"
by (induct xs, simp_all)

(* rubgonmar wilmorort pablucoto serrodcal migtermor paupeddeg
fraortmoy marpoldia1 danrodcha juacabsou *)
lemma length_borraDuplicados:
"length ( borraDuplicados xs ) ≤ length xs"
by (induct xs) auto

(* manmorjim1 *)
(* soy de ponerlo mejor por pasos *)
lemma length_borraDuplicados:
"length ( borraDuplicados xs ) ≤ length xs"
apply (induct xs)
apply auto
done

(* ivamenjim *)
(* Demostrando objetivo a objetivo *)
lemma length_borraDuplicados:
"length (borraDuplicados xs) ≤ length xs"
apply (induct xs)
apply simp
apply auto
done

(* bowma anaprarod *)
lemma length_borraDuplicados:
"length (borraDuplicados xs) ≤ length xs"
apply (induct xs)
apply (simp, simp) (* creo que es mejor poner aquí simp_all *)
done

(* ferrenseg *)

lemma length_borraDuplicados:
"length (borraDuplicados xs) ≤ length xs"
by (induct xs) simp_all (* Creo que se puede poner simp_all fuera de parentesis *)

text {*
---------------------------------------------------------------------
Ejercicio 4.2. Demostrar o refutar detalladamente
length (borraDuplicados xs) ≤ length xs
---------------------------------------------------------------------
*}

-- "La demostración estructurada es"
(* crigomgom *)
lemma length_borraDuplicados_2:
"length (borraDuplicados xs) ≤ length xs"
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix x xs
assume HI: "length (borraDuplicados xs) ≤ length xs"
show "length (borraDuplicados (x#xs)) ≤ length (x#xs)"
proof (cases)
assume "estaEn x xs"
then have "length (borraDuplicados (x#xs)) = length (borraDuplicados xs)" by simp
also have "... ≤ length xs" using HI by simp
also have "... ≤ length (x#xs)" by simp
finally show "length (borraDuplicados (x#xs)) ≤ length (x#xs)" by simp
next
assume "(¬ estaEn x xs)"
then have "length (borraDuplicados (x#xs)) = length (x#borraDuplicados xs)" by simp
also have "... = 1 + length (borraDuplicados xs)" by simp
also have "... ≤ 1 + length xs" using HI by simp
also have "... = length (x#xs)" by simp
finally show "length (borraDuplicados (x#xs)) ≤ length (x#xs)" by simp
qed
qed

(* ivamenjim wilmorort ferrenseg rubgonmar juacabsou *)
lemma length_borraDuplicados_2:
"length (borraDuplicados xs) ≤ length xs"
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix a xs
assume HI: "length (borraDuplicados xs) ≤ length xs"
have "length (borraDuplicados (a # xs)) ≤ 1+length (borraDuplicados xs)" by simp
also have "... ≤ 1+length xs" using HI by simp
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs)" by simp
qed

(* serrodcal anaprarod danrodcha *)
lemma length_borraDuplicados_2: "length (borraDuplicados xs) ≤ length xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "length (borraDuplicados (a # xs)) ≤ 1+length (borraDuplicados xs)" by simp
also have "... ≤ 1+length xs" using HI by simp
finally show "?P (a # xs)" by simp
qed

(* pablucoto *)
lemma length_borraDuplicados_2:
"length (borraDuplicados xs) ≤ length xs"
proof(induct xs)
show "length (borraDuplicados []) ≤ length [] " by simp
next
fix a xs
assume HI: " length (borraDuplicados xs) ≤ length xs "
have "length (borraDuplicados (a # xs)) ≤ 1 + length(borraDuplicados xs)" by simp
also have "... ≤ 1 + length xs" using HI by simp
also have "... ≤ length (a#xs)" by simp
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs) " by simp
qed

(* bowma *)
lemma length_borraDuplicados_3:
"length (borraDuplicados xs) ≤ length xs" (is "?p xs")
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix a xs
assume HI: "?p xs"
have "length (a#xs) = 1 + length xs" by simp
have "length(borraDuplicados (a#xs)) ≤ 1 + length(borraDuplicados xs)" by simp
also have "... ≤ 1+length xs" using HI by simp
also have "... ≤ length (a#xs)" by simp
finally show "?p (a # xs)" by simp
qed

(* migtermor *)

lemma length_borraDuplicados_2:
"length (borraDuplicados xs) ≤ length xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "length (borraDuplicados (a#xs)) ≤ (length (a#xs))"
proof (cases)
assume "(estaEn a xs)"
then have Aux: "length (borraDuplicados (a#xs)) = length (borraDuplicados xs)" by simp
also have "… ≤ length (a#xs)" using HI by simp
then show "length (borraDuplicados (a#xs)) ≤ (length (a#xs))" using Aux by simp
next
assume "¬ (estaEn a xs)"
then have Aux: "length (borraDuplicados (a#xs)) = 1+ length (borraDuplicados xs)" by simp
also have "… ≤ length (a#xs)" using HI by simp
then show "length (borraDuplicados (a#xs)) ≤ (length (a#xs))" using Aux by simp
qed
then show "?P (a#xs)" by simp
qed

(* paupeddeg marpoldia1*)
lemma length_borraDuplicados_4:
"length (borraDuplicados xs) ≤ length xs"
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix a xs
assume HI: "length (borraDuplicados xs) ≤ length xs"
have "length (borraDuplicados (a # xs)) ≤ length [a] + length (borraDuplicados xs) " by simp
also have "... ≤ 1 + length (borraDuplicados xs)" by simp
also have "... ≤ 1 + length xs" using HI by simp
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs)" by simp
qed

(* fraortmoy *)
(* muy parecida a alguna anterior, pero yo dí mas pasos *)
lemma length_borraDuplicados_2:
"length (borraDuplicados xs) ≤ length xs"
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix a xs
assume H1: "length (borraDuplicados xs) ≤ length xs"
have "length (borraDuplicados (a # xs)) ≤ length(borraDuplicados [a])+length (borraDuplicados xs)" by simp
also have "… ≤ 1 + length (borraDuplicados xs)" by simp
also have "… ≤ 1 + length xs" using H1 by simp
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 5.1. Demostrar o refutar automáticamente
estaEn a (borraDuplicados xs) = estaEn a xs
---------------------------------------------------------------------
*}

-- "La demostración automática es"
(* crigomgom rubgonmar wilmorort pablucoto serrodcal bowma
migtermor fraortmoy marpoldia1 ferrenseg danrodcha juacabsou paupeddeg*)
lemma estaEn_borraDuplicados:
"estaEn a (borraDuplicados xs) = estaEn a xs"
by (induct xs) auto

(* ivamenjim *)
lemma estaEn_borraDuplicados:
"estaEn a (borraDuplicados xs) = estaEn a xs"
apply (induct xs)
apply auto
done

(* anaprarod *)
lemma estaEn_borraDuplicados:
"estaEn a (borraDuplicados xs) = estaEn a xs"
by (induct xs, simp_all, blast)

(* anaprarod *)
lemma estaEn_borraDuplicados':
"estaEn a (borraDuplicados xs) = estaEn a xs"
apply (induct xs)
apply (cases "estaEn x xs")
apply (simp_all)
apply blast
done

(* bowma *)
lemma estaEn_borraDuplicados:
"estaEn a (borraDuplicados xs) = estaEn a xs"
apply (induct xs)
apply simp
apply (simp, blast)
done

text {*
---------------------------------------------------------------------
Ejercicio 5.2. Demostrar o refutar detalladamente
estaEn a (borraDuplicados xs) = estaEn a xs
Nota: Para la demostración de la equivalencia se puede usar
proof (rule iffI)
La regla iffI es
⟦P ⟹ Q ; Q ⟹ P⟧ ⟹ P = Q
---------------------------------------------------------------------
*}

-- "La demostración estructurada es"

(* wilmorort *)

lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs"
proof (induct xs)
show "estaEn a (borraDuplicados []) = estaEn a []" by simp
next
fix b xs
assume HI: "estaEn a (borraDuplicados xs) = estaEn a xs"
show "estaEn a (borraDuplicados (b#xs)) = estaEn a (b#xs)"
proof (rule iffI)
assume H1: "estaEn a (borraDuplicados (b#xs))"
show "estaEn a (b#xs)"
proof (cases)
assume "estaEn b xs"
then have "estaEn a (borraDuplicados xs)" using H1 by simp
then have "estaEn a xs" using HI by simp
then show "estaEn a (b#xs)" by simp
next
assume "¬ estaEn b xs"
then have "estaEn a (b#(borraDuplicados xs))" using H1 by simp
then have "a=b ∨ (estaEn a (borraDuplicados xs))" by simp
then have " a=b ∨ (estaEn a xs)" using HI by simp
then show "estaEn a (b#xs)" by simp
qed
next
assume H2: "estaEn a (b#xs)"
show "estaEn a (borraDuplicados (b#xs))"
proof (cases)
assume "a=b"
then have "estaEn b (borraDuplicados xs) = estaEn b xs" using HI by simp
then have "(estaEn b xs ⟶ estaEn b (borraDuplicados xs)) ∧
(¬ estaEn b xs ⟶ estaEn b (b # borraDuplicados xs))" by simp
then have "estaEn b (borraDuplicados (b#xs))" by simp
then show "estaEn a (borraDuplicados (b#xs))" using `a=b` by simp
next
assume "a≠b"
then have "estaEn a (b#xs)" using H2 by simp
then have "a = b ∨ estaEn a xs" by simp
then have "False ∨ estaEn a xs " using `a≠b` by simp
then have "estaEn a xs" by simp
then have "estaEn a (borraDuplicados xs)" using HI by simp
then show "estaEn a (borraDuplicados (b#xs))" using `a≠b` by simp
qed
qed
qed

(* anaprarod marpoldia1 ferrenseg juacabsou *)
lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix x xs
assume HI: "?P xs"
show "?P (x#xs)"
proof (cases)
assume "estaEn x xs"
then have "estaEn a (borraDuplicados (x#xs)) = estaEn a (borraDuplicados xs)" by simp
also have "...= estaEn a xs" using HI by simp
finally show "?P (x#xs)" by auto
next
assume "¬estaEn x xs"
then have "estaEn a (borraDuplicados (x#xs)) = estaEn a (x#borraDuplicados xs)" by simp
also have "...= (x = a ∨ estaEn a (borraDuplicados xs))" by simp
finally show "?P (x#xs)" using HI by simp
qed
qed

(* migtermor *)

lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix aa xs
assume HI: "?P xs"
have P1: "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)"
proof (cases)
assume C1: "(estaEn aa xs)"
have "estaEn a (borraDuplicados (aa#xs)) = estaEn a (borraDuplicados xs)"
using C1 by simp
also have P3: "… = estaEn a xs" using HI by simp
also have "… = estaEn a (aa#xs)"
proof (cases)
assume "(a=aa)"
then show "estaEn a xs = estaEn a (aa#xs)" using C1 by simp
next
assume "¬(a=aa)"
then show "estaEn a xs = estaEn a (aa#xs)" by simp
qed
then show "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)" using P3 by simp
next
assume C2: "¬(estaEn aa xs)"
then show "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)" using HI by simp
qed
also have Conc: "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)" using P1 by simp
finally show "?P (aa#xs)" using Conc by simp
qed

(* crigomgom *)
-- "La demostración estructurada es"
lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs"
proof (induct xs)
show "estaEn a (borraDuplicados []) = estaEn a []" by simp
next
fix x xs
assume HI: "estaEn a (borraDuplicados xs) = estaEn a xs"
show "estaEn a (borraDuplicados (x#xs)) = estaEn a (x#xs)"
proof (rule iffI)
assume a1: "estaEn a (borraDuplicados (x#xs))"
show "estaEn a (x#xs)"
proof (cases)
assume "estaEn x xs"
then have "estaEn a (borraDuplicados xs)" using a1 by simp
then have "estaEn a xs" using HI by simp
then show "estaEn a (x#xs)" by simp
next
assume "¬ estaEn x xs"
then have "estaEn a (x#(borraDuplicados xs))" using a1 by simp
then have " x=a ∨ (estaEn a (borraDuplicados xs))" by simp
then have " x=a ∨ (estaEn a xs)" using HI by simp
then show "estaEn a (x#xs)" by simp
qed
next
assume a2: "estaEn a (x#xs)"
show "estaEn a (borraDuplicados (x#xs))"
proof (cases)
assume "a=x"
then show "estaEn a (borraDuplicados (x#xs))" using HI by simp
next
assume b1: "a≠x"
then have "estaEn a (x#xs)" using a2 by simp
then have "x = a ∨ estaEn a xs" by simp
then have "estaEn a xs " using b1 by simp
then have "estaEn a (borraDuplicados xs)" using HI by simp
then show "estaEn a (borraDuplicados (x#xs))" using b1 by simp
qed
qed
qed

(* rubgonmar *)
lemma estaEn_borraDuplicados_2:
"estaEn a ( borraDuplicados xs ) = estaEn a xs"
proof (induct xs)
show "estaEn a (borraDuplicados []) = estaEn a []" by simp
next
fix x
fix xs
assume HI: "estaEn a (borraDuplicados xs) = estaEn a xs"
show "estaEn a (borraDuplicados (x#xs)) = estaEn a (x#xs)"
proof (rule iffI) (* usamos proof de la regla dada iffI*)
assume cprim: "estaEn a (borraDuplicados (x#xs))"
show "estaEn a (x#xs)"
proof (cases)
assume "estaEn x xs"
then show "estaEn a (x#xs)" using cprim HI by simp
next
assume "¬ estaEn x xs"
then show "estaEn a (x#xs)" using cprim HI by simp
qed
next
assume cseg: "estaEn a (x#xs)"
show "estaEn a (borraDuplicados (x#xs))"
proof (cases)
assume "a=x"
then show "estaEn a (borraDuplicados (x#xs))" using HI by auto
next
assume "a≠x"
then show "estaEn a (borraDuplicados (x#xs))" using `a≠x` cseg HI by simp
qed
qed
qed

(*
Aplico la regla iffI:
⟦P ⟹ Q ; Q ⟹ P⟧ ⟹ P = Q
Así:
[estaEn a (borraDuplicados (x # xs)) ⟹ estaEn a (x # xs); estaEn a (x # xs) ⟹ estaEn a (borraDuplicados (x # xs))]
⟹ estaEn a (borraDuplicados (x # xs)) = estaEn a (x # xs)
*)

(* bowma ivamenjim *)
lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?p xs")
proof (induct xs)
show "?p []" by simp
next
fix x xs
assume HI: "?p xs"
show "?p (x#xs)"
proof (cases)
assume H1:"estaEn x xs"
then have "estaEn a (borraDuplicados (x#xs)) = estaEn a (borraDuplicados xs)" by simp
also have "... = estaEn a xs" using HI by simp
also have "... = estaEn a (x#xs)"
proof(cases)
assume "x=a"
then show "estaEn a xs = estaEn a (x#xs)" using H1 by simp
next
assume "x≠a"
then show "estaEn a xs = estaEn a (x#xs)" by simp
qed
finally show "?p (x#xs)" by simp
next
assume H2:"¬estaEn x xs"
then have "estaEn a (borraDuplicados (x#xs)) = estaEn a (x#borraDuplicados xs)" by simp
also have "... = ((x=a) ∨ estaEn a (borraDuplicados xs))" by simp
also have "... = ((x=a) ∨ estaEn a xs)" using HI by simp
also have "... = estaEn a (x#xs)" by simp
finally show "?p (x#xs)" by simp
qed

(*danrodcha *)
(* es como la de ruben pero con diferencias de estilo *)
lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix x xs assume HI: "?P xs"
show "?P (x#xs)"
proof (rule iffI)
assume H1: "estaEn a (borraDuplicados (x # xs))"
show "estaEn a (x#xs)"
proof (cases "estaEn x xs")
case True
then show "estaEn a (x#xs)" using H1 HI by simp
next
case False
then show "estaEn a (x#xs)" using H1 HI by simp
qed
next
assume H2: "estaEn a (x#xs)"
show "estaEn a (borraDuplicados (x # xs))"
proof (cases "x=a")
case True
then show "estaEn a (borraDuplicados (x # xs))" using HI by simp
next
case False
then show "estaEn a (borraDuplicados (x # xs))" using H2 HI by simp
qed
qed
qed

text {*
---------------------------------------------------------------------
Ejercicio 6.1. Demostrar o refutar automáticamente
sinDuplicados (borraDuplicados xs)
---------------------------------------------------------------------
*}

-- "La demostración automática"

(* ivamenjim wilmorort serrodcal crigomgom anaprarod fraortmoy
marpoldia1 ferrenseg danrodcha juacabsou *)
lemma sinDuplicados_borraDuplicados:
"sinDuplicados (borraDuplicados xs)"
by (induct xs) (auto simp add: estaEn_borraDuplicados)

(* migtermor bowma *)
lemma sinDuplicados_borraDuplicados:
"sinDuplicados (borraDuplicados xs)"
by (induct xs, simp_all add: estaEn_borraDuplicados_2)

text {*
---------------------------------------------------------------------
Ejercicio 6.2. Demostrar o refutar detalladamente
sinDuplicados (borraDuplicados xs)
---------------------------------------------------------------------
*}

(*wilmorort*)

-- "La demostración estructurada es"
lemma sinDuplicados_borraDuplicados_2:
"sinDuplicados (borraDuplicados xs)"
proof (induct xs)
show " sinDuplicados (borraDuplicados [])" by simp
next
fix a xs
assume HI: "sinDuplicados (borraDuplicados xs)"
show "sinDuplicados (borraDuplicados (a # xs))"
proof (cases)
assume "estaEn a xs"
then show "sinDuplicados (borraDuplicados (a#xs))" using HI by simp
next
assume"¬ estaEn a xs"
then have "¬ (estaEn a xs) ∧ sinDuplicados (borraDuplicados xs)" using HI by simp
then have "¬ estaEn a (borraDuplicados xs) ∧ sinDuplicados (borraDuplicados xs)"
by (simp add: estaEn_borraDuplicados)
then have " sinDuplicados (a#borraDuplicados xs)" by simp
then show " sinDuplicados (borraDuplicados(a #xs))" by simp
qed
qed

(* ivamenjim migtermor crigomgom rubgonmar fraortmoy marpoldia1 ferrenseg bowma juacabsou *)

lemma sinDuplicados_borraDuplicados_2:
"sinDuplicados (borraDuplicados xs)"
proof (induct xs)
show "sinDuplicados (borraDuplicados [])" by simp
next
fix a xs
assume HI: "sinDuplicados (borraDuplicados xs)"
show "sinDuplicados (borraDuplicados (a # xs))"
proof (cases)
assume "estaEn a xs"
then show "sinDuplicados (borraDuplicados (a # xs))" using HI by simp
next
assume "¬(estaEn a xs)"
then show "sinDuplicados (borraDuplicados (a # xs))" using HI by (simp add: estaEn_borraDuplicados)
qed
qed

(* anaprarod *)
lemma sinDuplicados_borraDuplicados_2:
"sinDuplicados (borraDuplicados xs)" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix x xs
assume HI: "?P xs"
show "?P (x#xs)"
proof (cases)
assume c1: "estaEn x xs"
then show "sinDuplicados (borraDuplicados (x#xs))" using HI by simp
next
assume c2: "¬ estaEn x xs"
then have "sinDuplicados (borraDuplicados (x#xs)) =sinDuplicados (x#borraDuplicados xs)" by simp
also have "…= (¬estaEn x (borraDuplicados xs) ∧ sinDuplicados (borraDuplicados xs))" by simp
also have "… = (¬estaEn x (borraDuplicados xs))" using HI by simp
also have "… = (¬(estaEn x xs))" by (simp add:estaEn_borraDuplicados)
also have "… = True" using c2 by simp
finally show "?P (x#xs)" by simp
qed
qed

(* danrodcha *)
lemma sinDuplicados_borraDuplicados_2:
"sinDuplicados (borraDuplicados xs)" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix x xs assume HI: "?P xs"
show "?P (x#xs)"
proof (cases "estaEn x xs")
case True
then have 1: "sinDuplicados (borraDuplicados (x#xs))
= sinDuplicados (borraDuplicados xs)" by (simp add: estaEn_borraDuplicados_2)
show "?P (x#xs)" using HI 1 by simp
next
case False
then have "sinDuplicados (borraDuplicados (x#xs))
= sinDuplicados (x#borraDuplicados xs)" by simp
also have "… = (¬ (estaEn x (borraDuplicados xs)) ∧
sinDuplicados (borraDuplicados xs))" by simp
also have "… = True" using `¬ estaEn x xs` HI by (simp add:estaEn_borraDuplicados)
finally show "?P (x#xs)" by simp
qed
qed

text {*
---------------------------------------------------------------------
Ejercicio 7. Demostrar o refutar:
borraDuplicados (rev xs) = rev (borraDuplicados xs)
---------------------------------------------------------------------
*}

(*crigomgom rubgonmar ivamenjim wilmorort pablucoto migtermor
anaprarod fraortmoy ferrenseg marpoldia1 bowma danrodcha juacabsou paupeddeg *)
lemma "borraDuplicados (rev xs) = rev (borraDuplicados xs)"
quickcheck
oops

(* ivamenjim: Quickcheck encuentra el siguiente contraejemplo:
xs = [a1, a2, a1]
Por lo que:
· "borraDuplicados (rev xs) = [a2, a1]"
· "rev (borraDuplicados xs) = [a1, a2]" *)

end
</source>

Relación 5

2016-11-29T22:20:19Z

Manmorjim1:

<source lang="isar">
chapter {* R5: Eliminación de duplicados *}

theory R5_Eliminacion_de_duplicados
imports Main
begin

text {*
---------------------------------------------------------------------
Ejercicio 1. Definir la funcion primitiva recursiva
estaEn :: 'a ⇒ 'a list ⇒ bool
tal que (estaEn x xs) se verifica si el elemento x está en la lista
xs. Por ejemplo,
estaEn (2::nat) [3,2,4] = True
estaEn (1::nat) [3,2,4] = False
---------------------------------------------------------------------
*}

(* crigomgom rubgonmar bowma wilmorort pablucoto serrodcal
anaprarod migtermor paupeddeg fraortmoy marpoldia1
danrodcha manmorjim1 *)
fun estaEn :: "'a ⇒ 'a list ⇒ bool" where
"estaEn _ [] = False"
| "estaEn x (a#xs) = ((a = x) ∨ (estaEn x xs))"

value "estaEn (2::nat) [3,2,4] = True"
value "estaEn (1::nat) [3,2,4] = False"

(* ivamenjim, ferrenseg josgarsan juacabsou dancorgar *)
(* Igual que la anterior pero con x en lugar de _ en el caso base *)

fun estaEn1 :: "'a ⇒ 'a list ⇒ bool" where
"estaEn1 x [] = False"
| "estaEn1 x (a#xs) = ((x=a) ∨ estaEn1 x xs)"

value "estaEn1 (2::nat) [3,2,4] = True"
value "estaEn1 (1::nat) [3,2,4] = False"

(* wilmorort *)
(* reutilizando la funcion "algunos" de R4.thy*)
fun estaEn2 :: "'a ⇒ 'a list ⇒ bool" where
"estaEn2 a xs = algunos (λx. x = a) xs"

value "estaEn2 (2::nat) [3,2,4] = True"
value "estaEn2 (1::nat) [3,2,4] = False"

text {*
---------------------------------------------------------------------
Ejercicio 2. Definir la función primitiva recursiva
sinDuplicados :: 'a list ⇒ bool
tal que (sinDuplicados xs) se verifica si la lista xs no contiene
duplicados. Por ejemplo,
sinDuplicados [1::nat,4,2] = True
sinDuplicados [1::nat,4,2,4] = False
---------------------------------------------------------------------
*}

(* crigomgom rubgonmar ivamenjim wilmorort bowma pablucoto
serrodcal anaprarod migtermor paupeddeg fraortmoy marpoldia1
ferrenseg josgarsan danrodcha manmorjim1 juacabsou dancorgar *)
fun sinDuplicados :: "'a list ⇒ bool" where
"sinDuplicados [] = True"
| "sinDuplicados (x#xs) = (¬ estaEn x xs ∧ sinDuplicados xs)"

value "sinDuplicados [1::nat,4,2] = True"
value "sinDuplicados [1::nat,4,2,4] = False"

(* paupeddeg *)
(* Utilizando la función ∉ de Isabelle *)
fun sinDuplicados2 :: "'a list ⇒ bool" where
"sinDuplicados2 [] = True"
| "sinDuplicados2 (a#xs) = ((a ∉ set xs) ∧ sinDuplicados2 xs ) "

text {*
---------------------------------------------------------------------
Ejercicio 3. Definir la función primitiva recursiva
borraDuplicados :: 'a list ⇒ bool
tal que (borraDuplicados xs) es la lista obtenida eliminando los
elementos duplicados de la lista xs. Por ejemplo,
borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]

Nota: La función borraDuplicados es equivalente a la predefinida
remdups.
---------------------------------------------------------------------
*}

(* crigomgom rubgonmar wilmorort bowma pablucoto serrodcal
anaprarod migtermor paupeddeg fraortmoy marpoldia1 ferrenseg
josgarsan danrodcha juacabsou dancorgar manmorjim1 *)
fun borraDuplicados :: "'a list ⇒ 'a list" where
"borraDuplicados [] = []"
| "borraDuplicados (x#xs) =( if estaEn x xs then borraDuplicados xs else x#borraDuplicados xs)"

value "borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]"

(* ivamenjim *)
(* Utilizando la negación primero *)

fun borraDuplicados :: "'a list ⇒ 'a list" where
"borraDuplicados [] = []"
| "borraDuplicados (x#xs) = (if ¬(estaEn x xs) then (x#(borraDuplicados xs)) else borraDuplicados xs)"

(* rubgonmar *)
(* Otra forma Sin usar if
Utilizando case aunque se le sacaría más partido con más de 2 casos *)

fun borraDuplicados1 :: "'a list ⇒ 'a list" where
"borraDuplicados1 [] = []" |
"borraDuplicados1 (x#xs) = ( case estaEn x xs of False => x#borraDuplicados1 xs | True => borraDuplicados1 xs )"

(* rubgonmar *)
(*Otra forma utilizando let*)
fun borraDuplicados2 :: "'a list ⇒ 'a list" where
"borraDuplicados2 [] = []" |
"borraDuplicados2 (x#xs) = (let condicion = estaEn x xs::bool in
if condicion then borraDuplicados2 xs else x # borraDuplicados2 xs)"

value "borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]"

text {*
---------------------------------------------------------------------
Ejercicio 4.1. Demostrar o refutar automáticamente
length (borraDuplicados xs) ≤ length xs
---------------------------------------------------------------------
*}

-- "La demostración automática es"
(*crigomgom anaprarod ferrenseg*)
lemma length_borraDuplicados:
"length (borraDuplicados xs) ≤ length xs"
by (induct xs, simp_all)

(* rubgonmar wilmorort pablucoto serrodcal migtermor paupeddeg
fraortmoy marpoldia1 danrodcha juacabsou *)
lemma length_borraDuplicados:
"length ( borraDuplicados xs ) ≤ length xs"
by (induct xs) auto

(* ivamenjim *)
(* Demostrando objetivo a objetivo *)
lemma length_borraDuplicados:
"length (borraDuplicados xs) ≤ length xs"
apply (induct xs)
apply simp
apply auto
done

(* bowma anaprarod *)
lemma length_borraDuplicados:
"length (borraDuplicados xs) ≤ length xs"
apply (induct xs)
apply (simp, simp) (* creo que es mejor poner aquí simp_all *)
done

(* ferrenseg *)

lemma length_borraDuplicados:
"length (borraDuplicados xs) ≤ length xs"
by (induct xs) simp_all (* Creo que se puede poner simp_all fuera de parentesis *)

text {*
---------------------------------------------------------------------
Ejercicio 4.2. Demostrar o refutar detalladamente
length (borraDuplicados xs) ≤ length xs
---------------------------------------------------------------------
*}

-- "La demostración estructurada es"
(* crigomgom *)
lemma length_borraDuplicados_2:
"length (borraDuplicados xs) ≤ length xs"
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix x xs
assume HI: "length (borraDuplicados xs) ≤ length xs"
show "length (borraDuplicados (x#xs)) ≤ length (x#xs)"
proof (cases)
assume "estaEn x xs"
then have "length (borraDuplicados (x#xs)) = length (borraDuplicados xs)" by simp
also have "... ≤ length xs" using HI by simp
also have "... ≤ length (x#xs)" by simp
finally show "length (borraDuplicados (x#xs)) ≤ length (x#xs)" by simp
next
assume "(¬ estaEn x xs)"
then have "length (borraDuplicados (x#xs)) = length (x#borraDuplicados xs)" by simp
also have "... = 1 + length (borraDuplicados xs)" by simp
also have "... ≤ 1 + length xs" using HI by simp
also have "... = length (x#xs)" by simp
finally show "length (borraDuplicados (x#xs)) ≤ length (x#xs)" by simp
qed
qed

(* ivamenjim wilmorort ferrenseg rubgonmar juacabsou *)
lemma length_borraDuplicados_2:
"length (borraDuplicados xs) ≤ length xs"
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix a xs
assume HI: "length (borraDuplicados xs) ≤ length xs"
have "length (borraDuplicados (a # xs)) ≤ 1+length (borraDuplicados xs)" by simp
also have "... ≤ 1+length xs" using HI by simp
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs)" by simp
qed

(* serrodcal anaprarod danrodcha *)
lemma length_borraDuplicados_2: "length (borraDuplicados xs) ≤ length xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "length (borraDuplicados (a # xs)) ≤ 1+length (borraDuplicados xs)" by simp
also have "... ≤ 1+length xs" using HI by simp
finally show "?P (a # xs)" by simp
qed

(* pablucoto *)
lemma length_borraDuplicados_2:
"length (borraDuplicados xs) ≤ length xs"
proof(induct xs)
show "length (borraDuplicados []) ≤ length [] " by simp
next
fix a xs
assume HI: " length (borraDuplicados xs) ≤ length xs "
have "length (borraDuplicados (a # xs)) ≤ 1 + length(borraDuplicados xs)" by simp
also have "... ≤ 1 + length xs" using HI by simp
also have "... ≤ length (a#xs)" by simp
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs) " by simp
qed

(* bowma *)
lemma length_borraDuplicados_3:
"length (borraDuplicados xs) ≤ length xs" (is "?p xs")
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix a xs
assume HI: "?p xs"
have "length (a#xs) = 1 + length xs" by simp
have "length(borraDuplicados (a#xs)) ≤ 1 + length(borraDuplicados xs)" by simp
also have "... ≤ 1+length xs" using HI by simp
also have "... ≤ length (a#xs)" by simp
finally show "?p (a # xs)" by simp
qed

(* migtermor *)

lemma length_borraDuplicados_2:
"length (borraDuplicados xs) ≤ length xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "length (borraDuplicados (a#xs)) ≤ (length (a#xs))"
proof (cases)
assume "(estaEn a xs)"
then have Aux: "length (borraDuplicados (a#xs)) = length (borraDuplicados xs)" by simp
also have "… ≤ length (a#xs)" using HI by simp
then show "length (borraDuplicados (a#xs)) ≤ (length (a#xs))" using Aux by simp
next
assume "¬ (estaEn a xs)"
then have Aux: "length (borraDuplicados (a#xs)) = 1+ length (borraDuplicados xs)" by simp
also have "… ≤ length (a#xs)" using HI by simp
then show "length (borraDuplicados (a#xs)) ≤ (length (a#xs))" using Aux by simp
qed
then show "?P (a#xs)" by simp
qed

(* paupeddeg marpoldia1*)
lemma length_borraDuplicados_4:
"length (borraDuplicados xs) ≤ length xs"
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix a xs
assume HI: "length (borraDuplicados xs) ≤ length xs"
have "length (borraDuplicados (a # xs)) ≤ length [a] + length (borraDuplicados xs) " by simp
also have "... ≤ 1 + length (borraDuplicados xs)" by simp
also have "... ≤ 1 + length xs" using HI by simp
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs)" by simp
qed

(* fraortmoy *)
(* muy parecida a alguna anterior, pero yo dí mas pasos *)
lemma length_borraDuplicados_2:
"length (borraDuplicados xs) ≤ length xs"
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix a xs
assume H1: "length (borraDuplicados xs) ≤ length xs"
have "length (borraDuplicados (a # xs)) ≤ length(borraDuplicados [a])+length (borraDuplicados xs)" by simp
also have "… ≤ 1 + length (borraDuplicados xs)" by simp
also have "… ≤ 1 + length xs" using H1 by simp
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 5.1. Demostrar o refutar automáticamente
estaEn a (borraDuplicados xs) = estaEn a xs
---------------------------------------------------------------------
*}

-- "La demostración automática es"
(* crigomgom rubgonmar wilmorort pablucoto serrodcal bowma
migtermor fraortmoy marpoldia1 ferrenseg danrodcha juacabsou paupeddeg*)
lemma estaEn_borraDuplicados:
"estaEn a (borraDuplicados xs) = estaEn a xs"
by (induct xs) auto

(* ivamenjim *)
lemma estaEn_borraDuplicados:
"estaEn a (borraDuplicados xs) = estaEn a xs"
apply (induct xs)
apply auto
done

(* anaprarod *)
lemma estaEn_borraDuplicados:
"estaEn a (borraDuplicados xs) = estaEn a xs"
by (induct xs, simp_all, blast)

(* anaprarod *)
lemma estaEn_borraDuplicados':
"estaEn a (borraDuplicados xs) = estaEn a xs"
apply (induct xs)
apply (cases "estaEn x xs")
apply (simp_all)
apply blast
done

(* bowma *)
lemma estaEn_borraDuplicados:
"estaEn a (borraDuplicados xs) = estaEn a xs"
apply (induct xs)
apply simp
apply (simp, blast)
done

text {*
---------------------------------------------------------------------
Ejercicio 5.2. Demostrar o refutar detalladamente
estaEn a (borraDuplicados xs) = estaEn a xs
Nota: Para la demostración de la equivalencia se puede usar
proof (rule iffI)
La regla iffI es
⟦P ⟹ Q ; Q ⟹ P⟧ ⟹ P = Q
---------------------------------------------------------------------
*}

-- "La demostración estructurada es"

(* wilmorort *)

lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs"
proof (induct xs)
show "estaEn a (borraDuplicados []) = estaEn a []" by simp
next
fix b xs
assume HI: "estaEn a (borraDuplicados xs) = estaEn a xs"
show "estaEn a (borraDuplicados (b#xs)) = estaEn a (b#xs)"
proof (rule iffI)
assume H1: "estaEn a (borraDuplicados (b#xs))"
show "estaEn a (b#xs)"
proof (cases)
assume "estaEn b xs"
then have "estaEn a (borraDuplicados xs)" using H1 by simp
then have "estaEn a xs" using HI by simp
then show "estaEn a (b#xs)" by simp
next
assume "¬ estaEn b xs"
then have "estaEn a (b#(borraDuplicados xs))" using H1 by simp
then have "a=b ∨ (estaEn a (borraDuplicados xs))" by simp
then have " a=b ∨ (estaEn a xs)" using HI by simp
then show "estaEn a (b#xs)" by simp
qed
next
assume H2: "estaEn a (b#xs)"
show "estaEn a (borraDuplicados (b#xs))"
proof (cases)
assume "a=b"
then have "estaEn b (borraDuplicados xs) = estaEn b xs" using HI by simp
then have "(estaEn b xs ⟶ estaEn b (borraDuplicados xs)) ∧
(¬ estaEn b xs ⟶ estaEn b (b # borraDuplicados xs))" by simp
then have "estaEn b (borraDuplicados (b#xs))" by simp
then show "estaEn a (borraDuplicados (b#xs))" using `a=b` by simp
next
assume "a≠b"
then have "estaEn a (b#xs)" using H2 by simp
then have "a = b ∨ estaEn a xs" by simp
then have "False ∨ estaEn a xs " using `a≠b` by simp
then have "estaEn a xs" by simp
then have "estaEn a (borraDuplicados xs)" using HI by simp
then show "estaEn a (borraDuplicados (b#xs))" using `a≠b` by simp
qed
qed
qed

(* anaprarod marpoldia1 ferrenseg juacabsou *)
lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix x xs
assume HI: "?P xs"
show "?P (x#xs)"
proof (cases)
assume "estaEn x xs"
then have "estaEn a (borraDuplicados (x#xs)) = estaEn a (borraDuplicados xs)" by simp
also have "...= estaEn a xs" using HI by simp
finally show "?P (x#xs)" by auto
next
assume "¬estaEn x xs"
then have "estaEn a (borraDuplicados (x#xs)) = estaEn a (x#borraDuplicados xs)" by simp
also have "...= (x = a ∨ estaEn a (borraDuplicados xs))" by simp
finally show "?P (x#xs)" using HI by simp
qed
qed

(* migtermor *)

lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix aa xs
assume HI: "?P xs"
have P1: "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)"
proof (cases)
assume C1: "(estaEn aa xs)"
have "estaEn a (borraDuplicados (aa#xs)) = estaEn a (borraDuplicados xs)"
using C1 by simp
also have P3: "… = estaEn a xs" using HI by simp
also have "… = estaEn a (aa#xs)"
proof (cases)
assume "(a=aa)"
then show "estaEn a xs = estaEn a (aa#xs)" using C1 by simp
next
assume "¬(a=aa)"
then show "estaEn a xs = estaEn a (aa#xs)" by simp
qed
then show "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)" using P3 by simp
next
assume C2: "¬(estaEn aa xs)"
then show "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)" using HI by simp
qed
also have Conc: "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)" using P1 by simp
finally show "?P (aa#xs)" using Conc by simp
qed

(* crigomgom *)
-- "La demostración estructurada es"
lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs"
proof (induct xs)
show "estaEn a (borraDuplicados []) = estaEn a []" by simp
next
fix x xs
assume HI: "estaEn a (borraDuplicados xs) = estaEn a xs"
show "estaEn a (borraDuplicados (x#xs)) = estaEn a (x#xs)"
proof (rule iffI)
assume a1: "estaEn a (borraDuplicados (x#xs))"
show "estaEn a (x#xs)"
proof (cases)
assume "estaEn x xs"
then have "estaEn a (borraDuplicados xs)" using a1 by simp
then have "estaEn a xs" using HI by simp
then show "estaEn a (x#xs)" by simp
next
assume "¬ estaEn x xs"
then have "estaEn a (x#(borraDuplicados xs))" using a1 by simp
then have " x=a ∨ (estaEn a (borraDuplicados xs))" by simp
then have " x=a ∨ (estaEn a xs)" using HI by simp
then show "estaEn a (x#xs)" by simp
qed
next
assume a2: "estaEn a (x#xs)"
show "estaEn a (borraDuplicados (x#xs))"
proof (cases)
assume "a=x"
then show "estaEn a (borraDuplicados (x#xs))" using HI by simp
next
assume b1: "a≠x"
then have "estaEn a (x#xs)" using a2 by simp
then have "x = a ∨ estaEn a xs" by simp
then have "estaEn a xs " using b1 by simp
then have "estaEn a (borraDuplicados xs)" using HI by simp
then show "estaEn a (borraDuplicados (x#xs))" using b1 by simp
qed
qed
qed

(* rubgonmar *)
lemma estaEn_borraDuplicados_2:
"estaEn a ( borraDuplicados xs ) = estaEn a xs"
proof (induct xs)
show "estaEn a (borraDuplicados []) = estaEn a []" by simp
next
fix x
fix xs
assume HI: "estaEn a (borraDuplicados xs) = estaEn a xs"
show "estaEn a (borraDuplicados (x#xs)) = estaEn a (x#xs)"
proof (rule iffI) (* usamos proof de la regla dada iffI*)
assume cprim: "estaEn a (borraDuplicados (x#xs))"
show "estaEn a (x#xs)"
proof (cases)
assume "estaEn x xs"
then show "estaEn a (x#xs)" using cprim HI by simp
next
assume "¬ estaEn x xs"
then show "estaEn a (x#xs)" using cprim HI by simp
qed
next
assume cseg: "estaEn a (x#xs)"
show "estaEn a (borraDuplicados (x#xs))"
proof (cases)
assume "a=x"
then show "estaEn a (borraDuplicados (x#xs))" using HI by auto
next
assume "a≠x"
then show "estaEn a (borraDuplicados (x#xs))" using `a≠x` cseg HI by simp
qed
qed
qed

(*
Aplico la regla iffI:
⟦P ⟹ Q ; Q ⟹ P⟧ ⟹ P = Q
Así:
[estaEn a (borraDuplicados (x # xs)) ⟹ estaEn a (x # xs); estaEn a (x # xs) ⟹ estaEn a (borraDuplicados (x # xs))]
⟹ estaEn a (borraDuplicados (x # xs)) = estaEn a (x # xs)
*)

(* bowma ivamenjim *)
lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?p xs")
proof (induct xs)
show "?p []" by simp
next
fix x xs
assume HI: "?p xs"
show "?p (x#xs)"
proof (cases)
assume H1:"estaEn x xs"
then have "estaEn a (borraDuplicados (x#xs)) = estaEn a (borraDuplicados xs)" by simp
also have "... = estaEn a xs" using HI by simp
also have "... = estaEn a (x#xs)"
proof(cases)
assume "x=a"
then show "estaEn a xs = estaEn a (x#xs)" using H1 by simp
next
assume "x≠a"
then show "estaEn a xs = estaEn a (x#xs)" by simp
qed
finally show "?p (x#xs)" by simp
next
assume H2:"¬estaEn x xs"
then have "estaEn a (borraDuplicados (x#xs)) = estaEn a (x#borraDuplicados xs)" by simp
also have "... = ((x=a) ∨ estaEn a (borraDuplicados xs))" by simp
also have "... = ((x=a) ∨ estaEn a xs)" using HI by simp
also have "... = estaEn a (x#xs)" by simp
finally show "?p (x#xs)" by simp
qed

(*danrodcha *)
(* es como la de ruben pero con diferencias de estilo *)
lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix x xs assume HI: "?P xs"
show "?P (x#xs)"
proof (rule iffI)
assume H1: "estaEn a (borraDuplicados (x # xs))"
show "estaEn a (x#xs)"
proof (cases "estaEn x xs")
case True
then show "estaEn a (x#xs)" using H1 HI by simp
next
case False
then show "estaEn a (x#xs)" using H1 HI by simp
qed
next
assume H2: "estaEn a (x#xs)"
show "estaEn a (borraDuplicados (x # xs))"
proof (cases "x=a")
case True
then show "estaEn a (borraDuplicados (x # xs))" using HI by simp
next
case False
then show "estaEn a (borraDuplicados (x # xs))" using H2 HI by simp
qed
qed
qed

text {*
---------------------------------------------------------------------
Ejercicio 6.1. Demostrar o refutar automáticamente
sinDuplicados (borraDuplicados xs)
---------------------------------------------------------------------
*}

-- "La demostración automática"

(* ivamenjim wilmorort serrodcal crigomgom anaprarod fraortmoy
marpoldia1 ferrenseg danrodcha juacabsou *)
lemma sinDuplicados_borraDuplicados:
"sinDuplicados (borraDuplicados xs)"
by (induct xs) (auto simp add: estaEn_borraDuplicados)

(* migtermor bowma *)
lemma sinDuplicados_borraDuplicados:
"sinDuplicados (borraDuplicados xs)"
by (induct xs, simp_all add: estaEn_borraDuplicados_2)

text {*
---------------------------------------------------------------------
Ejercicio 6.2. Demostrar o refutar detalladamente
sinDuplicados (borraDuplicados xs)
---------------------------------------------------------------------
*}

(*wilmorort*)

-- "La demostración estructurada es"
lemma sinDuplicados_borraDuplicados_2:
"sinDuplicados (borraDuplicados xs)"
proof (induct xs)
show " sinDuplicados (borraDuplicados [])" by simp
next
fix a xs
assume HI: "sinDuplicados (borraDuplicados xs)"
show "sinDuplicados (borraDuplicados (a # xs))"
proof (cases)
assume "estaEn a xs"
then show "sinDuplicados (borraDuplicados (a#xs))" using HI by simp
next
assume"¬ estaEn a xs"
then have "¬ (estaEn a xs) ∧ sinDuplicados (borraDuplicados xs)" using HI by simp
then have "¬ estaEn a (borraDuplicados xs) ∧ sinDuplicados (borraDuplicados xs)"
by (simp add: estaEn_borraDuplicados)
then have " sinDuplicados (a#borraDuplicados xs)" by simp
then show " sinDuplicados (borraDuplicados(a #xs))" by simp
qed
qed

(* ivamenjim migtermor crigomgom rubgonmar fraortmoy marpoldia1 ferrenseg bowma juacabsou *)

lemma sinDuplicados_borraDuplicados_2:
"sinDuplicados (borraDuplicados xs)"
proof (induct xs)
show "sinDuplicados (borraDuplicados [])" by simp
next
fix a xs
assume HI: "sinDuplicados (borraDuplicados xs)"
show "sinDuplicados (borraDuplicados (a # xs))"
proof (cases)
assume "estaEn a xs"
then show "sinDuplicados (borraDuplicados (a # xs))" using HI by simp
next
assume "¬(estaEn a xs)"
then show "sinDuplicados (borraDuplicados (a # xs))" using HI by (simp add: estaEn_borraDuplicados)
qed
qed

(* anaprarod *)
lemma sinDuplicados_borraDuplicados_2:
"sinDuplicados (borraDuplicados xs)" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix x xs
assume HI: "?P xs"
show "?P (x#xs)"
proof (cases)
assume c1: "estaEn x xs"
then show "sinDuplicados (borraDuplicados (x#xs))" using HI by simp
next
assume c2: "¬ estaEn x xs"
then have "sinDuplicados (borraDuplicados (x#xs)) =sinDuplicados (x#borraDuplicados xs)" by simp
also have "…= (¬estaEn x (borraDuplicados xs) ∧ sinDuplicados (borraDuplicados xs))" by simp
also have "… = (¬estaEn x (borraDuplicados xs))" using HI by simp
also have "… = (¬(estaEn x xs))" by (simp add:estaEn_borraDuplicados)
also have "… = True" using c2 by simp
finally show "?P (x#xs)" by simp
qed
qed

(* danrodcha *)
lemma sinDuplicados_borraDuplicados_2:
"sinDuplicados (borraDuplicados xs)" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix x xs assume HI: "?P xs"
show "?P (x#xs)"
proof (cases "estaEn x xs")
case True
then have 1: "sinDuplicados (borraDuplicados (x#xs))
= sinDuplicados (borraDuplicados xs)" by (simp add: estaEn_borraDuplicados_2)
show "?P (x#xs)" using HI 1 by simp
next
case False
then have "sinDuplicados (borraDuplicados (x#xs))
= sinDuplicados (x#borraDuplicados xs)" by simp
also have "… = (¬ (estaEn x (borraDuplicados xs)) ∧
sinDuplicados (borraDuplicados xs))" by simp
also have "… = True" using `¬ estaEn x xs` HI by (simp add:estaEn_borraDuplicados)
finally show "?P (x#xs)" by simp
qed
qed

text {*
---------------------------------------------------------------------
Ejercicio 7. Demostrar o refutar:
borraDuplicados (rev xs) = rev (borraDuplicados xs)
---------------------------------------------------------------------
*}

(*crigomgom rubgonmar ivamenjim wilmorort pablucoto migtermor
anaprarod fraortmoy ferrenseg marpoldia1 bowma danrodcha juacabsou paupeddeg *)
lemma "borraDuplicados (rev xs) = rev (borraDuplicados xs)"
quickcheck
oops

(* ivamenjim: Quickcheck encuentra el siguiente contraejemplo:
xs = [a1, a2, a1]
Por lo que:
· "borraDuplicados (rev xs) = [a2, a1]"
· "rev (borraDuplicados xs) = [a1, a2]" *)

end
</source>

Relación 5

2016-11-29T20:23:52Z

Manmorjim1:

<source lang="isar">
chapter {* R5: Eliminación de duplicados *}

theory R5_Eliminacion_de_duplicados
imports Main
begin

text {*
---------------------------------------------------------------------
Ejercicio 1. Definir la funcion primitiva recursiva
estaEn :: 'a ⇒ 'a list ⇒ bool
tal que (estaEn x xs) se verifica si el elemento x está en la lista
xs. Por ejemplo,
estaEn (2::nat) [3,2,4] = True
estaEn (1::nat) [3,2,4] = False
---------------------------------------------------------------------
*}

(* crigomgom rubgonmar bowma wilmorort pablucoto serrodcal
anaprarod migtermor paupeddeg fraortmoy marpoldia1
danrodcha manmorjim1 *)
fun estaEn :: "'a ⇒ 'a list ⇒ bool" where
"estaEn _ [] = False"
| "estaEn x (a#xs) = ((a = x) ∨ (estaEn x xs))"

value "estaEn (2::nat) [3,2,4] = True"
value "estaEn (1::nat) [3,2,4] = False"

(* ivamenjim, ferrenseg josgarsan *)
(* Igual que la anterior pero con x en lugar de _ en el caso base *)

fun estaEn1 :: "'a ⇒ 'a list ⇒ bool" where
"estaEn1 x [] = False"
| "estaEn1 x (a#xs) = ((x=a) ∨ estaEn1 x xs)"

value "estaEn1 (2::nat) [3,2,4] = True"
value "estaEn1 (1::nat) [3,2,4] = False"

(* wilmorort *)
(* reutilizando la funcion "algunos" de R4.thy*)
fun estaEn2 :: "'a ⇒ 'a list ⇒ bool" where
"estaEn2 a xs = algunos (λx. x = a) xs"

value "estaEn2 (2::nat) [3,2,4] = True"
value "estaEn2 (1::nat) [3,2,4] = False"

text {*
---------------------------------------------------------------------
Ejercicio 2. Definir la función primitiva recursiva
sinDuplicados :: 'a list ⇒ bool
tal que (sinDuplicados xs) se verifica si la lista xs no contiene
duplicados. Por ejemplo,
sinDuplicados [1::nat,4,2] = True
sinDuplicados [1::nat,4,2,4] = False
---------------------------------------------------------------------
*}

(* crigomgom rubgonmar ivamenjim wilmorort bowma pablucoto
serrodcal anaprarod migtermor paupeddeg fraortmoy marpoldia1
ferrenseg josgarsan danrodcha manmorjim1 *)
fun sinDuplicados :: "'a list ⇒ bool" where
"sinDuplicados [] = True"
| "sinDuplicados (x#xs) = (¬ estaEn x xs ∧ sinDuplicados xs)"

value "sinDuplicados [1::nat,4,2] = True"
value "sinDuplicados [1::nat,4,2,4] = False"

(* paupeddeg *)
(* Utilizando la función ∉ de Isabelle *)
fun sinDuplicados2 :: "'a list ⇒ bool" where
"sinDuplicados2 [] = True"
| "sinDuplicados2 (a#xs) = ((a ∉ set xs) ∧ sinDuplicados2 xs ) "

text {*
---------------------------------------------------------------------
Ejercicio 3. Definir la función primitiva recursiva
borraDuplicados :: 'a list ⇒ bool
tal que (borraDuplicados xs) es la lista obtenida eliminando los
elementos duplicados de la lista xs. Por ejemplo,
borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]

Nota: La función borraDuplicados es equivalente a la predefinida
remdups.
---------------------------------------------------------------------
*}

(* crigomgom rubgonmar wilmorort bowma pablucoto serrodcal
anaprarod migtermor paupeddeg fraortmoy marpoldia1 ferrenseg
josgarsan danrodcha *)
fun borraDuplicados :: "'a list ⇒ 'a list" where
"borraDuplicados [] = []"
| "borraDuplicados (x#xs) =( if estaEn x xs then borraDuplicados xs else x#borraDuplicados xs)"

value "borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]"

(* ivamenjim *)
(* Utilizando la negación primero *)

fun borraDuplicados :: "'a list ⇒ 'a list" where
"borraDuplicados [] = []"
| "borraDuplicados (x#xs) = (if ¬(estaEn x xs) then (x#(borraDuplicados xs)) else borraDuplicados xs)"

(* rubgonmar *)
(* Otra forma Sin usar if
Utilizando case aunque se le sacaría más partido con más de 2 casos *)

fun borraDuplicados1 :: "'a list ⇒ 'a list" where
"borraDuplicados1 [] = []" |
"borraDuplicados1 (x#xs) = ( case estaEn x xs of False => x#borraDuplicados1 xs | True => borraDuplicados1 xs )"

(* rubgonmar *)
(*Otra forma utilizando let*)
fun borraDuplicados2 :: "'a list ⇒ 'a list" where
"borraDuplicados2 [] = []" |
"borraDuplicados2 (x#xs) = (let condicion = estaEn x xs::bool in
if condicion then borraDuplicados2 xs else x # borraDuplicados2 xs)"

value "borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]"

text {*
---------------------------------------------------------------------
Ejercicio 4.1. Demostrar o refutar automáticamente
length (borraDuplicados xs) ≤ length xs
---------------------------------------------------------------------
*}

-- "La demostración automática es"
(*crigomgom anaprarod ferrenseg*)
lemma length_borraDuplicados:
"length (borraDuplicados xs) ≤ length xs"
by (induct xs, simp_all)

(* rubgonmar wilmorort pablucoto serrodcal migtermor paupeddeg
fraortmoy marpoldia1 danrodcha *)
lemma length_borraDuplicados:
"length ( borraDuplicados xs ) ≤ length xs"
by (induct xs) auto

(* ivamenjim *)
(* Demostrando objetivo a objetivo *)
lemma length_borraDuplicados:
"length (borraDuplicados xs) ≤ length xs"
apply (induct xs)
apply simp
apply auto
done

(* bowma anaprarod *)
lemma length_borraDuplicados:
"length (borraDuplicados xs) ≤ length xs"
apply (induct xs)
apply (simp, simp) (* creo que es mejor poner aquí simp_all *)
done

(* ferrenseg *)

lemma length_borraDuplicados:
"length (borraDuplicados xs) ≤ length xs"
by (induct xs) simp_all (* Creo que se puede poner simp_all fuera de parentesis *)

text {*
---------------------------------------------------------------------
Ejercicio 4.2. Demostrar o refutar detalladamente
length (borraDuplicados xs) ≤ length xs
---------------------------------------------------------------------
*}

-- "La demostración estructurada es"
(* crigomgom *)
lemma length_borraDuplicados_2:
"length (borraDuplicados xs) ≤ length xs"
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix x xs
assume HI: "length (borraDuplicados xs) ≤ length xs"
show "length (borraDuplicados (x#xs)) ≤ length (x#xs)"
proof (cases)
assume "estaEn x xs"
then have "length (borraDuplicados (x#xs)) = length (borraDuplicados xs)" by simp
also have "... ≤ length xs" using HI by simp
also have "... ≤ length (x#xs)" by simp
finally show "length (borraDuplicados (x#xs)) ≤ length (x#xs)" by simp
next
assume "(¬ estaEn x xs)"
then have "length (borraDuplicados (x#xs)) = length (x#borraDuplicados xs)" by simp
also have "... = 1 + length (borraDuplicados xs)" by simp
also have "... ≤ 1 + length xs" using HI by simp
also have "... = length (x#xs)" by simp
finally show "length (borraDuplicados (x#xs)) ≤ length (x#xs)" by simp
qed
qed

(* ivamenjim wilmorort ferrenseg rubgonmar *)
lemma length_borraDuplicados_2:
"length (borraDuplicados xs) ≤ length xs"
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix a xs
assume HI: "length (borraDuplicados xs) ≤ length xs"
have "length (borraDuplicados (a # xs)) ≤ 1+length (borraDuplicados xs)" by simp
also have "... ≤ 1+length xs" using HI by simp
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs)" by simp
qed

(* serrodcal anaprarod danrodcha *)
lemma length_borraDuplicados_2: "length (borraDuplicados xs) ≤ length xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "length (borraDuplicados (a # xs)) ≤ 1+length (borraDuplicados xs)" by simp
also have "... ≤ 1+length xs" using HI by simp
finally show "?P (a # xs)" by simp
qed

(* pablucoto *)
lemma length_borraDuplicados_2:
"length (borraDuplicados xs) ≤ length xs"
proof(induct xs)
show "length (borraDuplicados []) ≤ length [] " by simp
next
fix a xs
assume HI: " length (borraDuplicados xs) ≤ length xs "
have "length (borraDuplicados (a # xs)) ≤ 1 + length(borraDuplicados xs)" by simp
also have "... ≤ 1 + length xs" using HI by simp
also have "... ≤ length (a#xs)" by simp
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs) " by simp
qed

(* bowma *)
lemma length_borraDuplicados_3:
"length (borraDuplicados xs) ≤ length xs" (is "?p xs")
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix a xs
assume HI: "?p xs"
have "length (a#xs) = 1 + length xs" by simp
have "length(borraDuplicados (a#xs)) ≤ 1 + length(borraDuplicados xs)" by simp
also have "... ≤ 1+length xs" using HI by simp
also have "... ≤ length (a#xs)" by simp
finally show "?p (a # xs)" by simp
qed

(* migtermor *)

lemma length_borraDuplicados_2:
"length (borraDuplicados xs) ≤ length xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "length (borraDuplicados (a#xs)) ≤ (length (a#xs))"
proof (cases)
assume "(estaEn a xs)"
then have Aux: "length (borraDuplicados (a#xs)) = length (borraDuplicados xs)" by simp
also have "… ≤ length (a#xs)" using HI by simp
then show "length (borraDuplicados (a#xs)) ≤ (length (a#xs))" using Aux by simp
next
assume "¬ (estaEn a xs)"
then have Aux: "length (borraDuplicados (a#xs)) = 1+ length (borraDuplicados xs)" by simp
also have "… ≤ length (a#xs)" using HI by simp
then show "length (borraDuplicados (a#xs)) ≤ (length (a#xs))" using Aux by simp
qed
then show "?P (a#xs)" by simp
qed

(* paupeddeg marpoldia1*)
lemma length_borraDuplicados_4:
"length (borraDuplicados xs) ≤ length xs"
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix a xs
assume HI: "length (borraDuplicados xs) ≤ length xs"
have "length (borraDuplicados (a # xs)) ≤ length [a] + length (borraDuplicados xs) " by simp
also have "... ≤ 1 + length (borraDuplicados xs)" by simp
also have "... ≤ 1 + length xs" using HI by simp
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs)" by simp
qed

(* fraortmoy *)
(* muy parecida a alguna anterior, pero yo dí mas pasos *)
lemma length_borraDuplicados_2:
"length (borraDuplicados xs) ≤ length xs"
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix a xs
assume H1: "length (borraDuplicados xs) ≤ length xs"
have "length (borraDuplicados (a # xs)) ≤ length(borraDuplicados [a])+length (borraDuplicados xs)" by simp
also have "… ≤ 1 + length (borraDuplicados xs)" by simp
also have "… ≤ 1 + length xs" using H1 by simp
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 5.1. Demostrar o refutar automáticamente
estaEn a (borraDuplicados xs) = estaEn a xs
---------------------------------------------------------------------
*}

-- "La demostración automática es"
(* crigomgom rubgonmar wilmorort pablucoto serrodcal bowma
migtermor fraortmoy marpoldia1 ferrenseg danrodcha *)
lemma estaEn_borraDuplicados:
"estaEn a (borraDuplicados xs) = estaEn a xs"
by (induct xs) auto

(* ivamenjim *)
lemma estaEn_borraDuplicados:
"estaEn a (borraDuplicados xs) = estaEn a xs"
apply (induct xs)
apply auto
done

(* anaprarod *)
lemma estaEn_borraDuplicados:
"estaEn a (borraDuplicados xs) = estaEn a xs"
by (induct xs, simp_all, blast)

(* anaprarod *)
lemma estaEn_borraDuplicados':
"estaEn a (borraDuplicados xs) = estaEn a xs"
apply (induct xs)
apply (cases "estaEn x xs")
apply (simp_all)
apply blast
done

(* bowma *)
lemma estaEn_borraDuplicados:
"estaEn a (borraDuplicados xs) = estaEn a xs"
apply (induct xs)
apply simp
apply (simp, blast)
done

text {*
---------------------------------------------------------------------
Ejercicio 5.2. Demostrar o refutar detalladamente
estaEn a (borraDuplicados xs) = estaEn a xs
Nota: Para la demostración de la equivalencia se puede usar
proof (rule iffI)
La regla iffI es
⟦P ⟹ Q ; Q ⟹ P⟧ ⟹ P = Q
---------------------------------------------------------------------
*}

-- "La demostración estructurada es"

(* wilmorort *)

lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs"
proof (induct xs)
show "estaEn a (borraDuplicados []) = estaEn a []" by simp
next
fix b xs
assume HI: "estaEn a (borraDuplicados xs) = estaEn a xs"
show "estaEn a (borraDuplicados (b#xs)) = estaEn a (b#xs)"
proof (rule iffI)
assume H1: "estaEn a (borraDuplicados (b#xs))"
show "estaEn a (b#xs)"
proof (cases)
assume "estaEn b xs"
then have "estaEn a (borraDuplicados xs)" using H1 by simp
then have "estaEn a xs" using HI by simp
then show "estaEn a (b#xs)" by simp
next
assume "¬ estaEn b xs"
then have "estaEn a (b#(borraDuplicados xs))" using H1 by simp
then have "a=b ∨ (estaEn a (borraDuplicados xs))" by simp
then have " a=b ∨ (estaEn a xs)" using HI by simp
then show "estaEn a (b#xs)" by simp
qed
next
assume H2: "estaEn a (b#xs)"
show "estaEn a (borraDuplicados (b#xs))"
proof (cases)
assume "a=b"
then have "estaEn b (borraDuplicados xs) = estaEn b xs" using HI by simp
then have "(estaEn b xs ⟶ estaEn b (borraDuplicados xs)) ∧
(¬ estaEn b xs ⟶ estaEn b (b # borraDuplicados xs))" by simp
then have "estaEn b (borraDuplicados (b#xs))" by simp
then show "estaEn a (borraDuplicados (b#xs))" using `a=b` by simp
next
assume "a≠b"
then have "estaEn a (b#xs)" using H2 by simp
then have "a = b ∨ estaEn a xs" by simp
then have "False ∨ estaEn a xs " using `a≠b` by simp
then have "estaEn a xs" by simp
then have "estaEn a (borraDuplicados xs)" using HI by simp
then show "estaEn a (borraDuplicados (b#xs))" using `a≠b` by simp
qed
qed
qed

(* anaprarod marpoldia1 ferrenseg *)
lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix x xs
assume HI: "?P xs"
show "?P (x#xs)"
proof (cases)
assume "estaEn x xs"
then have "estaEn a (borraDuplicados (x#xs)) = estaEn a (borraDuplicados xs)" by simp
also have "...= estaEn a xs" using HI by simp
finally show "?P (x#xs)" by auto
next
assume "¬estaEn x xs"
then have "estaEn a (borraDuplicados (x#xs)) = estaEn a (x#borraDuplicados xs)" by simp
also have "...= (x = a ∨ estaEn a (borraDuplicados xs))" by simp
finally show "?P (x#xs)" using HI by simp
qed
qed

(* migtermor *)

lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix aa xs
assume HI: "?P xs"
have P1: "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)"
proof (cases)
assume C1: "(estaEn aa xs)"
have "estaEn a (borraDuplicados (aa#xs)) = estaEn a (borraDuplicados xs)"
using C1 by simp
also have P3: "… = estaEn a xs" using HI by simp
also have "… = estaEn a (aa#xs)"
proof (cases)
assume "(a=aa)"
then show "estaEn a xs = estaEn a (aa#xs)" using C1 by simp
next
assume "¬(a=aa)"
then show "estaEn a xs = estaEn a (aa#xs)" by simp
qed
then show "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)" using P3 by simp
next
assume C2: "¬(estaEn aa xs)"
then show "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)" using HI by simp
qed
also have Conc: "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)" using P1 by simp
finally show "?P (aa#xs)" using Conc by simp
qed

(* crigomgom *)
-- "La demostración estructurada es"
lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs"
proof (induct xs)
show "estaEn a (borraDuplicados []) = estaEn a []" by simp
next
fix x xs
assume HI: "estaEn a (borraDuplicados xs) = estaEn a xs"
show "estaEn a (borraDuplicados (x#xs)) = estaEn a (x#xs)"
proof (rule iffI)
assume a1: "estaEn a (borraDuplicados (x#xs))"
show "estaEn a (x#xs)"
proof (cases)
assume "estaEn x xs"
then have "estaEn a (borraDuplicados xs)" using a1 by simp
then have "estaEn a xs" using HI by simp
then show "estaEn a (x#xs)" by simp
next
assume "¬ estaEn x xs"
then have "estaEn a (x#(borraDuplicados xs))" using a1 by simp
then have " x=a ∨ (estaEn a (borraDuplicados xs))" by simp
then have " x=a ∨ (estaEn a xs)" using HI by simp
then show "estaEn a (x#xs)" by simp
qed
next
assume a2: "estaEn a (x#xs)"
show "estaEn a (borraDuplicados (x#xs))"
proof (cases)
assume "a=x"
then show "estaEn a (borraDuplicados (x#xs))" using HI by simp
next
assume b1: "a≠x"
then have "estaEn a (x#xs)" using a2 by simp
then have "x = a ∨ estaEn a xs" by simp
then have "estaEn a xs " using b1 by simp
then have "estaEn a (borraDuplicados xs)" using HI by simp
then show "estaEn a (borraDuplicados (x#xs))" using b1 by simp
qed
qed
qed

(* rubgonmar *)
lemma estaEn_borraDuplicados_2:
"estaEn a ( borraDuplicados xs ) = estaEn a xs"
proof (induct xs)
show "estaEn a (borraDuplicados []) = estaEn a []" by simp
next
fix x
fix xs
assume HI: "estaEn a (borraDuplicados xs) = estaEn a xs"
show "estaEn a (borraDuplicados (x#xs)) = estaEn a (x#xs)"
proof (rule iffI) (* usamos proof de la regla dada iffI*)
assume cprim: "estaEn a (borraDuplicados (x#xs))"
show "estaEn a (x#xs)"
proof (cases)
assume "estaEn x xs"
then show "estaEn a (x#xs)" using cprim HI by simp
next
assume "¬ estaEn x xs"
then show "estaEn a (x#xs)" using cprim HI by simp
qed
next
assume cseg: "estaEn a (x#xs)"
show "estaEn a (borraDuplicados (x#xs))"
proof (cases)
assume "a=x"
then show "estaEn a (borraDuplicados (x#xs))" using HI by auto
next
assume "a≠x"
then show "estaEn a (borraDuplicados (x#xs))" using `a≠x` cseg HI by simp
qed
qed
qed

(*
Aplico la regla iffI:
⟦P ⟹ Q ; Q ⟹ P⟧ ⟹ P = Q
Así:
[estaEn a (borraDuplicados (x # xs)) ⟹ estaEn a (x # xs); estaEn a (x # xs) ⟹ estaEn a (borraDuplicados (x # xs))]
⟹ estaEn a (borraDuplicados (x # xs)) = estaEn a (x # xs)
*)

(* bowma ivamenjim *)
lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?p xs")
proof (induct xs)
show "?p []" by simp
next
fix x xs
assume HI: "?p xs"
show "?p (x#xs)"
proof (cases)
assume H1:"estaEn x xs"
then have "estaEn a (borraDuplicados (x#xs)) = estaEn a (borraDuplicados xs)" by simp
also have "... = estaEn a xs" using HI by simp
also have "... = estaEn a (x#xs)"
proof(cases)
assume "x=a"
then show "estaEn a xs = estaEn a (x#xs)" using H1 by simp
next
assume "x≠a"
then show "estaEn a xs = estaEn a (x#xs)" by simp
qed
finally show "?p (x#xs)" by simp
next
assume H2:"¬estaEn x xs"
then have "estaEn a (borraDuplicados (x#xs)) = estaEn a (x#borraDuplicados xs)" by simp
also have "... = ((x=a) ∨ estaEn a (borraDuplicados xs))" by simp
also have "... = ((x=a) ∨ estaEn a xs)" using HI by simp
also have "... = estaEn a (x#xs)" by simp
finally show "?p (x#xs)" by simp
qed

(*danrodcha *)
(* es como la de ruben pero con diferencias de estilo *)
lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix x xs assume HI: "?P xs"
show "?P (x#xs)"
proof (rule iffI)
assume H1: "estaEn a (borraDuplicados (x # xs))"
show "estaEn a (x#xs)"
proof (cases "estaEn x xs")
case True
then show "estaEn a (x#xs)" using H1 HI by simp
next
case False
then show "estaEn a (x#xs)" using H1 HI by simp
qed
next
assume H2: "estaEn a (x#xs)"
show "estaEn a (borraDuplicados (x # xs))"
proof (cases "x=a")
case True
then show "estaEn a (borraDuplicados (x # xs))" using HI by simp
next
case False
then show "estaEn a (borraDuplicados (x # xs))" using H2 HI by simp
qed
qed
qed

text {*
---------------------------------------------------------------------
Ejercicio 6.1. Demostrar o refutar automáticamente
sinDuplicados (borraDuplicados xs)
---------------------------------------------------------------------
*}

-- "La demostración automática"

(* ivamenjim wilmorort serrodcal crigomgom anaprarod fraortmoy
marpoldia1 ferrenseg danrodcha *)
lemma sinDuplicados_borraDuplicados:
"sinDuplicados (borraDuplicados xs)"
by (induct xs) (auto simp add: estaEn_borraDuplicados)

(* migtermor bowma *)
lemma sinDuplicados_borraDuplicados:
"sinDuplicados (borraDuplicados xs)"
by (induct xs, simp_all add: estaEn_borraDuplicados_2)

text {*
---------------------------------------------------------------------
Ejercicio 6.2. Demostrar o refutar detalladamente
sinDuplicados (borraDuplicados xs)
---------------------------------------------------------------------
*}

(*wilmorort*)

-- "La demostración estructurada es"
lemma sinDuplicados_borraDuplicados_2:
"sinDuplicados (borraDuplicados xs)"
proof (induct xs)
show " sinDuplicados (borraDuplicados [])" by simp
next
fix a xs
assume HI: "sinDuplicados (borraDuplicados xs)"
show "sinDuplicados (borraDuplicados (a # xs))"
proof (cases)
assume "estaEn a xs"
then show "sinDuplicados (borraDuplicados (a#xs))" using HI by simp
next
assume"¬ estaEn a xs"
then have "¬ (estaEn a xs) ∧ sinDuplicados (borraDuplicados xs)" using HI by simp
then have "¬ estaEn a (borraDuplicados xs) ∧ sinDuplicados (borraDuplicados xs)"
by (simp add: estaEn_borraDuplicados)
then have " sinDuplicados (a#borraDuplicados xs)" by simp
then show " sinDuplicados (borraDuplicados(a #xs))" by simp
qed
qed

(* ivamenjim migtermor crigomgom rubgonmar fraortmoy marpoldia1 ferrenseg bowma *)

lemma sinDuplicados_borraDuplicados_2:
"sinDuplicados (borraDuplicados xs)"
proof (induct xs)
show "sinDuplicados (borraDuplicados [])" by simp
next
fix a xs
assume HI: "sinDuplicados (borraDuplicados xs)"
show "sinDuplicados (borraDuplicados (a # xs))"
proof (cases)
assume "estaEn a xs"
then show "sinDuplicados (borraDuplicados (a # xs))" using HI by simp
next
assume "¬(estaEn a xs)"
then show "sinDuplicados (borraDuplicados (a # xs))" using HI by (simp add: estaEn_borraDuplicados)
qed
qed

(* anaprarod *)
lemma sinDuplicados_borraDuplicados_2:
"sinDuplicados (borraDuplicados xs)" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix x xs
assume HI: "?P xs"
show "?P (x#xs)"
proof (cases)
assume c1: "estaEn x xs"
then show "sinDuplicados (borraDuplicados (x#xs))" using HI by simp
next
assume c2: "¬ estaEn x xs"
then have "sinDuplicados (borraDuplicados (x#xs)) =sinDuplicados (x#borraDuplicados xs)" by simp
also have "…= (¬estaEn x (borraDuplicados xs) ∧ sinDuplicados (borraDuplicados xs))" by simp
also have "… = (¬estaEn x (borraDuplicados xs))" using HI by simp
also have "… = (¬(estaEn x xs))" by (simp add:estaEn_borraDuplicados)
also have "… = True" using c2 by simp
finally show "?P (x#xs)" by simp
qed
qed

(* danrodcha *)
lemma sinDuplicados_borraDuplicados_2:
"sinDuplicados (borraDuplicados xs)" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix x xs assume HI: "?P xs"
show "?P (x#xs)"
proof (cases "estaEn x xs")
case True
then have 1: "sinDuplicados (borraDuplicados (x#xs))
= sinDuplicados (borraDuplicados xs)" by (simp add: estaEn_borraDuplicados_2)
show "?P (x#xs)" using HI 1 by simp
next
case False
then have "sinDuplicados (borraDuplicados (x#xs))
= sinDuplicados (x#borraDuplicados xs)" by simp
also have "… = (¬ (estaEn x (borraDuplicados xs)) ∧
sinDuplicados (borraDuplicados xs))" by simp
also have "… = True" using `¬ estaEn x xs` HI by (simp add:estaEn_borraDuplicados)
finally show "?P (x#xs)" by simp
qed
qed

text {*
---------------------------------------------------------------------
Ejercicio 7. Demostrar o refutar:
borraDuplicados (rev xs) = rev (borraDuplicados xs)
---------------------------------------------------------------------
*}

(*crigomgom rubgonmar ivamenjim wilmorort pablucoto migtermor
anaprarod fraortmoy ferrenseg marpoldia1 bowma danrodcha *)
lemma "borraDuplicados (rev xs) = rev (borraDuplicados xs)"
quickcheck
oops

(* ivamenjim: Quickcheck encuentra el siguiente contraejemplo:
xs = [a1, a2, a1]
Por lo que:
· "borraDuplicados (rev xs) = [a2, a1]"
· "rev (borraDuplicados xs) = [a1, a2]" *)

end
</source>

Relación 5

2016-11-29T20:15:57Z

Manmorjim1:

<source lang="isar">
chapter {* R5: Eliminación de duplicados *}

theory R5_Eliminacion_de_duplicados
imports Main
begin

text {*
---------------------------------------------------------------------
Ejercicio 1. Definir la funcion primitiva recursiva
estaEn :: 'a ⇒ 'a list ⇒ bool
tal que (estaEn x xs) se verifica si el elemento x está en la lista
xs. Por ejemplo,
estaEn (2::nat) [3,2,4] = True
estaEn (1::nat) [3,2,4] = False
---------------------------------------------------------------------
*}

(* crigomgom rubgonmar bowma wilmorort pablucoto serrodcal
anaprarod migtermor paupeddeg fraortmoy marpoldia1
danrodcha manmorjim1 *)
fun estaEn :: "'a ⇒ 'a list ⇒ bool" where
"estaEn _ [] = False"
| "estaEn x (a#xs) = ((a = x) ∨ (estaEn x xs))"

value "estaEn (2::nat) [3,2,4] = True"
value "estaEn (1::nat) [3,2,4] = False"

(* ivamenjim, ferrenseg josgarsan *)
(* Igual que la anterior pero con x en lugar de _ en el caso base *)

fun estaEn1 :: "'a ⇒ 'a list ⇒ bool" where
"estaEn1 x [] = False"
| "estaEn1 x (a#xs) = ((x=a) ∨ estaEn1 x xs)"

value "estaEn1 (2::nat) [3,2,4] = True"
value "estaEn1 (1::nat) [3,2,4] = False"

(* wilmorort *)
(* reutilizando la funcion "algunos" de R4.thy*)
fun estaEn2 :: "'a ⇒ 'a list ⇒ bool" where
"estaEn2 a xs = algunos (λx. x = a) xs"

value "estaEn2 (2::nat) [3,2,4] = True"
value "estaEn2 (1::nat) [3,2,4] = False"

text {*
---------------------------------------------------------------------
Ejercicio 2. Definir la función primitiva recursiva
sinDuplicados :: 'a list ⇒ bool
tal que (sinDuplicados xs) se verifica si la lista xs no contiene
duplicados. Por ejemplo,
sinDuplicados [1::nat,4,2] = True
sinDuplicados [1::nat,4,2,4] = False
---------------------------------------------------------------------
*}

(* crigomgom rubgonmar ivamenjim wilmorort bowma pablucoto
serrodcal anaprarod migtermor paupeddeg fraortmoy marpoldia1
ferrenseg josgarsan danrodcha *)
fun sinDuplicados :: "'a list ⇒ bool" where
"sinDuplicados [] = True"
| "sinDuplicados (x#xs) = (¬ estaEn x xs ∧ sinDuplicados xs)"

value "sinDuplicados [1::nat,4,2] = True"
value "sinDuplicados [1::nat,4,2,4] = False"

(* paupeddeg *)
(* Utilizando la función ∉ de Isabelle *)
fun sinDuplicados2 :: "'a list ⇒ bool" where
"sinDuplicados2 [] = True"
| "sinDuplicados2 (a#xs) = ((a ∉ set xs) ∧ sinDuplicados2 xs ) "

text {*
---------------------------------------------------------------------
Ejercicio 3. Definir la función primitiva recursiva
borraDuplicados :: 'a list ⇒ bool
tal que (borraDuplicados xs) es la lista obtenida eliminando los
elementos duplicados de la lista xs. Por ejemplo,
borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]

Nota: La función borraDuplicados es equivalente a la predefinida
remdups.
---------------------------------------------------------------------
*}

(* crigomgom rubgonmar wilmorort bowma pablucoto serrodcal
anaprarod migtermor paupeddeg fraortmoy marpoldia1 ferrenseg
josgarsan danrodcha *)
fun borraDuplicados :: "'a list ⇒ 'a list" where
"borraDuplicados [] = []"
| "borraDuplicados (x#xs) =( if estaEn x xs then borraDuplicados xs else x#borraDuplicados xs)"

value "borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]"

(* ivamenjim *)
(* Utilizando la negación primero *)

fun borraDuplicados :: "'a list ⇒ 'a list" where
"borraDuplicados [] = []"
| "borraDuplicados (x#xs) = (if ¬(estaEn x xs) then (x#(borraDuplicados xs)) else borraDuplicados xs)"

(* rubgonmar *)
(* Otra forma Sin usar if
Utilizando case aunque se le sacaría más partido con más de 2 casos *)

fun borraDuplicados1 :: "'a list ⇒ 'a list" where
"borraDuplicados1 [] = []" |
"borraDuplicados1 (x#xs) = ( case estaEn x xs of False => x#borraDuplicados1 xs | True => borraDuplicados1 xs )"

(* rubgonmar *)
(*Otra forma utilizando let*)
fun borraDuplicados2 :: "'a list ⇒ 'a list" where
"borraDuplicados2 [] = []" |
"borraDuplicados2 (x#xs) = (let condicion = estaEn x xs::bool in
if condicion then borraDuplicados2 xs else x # borraDuplicados2 xs)"

value "borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]"

text {*
---------------------------------------------------------------------
Ejercicio 4.1. Demostrar o refutar automáticamente
length (borraDuplicados xs) ≤ length xs
---------------------------------------------------------------------
*}

-- "La demostración automática es"
(*crigomgom anaprarod ferrenseg*)
lemma length_borraDuplicados:
"length (borraDuplicados xs) ≤ length xs"
by (induct xs, simp_all)

(* rubgonmar wilmorort pablucoto serrodcal migtermor paupeddeg
fraortmoy marpoldia1 danrodcha *)
lemma length_borraDuplicados:
"length ( borraDuplicados xs ) ≤ length xs"
by (induct xs) auto

(* ivamenjim *)
(* Demostrando objetivo a objetivo *)
lemma length_borraDuplicados:
"length (borraDuplicados xs) ≤ length xs"
apply (induct xs)
apply simp
apply auto
done

(* bowma anaprarod *)
lemma length_borraDuplicados:
"length (borraDuplicados xs) ≤ length xs"
apply (induct xs)
apply (simp, simp) (* creo que es mejor poner aquí simp_all *)
done

(* ferrenseg *)

lemma length_borraDuplicados:
"length (borraDuplicados xs) ≤ length xs"
by (induct xs) simp_all (* Creo que se puede poner simp_all fuera de parentesis *)

text {*
---------------------------------------------------------------------
Ejercicio 4.2. Demostrar o refutar detalladamente
length (borraDuplicados xs) ≤ length xs
---------------------------------------------------------------------
*}

-- "La demostración estructurada es"
(* crigomgom *)
lemma length_borraDuplicados_2:
"length (borraDuplicados xs) ≤ length xs"
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix x xs
assume HI: "length (borraDuplicados xs) ≤ length xs"
show "length (borraDuplicados (x#xs)) ≤ length (x#xs)"
proof (cases)
assume "estaEn x xs"
then have "length (borraDuplicados (x#xs)) = length (borraDuplicados xs)" by simp
also have "... ≤ length xs" using HI by simp
also have "... ≤ length (x#xs)" by simp
finally show "length (borraDuplicados (x#xs)) ≤ length (x#xs)" by simp
next
assume "(¬ estaEn x xs)"
then have "length (borraDuplicados (x#xs)) = length (x#borraDuplicados xs)" by simp
also have "... = 1 + length (borraDuplicados xs)" by simp
also have "... ≤ 1 + length xs" using HI by simp
also have "... = length (x#xs)" by simp
finally show "length (borraDuplicados (x#xs)) ≤ length (x#xs)" by simp
qed
qed

(* ivamenjim wilmorort ferrenseg rubgonmar *)
lemma length_borraDuplicados_2:
"length (borraDuplicados xs) ≤ length xs"
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix a xs
assume HI: "length (borraDuplicados xs) ≤ length xs"
have "length (borraDuplicados (a # xs)) ≤ 1+length (borraDuplicados xs)" by simp
also have "... ≤ 1+length xs" using HI by simp
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs)" by simp
qed

(* serrodcal anaprarod danrodcha *)
lemma length_borraDuplicados_2: "length (borraDuplicados xs) ≤ length xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "length (borraDuplicados (a # xs)) ≤ 1+length (borraDuplicados xs)" by simp
also have "... ≤ 1+length xs" using HI by simp
finally show "?P (a # xs)" by simp
qed

(* pablucoto *)
lemma length_borraDuplicados_2:
"length (borraDuplicados xs) ≤ length xs"
proof(induct xs)
show "length (borraDuplicados []) ≤ length [] " by simp
next
fix a xs
assume HI: " length (borraDuplicados xs) ≤ length xs "
have "length (borraDuplicados (a # xs)) ≤ 1 + length(borraDuplicados xs)" by simp
also have "... ≤ 1 + length xs" using HI by simp
also have "... ≤ length (a#xs)" by simp
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs) " by simp
qed

(* bowma *)
lemma length_borraDuplicados_3:
"length (borraDuplicados xs) ≤ length xs" (is "?p xs")
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix a xs
assume HI: "?p xs"
have "length (a#xs) = 1 + length xs" by simp
have "length(borraDuplicados (a#xs)) ≤ 1 + length(borraDuplicados xs)" by simp
also have "... ≤ 1+length xs" using HI by simp
also have "... ≤ length (a#xs)" by simp
finally show "?p (a # xs)" by simp
qed

(* migtermor *)

lemma length_borraDuplicados_2:
"length (borraDuplicados xs) ≤ length xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "length (borraDuplicados (a#xs)) ≤ (length (a#xs))"
proof (cases)
assume "(estaEn a xs)"
then have Aux: "length (borraDuplicados (a#xs)) = length (borraDuplicados xs)" by simp
also have "… ≤ length (a#xs)" using HI by simp
then show "length (borraDuplicados (a#xs)) ≤ (length (a#xs))" using Aux by simp
next
assume "¬ (estaEn a xs)"
then have Aux: "length (borraDuplicados (a#xs)) = 1+ length (borraDuplicados xs)" by simp
also have "… ≤ length (a#xs)" using HI by simp
then show "length (borraDuplicados (a#xs)) ≤ (length (a#xs))" using Aux by simp
qed
then show "?P (a#xs)" by simp
qed

(* paupeddeg marpoldia1*)
lemma length_borraDuplicados_4:
"length (borraDuplicados xs) ≤ length xs"
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix a xs
assume HI: "length (borraDuplicados xs) ≤ length xs"
have "length (borraDuplicados (a # xs)) ≤ length [a] + length (borraDuplicados xs) " by simp
also have "... ≤ 1 + length (borraDuplicados xs)" by simp
also have "... ≤ 1 + length xs" using HI by simp
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs)" by simp
qed

(* fraortmoy *)
(* muy parecida a alguna anterior, pero yo dí mas pasos *)
lemma length_borraDuplicados_2:
"length (borraDuplicados xs) ≤ length xs"
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix a xs
assume H1: "length (borraDuplicados xs) ≤ length xs"
have "length (borraDuplicados (a # xs)) ≤ length(borraDuplicados [a])+length (borraDuplicados xs)" by simp
also have "… ≤ 1 + length (borraDuplicados xs)" by simp
also have "… ≤ 1 + length xs" using H1 by simp
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 5.1. Demostrar o refutar automáticamente
estaEn a (borraDuplicados xs) = estaEn a xs
---------------------------------------------------------------------
*}

-- "La demostración automática es"
(* crigomgom rubgonmar wilmorort pablucoto serrodcal bowma
migtermor fraortmoy marpoldia1 ferrenseg danrodcha *)
lemma estaEn_borraDuplicados:
"estaEn a (borraDuplicados xs) = estaEn a xs"
by (induct xs) auto

(* ivamenjim *)
lemma estaEn_borraDuplicados:
"estaEn a (borraDuplicados xs) = estaEn a xs"
apply (induct xs)
apply auto
done

(* anaprarod *)
lemma estaEn_borraDuplicados:
"estaEn a (borraDuplicados xs) = estaEn a xs"
by (induct xs, simp_all, blast)

(* anaprarod *)
lemma estaEn_borraDuplicados':
"estaEn a (borraDuplicados xs) = estaEn a xs"
apply (induct xs)
apply (cases "estaEn x xs")
apply (simp_all)
apply blast
done

(* bowma *)
lemma estaEn_borraDuplicados:
"estaEn a (borraDuplicados xs) = estaEn a xs"
apply (induct xs)
apply simp
apply (simp, blast)
done

text {*
---------------------------------------------------------------------
Ejercicio 5.2. Demostrar o refutar detalladamente
estaEn a (borraDuplicados xs) = estaEn a xs
Nota: Para la demostración de la equivalencia se puede usar
proof (rule iffI)
La regla iffI es
⟦P ⟹ Q ; Q ⟹ P⟧ ⟹ P = Q
---------------------------------------------------------------------
*}

-- "La demostración estructurada es"

(* wilmorort *)

lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs"
proof (induct xs)
show "estaEn a (borraDuplicados []) = estaEn a []" by simp
next
fix b xs
assume HI: "estaEn a (borraDuplicados xs) = estaEn a xs"
show "estaEn a (borraDuplicados (b#xs)) = estaEn a (b#xs)"
proof (rule iffI)
assume H1: "estaEn a (borraDuplicados (b#xs))"
show "estaEn a (b#xs)"
proof (cases)
assume "estaEn b xs"
then have "estaEn a (borraDuplicados xs)" using H1 by simp
then have "estaEn a xs" using HI by simp
then show "estaEn a (b#xs)" by simp
next
assume "¬ estaEn b xs"
then have "estaEn a (b#(borraDuplicados xs))" using H1 by simp
then have "a=b ∨ (estaEn a (borraDuplicados xs))" by simp
then have " a=b ∨ (estaEn a xs)" using HI by simp
then show "estaEn a (b#xs)" by simp
qed
next
assume H2: "estaEn a (b#xs)"
show "estaEn a (borraDuplicados (b#xs))"
proof (cases)
assume "a=b"
then have "estaEn b (borraDuplicados xs) = estaEn b xs" using HI by simp
then have "(estaEn b xs ⟶ estaEn b (borraDuplicados xs)) ∧
(¬ estaEn b xs ⟶ estaEn b (b # borraDuplicados xs))" by simp
then have "estaEn b (borraDuplicados (b#xs))" by simp
then show "estaEn a (borraDuplicados (b#xs))" using `a=b` by simp
next
assume "a≠b"
then have "estaEn a (b#xs)" using H2 by simp
then have "a = b ∨ estaEn a xs" by simp
then have "False ∨ estaEn a xs " using `a≠b` by simp
then have "estaEn a xs" by simp
then have "estaEn a (borraDuplicados xs)" using HI by simp
then show "estaEn a (borraDuplicados (b#xs))" using `a≠b` by simp
qed
qed
qed

(* anaprarod marpoldia1 ferrenseg *)
lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix x xs
assume HI: "?P xs"
show "?P (x#xs)"
proof (cases)
assume "estaEn x xs"
then have "estaEn a (borraDuplicados (x#xs)) = estaEn a (borraDuplicados xs)" by simp
also have "...= estaEn a xs" using HI by simp
finally show "?P (x#xs)" by auto
next
assume "¬estaEn x xs"
then have "estaEn a (borraDuplicados (x#xs)) = estaEn a (x#borraDuplicados xs)" by simp
also have "...= (x = a ∨ estaEn a (borraDuplicados xs))" by simp
finally show "?P (x#xs)" using HI by simp
qed
qed

(* migtermor *)

lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix aa xs
assume HI: "?P xs"
have P1: "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)"
proof (cases)
assume C1: "(estaEn aa xs)"
have "estaEn a (borraDuplicados (aa#xs)) = estaEn a (borraDuplicados xs)"
using C1 by simp
also have P3: "… = estaEn a xs" using HI by simp
also have "… = estaEn a (aa#xs)"
proof (cases)
assume "(a=aa)"
then show "estaEn a xs = estaEn a (aa#xs)" using C1 by simp
next
assume "¬(a=aa)"
then show "estaEn a xs = estaEn a (aa#xs)" by simp
qed
then show "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)" using P3 by simp
next
assume C2: "¬(estaEn aa xs)"
then show "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)" using HI by simp
qed
also have Conc: "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)" using P1 by simp
finally show "?P (aa#xs)" using Conc by simp
qed

(* crigomgom *)
-- "La demostración estructurada es"
lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs"
proof (induct xs)
show "estaEn a (borraDuplicados []) = estaEn a []" by simp
next
fix x xs
assume HI: "estaEn a (borraDuplicados xs) = estaEn a xs"
show "estaEn a (borraDuplicados (x#xs)) = estaEn a (x#xs)"
proof (rule iffI)
assume a1: "estaEn a (borraDuplicados (x#xs))"
show "estaEn a (x#xs)"
proof (cases)
assume "estaEn x xs"
then have "estaEn a (borraDuplicados xs)" using a1 by simp
then have "estaEn a xs" using HI by simp
then show "estaEn a (x#xs)" by simp
next
assume "¬ estaEn x xs"
then have "estaEn a (x#(borraDuplicados xs))" using a1 by simp
then have " x=a ∨ (estaEn a (borraDuplicados xs))" by simp
then have " x=a ∨ (estaEn a xs)" using HI by simp
then show "estaEn a (x#xs)" by simp
qed
next
assume a2: "estaEn a (x#xs)"
show "estaEn a (borraDuplicados (x#xs))"
proof (cases)
assume "a=x"
then show "estaEn a (borraDuplicados (x#xs))" using HI by simp
next
assume b1: "a≠x"
then have "estaEn a (x#xs)" using a2 by simp
then have "x = a ∨ estaEn a xs" by simp
then have "estaEn a xs " using b1 by simp
then have "estaEn a (borraDuplicados xs)" using HI by simp
then show "estaEn a (borraDuplicados (x#xs))" using b1 by simp
qed
qed
qed

(* rubgonmar *)
lemma estaEn_borraDuplicados_2:
"estaEn a ( borraDuplicados xs ) = estaEn a xs"
proof (induct xs)
show "estaEn a (borraDuplicados []) = estaEn a []" by simp
next
fix x
fix xs
assume HI: "estaEn a (borraDuplicados xs) = estaEn a xs"
show "estaEn a (borraDuplicados (x#xs)) = estaEn a (x#xs)"
proof (rule iffI) (* usamos proof de la regla dada iffI*)
assume cprim: "estaEn a (borraDuplicados (x#xs))"
show "estaEn a (x#xs)"
proof (cases)
assume "estaEn x xs"
then show "estaEn a (x#xs)" using cprim HI by simp
next
assume "¬ estaEn x xs"
then show "estaEn a (x#xs)" using cprim HI by simp
qed
next
assume cseg: "estaEn a (x#xs)"
show "estaEn a (borraDuplicados (x#xs))"
proof (cases)
assume "a=x"
then show "estaEn a (borraDuplicados (x#xs))" using HI by auto
next
assume "a≠x"
then show "estaEn a (borraDuplicados (x#xs))" using `a≠x` cseg HI by simp
qed
qed
qed

(*
Aplico la regla iffI:
⟦P ⟹ Q ; Q ⟹ P⟧ ⟹ P = Q
Así:
[estaEn a (borraDuplicados (x # xs)) ⟹ estaEn a (x # xs); estaEn a (x # xs) ⟹ estaEn a (borraDuplicados (x # xs))]
⟹ estaEn a (borraDuplicados (x # xs)) = estaEn a (x # xs)
*)

(* bowma ivamenjim *)
lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?p xs")
proof (induct xs)
show "?p []" by simp
next
fix x xs
assume HI: "?p xs"
show "?p (x#xs)"
proof (cases)
assume H1:"estaEn x xs"
then have "estaEn a (borraDuplicados (x#xs)) = estaEn a (borraDuplicados xs)" by simp
also have "... = estaEn a xs" using HI by simp
also have "... = estaEn a (x#xs)"
proof(cases)
assume "x=a"
then show "estaEn a xs = estaEn a (x#xs)" using H1 by simp
next
assume "x≠a"
then show "estaEn a xs = estaEn a (x#xs)" by simp
qed
finally show "?p (x#xs)" by simp
next
assume H2:"¬estaEn x xs"
then have "estaEn a (borraDuplicados (x#xs)) = estaEn a (x#borraDuplicados xs)" by simp
also have "... = ((x=a) ∨ estaEn a (borraDuplicados xs))" by simp
also have "... = ((x=a) ∨ estaEn a xs)" using HI by simp
also have "... = estaEn a (x#xs)" by simp
finally show "?p (x#xs)" by simp
qed

(*danrodcha *)
(* es como la de ruben pero con diferencias de estilo *)
lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix x xs assume HI: "?P xs"
show "?P (x#xs)"
proof (rule iffI)
assume H1: "estaEn a (borraDuplicados (x # xs))"
show "estaEn a (x#xs)"
proof (cases "estaEn x xs")
case True
then show "estaEn a (x#xs)" using H1 HI by simp
next
case False
then show "estaEn a (x#xs)" using H1 HI by simp
qed
next
assume H2: "estaEn a (x#xs)"
show "estaEn a (borraDuplicados (x # xs))"
proof (cases "x=a")
case True
then show "estaEn a (borraDuplicados (x # xs))" using HI by simp
next
case False
then show "estaEn a (borraDuplicados (x # xs))" using H2 HI by simp
qed
qed
qed

text {*
---------------------------------------------------------------------
Ejercicio 6.1. Demostrar o refutar automáticamente
sinDuplicados (borraDuplicados xs)
---------------------------------------------------------------------
*}

-- "La demostración automática"

(* ivamenjim wilmorort serrodcal crigomgom anaprarod fraortmoy
marpoldia1 ferrenseg danrodcha *)
lemma sinDuplicados_borraDuplicados:
"sinDuplicados (borraDuplicados xs)"
by (induct xs) (auto simp add: estaEn_borraDuplicados)

(* migtermor bowma *)
lemma sinDuplicados_borraDuplicados:
"sinDuplicados (borraDuplicados xs)"
by (induct xs, simp_all add: estaEn_borraDuplicados_2)

text {*
---------------------------------------------------------------------
Ejercicio 6.2. Demostrar o refutar detalladamente
sinDuplicados (borraDuplicados xs)
---------------------------------------------------------------------
*}

(*wilmorort*)

-- "La demostración estructurada es"
lemma sinDuplicados_borraDuplicados_2:
"sinDuplicados (borraDuplicados xs)"
proof (induct xs)
show " sinDuplicados (borraDuplicados [])" by simp
next
fix a xs
assume HI: "sinDuplicados (borraDuplicados xs)"
show "sinDuplicados (borraDuplicados (a # xs))"
proof (cases)
assume "estaEn a xs"
then show "sinDuplicados (borraDuplicados (a#xs))" using HI by simp
next
assume"¬ estaEn a xs"
then have "¬ (estaEn a xs) ∧ sinDuplicados (borraDuplicados xs)" using HI by simp
then have "¬ estaEn a (borraDuplicados xs) ∧ sinDuplicados (borraDuplicados xs)"
by (simp add: estaEn_borraDuplicados)
then have " sinDuplicados (a#borraDuplicados xs)" by simp
then show " sinDuplicados (borraDuplicados(a #xs))" by simp
qed
qed

(* ivamenjim migtermor crigomgom rubgonmar fraortmoy marpoldia1 ferrenseg bowma *)

lemma sinDuplicados_borraDuplicados_2:
"sinDuplicados (borraDuplicados xs)"
proof (induct xs)
show "sinDuplicados (borraDuplicados [])" by simp
next
fix a xs
assume HI: "sinDuplicados (borraDuplicados xs)"
show "sinDuplicados (borraDuplicados (a # xs))"
proof (cases)
assume "estaEn a xs"
then show "sinDuplicados (borraDuplicados (a # xs))" using HI by simp
next
assume "¬(estaEn a xs)"
then show "sinDuplicados (borraDuplicados (a # xs))" using HI by (simp add: estaEn_borraDuplicados)
qed
qed

(* anaprarod *)
lemma sinDuplicados_borraDuplicados_2:
"sinDuplicados (borraDuplicados xs)" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix x xs
assume HI: "?P xs"
show "?P (x#xs)"
proof (cases)
assume c1: "estaEn x xs"
then show "sinDuplicados (borraDuplicados (x#xs))" using HI by simp
next
assume c2: "¬ estaEn x xs"
then have "sinDuplicados (borraDuplicados (x#xs)) =sinDuplicados (x#borraDuplicados xs)" by simp
also have "…= (¬estaEn x (borraDuplicados xs) ∧ sinDuplicados (borraDuplicados xs))" by simp
also have "… = (¬estaEn x (borraDuplicados xs))" using HI by simp
also have "… = (¬(estaEn x xs))" by (simp add:estaEn_borraDuplicados)
also have "… = True" using c2 by simp
finally show "?P (x#xs)" by simp
qed
qed

(* danrodcha *)
lemma sinDuplicados_borraDuplicados_2:
"sinDuplicados (borraDuplicados xs)" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix x xs assume HI: "?P xs"
show "?P (x#xs)"
proof (cases "estaEn x xs")
case True
then have 1: "sinDuplicados (borraDuplicados (x#xs))
= sinDuplicados (borraDuplicados xs)" by (simp add: estaEn_borraDuplicados_2)
show "?P (x#xs)" using HI 1 by simp
next
case False
then have "sinDuplicados (borraDuplicados (x#xs))
= sinDuplicados (x#borraDuplicados xs)" by simp
also have "… = (¬ (estaEn x (borraDuplicados xs)) ∧
sinDuplicados (borraDuplicados xs))" by simp
also have "… = True" using `¬ estaEn x xs` HI by (simp add:estaEn_borraDuplicados)
finally show "?P (x#xs)" by simp
qed
qed

text {*
---------------------------------------------------------------------
Ejercicio 7. Demostrar o refutar:
borraDuplicados (rev xs) = rev (borraDuplicados xs)
---------------------------------------------------------------------
*}

(*crigomgom rubgonmar ivamenjim wilmorort pablucoto migtermor
anaprarod fraortmoy ferrenseg marpoldia1 bowma danrodcha *)
lemma "borraDuplicados (rev xs) = rev (borraDuplicados xs)"
quickcheck
oops

(* ivamenjim: Quickcheck encuentra el siguiente contraejemplo:
xs = [a1, a2, a1]
Por lo que:
· "borraDuplicados (rev xs) = [a2, a1]"
· "rev (borraDuplicados xs) = [a1, a2]" *)

end
</source>

Relación 2

2016-11-06T19:13:33Z

Manmorjim1:

<source lang="isar">
chapter {* R2: Razonamiento sobre programas en Isabelle/HOL *}

theory R2_Razonamiento_automatico_sobre_programas
imports Main
begin

declare [[names_short]]

text {* ---------------------------------------------------------------
Ejercicio 1.1. Definir la función
sumaImpares :: nat ⇒ nat
tal que (sumaImpares n) es la suma de los n primeros números
impares. Por ejemplo,
sumaImpares 5 = 25
------------------------------------------------------------------ *}

(* fracorjim1 *)
fun sumaImpares0 :: "nat ⇒ nat" where
"sumaImpares0 0 = 0"
| "sumaImpares0 n = (n mod 2)*n + sumaImpares(n-1)"

(* ivamenjim wilmorort*)
fun sumaImpares :: "nat ⇒ nat" where
"sumaImpares 0 = 0"
| "sumaImpares n = 2*(n-1) + 1 + sumaImpares (n-1)"

value "sumaImpares 5 = 25"

(* wilmorort *)

(*Notar: Por la propiedad de Gauss se puede deducir que:
la suma de los n números impares es igual a la suma de los
n y n-1 números consecutivos*)
fun suma :: "nat ⇒ nat" where
"suma 0 = 0"
| "suma n = n + suma(n-1)"

fun sumaImpares1 :: "nat ⇒ nat" where
"sumaImpares1 0 = 0"
| "sumaImpares1 n = suma(n) + suma(n-1)"

value "sumaImpares1 5 = 25"

(* Comentario: La definición sumaImpares1 no es recursiva y, por tanto,
no es apropiada para demostraciones por inducción.

Comentario: La definición sumaImpares1 no es recursiva pero la
demostración, está realizada por inducción, por lo tanto no se puede
concluir que: "Si una def. no es recuriva no se puede demostrar por
inducción"

Comentario: Las definiciones no recursivas no generan esquemas de
inducción.
*)

(* fraortmoy *)
fun sumaImpares2 :: "nat ⇒ nat" where
"sumaImpares2 0 = 0"
| "sumaImpares2 (Suc(n)) = ((Suc(n)*2)-1) + sumaImpares2 n"

value "sumaImpares2 5 = 25"

(* anaprarod danrodcha crigomgom migtermor*)
fun sumaImpares3 :: "nat ⇒ nat" where
"sumaImpares3 0 = 0"
|"sumaImpares3 (Suc n) = (2*n +1) + (sumaImpares3 n)"

value "sumaImpares3 5 = 25"

(* manmorjim1 *)
fun sumaImpares4 :: "nat ⇒ nat" where
"sumaImpares4 0 = 0"
| "sumaImpares4 (Suc n) = (Suc (2 * n)) + (sumaImpares4 n)"

value "sumaImpares4 5 = 25"

(* pablucoto serrodcal *)
fun sumaImpares5 :: "nat ⇒ nat" where
"sumaImpares5 0 = 0"
| "sumaImpares5 n = (2*n-1) + sumaImpares5 (n-1) "

value "sumaImpares5 5 = 25"

(* ferrenseg *)

fun sumaImpares6 :: "nat ⇒ nat" where
"sumaImpares6 0 = 0"
|"sumaImpares6 (Suc n) = (2 * (Suc n) - 1) + sumaImpares6 n"

value "sumaImpares6 5 = 25"

text {* ---------------------------------------------------------------
Ejercicio 1.2. Demostrar que
sumaImpares n = n*n
------------------------------------------------------------------- *}

(* ivamenjim wilmorort anaprarod crigomgom manmorjim1 pablucoto serrodcal *)
lemma "sumaImpares n = n*n"
apply (induct n)
apply auto
done

(* wilmorort *)
(* Demostración estructurada *)

lemma aux1 : "suma(Suc n) + suma(n) = (n+1) + suma(n) + n + suma(n-1)"
apply (induct n)
apply auto
done

lemma aux2: "(n+1) + suma(n) + n + suma(n-1) = (n+1) + n + sumaImpares1 n "
apply (induct n)
apply auto
done

lemma "sumaImpares1 n = n*n"
proof (induct n)
show "sumaImpares1 0 = 0*0" by simp
next
fix n
assume HI:"sumaImpares1 n = n*n"
have "sumaImpares1 (Suc n) = suma(Suc n) + suma(n)" by simp
have "suma(Suc n) + suma(n) = (n+1) + suma(n) + n + suma(n-1)" using
"aux1" by simp
also have "... = (n+1) + n + sumaImpares1 n " using "aux2" by simp
also have "... = 2*n +1 + n*n" using HI by simp
also have "... = (n+1)*(n+1)" by simp
finally show "sumaImpares1 (Suc n) = Suc n * Suc n" by simp
qed

(* fraortmoy *)
(* La demostración para la función sumaImpares2 es la misma que la de
sumaImpares*)

lemma "sumaImpares2 n = n*n"
apply (induct n)
apply auto
done

(* danrodcha migtermor*)
lemma "sumaImpares n = n*n"
by (induct n) auto

(* danrodcha *)
lemma "sumaImpares n = n*n" (is "?P n")
proof (induct n)
show "?P 0" by simp
next
fix n assume "?P n"
thus "?P (Suc n)" by simp
qed

(* ferrenseg *)

lemma "sumaImpares3 n = n*n"
by (inductive n) simp_all

text {* ---------------------------------------------------------------
Ejercicio 2.1. Definir la función
sumaPotenciasDeDosMasUno :: nat ⇒ nat
tal que
(sumaPotenciasDeDosMasUno n) = 1 + 2^0 + 2^1 + 2^2 + ... + 2^n.
Por ejemplo,
sumaPotenciasDeDosMasUno 3 = 16
------------------------------------------------------------------ *}

(* wilmorort ivamenjim pablucoto serrodcal*)
fun sumaPotenciasDeDosMasUno :: "nat ⇒ nat" where
"sumaPotenciasDeDosMasUno 0 = 2"
| "sumaPotenciasDeDosMasUno n = 2^n + sumaPotenciasDeDosMasUno (n-1)"

value "sumaPotenciasDeDosMasUno 3 = 16"

(* fraortmoy anaprarod danrodcha crigomgom migtermor manmorjim1 *)
fun sumaPotenciasDeDosMasUno2 :: "nat ⇒ nat" where
"sumaPotenciasDeDosMasUno2 0 = 2"
| "sumaPotenciasDeDosMasUno2 (Suc(n)) =
(2^(Suc n)) + sumaPotenciasDeDosMasUno2 n"

value "sumaPotenciasDeDosMasUno2 3 = 16"

text {* ---------------------------------------------------------------
Ejercicio 2.2. Demostrar que
sumaPotenciasDeDosMasUno n = 2^(n+1)
------------------------------------------------------------------- *}

(* wilmorort ivamenjim fraortmoy anaprarod crigomgom pablucoto serrodcal manmorjim1 *)
(* esta demostración funciona con sumaPotenciasDeDosMasUno2 *)

lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"
apply (induct n)
apply auto
done

(* danrodcha migtermor*)
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"
by (induct n) auto

(* danrodcha *)
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)" (is "?P n")
proof (induct n)
show "?P 0" by simp
next
fix n assume "?P n"
thus "?P (Suc n)"by simp
qed

text {* ---------------------------------------------------------------
Ejercicio 3.1. Definir la función
copia :: nat ⇒ 'a ⇒ 'a list
tal que (copia n x) es la lista formado por n copias del elemento
x. Por ejemplo,
copia 3 x = [x,x,x]
------------------------------------------------------------------ *}

(* ivamenjim wilmorort pablucoto*)
fun copia :: "nat ⇒ 'a ⇒ 'a list" where
"copia 0 x = [] "
| "copia n x = x # (copia (n-1) x)"

(* wilmorort anaprarod *)
fun copia1 :: "nat ⇒ 'a ⇒ 'a list" where
"copia1 0 x = []"
| "copia1 (Suc n) x = [x] @ (copia1 n x) "

(* Comentario: La dedinición copia1 se puede simplificar eliminando @ *)

(* danrodcha crigomgom migtermor manmorjim1 *)
fun copia2 :: "nat ⇒ 'a ⇒ 'a list" where
"copia2 0 x = []"
| "copia2 (Suc n) x = x # copia2 n x"

(* serrodcal *)
fun copia3 :: "nat ⇒ 'a ⇒ 'a list" where
"copia3 0 x = []"
| "copia3 n x = [x] @ copia3 n-1 x"

value "copia 3 x = [x,x,x]"

text {* ---------------------------------------------------------------
Ejercicio 3.2. Definir la función
todos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool
tal que (todos p xs) se verifica si todos los elementos de xs cumplen
la propiedad p. Por ejemplo,
todos (λx. x>(1::nat)) [2,6,4] = True
todos (λx. x>(2::nat)) [2,6,4] = False
Nota: La conjunción se representa por ∧
----------------------------------------------------------------- *}

(* wilmorort *)
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos p [] = True"
| "todos p (x#xs) = (p x = True ∧ todos p xs) "

(* Comentario: La definición todos se puede simplificar eliminando True
*)

value "todos (λx. x>(1::nat)) [2,6,4] = True"
value "todos (λx. x>(2::nat)) [2,6,4] = False"

(* fraortmoy anaprarod danrodcha crigomgom migtermor ivamenjim serrodcal *)
fun todos1 :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos1 p [] = True"
| "todos1 p (x#xs) = (p x ∧ todos1 p xs )"

value "todos1 (λx. x>(1::nat)) [2,6,4] = True"
value "todos1 (λx. x>(2::nat)) [2,6,4] = False"

(* manmorjim1 *)
fun todos2 :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos2 p xs = ((filter p xs) = xs)"

text {* ---------------------------------------------------------------
Ejercicio 3.3. Demostrar que todos los elementos de (copia n x) son
iguales a x.
------------------------------------------------------------------- *}

(* wilmorort fraortmoy anaprarod crigomgom ivamenjim serrodcal manmorjim1 *)
lemma "todos (λy. y=x) (copia n x)"
apply (induct n)
apply auto
done

(* danrodcha*)
lemma "todos (λy. y=x) (copia n x)"
by (induct n) auto

(* danrodcha *)
lemma "todos (λy. y=x) (copia n x)" (is "?P n")
proof (induct n)
show "?P 0" by simp
next
fix n assume "?P n"
thus "?P (Suc n)" by simp
qed

text {* ---------------------------------------------------------------
Ejercicio 4.1. Definir, recursivamente y sin usar (@), la función
amplia :: 'a list ⇒ 'a ⇒ 'a list
tal que (amplia xs y) es la lista obtenida añadiendo el elemento y al
final de la lista xs. Por ejemplo,
amplia [d,a] t = [d,a,t]
------------------------------------------------------------------ *}

(* wilmorort fraortmoy anaprarod crigomgom migtermor ivamenjim serrodcal manmorjim1 *)
fun amplia :: "'a list ⇒ 'a ⇒ 'a list" where
"amplia [] y = [y] "
| "amplia (x#xs) y = x # amplia xs y"

value "amplia [d,a] t = [d,a,t]"

text {* ---------------------------------------------------------------
Ejercicio 4.2. Demostrar que
amplia xs y = xs @ [y]
------------------------------------------------------------------- *}

(* wilmorort fraortmoy anaprarod crigomgom ivamenjim serrodcal manmorjim1 *)
lemma "amplia xs y = xs @ [y]"
apply (induct xs)
apply auto
done

(* danrodcha migtermor*)
lemma "amplia xs y = xs @ [y]"
by (induct xs) auto

(* danrodcha *)
lemma "amplia xs y = xs @ [y]" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs assume "?P xs"
thus "?P (a#xs)" by simp
qed

end
</source>

Relación 2

2016-11-06T19:08:43Z

Manmorjim1:

<source lang="isar">
chapter {* R2: Razonamiento sobre programas en Isabelle/HOL *}

theory R2_Razonamiento_automatico_sobre_programas
imports Main
begin

declare [[names_short]]

text {* ---------------------------------------------------------------
Ejercicio 1.1. Definir la función
sumaImpares :: nat ⇒ nat
tal que (sumaImpares n) es la suma de los n primeros números
impares. Por ejemplo,
sumaImpares 5 = 25
------------------------------------------------------------------ *}

(* fracorjim1 *)
fun sumaImpares0 :: "nat ⇒ nat" where
"sumaImpares0 0 = 0"
| "sumaImpares0 n = (n mod 2)*n + sumaImpares(n-1)"

(* ivamenjim wilmorort*)
fun sumaImpares :: "nat ⇒ nat" where
"sumaImpares 0 = 0"
| "sumaImpares n = 2*(n-1) + 1 + sumaImpares (n-1)"

value "sumaImpares 5 = 25"

(* wilmorort *)

(*Notar: Por la propiedad de Gauss se puede deducir que:
la suma de los n números impares es igual a la suma de los
n y n-1 números consecutivos*)
fun suma :: "nat ⇒ nat" where
"suma 0 = 0"
| "suma n = n + suma(n-1)"

fun sumaImpares1 :: "nat ⇒ nat" where
"sumaImpares1 0 = 0"
| "sumaImpares1 n = suma(n) + suma(n-1)"

value "sumaImpares1 5 = 25"

(* Comentario: La definición sumaImpares1 no es recursiva y, por tanto,
no es apropiada para demostraciones por inducción.

Comentario: La definición sumaImpares1 no es recursiva pero la
demostración, está realizada por inducción, por lo tanto no se puede
concluir que: "Si una def. no es recuriva no se puede demostrar por
inducción"

Comentario: Las definiciones no recursivas no generan esquemas de
inducción.
*)

(* fraortmoy *)
fun sumaImpares2 :: "nat ⇒ nat" where
"sumaImpares2 0 = 0"
| "sumaImpares2 (Suc(n)) = ((Suc(n)*2)-1) + sumaImpares2 n"

value "sumaImpares2 5 = 25"

(* anaprarod danrodcha crigomgom migtermor*)
fun sumaImpares3 :: "nat ⇒ nat" where
"sumaImpares3 0 = 0"
|"sumaImpares3 (Suc n) = (2*n +1) + (sumaImpares3 n)"

value "sumaImpares3 5 = 25"

(* manmorjim1 *)
fun sumaImpares4 :: "nat ⇒ nat" where
"sumaImpares4 0 = 0"
| "sumaImpares4 (Suc n) = (Suc (2 * n)) + (sumaImpares4 n)"

value "sumaImpares4 5 = 25"

(* pablucoto serrodcal *)
fun sumaImpares5 :: "nat ⇒ nat" where
"sumaImpares5 0 = 0"
| "sumaImpares5 n = (2*n-1) + sumaImpares5 (n-1) "

value "sumaImpares5 5 = 25"

(* ferrenseg *)

fun sumaImpares6 :: "nat ⇒ nat" where
"sumaImpares6 0 = 0"
|"sumaImpares6 (Suc n) = (2 * (Suc n) - 1) + sumaImpares6 n"

value "sumaImpares6 5 = 25"

text {* ---------------------------------------------------------------
Ejercicio 1.2. Demostrar que
sumaImpares n = n*n
------------------------------------------------------------------- *}

(* ivamenjim wilmorort anaprarod crigomgom manmorjim1 pablucoto serrodcal *)
lemma "sumaImpares n = n*n"
apply (induct n)
apply auto
done

(* wilmorort *)
(* Demostración estructurada *)

lemma aux1 : "suma(Suc n) + suma(n) = (n+1) + suma(n) + n + suma(n-1)"
apply (induct n)
apply auto
done

lemma aux2: "(n+1) + suma(n) + n + suma(n-1) = (n+1) + n + sumaImpares1 n "
apply (induct n)
apply auto
done

lemma "sumaImpares1 n = n*n"
proof (induct n)
show "sumaImpares1 0 = 0*0" by simp
next
fix n
assume HI:"sumaImpares1 n = n*n"
have "sumaImpares1 (Suc n) = suma(Suc n) + suma(n)" by simp
have "suma(Suc n) + suma(n) = (n+1) + suma(n) + n + suma(n-1)" using
"aux1" by simp
also have "... = (n+1) + n + sumaImpares1 n " using "aux2" by simp
also have "... = 2*n +1 + n*n" using HI by simp
also have "... = (n+1)*(n+1)" by simp
finally show "sumaImpares1 (Suc n) = Suc n * Suc n" by simp
qed

(* fraortmoy *)
(* La demostración para la función sumaImpares2 es la misma que la de
sumaImpares*)

lemma "sumaImpares2 n = n*n"
apply (induct n)
apply auto
done

(* danrodcha migtermor*)
lemma "sumaImpares n = n*n"
by (induct n) auto

(* danrodcha *)
lemma "sumaImpares n = n*n" (is "?P n")
proof (induct n)
show "?P 0" by simp
next
fix n assume "?P n"
thus "?P (Suc n)" by simp
qed

(* ferrenseg *)

lemma "sumaImpares3 n = n*n"
by (inductive n) simp_all

text {* ---------------------------------------------------------------
Ejercicio 2.1. Definir la función
sumaPotenciasDeDosMasUno :: nat ⇒ nat
tal que
(sumaPotenciasDeDosMasUno n) = 1 + 2^0 + 2^1 + 2^2 + ... + 2^n.
Por ejemplo,
sumaPotenciasDeDosMasUno 3 = 16
------------------------------------------------------------------ *}

(* wilmorort ivamenjim pablucoto serrodcal*)
fun sumaPotenciasDeDosMasUno :: "nat ⇒ nat" where
"sumaPotenciasDeDosMasUno 0 = 2"
| "sumaPotenciasDeDosMasUno n = 2^n + sumaPotenciasDeDosMasUno (n-1)"

value "sumaPotenciasDeDosMasUno 3 = 16"

(* fraortmoy anaprarod danrodcha crigomgom migtermor manmorjim1 *)
fun sumaPotenciasDeDosMasUno2 :: "nat ⇒ nat" where
"sumaPotenciasDeDosMasUno2 0 = 2"
| "sumaPotenciasDeDosMasUno2 (Suc(n)) =
(2^(Suc n)) + sumaPotenciasDeDosMasUno2 n"

value "sumaPotenciasDeDosMasUno2 3 = 16"

text {* ---------------------------------------------------------------
Ejercicio 2.2. Demostrar que
sumaPotenciasDeDosMasUno n = 2^(n+1)
------------------------------------------------------------------- *}

(* wilmorort ivamenjim fraortmoy anaprarod crigomgom pablucoto serrodcal manmorjim1 *)
(* esta demostración funciona con sumaPotenciasDeDosMasUno2 *)

lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"
apply (induct n)
apply auto
done

(* danrodcha migtermor*)
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"
by (induct n) auto

(* danrodcha *)
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)" (is "?P n")
proof (induct n)
show "?P 0" by simp
next
fix n assume "?P n"
thus "?P (Suc n)"by simp
qed

text {* ---------------------------------------------------------------
Ejercicio 3.1. Definir la función
copia :: nat ⇒ 'a ⇒ 'a list
tal que (copia n x) es la lista formado por n copias del elemento
x. Por ejemplo,
copia 3 x = [x,x,x]
------------------------------------------------------------------ *}

(* ivamenjim wilmorort pablucoto*)
fun copia :: "nat ⇒ 'a ⇒ 'a list" where
"copia 0 x = [] "
| "copia n x = x # (copia (n-1) x)"

(* wilmorort anaprarod *)
fun copia1 :: "nat ⇒ 'a ⇒ 'a list" where
"copia1 0 x = []"
| "copia1 (Suc n) x = [x] @ (copia1 n x) "

(* Comentario: La dedinición copia1 se puede simplificar eliminando @ *)

(* danrodcha crigomgom migtermor manmorjim1 *)
fun copia2 :: "nat ⇒ 'a ⇒ 'a list" where
"copia2 0 x = []"
| "copia2 (Suc n) x = x # copia2 n x"

(* serrodcal *)
fun copia3 :: "nat ⇒ 'a ⇒ 'a list" where
"copia3 0 x = []"
| "copia3 n x = [x] @ copia3 n-1 x"

value "copia 3 x = [x,x,x]"

text {* ---------------------------------------------------------------
Ejercicio 3.2. Definir la función
todos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool
tal que (todos p xs) se verifica si todos los elementos de xs cumplen
la propiedad p. Por ejemplo,
todos (λx. x>(1::nat)) [2,6,4] = True
todos (λx. x>(2::nat)) [2,6,4] = False
Nota: La conjunción se representa por ∧
----------------------------------------------------------------- *}

(* wilmorort *)
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos p [] = True"
| "todos p (x#xs) = (p x = True ∧ todos p xs) "

(* Comentario: La definición todos se puede simplificar eliminando True
*)

value "todos (λx. x>(1::nat)) [2,6,4] = True"
value "todos (λx. x>(2::nat)) [2,6,4] = False"

(* fraortmoy anaprarod danrodcha crigomgom migtermor ivamenjim serrodcal *)
fun todos1 :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos1 p [] = True"
| "todos1 p (x#xs) = (p x ∧ todos1 p xs )"

value "todos1 (λx. x>(1::nat)) [2,6,4] = True"
value "todos1 (λx. x>(2::nat)) [2,6,4] = False"

(* manmorjim1 *)
fun todos2 :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos2 p xs = ((filter p xs) = xs)"

text {* ---------------------------------------------------------------
Ejercicio 3.3. Demostrar que todos los elementos de (copia n x) son
iguales a x.
------------------------------------------------------------------- *}

(* wilmorort fraortmoy anaprarod crigomgom ivamenjim serrodcal manmorjim1 *)
lemma "todos (λy. y=x) (copia n x)"
apply (induct n)
apply auto
done

(* danrodcha*)
lemma "todos (λy. y=x) (copia n x)"
by (induct n) auto

(* danrodcha *)
lemma "todos (λy. y=x) (copia n x)" (is "?P n")
proof (induct n)
show "?P 0" by simp
next
fix n assume "?P n"
thus "?P (Suc n)" by simp
qed

text {* ---------------------------------------------------------------
Ejercicio 4.1. Definir, recursivamente y sin usar (@), la función
amplia :: 'a list ⇒ 'a ⇒ 'a list
tal que (amplia xs y) es la lista obtenida añadiendo el elemento y al
final de la lista xs. Por ejemplo,
amplia [d,a] t = [d,a,t]
------------------------------------------------------------------ *}

(* wilmorort fraortmoy anaprarod crigomgom migtermor ivamenjim serrodcal *)
fun amplia :: "'a list ⇒ 'a ⇒ 'a list" where
"amplia [] y = [y] "
| "amplia (x#xs) y = x # amplia xs y"

value "amplia [d,a] t = [d,a,t]"

text {* ---------------------------------------------------------------
Ejercicio 4.2. Demostrar que
amplia xs y = xs @ [y]
------------------------------------------------------------------- *}

(* wilmorort fraortmoy anaprarod crigomgom ivamenjim serrodcal *)
lemma "amplia xs y = xs @ [y]"
apply (induct xs)
apply auto
done

(* danrodcha migtermor*)
lemma "amplia xs y = xs @ [y]"
by (induct xs) auto

(* danrodcha *)
lemma "amplia xs y = xs @ [y]" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs assume "?P xs"
thus "?P (a#xs)" by simp
qed

end
</source>

Relación 2

2016-11-06T18:59:07Z

Manmorjim1:

<source lang="isar">
chapter {* R2: Razonamiento sobre programas en Isabelle/HOL *}

theory R2_Razonamiento_automatico_sobre_programas
imports Main
begin

declare [[names_short]]

text {* ---------------------------------------------------------------
Ejercicio 1.1. Definir la función
sumaImpares :: nat ⇒ nat
tal que (sumaImpares n) es la suma de los n primeros números
impares. Por ejemplo,
sumaImpares 5 = 25
------------------------------------------------------------------ *}

(* fracorjim1 *)
fun sumaImpares0 :: "nat ⇒ nat" where
"sumaImpares0 0 = 0"
| "sumaImpares0 n = (n mod 2)*n + sumaImpares(n-1)"

(* ivamenjim wilmorort*)
fun sumaImpares :: "nat ⇒ nat" where
"sumaImpares 0 = 0"
| "sumaImpares n = 2*(n-1) + 1 + sumaImpares (n-1)"

value "sumaImpares 5 = 25"

(* wilmorort *)

(*Notar: Por la propiedad de Gauss se puede deducir que:
la suma de los n números impares es igual a la suma de los
n y n-1 números consecutivos*)
fun suma :: "nat ⇒ nat" where
"suma 0 = 0"
| "suma n = n + suma(n-1)"

fun sumaImpares1 :: "nat ⇒ nat" where
"sumaImpares1 0 = 0"
| "sumaImpares1 n = suma(n) + suma(n-1)"

value "sumaImpares1 5 = 25"

(* Comentario: La definición sumaImpares1 no es recursiva y, por tanto,
no es apropiada para demostraciones por inducción.

Comentario: La definición sumaImpares1 no es recursiva pero la
demostración, está realizada por inducción, por lo tanto no se puede
concluir que: "Si una def. no es recuriva no se puede demostrar por
inducción"

Comentario: Las definiciones no recursivas no generan esquemas de
inducción.
*)

(* fraortmoy *)
fun sumaImpares2 :: "nat ⇒ nat" where
"sumaImpares2 0 = 0"
| "sumaImpares2 (Suc(n)) = ((Suc(n)*2)-1) + sumaImpares2 n"

value "sumaImpares2 5 = 25"

(* anaprarod danrodcha crigomgom migtermor*)
fun sumaImpares3 :: "nat ⇒ nat" where
"sumaImpares3 0 = 0"
|"sumaImpares3 (Suc n) = (2*n +1) + (sumaImpares3 n)"

value "sumaImpares3 5 = 25"

(* manmorjim1 *)
fun sumaImpares4 :: "nat ⇒ nat" where
"sumaImpares4 0 = 0"
| "sumaImpares4 (Suc n) = (Suc (2 * n)) + (sumaImpares4 n)"

value "sumaImpares4 5 = 25"

(* pablucoto serrodcal *)
fun sumaImpares5 :: "nat ⇒ nat" where
"sumaImpares5 0 = 0"
| "sumaImpares5 n = (2*n-1) + sumaImpares5 (n-1) "

value "sumaImpares5 5 = 25"

(* ferrenseg *)

fun sumaImpares6 :: "nat ⇒ nat" where
"sumaImpares6 0 = 0"
|"sumaImpares6 (Suc n) = (2 * (Suc n) - 1) + sumaImpares6 n"

value "sumaImpares6 5 = 25"

text {* ---------------------------------------------------------------
Ejercicio 1.2. Demostrar que
sumaImpares n = n*n
------------------------------------------------------------------- *}

(* ivamenjim wilmorort anaprarod crigomgom manmorjim1 pablucoto serrodcal *)
lemma "sumaImpares n = n*n"
apply (induct n)
apply auto
done

(* wilmorort *)
(* Demostración estructurada *)

lemma aux1 : "suma(Suc n) + suma(n) = (n+1) + suma(n) + n + suma(n-1)"
apply (induct n)
apply auto
done

lemma aux2: "(n+1) + suma(n) + n + suma(n-1) = (n+1) + n + sumaImpares1 n "
apply (induct n)
apply auto
done

lemma "sumaImpares1 n = n*n"
proof (induct n)
show "sumaImpares1 0 = 0*0" by simp
next
fix n
assume HI:"sumaImpares1 n = n*n"
have "sumaImpares1 (Suc n) = suma(Suc n) + suma(n)" by simp
have "suma(Suc n) + suma(n) = (n+1) + suma(n) + n + suma(n-1)" using
"aux1" by simp
also have "... = (n+1) + n + sumaImpares1 n " using "aux2" by simp
also have "... = 2*n +1 + n*n" using HI by simp
also have "... = (n+1)*(n+1)" by simp
finally show "sumaImpares1 (Suc n) = Suc n * Suc n" by simp
qed

(* fraortmoy *)
(* La demostración para la función sumaImpares2 es la misma que la de
sumaImpares*)

lemma "sumaImpares2 n = n*n"
apply (induct n)
apply auto
done

(* danrodcha migtermor*)
lemma "sumaImpares n = n*n"
by (induct n) auto

(* danrodcha *)
lemma "sumaImpares n = n*n" (is "?P n")
proof (induct n)
show "?P 0" by simp
next
fix n assume "?P n"
thus "?P (Suc n)" by simp
qed

(* ferrenseg *)

lemma "sumaImpares3 n = n*n"
by (inductive n) simp_all

text {* ---------------------------------------------------------------
Ejercicio 2.1. Definir la función
sumaPotenciasDeDosMasUno :: nat ⇒ nat
tal que
(sumaPotenciasDeDosMasUno n) = 1 + 2^0 + 2^1 + 2^2 + ... + 2^n.
Por ejemplo,
sumaPotenciasDeDosMasUno 3 = 16
------------------------------------------------------------------ *}

(* wilmorort ivamenjim pablucoto serrodcal*)
fun sumaPotenciasDeDosMasUno :: "nat ⇒ nat" where
"sumaPotenciasDeDosMasUno 0 = 2"
| "sumaPotenciasDeDosMasUno n = 2^n + sumaPotenciasDeDosMasUno (n-1)"

value "sumaPotenciasDeDosMasUno 3 = 16"

(* fraortmoy anaprarod danrodcha crigomgom migtermor manmorjim1 *)
fun sumaPotenciasDeDosMasUno2 :: "nat ⇒ nat" where
"sumaPotenciasDeDosMasUno2 0 = 2"
| "sumaPotenciasDeDosMasUno2 (Suc(n)) =
(2^(Suc n)) + sumaPotenciasDeDosMasUno2 n"

value "sumaPotenciasDeDosMasUno2 3 = 16"

text {* ---------------------------------------------------------------
Ejercicio 2.2. Demostrar que
sumaPotenciasDeDosMasUno n = 2^(n+1)
------------------------------------------------------------------- *}

(* wilmorort ivamenjim fraortmoy anaprarod crigomgom pablucoto serrodcal manmorjim1 *)
(* esta demostración funciona con sumaPotenciasDeDosMasUno2 *)

lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"
apply (induct n)
apply auto
done

(* danrodcha migtermor*)
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"
by (induct n) auto

(* danrodcha *)
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)" (is "?P n")
proof (induct n)
show "?P 0" by simp
next
fix n assume "?P n"
thus "?P (Suc n)"by simp
qed

text {* ---------------------------------------------------------------
Ejercicio 3.1. Definir la función
copia :: nat ⇒ 'a ⇒ 'a list
tal que (copia n x) es la lista formado por n copias del elemento
x. Por ejemplo,
copia 3 x = [x,x,x]
------------------------------------------------------------------ *}

(* ivamenjim wilmorort pablucoto*)
fun copia :: "nat ⇒ 'a ⇒ 'a list" where
"copia 0 x = [] "
| "copia n x = x # (copia (n-1) x)"

(* wilmorort anaprarod *)
fun copia1 :: "nat ⇒ 'a ⇒ 'a list" where
"copia1 0 x = []"
| "copia1 (Suc n) x = [x] @ (copia1 n x) "

(* Comentario: La dedinición copia1 se puede simplificar eliminando @ *)

(* danrodcha crigomgom migtermor manmorjim1 *)
fun copia2 :: "nat ⇒ 'a ⇒ 'a list" where
"copia2 0 x = []"
| "copia2 (Suc n) x = x # copia2 n x"

(* serrodcal *)
fun copia3 :: "nat ⇒ 'a ⇒ 'a list" where
"copia3 0 x = []"
| "copia3 n x = [x] @ copia3 n-1 x"

value "copia 3 x = [x,x,x]"

text {* ---------------------------------------------------------------
Ejercicio 3.2. Definir la función
todos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool
tal que (todos p xs) se verifica si todos los elementos de xs cumplen
la propiedad p. Por ejemplo,
todos (λx. x>(1::nat)) [2,6,4] = True
todos (λx. x>(2::nat)) [2,6,4] = False
Nota: La conjunción se representa por ∧
----------------------------------------------------------------- *}

(* wilmorort *)
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos p [] = True"
| "todos p (x#xs) = (p x = True ∧ todos p xs) "

(* Comentario: La definición todos se puede simplificar eliminando True
*)

value "todos (λx. x>(1::nat)) [2,6,4] = True"
value "todos (λx. x>(2::nat)) [2,6,4] = False"

(* fraortmoy anaprarod danrodcha crigomgom migtermor ivamenjim serrodcal *)
fun todos1 :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos1 p [] = True"
| "todos1 p (x#xs) = (p x ∧ todos1 p xs )"

value "todos1 (λx. x>(1::nat)) [2,6,4] = True"
value "todos1 (λx. x>(2::nat)) [2,6,4] = False"

text {* ---------------------------------------------------------------
Ejercicio 3.3. Demostrar que todos los elementos de (copia n x) son
iguales a x.
------------------------------------------------------------------- *}

(* wilmorort fraortmoy anaprarod crigomgom ivamenjim serrodcal *)
lemma "todos (λy. y=x) (copia n x)"
apply (induct n)
apply auto
done

(* danrodcha*)
lemma "todos (λy. y=x) (copia n x)"
by (induct n) auto

(* danrodcha *)
lemma "todos (λy. y=x) (copia n x)" (is "?P n")
proof (induct n)
show "?P 0" by simp
next
fix n assume "?P n"
thus "?P (Suc n)" by simp
qed

text {* ---------------------------------------------------------------
Ejercicio 4.1. Definir, recursivamente y sin usar (@), la función
amplia :: 'a list ⇒ 'a ⇒ 'a list
tal que (amplia xs y) es la lista obtenida añadiendo el elemento y al
final de la lista xs. Por ejemplo,
amplia [d,a] t = [d,a,t]
------------------------------------------------------------------ *}

(* wilmorort fraortmoy anaprarod crigomgom migtermor ivamenjim serrodcal *)
fun amplia :: "'a list ⇒ 'a ⇒ 'a list" where
"amplia [] y = [y] "
| "amplia (x#xs) y = x # amplia xs y"

value "amplia [d,a] t = [d,a,t]"

text {* ---------------------------------------------------------------
Ejercicio 4.2. Demostrar que
amplia xs y = xs @ [y]
------------------------------------------------------------------- *}

(* wilmorort fraortmoy anaprarod crigomgom ivamenjim serrodcal *)
lemma "amplia xs y = xs @ [y]"
apply (induct xs)
apply auto
done

(* danrodcha migtermor*)
lemma "amplia xs y = xs @ [y]"
by (induct xs) auto

(* danrodcha *)
lemma "amplia xs y = xs @ [y]" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs assume "?P xs"
thus "?P (a#xs)" by simp
qed

end
</source>

Relación 2

2016-11-06T14:33:27Z

Manmorjim1:

<source lang="isar">
chapter {* R2: Razonamiento sobre programas en Isabelle/HOL *}

theory R2_Razonamiento_automatico_sobre_programas
imports Main
begin

declare [[names_short]]

text {* ---------------------------------------------------------------
Ejercicio 1.1. Definir la función
sumaImpares :: nat ⇒ nat
tal que (sumaImpares n) es la suma de los n primeros números
impares. Por ejemplo,
sumaImpares 5 = 25
------------------------------------------------------------------ *}

(* ivamenjim wilmorort*)
fun sumaImpares :: "nat ⇒ nat" where
"sumaImpares 0 = 0"
| "sumaImpares n = 2*(n-1) + 1 + sumaImpares (n-1)"

value "sumaImpares 5 = 25"

(* wilmorort *)

(*Notar: Por la propiedad de Gauss se puede deducir que:
la suma de los n números impares es igual a la suma de los
n y n-1 números consecutivos*)
fun suma :: "nat ⇒ nat" where
"suma 0 = 0"
| "suma n = n + suma(n-1)"

fun sumaImpares1 :: "nat ⇒ nat" where
"sumaImpares1 0 = 0"
| "sumaImpares1 n = suma(n) + suma(n-1)"

value "sumaImpares1 5 = 25"

(* Comentario: La definición sumaImpares1 no es recursiva y, por tanto,
no es apropiada para demostraciones por inducción.

Comentario: La definición sumaImpares1 no es recursiva pero la
demostración, está realizada por inducción, por lo tanto no se puede
concluir que: "Si una def. no es recuriva no se puede demostrar por
inducción"

Comentario: Las definiciones no recursivas no generan esquemas de
inducción.
*)

(* fraortmoy *)
fun sumaImpares2 :: "nat ⇒ nat" where
"sumaImpares2 0 = 0"
| "sumaImpares2 (Suc(n)) = ((Suc(n)*2)-1) + sumaImpares2 n"

value "sumaImpares2 5 = 25"

(* anaprarod danrodcha crigomgom migtermor*)
fun sumaImpares3 :: "nat ⇒ nat" where
"sumaImpares3 0 = 0"
|"sumaImpares3 (Suc n) = (2*n +1) + (sumaImpares3 n)"

value "sumaImpares3 5 = 25"

(* manmorjim1 *)
fun sumaImpares4 :: "nat ⇒ nat" where
"sumaImpares4 0 = 0"
| "sumaImpares4 (Suc n) = (Suc (2 * n)) + (sumaImpares4 n)"

value "sumaImpares4 5 = 25"

(* pablucoto serrodcal *)
fun sumaImpares5 :: "nat ⇒ nat" where
"sumaImpares5 0 = 0"
| "sumaImpares5 n = (2*n-1) + sumaImpares5 (n-1) "

value "sumaImpares5 5 = 25"

text {* ---------------------------------------------------------------
Ejercicio 1.2. Demostrar que
sumaImpares n = n*n
------------------------------------------------------------------- *}

(* ivamenjim wilmorort anaprarod crigomgom manmorjim1 pablucoto serrodcal *)
lemma "sumaImpares n = n*n"
apply (induct n)
apply auto
done

(* wilmorort *)
(* Demostración estructurada *)

lemma aux1 : "suma(Suc n) + suma(n) = (n+1) + suma(n) + n + suma(n-1)"
apply (induct n)
apply auto
done

lemma aux2: "(n+1) + suma(n) + n + suma(n-1) = (n+1) + n + sumaImpares1 n "
apply (induct n)
apply auto
done

lemma "sumaImpares1 n = n*n"
proof (induct n)
show "sumaImpares1 0 = 0*0" by simp
next
fix n
assume HI:"sumaImpares1 n = n*n"
have "sumaImpares1 (Suc n) = suma(Suc n) + suma(n)" by simp
have "suma(Suc n) + suma(n) = (n+1) + suma(n) + n + suma(n-1)" using
"aux1" by simp
also have "... = (n+1) + n + sumaImpares1 n " using "aux2" by simp
also have "... = 2*n +1 + n*n" using HI by simp
also have "... = (n+1)*(n+1)" by simp
finally show "sumaImpares1 (Suc n) = Suc n * Suc n" by simp
qed

(* fraortmoy *)
(* La demostración para la función sumaImpares2 es la misma que la de
sumaImpares*)

lemma "sumaImpares2 n = n*n"
apply (induct n)
apply auto
done

(* danrodcha migtermor*)
lemma "sumaImpares n = n*n"
by (induct n) auto

(* danrodcha *)
lemma "sumaImpares n = n*n" (is "?P n")
proof (induct n)
show "?P 0" by simp
next
fix n assume "?P n"
thus "?P (Suc n)" by simp
qed

text {* ---------------------------------------------------------------
Ejercicio 2.1. Definir la función
sumaPotenciasDeDosMasUno :: nat ⇒ nat
tal que
(sumaPotenciasDeDosMasUno n) = 1 + 2^0 + 2^1 + 2^2 + ... + 2^n.
Por ejemplo,
sumaPotenciasDeDosMasUno 3 = 16
------------------------------------------------------------------ *}

(* wilmorort ivamenjim pablucoto serrodcal*)
fun sumaPotenciasDeDosMasUno :: "nat ⇒ nat" where
"sumaPotenciasDeDosMasUno 0 = 2"
| "sumaPotenciasDeDosMasUno n = 2^n + sumaPotenciasDeDosMasUno (n-1)"

value "sumaPotenciasDeDosMasUno 3 = 16"

(* fraortmoy anaprarod danrodcha crigomgom migtermor manmorjim1 *)
fun sumaPotenciasDeDosMasUno2 :: "nat ⇒ nat" where
"sumaPotenciasDeDosMasUno2 0 = 2"
| "sumaPotenciasDeDosMasUno2 (Suc(n)) =
(2^(Suc n)) + sumaPotenciasDeDosMasUno2 n"

value "sumaPotenciasDeDosMasUno2 3 = 16"

text {* ---------------------------------------------------------------
Ejercicio 2.2. Demostrar que
sumaPotenciasDeDosMasUno n = 2^(n+1)
------------------------------------------------------------------- *}

(* wilmorort ivamenjim fraortmoy anaprarod crigomgom pablucoto serrodcal manmorjim1 *)
(* esta demostración funciona con sumaPotenciasDeDosMasUno2 *)

lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"
apply (induct n)
apply auto
done

(* danrodcha migtermor*)
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"
by (induct n) auto

(* danrodcha *)
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)" (is "?P n")
proof (induct n)
show "?P 0" by simp
next
fix n assume "?P n"
thus "?P (Suc n)"by simp
qed

text {* ---------------------------------------------------------------
Ejercicio 3.1. Definir la función
copia :: nat ⇒ 'a ⇒ 'a list
tal que (copia n x) es la lista formado por n copias del elemento
x. Por ejemplo,
copia 3 x = [x,x,x]
------------------------------------------------------------------ *}

(* ivamenjim wilmorort pablucoto*)
fun copia :: "nat ⇒ 'a ⇒ 'a list" where
"copia 0 x = [] "
| "copia n x = x # (copia (n-1) x)"

(* wilmorort anaprarod *)
fun copia1 :: "nat ⇒ 'a ⇒ 'a list" where
"copia1 0 x = []"
| "copia1 (Suc n) x = [x] @ (copia1 n x) "

(* Comentario: La dedinición copia1 se puede simplificar eliminando @ *)

(* danrodcha crigomgom migtermor*)
fun copia2 :: "nat ⇒ 'a ⇒ 'a list" where
"copia2 0 x = []"
| "copia2 (Suc n) x = x # copia2 n x"

(* serrodcal *)
fun copia3 :: "nat ⇒ 'a ⇒ 'a list" where
"copia3 0 x = []"
| "copia3 n x = [x] @ copia3 n-1 x"

value "copia 3 x = [x,x,x]"

text {* ---------------------------------------------------------------
Ejercicio 3.2. Definir la función
todos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool
tal que (todos p xs) se verifica si todos los elementos de xs cumplen
la propiedad p. Por ejemplo,
todos (λx. x>(1::nat)) [2,6,4] = True
todos (λx. x>(2::nat)) [2,6,4] = False
Nota: La conjunción se representa por ∧
----------------------------------------------------------------- *}

(* wilmorort *)
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos p [] = True"
| "todos p (x#xs) = (p x = True ∧ todos p xs) "

(* Comentario: La definición todos se puede simplificar eliminando True
*)

value "todos (λx. x>(1::nat)) [2,6,4] = True"
value "todos (λx. x>(2::nat)) [2,6,4] = False"

(* fraortmoy anaprarod danrodcha crigomgom migtermor ivamenjim serrodcal *)
fun todos1 :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos1 p [] = True"
| "todos1 p (x#xs) = (p x ∧ todos1 p xs )"

value "todos1 (λx. x>(1::nat)) [2,6,4] = True"
value "todos1 (λx. x>(2::nat)) [2,6,4] = False"

text {* ---------------------------------------------------------------
Ejercicio 3.3. Demostrar que todos los elementos de (copia n x) son
iguales a x.
------------------------------------------------------------------- *}

(* wilmorort fraortmoy anaprarod crigomgom ivamenjim serrodcal *)
lemma "todos (λy. y=x) (copia n x)"
apply (induct n)
apply auto
done

(* danrodcha*)
lemma "todos (λy. y=x) (copia n x)"
by (induct n) auto

(* danrodcha *)
lemma "todos (λy. y=x) (copia n x)" (is "?P n")
proof (induct n)
show "?P 0" by simp
next
fix n assume "?P n"
thus "?P (Suc n)" by simp
qed

text {* ---------------------------------------------------------------
Ejercicio 4.1. Definir, recursivamente y sin usar (@), la función
amplia :: 'a list ⇒ 'a ⇒ 'a list
tal que (amplia xs y) es la lista obtenida añadiendo el elemento y al
final de la lista xs. Por ejemplo,
amplia [d,a] t = [d,a,t]
------------------------------------------------------------------ *}

(* wilmorort fraortmoy anaprarod crigomgom migtermor ivamenjim serrodcal *)
fun amplia :: "'a list ⇒ 'a ⇒ 'a list" where
"amplia [] y = [y] "
| "amplia (x#xs) y = x # amplia xs y"

value "amplia [d,a] t = [d,a,t]"

text {* ---------------------------------------------------------------
Ejercicio 4.2. Demostrar que
amplia xs y = xs @ [y]
------------------------------------------------------------------- *}

(* wilmorort fraortmoy anaprarod crigomgom ivamenjim serrodcal *)
lemma "amplia xs y = xs @ [y]"
apply (induct xs)
apply auto
done

(* danrodcha migtermor*)
lemma "amplia xs y = xs @ [y]"
by (induct xs) auto

(* danrodcha *)
lemma "amplia xs y = xs @ [y]" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs assume "?P xs"
thus "?P (a#xs)" by simp
qed

end
</source>

Discusión:Relación 2

2016-11-05T17:32:56Z

Manmorjim1:

== Pregunta de Wilmorort ==

<source lang="isar">
chapter {* R2: Razonamiento sobre programas en Isabelle/HOL *}

theory R2
imports Main
begin

declare [[names_short]]

text {* ---------------------------------------------------------------
Ejercicio 1.1. Definir la función
sumaImpares :: nat ⇒ nat
tal que (sumaImpares n) es la suma de los n primeros números
impares. Por ejemplo,
sumaImpares 5 = 25
------------------------------------------------------------------ *}

(*Notar: Por la propiedad de Gauss se puede deducir que:
la suma de los n numeros impares es igual a la suma de los
n y n-1 numeros consecutivos*)
fun suma :: "nat ⇒ nat" where
"suma 0 = 0"|
"suma n = n + suma(n-1)"

fun sumaImpares1 :: "nat ⇒ nat" where
"sumaImpares1 0 = 0" |
"sumaImpares1 n = suma(n) + suma(n-1)"

value "sumaImpares1 3 = 9 "

lemma aux1 : "suma(Suc n) + suma(n) = (n+1) + suma(n) + n + suma(n-1)"
apply (induct n)
apply auto
done

lemma aux2: "(n+1) + suma(n) + n + suma(n-1) = (n+1) + n + sumaImpares1 n "
apply (induct n)
apply auto
done

text {* ---------------------------------------------------------------
Ejercicio 1.2. Demostrar que
sumaImpares n = n*n
------------------------------------------------------------------- *}
text {* ---------------------------------------------------------------
Por que no funciona demostrar con el siguiente algoritmo by(induct n) auto
y se tiene que hacer una demostración detallada ???
------------------------------------------------------------------- *}

text {* ---------------------------------------------------------------
Creo que no funcionaba porque al usar una funcion auxiliar en tu definicion
de sumaImpares te queda por demostrar la propiedad de esa funcion. Es decir, si aplicas
induccion en "n+1" te quedaría por demostrar:
(n+1)+(suma n)+(suma n) = (n+1)(n+1)
(suma n) + (suma n) = n*n + n
Asi que lo he demostrado anteriormente y he usado ese teorema en la demostracion

~~~~
------------------------------------------------------------------- *}

(* ~~~~ *)
lemma sumAux: "(suma n) + (suma n) = n + n * n"
apply (induct n)
apply auto
done

(*----------------------ERROR ----------------------------------------*)
lemma "sumaImpares1 n = n*n"
apply (induct n)
apply auto
apply (simp add: sumAux) (* ~~~~ *)
done
(*--------------------------------------------------------------------*)

lemma "sumaImpares1 n = n*n"
proof (induct n)
show "sumaImpares1 0 = 0*0" by simp
next
fix n
assume HI:"sumaImpares1 n = n*n"
have "sumaImpares1 (Suc n) = suma(Suc n) + suma(n)" by simp
have "suma(Suc n) + suma(n) = (n+1) + suma(n) + n + suma(n-1)" using
"aux1" by simp
also have "... = (n+1) + n + sumaImpares1 n " using "aux2" by simp
also have "... = 2*n +1 + n*n" using HI by simp
also have "... = (n+1)*(n+1)" by simp
finally show "sumaImpares1 (Suc n) = Suc n * Suc n" by simp
qed

end
</source>

Relación 2

2016-11-05T16:28:55Z

Manmorjim1:

<source lang="isar">
chapter {* R2: Razonamiento sobre programas en Isabelle/HOL *}

theory R2_Razonamiento_automatico_sobre_programas
imports Main
begin

declare [[names_short]]

text {* ---------------------------------------------------------------
Ejercicio 1.1. Definir la función
sumaImpares :: nat ⇒ nat
tal que (sumaImpares n) es la suma de los n primeros números
impares. Por ejemplo,
sumaImpares 5 = 25
------------------------------------------------------------------ *}

(* ivamenjim wilmorort*)
fun sumaImpares :: "nat ⇒ nat" where
"sumaImpares 0 = 0"
| "sumaImpares n = 2*(n-1) + 1 + sumaImpares (n-1)"

value "sumaImpares 5 = 25"

(* wilmorort *)

(*Notar: Por la propiedad de Gauss se puede deducir que:
la suma de los n números impares es igual a la suma de los
n y n-1 números consecutivos*)
fun suma :: "nat ⇒ nat" where
"suma 0 = 0"
| "suma n = n + suma(n-1)"

fun sumaImpares1 :: "nat ⇒ nat" where
"sumaImpares1 0 = 0"
| "sumaImpares1 n = suma(n) + suma(n-1)"

value "sumaImpares1 5 = 25"

(* Comentario: La definición sumaImpares1 no es recursiva y, por tanto,
no es apropiada para demostraciones por inducción. *)

(* fraortmoy *)
fun sumaImpares2 :: "nat ⇒ nat" where
"sumaImpares2 0 = 0"
| "sumaImpares2 (Suc(n)) = ((Suc(n)*2)-1) + sumaImpares2 n"

value "sumaImpares2 5 = 25"

(* anaprarod danrodcha crigomgom migtermor*)
fun sumaImpares3 :: "nat ⇒ nat" where
"sumaImpares3 0 = 0"
|"sumaImpares3 (Suc n) = (2*n +1) + (sumaImpares3 n)"

value "sumaImpares3 5 = 25"

(* manmorjim1 *)
fun sumaImpares4 :: "nat ⇒ nat" where
"sumaImpares4 0 = 0"
| "sumaImpares4 (Suc n) = (Suc (2 * n)) + (sumaImpares4 n)"

value "sumaImpares4 5 = 25"

text {* ---------------------------------------------------------------
Ejercicio 1.2. Demostrar que
sumaImpares n = n*n
------------------------------------------------------------------- *}

(* ivamenjim wilmorort anaprarod crigomgom manmorjim1 *)
lemma "sumaImpares n = n*n"
apply (induct n)
apply auto
done

(* wilmorort *)
(* Demostración estructurada *)

lemma aux1 : "suma(Suc n) + suma(n) = (n+1) + suma(n) + n + suma(n-1)"
apply (induct n)
apply auto
done

lemma aux2: "(n+1) + suma(n) + n + suma(n-1) = (n+1) + n + sumaImpares1 n "
apply (induct n)
apply auto
done

lemma "sumaImpares1 n = n*n"
proof (induct n)
show "sumaImpares1 0 = 0*0" by simp
next
fix n
assume HI:"sumaImpares1 n = n*n"
have "sumaImpares1 (Suc n) = suma(Suc n) + suma(n)" by simp
have "suma(Suc n) + suma(n) = (n+1) + suma(n) + n + suma(n-1)" using
"aux1" by simp
also have "... = (n+1) + n + sumaImpares1 n " using "aux2" by simp
also have "... = 2*n +1 + n*n" using HI by simp
also have "... = (n+1)*(n+1)" by simp
finally show "sumaImpares1 (Suc n) = Suc n * Suc n" by simp
qed

(* fraortmoy *)
(* La demostración para la función sumaImpares2 es la misma que la de
sumaImpares*)

lemma "sumaImpares2 n = n*n"
apply (induct n)
apply auto
done

(* danrodcha migtermor*)
lemma "sumaImpares n = n*n"
by (induct n) auto

(* danrodcha *)
lemma "sumaImpares n = n*n" (is "?P n")
proof (induct n)
show "?P 0" by simp
next
fix n assume "?P n"
thus "?P (Suc n)" by simp
qed

text {* ---------------------------------------------------------------
Ejercicio 2.1. Definir la función
sumaPotenciasDeDosMasUno :: nat ⇒ nat
tal que
(sumaPotenciasDeDosMasUno n) = 1 + 2^0 + 2^1 + 2^2 + ... + 2^n.
Por ejemplo,
sumaPotenciasDeDosMasUno 3 = 16
------------------------------------------------------------------ *}

(* wilmorort ivamenjim *)
fun sumaPotenciasDeDosMasUno :: "nat ⇒ nat" where
"sumaPotenciasDeDosMasUno 0 = 2"
| "sumaPotenciasDeDosMasUno n = 2^n + sumaPotenciasDeDosMasUno (n-1)"

value "sumaPotenciasDeDosMasUno 3 = 16"

(* fraortmoy anaprarod danrodcha crigomgom migtermor*)
fun sumaPotenciasDeDosMasUno2 :: "nat ⇒ nat" where
"sumaPotenciasDeDosMasUno2 0 = 2"
| "sumaPotenciasDeDosMasUno2 (Suc(n)) =
(2^(Suc n)) + sumaPotenciasDeDosMasUno2 n"

value "sumaPotenciasDeDosMasUno2 3 = 16"

text {* ---------------------------------------------------------------
Ejercicio 2.2. Demostrar que
sumaPotenciasDeDosMasUno n = 2^(n+1)
------------------------------------------------------------------- *}

(* wilmorort ivamenjim fraortmoy anaprarod crigomgom*)
(* esta demostración funciona con sumaPotenciasDeDosMasUno2 *)

lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"
apply (induct n)
apply auto
done

(* danrodcha migtermor*)
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"
by (induct n) auto

(* danrodcha *)
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)" (is "?P n")
proof (induct n)
show "?P 0" by simp
next
fix n assume "?P n"
thus "?P (Suc n)"by simp
qed

text {* ---------------------------------------------------------------
Ejercicio 3.1. Definir la función
copia :: nat ⇒ 'a ⇒ 'a list
tal que (copia n x) es la lista formado por n copias del elemento
x. Por ejemplo,
copia 3 x = [x,x,x]
------------------------------------------------------------------ *}

(* ivamenjim wilmorort*)
fun copia :: "nat ⇒ 'a ⇒ 'a list" where
"copia 0 x = [] "
| "copia n x = x # (copia (n-1) x)"

(* wilmorort anaprarod *)
fun copia1 :: "nat ⇒ 'a ⇒ 'a list" where
"copia1 0 x = []"
| "copia1 (Suc n) x = [x] @ (copia1 n x) "

(* Comentario: La dedinición copia1 se puede simplificar eliminando @ *)

(* danrodcha crigomgom migtermor*)
fun copia2 :: "nat ⇒ 'a ⇒ 'a list" where
"copia2 0 x = []"
| "copia2 (Suc n) x = x # copia2 n x"

value "copia 3 x = [x,x,x]"

text {* ---------------------------------------------------------------
Ejercicio 3.2. Definir la función
todos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool
tal que (todos p xs) se verifica si todos los elementos de xs cumplen
la propiedad p. Por ejemplo,
todos (λx. x>(1::nat)) [2,6,4] = True
todos (λx. x>(2::nat)) [2,6,4] = False
Nota: La conjunción se representa por ∧
----------------------------------------------------------------- *}

(* wilmorort *)
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos p [] = True"
| "todos p (x#xs) = (p x = True ∧ todos p xs) "

(* Comentario: La definición todos se puede simplificar eliminando True
*)

value "todos (λx. x>(1::nat)) [2,6,4] = True"
value "todos (λx. x>(2::nat)) [2,6,4] = False"

(* fraortmoy anaprarod danrodcha crigomgom migtermor ivamenjim *)
fun todos1 :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos1 p [] = True"
| "todos1 p (x#xs) = (p x ∧ todos1 p xs )"

value "todos1 (λx. x>(1::nat)) [2,6,4] = True"
value "todos1 (λx. x>(2::nat)) [2,6,4] = False"

text {* ---------------------------------------------------------------
Ejercicio 3.3. Demostrar que todos los elementos de (copia n x) son
iguales a x.
------------------------------------------------------------------- *}
(*wilmorort fraortmoy anaprarod crigomgom ivamenjim *)

lemma "todos (λy. y=x) (copia n x)"
apply (induct n)
apply auto
done

(* danrodcha*)
lemma "todos (λy. y=x) (copia n x)"
by (induct n) auto

(* danrodcha *)
lemma "todos (λy. y=x) (copia n x)" (is "?P n")
proof (induct n)
show "?P 0" by simp
next
fix n assume "?P n"
thus "?P (Suc n)" by simp
qed

text {* ---------------------------------------------------------------
Ejercicio 4.1. Definir, recursivamente y sin usar (@), la función
amplia :: 'a list ⇒ 'a ⇒ 'a list
tal que (amplia xs y) es la lista obtenida añadiendo el elemento y al
final de la lista xs. Por ejemplo,
amplia [d,a] t = [d,a,t]
------------------------------------------------------------------ *}

(* wilmorort fraortmoy anaprarod crigomgom migtermor ivamenjim *)
fun amplia :: "'a list ⇒ 'a ⇒ 'a list" where
"amplia [] y = [y] "
| "amplia (x#xs) y = x # amplia xs y"

value "amplia [d,a] t = [d,a,t]"

text {* ---------------------------------------------------------------
Ejercicio 4.2. Demostrar que
amplia xs y = xs @ [y]
------------------------------------------------------------------- *}

(* wilmorort fraortmoy anaprarod crigomgom ivamenjim *)
lemma "amplia xs y = xs @ [y]"
apply (induct xs)
apply auto
done

(* danrodcha migtermor*)
lemma "amplia xs y = xs @ [y]"
by (induct xs) auto

(* danrodcha *)
lemma "amplia xs y = xs @ [y]" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs assume "?P xs"
thus "?P (a#xs)" by simp
qed

end
</source>

Relación 2

2016-11-05T16:20:22Z

Manmorjim1:

<source lang="isar">
chapter {* R2: Razonamiento sobre programas en Isabelle/HOL *}

theory R2_Razonamiento_automatico_sobre_programas
imports Main
begin

declare [[names_short]]

text {* ---------------------------------------------------------------
Ejercicio 1.1. Definir la función
sumaImpares :: nat ⇒ nat
tal que (sumaImpares n) es la suma de los n primeros números
impares. Por ejemplo,
sumaImpares 5 = 25
------------------------------------------------------------------ *}

(* ivamenjim wilmorort*)
fun sumaImpares :: "nat ⇒ nat" where
"sumaImpares 0 = 0"
| "sumaImpares n = 2*(n-1) + 1 + sumaImpares (n-1)"

value "sumaImpares 5 = 25"

(* wilmorort *)

(*Notar: Por la propiedad de Gauss se puede deducir que:
la suma de los n números impares es igual a la suma de los
n y n-1 números consecutivos*)
fun suma :: "nat ⇒ nat" where
"suma 0 = 0"
| "suma n = n + suma(n-1)"

fun sumaImpares1 :: "nat ⇒ nat" where
"sumaImpares1 0 = 0"
| "sumaImpares1 n = suma(n) + suma(n-1)"

value "sumaImpares1 5 = 25"

(* Comentario: La definición sumaImpares1 no es recursiva y, por tanto,
no es apropiada para demostraciones por inducción. *)

(* fraortmoy *)
fun sumaImpares2 :: "nat ⇒ nat" where
"sumaImpares2 0 = 0"
| "sumaImpares2 (Suc(n)) = ((Suc(n)*2)-1) + sumaImpares2 n"

value "sumaImpares2 5 = 25"

(* anaprarod danrodcha crigomgom migtermor*)
fun sumaImpares3 :: "nat ⇒ nat" where
"sumaImpares3 0 = 0"
|"sumaImpares3 (Suc n) = (2*n +1) + (sumaImpares3 n)"

value "sumaImpares3 5 = 25"

(* manmorjim1 *)
fun sumaImpares4 :: "nat ⇒ nat" where
"sumaImpares4 0 = 0"
| "sumaImpares4 (Suc n) = (Suc (2 * n)) + (sumaImpares4 n)"

value "sumaImpares4 5 = 25"

text {* ---------------------------------------------------------------
Ejercicio 1.2. Demostrar que
sumaImpares n = n*n
------------------------------------------------------------------- *}

(* ivamenjim wilmorort anaprarod crigomgom *)
lemma "sumaImpares n = n*n"
apply (induct n)
apply auto
done

(* wilmorort *)
(* Demostración estructurada *)

lemma aux1 : "suma(Suc n) + suma(n) = (n+1) + suma(n) + n + suma(n-1)"
apply (induct n)
apply auto
done

lemma aux2: "(n+1) + suma(n) + n + suma(n-1) = (n+1) + n + sumaImpares1 n "
apply (induct n)
apply auto
done

lemma "sumaImpares1 n = n*n"
proof (induct n)
show "sumaImpares1 0 = 0*0" by simp
next
fix n
assume HI:"sumaImpares1 n = n*n"
have "sumaImpares1 (Suc n) = suma(Suc n) + suma(n)" by simp
have "suma(Suc n) + suma(n) = (n+1) + suma(n) + n + suma(n-1)" using
"aux1" by simp
also have "... = (n+1) + n + sumaImpares1 n " using "aux2" by simp
also have "... = 2*n +1 + n*n" using HI by simp
also have "... = (n+1)*(n+1)" by simp
finally show "sumaImpares1 (Suc n) = Suc n * Suc n" by simp
qed

(* fraortmoy *)
(* La demostración para la función sumaImpares2 es la misma que la de
sumaImpares*)

lemma "sumaImpares2 n = n*n"
apply (induct n)
apply auto
done

(* danrodcha migtermor*)
lemma "sumaImpares n = n*n"
by (induct n) auto

(* danrodcha *)
lemma "sumaImpares n = n*n" (is "?P n")
proof (induct n)
show "?P 0" by simp
next
fix n assume "?P n"
thus "?P (Suc n)" by simp
qed

text {* ---------------------------------------------------------------
Ejercicio 2.1. Definir la función
sumaPotenciasDeDosMasUno :: nat ⇒ nat
tal que
(sumaPotenciasDeDosMasUno n) = 1 + 2^0 + 2^1 + 2^2 + ... + 2^n.
Por ejemplo,
sumaPotenciasDeDosMasUno 3 = 16
------------------------------------------------------------------ *}

(* wilmorort ivamenjim *)
fun sumaPotenciasDeDosMasUno :: "nat ⇒ nat" where
"sumaPotenciasDeDosMasUno 0 = 2"
| "sumaPotenciasDeDosMasUno n = 2^n + sumaPotenciasDeDosMasUno (n-1)"

value "sumaPotenciasDeDosMasUno 3 = 16"

(* fraortmoy anaprarod danrodcha crigomgom migtermor*)
fun sumaPotenciasDeDosMasUno2 :: "nat ⇒ nat" where
"sumaPotenciasDeDosMasUno2 0 = 2"
| "sumaPotenciasDeDosMasUno2 (Suc(n)) =
(2^(Suc n)) + sumaPotenciasDeDosMasUno2 n"

value "sumaPotenciasDeDosMasUno2 3 = 16"

text {* ---------------------------------------------------------------
Ejercicio 2.2. Demostrar que
sumaPotenciasDeDosMasUno n = 2^(n+1)
------------------------------------------------------------------- *}

(* wilmorort ivamenjim fraortmoy anaprarod crigomgom*)
(* esta demostración funciona con sumaPotenciasDeDosMasUno2 *)

lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"
apply (induct n)
apply auto
done

(* danrodcha migtermor*)
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"
by (induct n) auto

(* danrodcha *)
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)" (is "?P n")
proof (induct n)
show "?P 0" by simp
next
fix n assume "?P n"
thus "?P (Suc n)"by simp
qed

text {* ---------------------------------------------------------------
Ejercicio 3.1. Definir la función
copia :: nat ⇒ 'a ⇒ 'a list
tal que (copia n x) es la lista formado por n copias del elemento
x. Por ejemplo,
copia 3 x = [x,x,x]
------------------------------------------------------------------ *}

(* ivamenjim wilmorort*)
fun copia :: "nat ⇒ 'a ⇒ 'a list" where
"copia 0 x = [] "
| "copia n x = x # (copia (n-1) x)"

(* wilmorort anaprarod *)
fun copia1 :: "nat ⇒ 'a ⇒ 'a list" where
"copia1 0 x = []"
| "copia1 (Suc n) x = [x] @ (copia1 n x) "

(* Comentario: La dedinición copia1 se puede simplificar eliminando @ *)

(* danrodcha crigomgom migtermor*)
fun copia2 :: "nat ⇒ 'a ⇒ 'a list" where
"copia2 0 x = []"
| "copia2 (Suc n) x = x # copia2 n x"

value "copia 3 x = [x,x,x]"

text {* ---------------------------------------------------------------
Ejercicio 3.2. Definir la función
todos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool
tal que (todos p xs) se verifica si todos los elementos de xs cumplen
la propiedad p. Por ejemplo,
todos (λx. x>(1::nat)) [2,6,4] = True
todos (λx. x>(2::nat)) [2,6,4] = False
Nota: La conjunción se representa por ∧
----------------------------------------------------------------- *}

(* wilmorort *)
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos p [] = True"
| "todos p (x#xs) = (p x = True ∧ todos p xs) "

(* Comentario: La definición todos se puede simplificar eliminando True
*)

value "todos (λx. x>(1::nat)) [2,6,4] = True"
value "todos (λx. x>(2::nat)) [2,6,4] = False"

(* fraortmoy anaprarod danrodcha crigomgom migtermor ivamenjim *)
fun todos1 :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos1 p [] = True"
| "todos1 p (x#xs) = (p x ∧ todos1 p xs )"

value "todos1 (λx. x>(1::nat)) [2,6,4] = True"
value "todos1 (λx. x>(2::nat)) [2,6,4] = False"

text {* ---------------------------------------------------------------
Ejercicio 3.3. Demostrar que todos los elementos de (copia n x) son
iguales a x.
------------------------------------------------------------------- *}
(*wilmorort fraortmoy anaprarod crigomgom ivamenjim *)

lemma "todos (λy. y=x) (copia n x)"
apply (induct n)
apply auto
done

(* danrodcha*)
lemma "todos (λy. y=x) (copia n x)"
by (induct n) auto

(* danrodcha *)
lemma "todos (λy. y=x) (copia n x)" (is "?P n")
proof (induct n)
show "?P 0" by simp
next
fix n assume "?P n"
thus "?P (Suc n)" by simp
qed

text {* ---------------------------------------------------------------
Ejercicio 4.1. Definir, recursivamente y sin usar (@), la función
amplia :: 'a list ⇒ 'a ⇒ 'a list
tal que (amplia xs y) es la lista obtenida añadiendo el elemento y al
final de la lista xs. Por ejemplo,
amplia [d,a] t = [d,a,t]
------------------------------------------------------------------ *}

(* wilmorort fraortmoy anaprarod crigomgom migtermor ivamenjim *)
fun amplia :: "'a list ⇒ 'a ⇒ 'a list" where
"amplia [] y = [y] "
| "amplia (x#xs) y = x # amplia xs y"

value "amplia [d,a] t = [d,a,t]"

text {* ---------------------------------------------------------------
Ejercicio 4.2. Demostrar que
amplia xs y = xs @ [y]
------------------------------------------------------------------- *}

(* wilmorort fraortmoy anaprarod crigomgom ivamenjim *)
lemma "amplia xs y = xs @ [y]"
apply (induct xs)
apply auto
done

(* danrodcha migtermor*)
lemma "amplia xs y = xs @ [y]"
by (induct xs) auto

(* danrodcha *)
lemma "amplia xs y = xs @ [y]" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs assume "?P xs"
thus "?P (a#xs)" by simp
qed

end
</source>

Relación 1

2016-10-30T15:43:34Z

Manmorjim1:

<source lang="isar">
chapter {* R1: Programación funcional en Isabelle *}

theory R1
imports Main
begin

text {* ----------------------------------------------------------------
Ejercicio 0. Definir, por recursión, la función
factorial :: nat ⇒ nat
tal que (factorial n) es el factorial de n. Por ejemplo,
factorial 4 = 24
------------------------------------------------------------------- *}

(* danrodcha, anaprarod, ivamenjim, serrodcal, manmorjim1 *)
fun factorial :: "nat ⇒ nat" where
"factorial 0 = 1"
| "factorial (Suc n) = (Suc n) * factorial n"

(*wilmorort, pablucoto,marcarmor13, crigomgom, rubgonmar*)
fun factorial1 :: "nat ⇒ nat" where
"factorial1 0 = 1 "
| "factorial1 n = n * factorial1(n-1)"

value "factorial 4" -- "24"

text {* ----------------------------------------------------------------
Ejercicio 1. Definir, por recursión, la función
longitud :: 'a list ⇒ nat
tal que (longitud xs) es la longitud de la listas xs. Por ejemplo,
longitud [4,2,5] = 3
------------------------------------------------------------------- *}

(*wilmorort*)
(* Para usar las lista en forma de [a,b,c] *)

translations
"[x, xs]" == "x#[xs]"
"[x]" == "x#[]"

(*wilmorort, serrodcal,crigomgom,rubgonmar*)
fun longitud :: "'a list ⇒ nat" where
"longitud [] = 0" |
"longitud (x # xs) = 1 + longitud xs "

(*pablucoto*)
fun longitud0 :: "'a list ⇒ nat " where
" longitud0 [] = 0"
|"longitud0 xs = 1 + longitud0 ((butlast xs)) "

fun longitud0_1 :: "'a list ⇒ nat " where
"longitud0_1 xs = (if xs =[] then 0 else 1 + longitud0_1((butlast xs))) "

(*marcarmor13, manmorjim1*)
fun longitud1 :: "'a list ⇒ nat" where
"longitud1 [] = 0 "
| "longitud1 xs = (1+ longitud2 (tl xs))"

(*danrodcha*)
fun longitud2 :: "'a list ⇒ nat" where
"longitud2 [] = 0"
| "longitud2 (x#xs) = Suc (longitud2 xs)"

(*serrodcal*)
fun longitud3 :: "'a list ⇒ nat" where
"longitud xs = length (xs)"

value "longitud [4,2,5] " -- "= 3"

text {* ---------------------------------------------------------------
Ejercicio 2. Definir la función
fun intercambia :: 'a × 'b ⇒ 'b × 'a
tal que (intercambia p) es el par obtenido intercambiando las
componentes del par p. Por ejemplo,
intercambia (u,v) = (v,u)
------------------------------------------------------------------ *}

(*wilmorort,marcarmor13,danrodcha,crigomgom,pablucoto, rubgonmar, manmorjim1*)
fun intercambia :: "'a × 'b ⇒ 'b × 'a" where
"intercambia (x,y) = (y,x) "

value "intercambia (u,v)"-- "= (v,u)"

text {* ---------------------------------------------------------------
Ejercicio 3. Definir, por recursión, la función
inversa :: 'a list ⇒ 'a list
tal que (inversa xs) es la lista obtenida invirtiendo el orden de los
elementos de xs. Por ejemplo,
inversa [a,d,c] = [c,d,a]
------------------------------------------------------------------ *}

(*wilmorort*)
(* @ :: "'a list => 'a list => 'a list", función agregación definida
en Theory Main, concatena dos listas: [a,b] @ [c,d] = [a,b,c,d] *)
fun inversa :: "'a list ⇒ 'a list" where
"inversa [] = []" |
"inversa (x # xs) = (inversa xs)@(x#[]) "

(*marcarmor13, manmorjim1*)
fun inversa1 :: "'a list ⇒ 'a list" where
"inversa1 [] = []"
| "inversa1 xs = inversa1 (tl xs)@ ((hd xs)#[])"

(*danrodcha, pablucoto*)
(* es igual que inversa sustituyendo x#[] por [x] *)
fun inversa2 :: "'a list ⇒ 'a list" where
"inversa2 [] = []"
| "inversa2 (x#xs) = (inversa2 xs)@[x] "

(*danrodcha*)
fun inversa3 :: "'a list ⇒ 'a list" where
"inversa3 [] = []"
| "inversa3 (x#xs) = concat [(inversa3 xs),[x]] "

(*crigomgom*)
fun inversa4 :: "'a list ⇒ 'a list" where
"inversa4 [] = []" |
"inversa4 xs = (last xs)#(inversa4(butlast xs)) "

value "inversa [a,d,c]" -- "= [c,d,a]"

text {* ---------------------------------------------------------------
Ejercicio 4. Definir la función
repite :: nat ⇒ 'a ⇒ 'a list
tal que (repite n x) es la lista formada por n copias del elemento
x. Por ejemplo,
repite 3 a = [a,a,a]
------------------------------------------------------------------ *}

(*wilmorort,marcarmor13, crigomgom, pablucoto, rubgonmar, manmorjim1*)
fun repite :: "nat ⇒ 'a ⇒ 'a list" where
"repite 0 x = [] " |
"repite n x = x # (repite(n-1) x) "

(*danrodcha*)
fun repite1 :: "nat ⇒ 'a ⇒ 'a list" where
"repite1 0 x = []"
| "repite1 (Suc n) x = x#(repite1 n x)"

value "repite 3 a" -- "= [a,a,a]"

text {* ---------------------------------------------------------------
Ejercicio 5. Definir la función
conc :: 'a list ⇒ 'a list ⇒ 'a list
tal que (conc xs ys) es la concatención de las listas xs e ys. Por
ejemplo,
conc [a,d] [b,d,a,c] = [a,d,b,d,a,c]
------------------------------------------------------------------ *}

(*marcarmor13*)
fun conc :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc xs ys = xs@ys"

(*danrodcha, crigomgom, rubgonmar*)
fun conc1 :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc1 [] ys = ys"
| "conc1 xs [] = xs" (*esta no hace falta*)
| "conc1 (x#xs) ys = x#(conc1 xs ys)"

(*pablucoto*)
fun conc2 :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc2 [] ys = ys" |
"conc2 (x#xs) ys = x # (conc2 xs ys)"

(* manmorjim1 *)
fun conc3 :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc3 [] ys = ys"
| "conc3 xs ys = (hd xs)#[] @ (conc3 (tl xs) ys)"

value "conc [a,d] [b,d,a,c]" -- "= [a,d,b,d,a,c]"

text {* ---------------------------------------------------------------
Ejercicio 6. Definir la función
coge :: nat ⇒ 'a list ⇒ 'a list
tal que (coge n xs) es la lista de los n primeros elementos de xs. Por
ejemplo,
coge 2 [a,c,d,b,e] = [a,c]
------------------------------------------------------------------ *}
(*marcarmor13, manmorjim1*)
fun coge :: "nat ⇒ 'a list ⇒ 'a list" where
"coge 0 xs = []"
| "coge n xs = (hd xs)#(coge (n-1) (tl xs)) "

(*danrodcha*)
fun coge1 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge1 0 _ = []"
| "coge1 _ [] = []"
| "coge1 (Suc n) (x#xs) = x#(coge1 n xs)"

(*crimgomgom*)
fun coge2 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge2 0 _ = []"|
"coge2 _ [] = []" |
"coge2 n (x#xs) = x#(coge2 (n-1) xs) "

(*pablucoto*)
fun coge3 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge3 0 xs = []" |
"coge3 n (x#xs) = (if n>length(x#xs) then (x#xs) else x # (coge3 (n-1) xs))"

value "coge 2 [a,c,d,b,e]" -- "= [a,c]"

text {* ---------------------------------------------------------------
Ejercicio 7. Definir la función
elimina :: nat ⇒ 'a list ⇒ 'a list
tal que (elimina n xs) es la lista obtenida eliminando los n primeros
elementos de xs. Por ejemplo,
elimina 2 [a,c,d,b,e] = [d,b,e]
------------------------------------------------------------------ *}
(*marcarmor13, manmorjim1*)
fun elimina :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina 0 xs = xs"
| "elimina n xs = (elimina (n-1) (tl xs ))"

(*danrodcha, crigomgom*)
fun elimina1 :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina1 0 xs = xs"
| "elimina1 _ [] = []"
| "elimina1 (Suc n) (x#xs) = elimina1 n xs"

(*pablucoto*)
fun elimina2 :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina2 0 xs = xs" |
"elimina2 n (x#xs) = (if n>length(x#xs) then [] else (elimina2 (n-1) xs))"

value "elimina 2 [a,c,d,b,e]" -- "= [d,b,e]"

text {* ---------------------------------------------------------------
Ejercicio 8. Definir la función
esVacia :: 'a list ⇒ bool
tal que (esVacia xs) se verifica si xs es la lista vacía. Por ejemplo,
esVacia [] = True
esVacia [1] = False
------------------------------------------------------------------ *}

(*marcarmor13, rubgonmar, danrodcha, crigomgom*)
fun esVacia :: "'a list ⇒ bool" where
"esVacia [] = True "
| "esVacia xs = False"

(*danrodcha*)
fun esVacia1 :: "'a list ⇒ bool" where
"esVacia1 xs = (xs = [])"

(*pablucoto*)
fun esVacia2 :: "'a list ⇒ bool" where
"esVacia2 xs = (if xs=[] then True else False)"

(*manmorjim1*)
fun esVacia3 :: "'a list ⇒ bool" where
"esVacia3 xs = (length xs = 0)"

value "esVacia []" -- "= True"
value "esVacia [1]" -- "= False"

text {* ---------------------------------------------------------------
Ejercicio 9. Definir la función
inversaAc :: 'a list ⇒ 'a list
tal que (inversaAc xs) es a inversa de xs calculada usando
acumuladores. Por ejemplo,
inversaAc [a,c,b,e] = [e,b,c,a]
------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13, mamnorjim1*)
fun inversaAcAux :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux [] ys = ys"
| "inversaAcAux xs ys = inversaAcAux (tl xs) (hd xs#ys) "

(*danrodcha, crigomgom*)
fun inversaAcAux1 :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux1 [] ys = ys"
| "inversaAcAux1 (x#xs) ys = inversaAcAux1 xs (x#ys)"

(*rubgonmar, danrodcha, crigomgom, manmorjim1*)
fun inversaAc :: "'a list ⇒ 'a list" where
"inversaAc xs = inversaAcAux xs []"

(*pablucoto*)
fun inversaAcAux2 :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux2 [] [] = []"|
"inversaAcAux2 xs (y#ys) = (inversaAcAux2 [] ys) @ [y]"
fun inversaAc2 :: "'a list ⇒ 'a list" where
"inversaAc2 xs = inversaAcAux2 [] xs "

value "inversaAc [a,c,b,e]" -- "= [e,b,c,a]"

text {* ---------------------------------------------------------------
Ejercicio 10. Definir la función
sum :: nat list ⇒ nat
tal que (sum xs) es la suma de los elementos de xs. Por ejemplo,
sum [3,2,5] = 10
------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13, manmorjim1*)
fun sum :: "nat list ⇒ nat" where
"sum [] = 0"
|"sum xs = hd xs + sum (tl xs)"

(*danrodcha, crigomgom, pablucoto*)
fun sum1 :: "nat list ⇒ nat" where
"sum1 [] = 0"
| "sum1 (x#xs) = x + sum1 xs"

(*danrodcha*)
fun sum2 :: "nat list ⇒ nat" where
"sum2 xs = fold (op +) xs 0"

value "sum [3,2,5]" -- "= 10"

text {* ---------------------------------------------------------------
Ejercicio 11. Definir la función
map :: ('a ⇒ 'b) ⇒ 'a list ⇒ 'b list
tal que (map f xs) es la lista obtenida aplicando la función f a los
elementos de xs. Por ejemplo,
map (λx. 2*x) [3,2,5] = [6,4,10]
------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13,manmorjim1*)
fun map :: "('a ⇒ 'b) ⇒ 'a list ⇒ 'b list" where
"map f [] = []"
| "map f xs = f(hd xs)#map f (tl xs)"

(*wilmorort, danrodcha, crigomgom, pablucoto*)
fun map :: "('a ⇒ 'b) ⇒ 'a list ⇒ 'b list" where
"map f [] = []"
|"map f (x # xs) = f x # map f xs" (*yo pondría paréntesis, pero sin
ellos lo entiende*)

value "map (λx. 2*x) [3::nat,2,5]" -- "= [6,4,10]"

end
</source>

Relación 1

2016-10-30T15:40:05Z

Manmorjim1:

<source lang="isar">
chapter {* R1: Programación funcional en Isabelle *}

theory R1
imports Main
begin

text {* ----------------------------------------------------------------
Ejercicio 0. Definir, por recursión, la función
factorial :: nat ⇒ nat
tal que (factorial n) es el factorial de n. Por ejemplo,
factorial 4 = 24
------------------------------------------------------------------- *}

(* danrodcha, anaprarod, ivamenjim, serrodcal, manmorjim1 *)
fun factorial :: "nat ⇒ nat" where
"factorial 0 = 1"
| "factorial (Suc n) = (Suc n) * factorial n"

(*wilmorort, pablucoto,marcarmor13, crigomgom, rubgonmar*)
fun factorial1 :: "nat ⇒ nat" where
"factorial1 0 = 1 "
| "factorial1 n = n * factorial1(n-1)"

value "factorial 4" -- "24"

text {* ----------------------------------------------------------------
Ejercicio 1. Definir, por recursión, la función
longitud :: 'a list ⇒ nat
tal que (longitud xs) es la longitud de la listas xs. Por ejemplo,
longitud [4,2,5] = 3
------------------------------------------------------------------- *}

(*wilmorort*)
(* Para usar las lista en forma de [a,b,c] *)

translations
"[x, xs]" == "x#[xs]"
"[x]" == "x#[]"

(*wilmorort, serrodcal,crigomgom,rubgonmar*)
fun longitud :: "'a list ⇒ nat" where
"longitud [] = 0" |
"longitud (x # xs) = 1 + longitud xs "

(*pablucoto*)
fun longitud0 :: "'a list ⇒ nat " where
" longitud0 [] = 0"
|"longitud0 xs = 1 + longitud0 ((butlast xs)) "

fun longitud0_1 :: "'a list ⇒ nat " where
"longitud0_1 xs = (if xs =[] then 0 else 1 + longitud0_1((butlast xs))) "

(*marcarmor13, manmorjim1*)
fun longitud1 :: "'a list ⇒ nat" where
"longitud1 [] = 0 "
| "longitud1 xs = (1+ longitud2 (tl xs))"

(*danrodcha*)
fun longitud2 :: "'a list ⇒ nat" where
"longitud2 [] = 0"
| "longitud2 (x#xs) = Suc (longitud2 xs)"

(*serrodcal*)
fun longitud3 :: "'a list ⇒ nat" where
"longitud xs = length (xs)"

value "longitud [4,2,5] " -- "= 3"

text {* ---------------------------------------------------------------
Ejercicio 2. Definir la función
fun intercambia :: 'a × 'b ⇒ 'b × 'a
tal que (intercambia p) es el par obtenido intercambiando las
componentes del par p. Por ejemplo,
intercambia (u,v) = (v,u)
------------------------------------------------------------------ *}

(*wilmorort,marcarmor13,danrodcha,crigomgom,pablucoto, rubgonmar, manmorjim1*)
fun intercambia :: "'a × 'b ⇒ 'b × 'a" where
"intercambia (x,y) = (y,x) "

value "intercambia (u,v)"-- "= (v,u)"

text {* ---------------------------------------------------------------
Ejercicio 3. Definir, por recursión, la función
inversa :: 'a list ⇒ 'a list
tal que (inversa xs) es la lista obtenida invirtiendo el orden de los
elementos de xs. Por ejemplo,
inversa [a,d,c] = [c,d,a]
------------------------------------------------------------------ *}

(*wilmorort*)
(* @ :: "'a list => 'a list => 'a list", función agregación definida
en Theory Main, concatena dos listas: [a,b] @ [c,d] = [a,b,c,d] *)
fun inversa :: "'a list ⇒ 'a list" where
"inversa [] = []" |
"inversa (x # xs) = (inversa xs)@(x#[]) "

(*marcarmor13, manmorjim1*)
fun inversa1 :: "'a list ⇒ 'a list" where
"inversa1 [] = []"
| "inversa1 xs = inversa1 (tl xs)@ ((hd xs)#[])"

(*danrodcha, pablucoto*)
(* es igual que inversa sustituyendo x#[] por [x] *)
fun inversa2 :: "'a list ⇒ 'a list" where
"inversa2 [] = []"
| "inversa2 (x#xs) = (inversa2 xs)@[x] "

(*danrodcha*)
fun inversa3 :: "'a list ⇒ 'a list" where
"inversa3 [] = []"
| "inversa3 (x#xs) = concat [(inversa3 xs),[x]] "

(*crigomgom*)
fun inversa4 :: "'a list ⇒ 'a list" where
"inversa4 [] = []" |
"inversa4 xs = (last xs)#(inversa4(butlast xs)) "

value "inversa [a,d,c]" -- "= [c,d,a]"

text {* ---------------------------------------------------------------
Ejercicio 4. Definir la función
repite :: nat ⇒ 'a ⇒ 'a list
tal que (repite n x) es la lista formada por n copias del elemento
x. Por ejemplo,
repite 3 a = [a,a,a]
------------------------------------------------------------------ *}

(*wilmorort,marcarmor13, crigomgom, pablucoto, rubgonmar, manmorjim1*)
fun repite :: "nat ⇒ 'a ⇒ 'a list" where
"repite 0 x = [] " |
"repite n x = x # (repite(n-1) x) "

(*danrodcha*)
fun repite1 :: "nat ⇒ 'a ⇒ 'a list" where
"repite1 0 x = []"
| "repite1 (Suc n) x = x#(repite1 n x)"

value "repite 3 a" -- "= [a,a,a]"

text {* ---------------------------------------------------------------
Ejercicio 5. Definir la función
conc :: 'a list ⇒ 'a list ⇒ 'a list
tal que (conc xs ys) es la concatención de las listas xs e ys. Por
ejemplo,
conc [a,d] [b,d,a,c] = [a,d,b,d,a,c]
------------------------------------------------------------------ *}

(*marcarmor13*)
fun conc :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc xs ys = xs@ys"

(*danrodcha, crigomgom, rubgonmar*)
fun conc1 :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc1 [] ys = ys"
| "conc1 xs [] = xs" (*esta no hace falta*)
| "conc1 (x#xs) ys = x#(conc1 xs ys)"

(*pablucoto*)
fun conc2 :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc2 [] ys = ys" |
"conc2 (x#xs) ys = x # (conc2 xs ys)"

(* manmorjim1 *)
fun conc3 :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc3 [] ys = ys"
| "conc3 xs ys = (hd xs)#[] @ (conc3 (tl xs) ys)"

value "conc [a,d] [b,d,a,c]" -- "= [a,d,b,d,a,c]"

text {* ---------------------------------------------------------------
Ejercicio 6. Definir la función
coge :: nat ⇒ 'a list ⇒ 'a list
tal que (coge n xs) es la lista de los n primeros elementos de xs. Por
ejemplo,
coge 2 [a,c,d,b,e] = [a,c]
------------------------------------------------------------------ *}
(*marcarmor13, manmorjim1*)
fun coge :: "nat ⇒ 'a list ⇒ 'a list" where
"coge 0 xs = []"
| "coge n xs = (hd xs)#(coge (n-1) (tl xs)) "

(*danrodcha*)
fun coge1 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge1 0 _ = []"
| "coge1 _ [] = []"
| "coge1 (Suc n) (x#xs) = x#(coge1 n xs)"

(*crimgomgom*)
fun coge2 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge2 0 _ = []"|
"coge2 _ [] = []" |
"coge2 n (x#xs) = x#(coge2 (n-1) xs) "

(*pablucoto*)
fun coge3 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge3 0 xs = []" |
"coge3 n (x#xs) = (if n>length(x#xs) then (x#xs) else x # (coge3 (n-1) xs))"

value "coge 2 [a,c,d,b,e]" -- "= [a,c]"

text {* ---------------------------------------------------------------
Ejercicio 7. Definir la función
elimina :: nat ⇒ 'a list ⇒ 'a list
tal que (elimina n xs) es la lista obtenida eliminando los n primeros
elementos de xs. Por ejemplo,
elimina 2 [a,c,d,b,e] = [d,b,e]
------------------------------------------------------------------ *}
(*marcarmor13, manmorjim1*)
fun elimina :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina 0 xs = xs"
| "elimina n xs = (elimina (n-1) (tl xs ))"

(*danrodcha, crigomgom*)
fun elimina1 :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina1 0 xs = xs"
| "elimina1 _ [] = []"
| "elimina1 (Suc n) (x#xs) = elimina1 n xs"

(*pablucoto*)
fun elimina2 :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina2 0 xs = xs" |
"elimina2 n (x#xs) = (if n>length(x#xs) then [] else (elimina2 (n-1) xs))"

value "elimina 2 [a,c,d,b,e]" -- "= [d,b,e]"

text {* ---------------------------------------------------------------
Ejercicio 8. Definir la función
esVacia :: 'a list ⇒ bool
tal que (esVacia xs) se verifica si xs es la lista vacía. Por ejemplo,
esVacia [] = True
esVacia [1] = False
------------------------------------------------------------------ *}

(*marcarmor13, rubgonmar, danrodcha, crigomgom*)
fun esVacia :: "'a list ⇒ bool" where
"esVacia [] = True "
| "esVacia xs = False"

(*danrodcha*)
fun esVacia1 :: "'a list ⇒ bool" where
"esVacia1 xs = (xs = [])"

(*pablucoto*)
fun esVacia2 :: "'a list ⇒ bool" where
"esVacia2 xs = (if xs=[] then True else False)"

(*manmorjim1*)
fun esVacia3 :: "'a list ⇒ bool" where
"esVacia3 xs = (length xs = 0)"

value "esVacia []" -- "= True"
value "esVacia [1]" -- "= False"

text {* ---------------------------------------------------------------
Ejercicio 9. Definir la función
inversaAc :: 'a list ⇒ 'a list
tal que (inversaAc xs) es a inversa de xs calculada usando
acumuladores. Por ejemplo,
inversaAc [a,c,b,e] = [e,b,c,a]
------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13, mamnorjim1*)
fun inversaAcAux :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux [] ys = ys"
| "inversaAcAux xs ys = inversaAcAux (tl xs) (hd xs#ys) "

(*danrodcha, crigomgom*)
fun inversaAcAux1 :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux1 [] ys = ys"
| "inversaAcAux1 (x#xs) ys = inversaAcAux1 xs (x#ys)"

(*rubgonmar, danrodcha, crigomgom, manmorjim1*)
fun inversaAc :: "'a list ⇒ 'a list" where
"inversaAc xs = inversaAcAux xs []"

(*pablucoto*)
fun inversaAcAux2 :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux2 [] [] = []"|
"inversaAcAux2 xs (y#ys) = (inversaAcAux2 [] ys) @ [y]"
fun inversaAc2 :: "'a list ⇒ 'a list" where
"inversaAc2 xs = inversaAcAux2 [] xs "

value "inversaAc [a,c,b,e]" -- "= [e,b,c,a]"

text {* ---------------------------------------------------------------
Ejercicio 10. Definir la función
sum :: nat list ⇒ nat
tal que (sum xs) es la suma de los elementos de xs. Por ejemplo,
sum [3,2,5] = 10
------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13, manmorjim1*)
fun sum :: "nat list ⇒ nat" where
"sum [] = 0"
|"sum xs = hd xs + sum (tl xs)"

(*danrodcha, crigomgom, pablucoto*)
fun sum1 :: "nat list ⇒ nat" where
"sum1 [] = 0"
| "sum1 (x#xs) = x + sum1 xs"

(*danrodcha*)
fun sum2 :: "nat list ⇒ nat" where
"sum2 xs = fold (op +) xs 0"

value "sum [3,2,5]" -- "= 10"

text {* ---------------------------------------------------------------
Ejercicio 11. Definir la función
map :: ('a ⇒ 'b) ⇒ 'a list ⇒ 'b list
tal que (map f xs) es la lista obtenida aplicando la función f a los
elementos de xs. Por ejemplo,
map (λx. 2*x) [3,2,5] = [6,4,10]
------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13*)
fun map :: "('a ⇒ 'b) ⇒ 'a list ⇒ 'b list" where
"map f [] = []"
| "map f xs = f(hd xs)#map f (tl xs)"

(*wilmorort, danrodcha, crigomgom, pablucoto*)
fun map :: "('a ⇒ 'b) ⇒ 'a list ⇒ 'b list" where
"map f [] = []"
|"map f (x # xs) = f x # map f xs" (*yo pondría paréntesis, pero sin
ellos lo entiende*)

value "map (λx. 2*x) [3::nat,2,5]" -- "= [6,4,10]"

end
</source>

Relación 1

2016-10-30T15:37:28Z

Manmorjim1:

<source lang="isar">
chapter {* R1: Programación funcional en Isabelle *}

theory R1
imports Main
begin

text {* ----------------------------------------------------------------
Ejercicio 0. Definir, por recursión, la función
factorial :: nat ⇒ nat
tal que (factorial n) es el factorial de n. Por ejemplo,
factorial 4 = 24
------------------------------------------------------------------- *}

(* danrodcha, anaprarod, ivamenjim, serrodcal, manmorjim1 *)
fun factorial :: "nat ⇒ nat" where
"factorial 0 = 1"
| "factorial (Suc n) = (Suc n) * factorial n"

(*wilmorort, pablucoto,marcarmor13, crigomgom, rubgonmar*)
fun factorial1 :: "nat ⇒ nat" where
"factorial1 0 = 1 "
| "factorial1 n = n * factorial1(n-1)"

value "factorial 4" -- "24"

text {* ----------------------------------------------------------------
Ejercicio 1. Definir, por recursión, la función
longitud :: 'a list ⇒ nat
tal que (longitud xs) es la longitud de la listas xs. Por ejemplo,
longitud [4,2,5] = 3
------------------------------------------------------------------- *}

(*wilmorort*)
(* Para usar las lista en forma de [a,b,c] *)

translations
"[x, xs]" == "x#[xs]"
"[x]" == "x#[]"

(*wilmorort, serrodcal,crigomgom,rubgonmar*)
fun longitud :: "'a list ⇒ nat" where
"longitud [] = 0" |
"longitud (x # xs) = 1 + longitud xs "

(*pablucoto*)
fun longitud0 :: "'a list ⇒ nat " where
" longitud0 [] = 0"
|"longitud0 xs = 1 + longitud0 ((butlast xs)) "

fun longitud0_1 :: "'a list ⇒ nat " where
"longitud0_1 xs = (if xs =[] then 0 else 1 + longitud0_1((butlast xs))) "

(*marcarmor13, manmorjim1*)
fun longitud1 :: "'a list ⇒ nat" where
"longitud1 [] = 0 "
| "longitud1 xs = (1+ longitud2 (tl xs))"

(*danrodcha*)
fun longitud2 :: "'a list ⇒ nat" where
"longitud2 [] = 0"
| "longitud2 (x#xs) = Suc (longitud2 xs)"

(*serrodcal*)
fun longitud3 :: "'a list ⇒ nat" where
"longitud xs = length (xs)"

value "longitud [4,2,5] " -- "= 3"

text {* ---------------------------------------------------------------
Ejercicio 2. Definir la función
fun intercambia :: 'a × 'b ⇒ 'b × 'a
tal que (intercambia p) es el par obtenido intercambiando las
componentes del par p. Por ejemplo,
intercambia (u,v) = (v,u)
------------------------------------------------------------------ *}

(*wilmorort,marcarmor13,danrodcha,crigomgom,pablucoto, rubgonmar, manmorjim1*)
fun intercambia :: "'a × 'b ⇒ 'b × 'a" where
"intercambia (x,y) = (y,x) "

value "intercambia (u,v)"-- "= (v,u)"

text {* ---------------------------------------------------------------
Ejercicio 3. Definir, por recursión, la función
inversa :: 'a list ⇒ 'a list
tal que (inversa xs) es la lista obtenida invirtiendo el orden de los
elementos de xs. Por ejemplo,
inversa [a,d,c] = [c,d,a]
------------------------------------------------------------------ *}

(*wilmorort*)
(* @ :: "'a list => 'a list => 'a list", función agregación definida
en Theory Main, concatena dos listas: [a,b] @ [c,d] = [a,b,c,d] *)
fun inversa :: "'a list ⇒ 'a list" where
"inversa [] = []" |
"inversa (x # xs) = (inversa xs)@(x#[]) "

(*marcarmor13, manmorjim1*)
fun inversa1 :: "'a list ⇒ 'a list" where
"inversa1 [] = []"
| "inversa1 xs = inversa1 (tl xs)@ ((hd xs)#[])"

(*danrodcha, pablucoto*)
(* es igual que inversa sustituyendo x#[] por [x] *)
fun inversa2 :: "'a list ⇒ 'a list" where
"inversa2 [] = []"
| "inversa2 (x#xs) = (inversa2 xs)@[x] "

(*danrodcha*)
fun inversa3 :: "'a list ⇒ 'a list" where
"inversa3 [] = []"
| "inversa3 (x#xs) = concat [(inversa3 xs),[x]] "

(*crigomgom*)
fun inversa4 :: "'a list ⇒ 'a list" where
"inversa4 [] = []" |
"inversa4 xs = (last xs)#(inversa4(butlast xs)) "

value "inversa [a,d,c]" -- "= [c,d,a]"

text {* ---------------------------------------------------------------
Ejercicio 4. Definir la función
repite :: nat ⇒ 'a ⇒ 'a list
tal que (repite n x) es la lista formada por n copias del elemento
x. Por ejemplo,
repite 3 a = [a,a,a]
------------------------------------------------------------------ *}

(*wilmorort,marcarmor13, crigomgom, pablucoto, rubgonmar, manmorjim1*)
fun repite :: "nat ⇒ 'a ⇒ 'a list" where
"repite 0 x = [] " |
"repite n x = x # (repite(n-1) x) "

(*danrodcha*)
fun repite1 :: "nat ⇒ 'a ⇒ 'a list" where
"repite1 0 x = []"
| "repite1 (Suc n) x = x#(repite1 n x)"

value "repite 3 a" -- "= [a,a,a]"

text {* ---------------------------------------------------------------
Ejercicio 5. Definir la función
conc :: 'a list ⇒ 'a list ⇒ 'a list
tal que (conc xs ys) es la concatención de las listas xs e ys. Por
ejemplo,
conc [a,d] [b,d,a,c] = [a,d,b,d,a,c]
------------------------------------------------------------------ *}

(*marcarmor13*)
fun conc :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc xs ys = xs@ys"

(*danrodcha, crigomgom, rubgonmar*)
fun conc1 :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc1 [] ys = ys"
| "conc1 xs [] = xs" (*esta no hace falta*)
| "conc1 (x#xs) ys = x#(conc1 xs ys)"

(*pablucoto*)
fun conc2 :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc2 [] ys = ys" |
"conc2 (x#xs) ys = x # (conc2 xs ys)"

(* manmorjim1 *)
fun conc3 :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc3 [] ys = ys"
| "conc3 xs ys = (hd xs)#[] @ (conc3 (tl xs) ys)"

value "conc [a,d] [b,d,a,c]" -- "= [a,d,b,d,a,c]"

text {* ---------------------------------------------------------------
Ejercicio 6. Definir la función
coge :: nat ⇒ 'a list ⇒ 'a list
tal que (coge n xs) es la lista de los n primeros elementos de xs. Por
ejemplo,
coge 2 [a,c,d,b,e] = [a,c]
------------------------------------------------------------------ *}
(*marcarmor13, manmorjim1*)
fun coge :: "nat ⇒ 'a list ⇒ 'a list" where
"coge 0 xs = []"
| "coge n xs = (hd xs)#(coge (n-1) (tl xs)) "

(*danrodcha*)
fun coge1 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge1 0 _ = []"
| "coge1 _ [] = []"
| "coge1 (Suc n) (x#xs) = x#(coge1 n xs)"

(*crimgomgom*)
fun coge2 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge2 0 _ = []"|
"coge2 _ [] = []" |
"coge2 n (x#xs) = x#(coge2 (n-1) xs) "

(*pablucoto*)
fun coge3 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge3 0 xs = []" |
"coge3 n (x#xs) = (if n>length(x#xs) then (x#xs) else x # (coge3 (n-1) xs))"

value "coge 2 [a,c,d,b,e]" -- "= [a,c]"

text {* ---------------------------------------------------------------
Ejercicio 7. Definir la función
elimina :: nat ⇒ 'a list ⇒ 'a list
tal que (elimina n xs) es la lista obtenida eliminando los n primeros
elementos de xs. Por ejemplo,
elimina 2 [a,c,d,b,e] = [d,b,e]
------------------------------------------------------------------ *}
(*marcarmor13, manmorjim1*)
fun elimina :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina 0 xs = xs"
| "elimina n xs = (elimina (n-1) (tl xs ))"

(*danrodcha, crigomgom*)
fun elimina1 :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina1 0 xs = xs"
| "elimina1 _ [] = []"
| "elimina1 (Suc n) (x#xs) = elimina1 n xs"

(*pablucoto*)
fun elimina2 :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina2 0 xs = xs" |
"elimina2 n (x#xs) = (if n>length(x#xs) then [] else (elimina2 (n-1) xs))"

value "elimina 2 [a,c,d,b,e]" -- "= [d,b,e]"

text {* ---------------------------------------------------------------
Ejercicio 8. Definir la función
esVacia :: 'a list ⇒ bool
tal que (esVacia xs) se verifica si xs es la lista vacía. Por ejemplo,
esVacia [] = True
esVacia [1] = False
------------------------------------------------------------------ *}

(*marcarmor13, rubgonmar, danrodcha, crigomgom*)
fun esVacia :: "'a list ⇒ bool" where
"esVacia [] = True "
| "esVacia xs = False"

(*danrodcha*)
fun esVacia1 :: "'a list ⇒ bool" where
"esVacia1 xs = (xs = [])"

(*pablucoto*)
fun esVacia2 :: "'a list ⇒ bool" where
"esVacia2 xs = (if xs=[] then True else False)"

(*manmorjim1*)
fun esVacia3 :: "'a list ⇒ bool" where
"esVacia3 xs = (length xs = 0)"

value "esVacia []" -- "= True"
value "esVacia [1]" -- "= False"

text {* ---------------------------------------------------------------
Ejercicio 9. Definir la función
inversaAc :: 'a list ⇒ 'a list
tal que (inversaAc xs) es a inversa de xs calculada usando
acumuladores. Por ejemplo,
inversaAc [a,c,b,e] = [e,b,c,a]
------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13, mamnorjim1*)
fun inversaAcAux :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux [] ys = ys"
| "inversaAcAux xs ys = inversaAcAux (tl xs) (hd xs#ys) "

(*danrodcha, crigomgom*)
fun inversaAcAux1 :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux1 [] ys = ys"
| "inversaAcAux1 (x#xs) ys = inversaAcAux1 xs (x#ys)"

(*rubgonmar, danrodcha, crigomgom, manmorjim1*)
fun inversaAc :: "'a list ⇒ 'a list" where
"inversaAc xs = inversaAcAux xs []"

(*pablucoto*)
fun inversaAcAux2 :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux2 [] [] = []"|
"inversaAcAux2 xs (y#ys) = (inversaAcAux2 [] ys) @ [y]"
fun inversaAc2 :: "'a list ⇒ 'a list" where
"inversaAc2 xs = inversaAcAux2 [] xs "

value "inversaAc [a,c,b,e]" -- "= [e,b,c,a]"

text {* ---------------------------------------------------------------
Ejercicio 10. Definir la función
sum :: nat list ⇒ nat
tal que (sum xs) es la suma de los elementos de xs. Por ejemplo,
sum [3,2,5] = 10
------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13*)
fun sum :: "nat list ⇒ nat" where
"sum [] = 0"
|"sum xs = hd xs + sum (tl xs)"

(*danrodcha, crigomgom, pablucoto*)
fun sum1 :: "nat list ⇒ nat" where
"sum1 [] = 0"
| "sum1 (x#xs) = x + sum1 xs"

(*danrodcha*)
fun sum2 :: "nat list ⇒ nat" where
"sum2 xs = fold (op +) xs 0"

value "sum [3,2,5]" -- "= 10"

text {* ---------------------------------------------------------------
Ejercicio 11. Definir la función
map :: ('a ⇒ 'b) ⇒ 'a list ⇒ 'b list
tal que (map f xs) es la lista obtenida aplicando la función f a los
elementos de xs. Por ejemplo,
map (λx. 2*x) [3,2,5] = [6,4,10]
------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13*)
fun map :: "('a ⇒ 'b) ⇒ 'a list ⇒ 'b list" where
"map f [] = []"
| "map f xs = f(hd xs)#map f (tl xs)"

(*wilmorort, danrodcha, crigomgom, pablucoto*)
fun map :: "('a ⇒ 'b) ⇒ 'a list ⇒ 'b list" where
"map f [] = []"
|"map f (x # xs) = f x # map f xs" (*yo pondría paréntesis, pero sin
ellos lo entiende*)

value "map (λx. 2*x) [3::nat,2,5]" -- "= [6,4,10]"

end
</source>

Relación 1

2016-10-30T14:33:22Z

Manmorjim1:

<source lang="isar">
chapter {* R1: Programación funcional en Isabelle *}

theory R1
imports Main
begin

text {* ----------------------------------------------------------------
Ejercicio 0. Definir, por recursión, la función
factorial :: nat ⇒ nat
tal que (factorial n) es el factorial de n. Por ejemplo,
factorial 4 = 24
------------------------------------------------------------------- *}

(* danrodcha, anaprarod, ivamenjim, serrodcal, manmorjim1 *)
fun factorial :: "nat ⇒ nat" where
"factorial 0 = 1"
| "factorial (Suc n) = (Suc n) * factorial n"

(*wilmorort, pablucoto,marcarmor13, crigomgom, rubgonmar*)
fun factorial1 :: "nat ⇒ nat" where
"factorial1 0 = 1 "
| "factorial1 n = n * factorial1(n-1)"

value "factorial 4" -- "24"

text {* ----------------------------------------------------------------
Ejercicio 1. Definir, por recursión, la función
longitud :: 'a list ⇒ nat
tal que (longitud xs) es la longitud de la listas xs. Por ejemplo,
longitud [4,2,5] = 3
------------------------------------------------------------------- *}

(*wilmorort*)
(* Para usar las lista en forma de [a,b,c] *)

translations
"[x, xs]" == "x#[xs]"
"[x]" == "x#[]"

(*wilmorort, serrodcal,crigomgom,rubgonmar*)
fun longitud :: "'a list ⇒ nat" where
"longitud [] = 0" |
"longitud (x # xs) = 1 + longitud xs "

(*pablucoto*)
fun longitud0 :: "'a list ⇒ nat " where
" longitud0 [] = 0"
|"longitud0 xs = 1 + longitud0 ((butlast xs)) "

fun longitud0_1 :: "'a list ⇒ nat " where
"longitud0_1 xs = (if xs =[] then 0 else 1 + longitud0_1((butlast xs))) "

(*marcarmor13, manmorjim1*)
fun longitud1 :: "'a list ⇒ nat" where
"longitud1 [] = 0 "
| "longitud1 xs = (1+ longitud2 (tl xs))"

(*danrodcha*)
fun longitud2 :: "'a list ⇒ nat" where
"longitud2 [] = 0"
| "longitud2 (x#xs) = Suc (longitud2 xs)"

(*serrodcal*)
fun longitud3 :: "'a list ⇒ nat" where
"longitud xs = length (xs)"

value "longitud [4,2,5] " -- "= 3"

text {* ---------------------------------------------------------------
Ejercicio 2. Definir la función
fun intercambia :: 'a × 'b ⇒ 'b × 'a
tal que (intercambia p) es el par obtenido intercambiando las
componentes del par p. Por ejemplo,
intercambia (u,v) = (v,u)
------------------------------------------------------------------ *}

(*wilmorort,marcarmor13,danrodcha,crigomgom,pablucoto, rubgonmar, manmorjim1*)
fun intercambia :: "'a × 'b ⇒ 'b × 'a" where
"intercambia (x,y) = (y,x) "

value "intercambia (u,v)"-- "= (v,u)"

text {* ---------------------------------------------------------------
Ejercicio 3. Definir, por recursión, la función
inversa :: 'a list ⇒ 'a list
tal que (inversa xs) es la lista obtenida invirtiendo el orden de los
elementos de xs. Por ejemplo,
inversa [a,d,c] = [c,d,a]
------------------------------------------------------------------ *}

(*wilmorort*)
(* @ :: "'a list => 'a list => 'a list", función agregación definida
en Theory Main, concatena dos listas: [a,b] @ [c,d] = [a,b,c,d] *)
fun inversa :: "'a list ⇒ 'a list" where
"inversa [] = []" |
"inversa (x # xs) = (inversa xs)@(x#[]) "

(*marcarmor13, manmorjim1*)
fun inversa1 :: "'a list ⇒ 'a list" where
"inversa1 [] = []"
| "inversa1 xs = inversa1 (tl xs)@ ((hd xs)#[])"

(*danrodcha, pablucoto*)
(* es igual que inversa sustituyendo x#[] por [x] *)
fun inversa2 :: "'a list ⇒ 'a list" where
"inversa2 [] = []"
| "inversa2 (x#xs) = (inversa2 xs)@[x] "

(*danrodcha*)
fun inversa3 :: "'a list ⇒ 'a list" where
"inversa3 [] = []"
| "inversa3 (x#xs) = concat [(inversa3 xs),[x]] "

(*crigomgom*)
fun inversa4 :: "'a list ⇒ 'a list" where
"inversa4 [] = []" |
"inversa4 xs = (last xs)#(inversa4(butlast xs)) "

value "inversa [a,d,c]" -- "= [c,d,a]"

text {* ---------------------------------------------------------------
Ejercicio 4. Definir la función
repite :: nat ⇒ 'a ⇒ 'a list
tal que (repite n x) es la lista formada por n copias del elemento
x. Por ejemplo,
repite 3 a = [a,a,a]
------------------------------------------------------------------ *}

(*wilmorort,marcarmor13, crigomgom, pablucoto, rubgonmar, manmorjim1*)
fun repite :: "nat ⇒ 'a ⇒ 'a list" where
"repite 0 x = [] " |
"repite n x = x # (repite(n-1) x) "

(*danrodcha*)
fun repite1 :: "nat ⇒ 'a ⇒ 'a list" where
"repite1 0 x = []"
| "repite1 (Suc n) x = x#(repite1 n x)"

value "repite 3 a" -- "= [a,a,a]"

text {* ---------------------------------------------------------------
Ejercicio 5. Definir la función
conc :: 'a list ⇒ 'a list ⇒ 'a list
tal que (conc xs ys) es la concatención de las listas xs e ys. Por
ejemplo,
conc [a,d] [b,d,a,c] = [a,d,b,d,a,c]
------------------------------------------------------------------ *}

(*marcarmor13*)
fun conc :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc xs ys = xs@ys"

(*danrodcha, crigomgom, rubgonmar*)
fun conc1 :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc1 [] ys = ys"
| "conc1 xs [] = xs" (*esta no hace falta*)
| "conc1 (x#xs) ys = x#(conc1 xs ys)"

(*pablucoto*)
fun conc2 :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc2 [] ys = ys" |
"conc2 (x#xs) ys = x # (conc2 xs ys)"

(* manmorjim1 *)
fun conc3 :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc3 [] ys = ys"
| "conc3 xs ys = (hd xs)#[] @ (conc3 (tl xs) ys)"

value "conc [a,d] [b,d,a,c]" -- "= [a,d,b,d,a,c]"

text {* ---------------------------------------------------------------
Ejercicio 6. Definir la función
coge :: nat ⇒ 'a list ⇒ 'a list
tal que (coge n xs) es la lista de los n primeros elementos de xs. Por
ejemplo,
coge 2 [a,c,d,b,e] = [a,c]
------------------------------------------------------------------ *}
(*marcarmor13, manmorjim1*)
fun coge :: "nat ⇒ 'a list ⇒ 'a list" where
"coge 0 xs = []"
| "coge n xs = (hd xs)#(coge (n-1) (tl xs)) "

(*danrodcha*)
fun coge1 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge1 0 _ = []"
| "coge1 _ [] = []"
| "coge1 (Suc n) (x#xs) = x#(coge1 n xs)"

(*crimgomgom*)
fun coge2 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge2 0 _ = []"|
"coge2 _ [] = []" |
"coge2 n (x#xs) = x#(coge2 (n-1) xs) "

(*pablucoto*)
fun coge3 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge3 0 xs = []" |
"coge3 n (x#xs) = (if n>length(x#xs) then (x#xs) else x # (coge3 (n-1) xs))"

value "coge 2 [a,c,d,b,e]" -- "= [a,c]"

text {* ---------------------------------------------------------------
Ejercicio 7. Definir la función
elimina :: nat ⇒ 'a list ⇒ 'a list
tal que (elimina n xs) es la lista obtenida eliminando los n primeros
elementos de xs. Por ejemplo,
elimina 2 [a,c,d,b,e] = [d,b,e]
------------------------------------------------------------------ *}
(*marcarmor13, manmorjim1*)
fun elimina :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina 0 xs = xs"
| "elimina n xs = (elimina (n-1) (tl xs ))"

(*danrodcha, crigomgom*)
fun elimina1 :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina1 0 xs = xs"
| "elimina1 _ [] = []"
| "elimina1 (Suc n) (x#xs) = elimina1 n xs"

(*pablucoto*)
fun elimina2 :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina2 0 xs = xs" |
"elimina2 n (x#xs) = (if n>length(x#xs) then [] else (elimina2 (n-1) xs))"

value "elimina 2 [a,c,d,b,e]" -- "= [d,b,e]"

text {* ---------------------------------------------------------------
Ejercicio 8. Definir la función
esVacia :: 'a list ⇒ bool
tal que (esVacia xs) se verifica si xs es la lista vacía. Por ejemplo,
esVacia [] = True
esVacia [1] = False
------------------------------------------------------------------ *}

(*marcarmor13, rubgonmar, danrodcha, crigomgom*)
fun esVacia :: "'a list ⇒ bool" where
"esVacia [] = True "
| "esVacia xs = False"

(*danrodcha*)
fun esVacia1 :: "'a list ⇒ bool" where
"esVacia1 xs = (xs = [])"

(*pablucoto*)
fun esVacia2 :: "'a list ⇒ bool" where
"esVacia2 xs = (if xs=[] then True else False)"

(*manmorjim1*)
fun esVacia3 :: "'a list ⇒ bool" where
"esVacia3 xs = (length xs = 0)"

value "esVacia []" -- "= True"
value "esVacia [1]" -- "= False"

text {* ---------------------------------------------------------------
Ejercicio 9. Definir la función
inversaAc :: 'a list ⇒ 'a list
tal que (inversaAc xs) es a inversa de xs calculada usando
acumuladores. Por ejemplo,
inversaAc [a,c,b,e] = [e,b,c,a]
------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13*)
fun inversaAcAux :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux [] ys = ys"
| "inversaAcAux xs ys = inversaAcAux (tl xs) (hd xs#ys) "

(*danrodcha, crigomgom*)
fun inversaAcAux1 :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux1 [] ys = ys"
| "inversaAcAux1 (x#xs) ys = inversaAcAux1 xs (x#ys)"

(*rubgonmar, danrodcha, crigomgom*)
fun inversaAc :: "'a list ⇒ 'a list" where
"inversaAc xs = inversaAcAux xs []"

(*pablucoto*)
fun inversaAcAux2 :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux2 [] [] = []"|
"inversaAcAux2 xs (y#ys) = (inversaAcAux2 [] ys) @ [y]"
fun inversaAc2 :: "'a list ⇒ 'a list" where
"inversaAc2 xs = inversaAcAux2 [] xs "

value "inversaAc [a,c,b,e]" -- "= [e,b,c,a]"

text {* ---------------------------------------------------------------
Ejercicio 10. Definir la función
sum :: nat list ⇒ nat
tal que (sum xs) es la suma de los elementos de xs. Por ejemplo,
sum [3,2,5] = 10
------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13*)
fun sum :: "nat list ⇒ nat" where
"sum [] = 0"
|"sum xs = hd xs + sum (tl xs)"

(*danrodcha, crigomgom, pablucoto*)
fun sum1 :: "nat list ⇒ nat" where
"sum1 [] = 0"
| "sum1 (x#xs) = x + sum1 xs"

(*danrodcha*)
fun sum2 :: "nat list ⇒ nat" where
"sum2 xs = fold (op +) xs 0"

value "sum [3,2,5]" -- "= 10"

text {* ---------------------------------------------------------------
Ejercicio 11. Definir la función
map :: ('a ⇒ 'b) ⇒ 'a list ⇒ 'b list
tal que (map f xs) es la lista obtenida aplicando la función f a los
elementos de xs. Por ejemplo,
map (λx. 2*x) [3,2,5] = [6,4,10]
------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13*)
fun map :: "('a ⇒ 'b) ⇒ 'a list ⇒ 'b list" where
"map f [] = []"
| "map f xs = f(hd xs)#map f (tl xs)"

(*wilmorort, danrodcha, crigomgom, pablucoto*)
fun map :: "('a ⇒ 'b) ⇒ 'a list ⇒ 'b list" where
"map f [] = []"
|"map f (x # xs) = f x # map f xs" (*yo pondría paréntesis, pero sin
ellos lo entiende*)

value "map (λx. 2*x) [3::nat,2,5]" -- "= [6,4,10]"

end
</source>

Relación 1

2016-10-30T14:31:08Z

Manmorjim1:

<source lang="isar">
chapter {* R1: Programación funcional en Isabelle *}

theory R1
imports Main
begin

text {* ----------------------------------------------------------------
Ejercicio 0. Definir, por recursión, la función
factorial :: nat ⇒ nat
tal que (factorial n) es el factorial de n. Por ejemplo,
factorial 4 = 24
------------------------------------------------------------------- *}

(* danrodcha, anaprarod, ivamenjim, serrodcal, manmorjim1 *)
fun factorial :: "nat ⇒ nat" where
"factorial 0 = 1"
| "factorial (Suc n) = (Suc n) * factorial n"

(*wilmorort, pablucoto,marcarmor13, crigomgom, rubgonmar*)
fun factorial1 :: "nat ⇒ nat" where
"factorial1 0 = 1 "
| "factorial1 n = n * factorial1(n-1)"

value "factorial 4" -- "24"

text {* ----------------------------------------------------------------
Ejercicio 1. Definir, por recursión, la función
longitud :: 'a list ⇒ nat
tal que (longitud xs) es la longitud de la listas xs. Por ejemplo,
longitud [4,2,5] = 3
------------------------------------------------------------------- *}

(*wilmorort*)
(* Para usar las lista en forma de [a,b,c] *)

translations
"[x, xs]" == "x#[xs]"
"[x]" == "x#[]"

(*wilmorort, serrodcal,crigomgom,rubgonmar*)
fun longitud :: "'a list ⇒ nat" where
"longitud [] = 0" |
"longitud (x # xs) = 1 + longitud xs "

(*pablucoto*)
fun longitud0 :: "'a list ⇒ nat " where
" longitud0 [] = 0"
|"longitud0 xs = 1 + longitud0 ((butlast xs)) "

fun longitud0_1 :: "'a list ⇒ nat " where
"longitud0_1 xs = (if xs =[] then 0 else 1 + longitud0_1((butlast xs))) "

(*marcarmor13, manmorjim1*)
fun longitud1 :: "'a list ⇒ nat" where
"longitud1 [] = 0 "
| "longitud1 xs = (1+ longitud2 (tl xs))"

(*danrodcha*)
fun longitud2 :: "'a list ⇒ nat" where
"longitud2 [] = 0"
| "longitud2 (x#xs) = Suc (longitud2 xs)"

(*serrodcal*)
fun longitud3 :: "'a list ⇒ nat" where
"longitud xs = length (xs)"

value "longitud [4,2,5] " -- "= 3"

text {* ---------------------------------------------------------------
Ejercicio 2. Definir la función
fun intercambia :: 'a × 'b ⇒ 'b × 'a
tal que (intercambia p) es el par obtenido intercambiando las
componentes del par p. Por ejemplo,
intercambia (u,v) = (v,u)
------------------------------------------------------------------ *}

(*wilmorort,marcarmor13,danrodcha,crigomgom,pablucoto, rubgonmar, manmorjim1*)
fun intercambia :: "'a × 'b ⇒ 'b × 'a" where
"intercambia (x,y) = (y,x) "

value "intercambia (u,v)"-- "= (v,u)"

text {* ---------------------------------------------------------------
Ejercicio 3. Definir, por recursión, la función
inversa :: 'a list ⇒ 'a list
tal que (inversa xs) es la lista obtenida invirtiendo el orden de los
elementos de xs. Por ejemplo,
inversa [a,d,c] = [c,d,a]
------------------------------------------------------------------ *}

(*wilmorort*)
(* @ :: "'a list => 'a list => 'a list", función agregación definida
en Theory Main, concatena dos listas: [a,b] @ [c,d] = [a,b,c,d] *)
fun inversa :: "'a list ⇒ 'a list" where
"inversa [] = []" |
"inversa (x # xs) = (inversa xs)@(x#[]) "

(*marcarmor13, manmorjim1*)
fun inversa1 :: "'a list ⇒ 'a list" where
"inversa1 [] = []"
| "inversa1 xs = inversa1 (tl xs)@ ((hd xs)#[])"

(*danrodcha, pablucoto*)
(* es igual que inversa sustituyendo x#[] por [x] *)
fun inversa2 :: "'a list ⇒ 'a list" where
"inversa2 [] = []"
| "inversa2 (x#xs) = (inversa2 xs)@[x] "

(*danrodcha*)
fun inversa3 :: "'a list ⇒ 'a list" where
"inversa3 [] = []"
| "inversa3 (x#xs) = concat [(inversa3 xs),[x]] "

(*crigomgom*)
fun inversa4 :: "'a list ⇒ 'a list" where
"inversa4 [] = []" |
"inversa4 xs = (last xs)#(inversa4(butlast xs)) "

value "inversa [a,d,c]" -- "= [c,d,a]"

text {* ---------------------------------------------------------------
Ejercicio 4. Definir la función
repite :: nat ⇒ 'a ⇒ 'a list
tal que (repite n x) es la lista formada por n copias del elemento
x. Por ejemplo,
repite 3 a = [a,a,a]
------------------------------------------------------------------ *}

(*wilmorort,marcarmor13, crigomgom, pablucoto, rubgonmar, manmorjim1*)
fun repite :: "nat ⇒ 'a ⇒ 'a list" where
"repite 0 x = [] " |
"repite n x = x # (repite(n-1) x) "

(*danrodcha*)
fun repite1 :: "nat ⇒ 'a ⇒ 'a list" where
"repite1 0 x = []"
| "repite1 (Suc n) x = x#(repite1 n x)"

value "repite 3 a" -- "= [a,a,a]"

text {* ---------------------------------------------------------------
Ejercicio 5. Definir la función
conc :: 'a list ⇒ 'a list ⇒ 'a list
tal que (conc xs ys) es la concatención de las listas xs e ys. Por
ejemplo,
conc [a,d] [b,d,a,c] = [a,d,b,d,a,c]
------------------------------------------------------------------ *}

(*marcarmor13*)
fun conc :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc xs ys = xs@ys"

(*danrodcha, crigomgom, rubgonmar*)
fun conc1 :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc1 [] ys = ys"
| "conc1 xs [] = xs" (*esta no hace falta*)
| "conc1 (x#xs) ys = x#(conc1 xs ys)"

(*pablucoto*)
fun conc2 :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc2 [] ys = ys" |
"conc2 (x#xs) ys = x # (conc2 xs ys)"

(* manmorjim1 *)
fun conc3 :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc3 [] ys = ys"
| "conc3 xs ys = (hd xs)#[] @ (conc3 (tl xs) ys)"

value "conc [a,d] [b,d,a,c]" -- "= [a,d,b,d,a,c]"

text {* ---------------------------------------------------------------
Ejercicio 6. Definir la función
coge :: nat ⇒ 'a list ⇒ 'a list
tal que (coge n xs) es la lista de los n primeros elementos de xs. Por
ejemplo,
coge 2 [a,c,d,b,e] = [a,c]
------------------------------------------------------------------ *}
(*marcarmor13, manmorjim1*)
fun coge :: "nat ⇒ 'a list ⇒ 'a list" where
"coge 0 xs = []"
| "coge n xs = (hd xs)#(coge (n-1) (tl xs)) "

(*danrodcha*)
fun coge1 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge1 0 _ = []"
| "coge1 _ [] = []"
| "coge1 (Suc n) (x#xs) = x#(coge1 n xs)"

(*crimgomgom*)
fun coge2 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge2 0 _ = []"|
"coge2 _ [] = []" |
"coge2 n (x#xs) = x#(coge2 (n-1) xs) "

(*pablucoto*)
fun coge3 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge3 0 xs = []" |
"coge3 n (x#xs) = (if n>length(x#xs) then (x#xs) else x # (coge3 (n-1) xs))"

value "coge 2 [a,c,d,b,e]" -- "= [a,c]"

text {* ---------------------------------------------------------------
Ejercicio 7. Definir la función
elimina :: nat ⇒ 'a list ⇒ 'a list
tal que (elimina n xs) es la lista obtenida eliminando los n primeros
elementos de xs. Por ejemplo,
elimina 2 [a,c,d,b,e] = [d,b,e]
------------------------------------------------------------------ *}
(*marcarmor13, manmorjim1*)
fun elimina :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina 0 xs = xs"
| "elimina n xs = (elimina (n-1) (tl xs ))"

(*danrodcha, crigomgom*)
fun elimina1 :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina1 0 xs = xs"
| "elimina1 _ [] = []"
| "elimina1 (Suc n) (x#xs) = elimina1 n xs"

(*pablucoto*)
fun elimina2 :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina2 0 xs = xs" |
"elimina2 n (x#xs) = (if n>length(x#xs) then [] else (elimina2 (n-1) xs))"

value "elimina 2 [a,c,d,b,e]" -- "= [d,b,e]"

text {* ---------------------------------------------------------------
Ejercicio 8. Definir la función
esVacia :: 'a list ⇒ bool
tal que (esVacia xs) se verifica si xs es la lista vacía. Por ejemplo,
esVacia [] = True
esVacia [1] = False
------------------------------------------------------------------ *}

(*marcarmor13, rubgonmar, danrodcha, crigomgom*)
fun esVacia :: "'a list ⇒ bool" where
"esVacia [] = True "
| "esVacia xs = False"

(*danrodcha*)
fun esVacia1 :: "'a list ⇒ bool" where
"esVacia1 xs = (xs = [])"

(*pablucoto*)
fun esVacia2 :: "'a list ⇒ bool" where
"esVacia2 xs = (if xs=[] then True else False)"

value "esVacia []" -- "= True"
value "esVacia [1]" -- "= False"

text {* ---------------------------------------------------------------
Ejercicio 9. Definir la función
inversaAc :: 'a list ⇒ 'a list
tal que (inversaAc xs) es a inversa de xs calculada usando
acumuladores. Por ejemplo,
inversaAc [a,c,b,e] = [e,b,c,a]
------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13*)
fun inversaAcAux :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux [] ys = ys"
| "inversaAcAux xs ys = inversaAcAux (tl xs) (hd xs#ys) "

(*danrodcha, crigomgom*)
fun inversaAcAux1 :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux1 [] ys = ys"
| "inversaAcAux1 (x#xs) ys = inversaAcAux1 xs (x#ys)"

(*rubgonmar, danrodcha, crigomgom*)
fun inversaAc :: "'a list ⇒ 'a list" where
"inversaAc xs = inversaAcAux xs []"

(*pablucoto*)
fun inversaAcAux2 :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux2 [] [] = []"|
"inversaAcAux2 xs (y#ys) = (inversaAcAux2 [] ys) @ [y]"
fun inversaAc2 :: "'a list ⇒ 'a list" where
"inversaAc2 xs = inversaAcAux2 [] xs "

value "inversaAc [a,c,b,e]" -- "= [e,b,c,a]"

text {* ---------------------------------------------------------------
Ejercicio 10. Definir la función
sum :: nat list ⇒ nat
tal que (sum xs) es la suma de los elementos de xs. Por ejemplo,
sum [3,2,5] = 10
------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13*)
fun sum :: "nat list ⇒ nat" where
"sum [] = 0"
|"sum xs = hd xs + sum (tl xs)"

(*danrodcha, crigomgom, pablucoto*)
fun sum1 :: "nat list ⇒ nat" where
"sum1 [] = 0"
| "sum1 (x#xs) = x + sum1 xs"

(*danrodcha*)
fun sum2 :: "nat list ⇒ nat" where
"sum2 xs = fold (op +) xs 0"

value "sum [3,2,5]" -- "= 10"

text {* ---------------------------------------------------------------
Ejercicio 11. Definir la función
map :: ('a ⇒ 'b) ⇒ 'a list ⇒ 'b list
tal que (map f xs) es la lista obtenida aplicando la función f a los
elementos de xs. Por ejemplo,
map (λx. 2*x) [3,2,5] = [6,4,10]
------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13*)
fun map :: "('a ⇒ 'b) ⇒ 'a list ⇒ 'b list" where
"map f [] = []"
| "map f xs = f(hd xs)#map f (tl xs)"

(*wilmorort, danrodcha, crigomgom, pablucoto*)
fun map :: "('a ⇒ 'b) ⇒ 'a list ⇒ 'b list" where
"map f [] = []"
|"map f (x # xs) = f x # map f xs" (*yo pondría paréntesis, pero sin
ellos lo entiende*)

value "map (λx. 2*x) [3::nat,2,5]" -- "= [6,4,10]"

end
</source>

Relación 1

2016-10-30T13:44:34Z

Manmorjim1:

<source lang="isar">
chapter {* R1: Programación funcional en Isabelle *}

theory R1
imports Main
begin

text {* ----------------------------------------------------------------
Ejercicio 0. Definir, por recursión, la función
factorial :: nat ⇒ nat
tal que (factorial n) es el factorial de n. Por ejemplo,
factorial 4 = 24
------------------------------------------------------------------- *}

(* danrodcha, anaprarod, ivamenjim, serrodcal, manmorjim1 *)
fun factorial :: "nat ⇒ nat" where
"factorial 0 = 1"
| "factorial (Suc n) = (Suc n) * factorial n"

(*wilmorort, pablucoto,marcarmor13, crigomgom, rubgonmar*)
fun factorial1 :: "nat ⇒ nat" where
"factorial1 0 = 1 "
| "factorial1 n = n * factorial1(n-1)"

value "factorial 4" -- "24"

text {* ----------------------------------------------------------------
Ejercicio 1. Definir, por recursión, la función
longitud :: 'a list ⇒ nat
tal que (longitud xs) es la longitud de la listas xs. Por ejemplo,
longitud [4,2,5] = 3
------------------------------------------------------------------- *}

(*wilmorort*)
(* Para usar las lista en forma de [a,b,c] *)

translations
"[x, xs]" == "x#[xs]"
"[x]" == "x#[]"

(*wilmorort, serrodcal,crigomgom,rubgonmar*)
fun longitud :: "'a list ⇒ nat" where
"longitud [] = 0" |
"longitud (x # xs) = 1 + longitud xs "

(*pablucoto*)
fun longitud0 :: "'a list ⇒ nat " where
" longitud0 [] = 0"
|"longitud0 xs = 1 + longitud0 ((butlast xs)) "

fun longitud0_1 :: "'a list ⇒ nat " where
"longitud0_1 xs = (if xs =[] then 0 else 1 + longitud0_1((butlast xs))) "

(*marcarmor13, manmorjim1*)
fun longitud1 :: "'a list ⇒ nat" where
"longitud1 [] = 0 "
| "longitud1 xs = (1+ longitud2 (tl xs))"

(*danrodcha*)
fun longitud2 :: "'a list ⇒ nat" where
"longitud2 [] = 0"
| "longitud2 (x#xs) = Suc (longitud2 xs)"

(*serrodcal*)
fun longitud3 :: "'a list ⇒ nat" where
"longitud xs = length (xs)"

value "longitud [4,2,5] " -- "= 3"

text {* ---------------------------------------------------------------
Ejercicio 2. Definir la función
fun intercambia :: 'a × 'b ⇒ 'b × 'a
tal que (intercambia p) es el par obtenido intercambiando las
componentes del par p. Por ejemplo,
intercambia (u,v) = (v,u)
------------------------------------------------------------------ *}

(*wilmorort,marcarmor13,danrodcha,crigomgom,pablucoto, rubgonmar, manmorjim1*)
fun intercambia :: "'a × 'b ⇒ 'b × 'a" where
"intercambia (x,y) = (y,x) "

value "intercambia (u,v)"-- "= (v,u)"

text {* ---------------------------------------------------------------
Ejercicio 3. Definir, por recursión, la función
inversa :: 'a list ⇒ 'a list
tal que (inversa xs) es la lista obtenida invirtiendo el orden de los
elementos de xs. Por ejemplo,
inversa [a,d,c] = [c,d,a]
------------------------------------------------------------------ *}

(*wilmorort*)
(* @ :: "'a list => 'a list => 'a list", función agregación definida
en Theory Main, concatena dos listas: [a,b] @ [c,d] = [a,b,c,d] *)
fun inversa :: "'a list ⇒ 'a list" where
"inversa [] = []" |
"inversa (x # xs) = (inversa xs)@(x#[]) "

(*marcarmor13, manmorjim1*)
fun inversa1 :: "'a list ⇒ 'a list" where
"inversa1 [] = []"
| "inversa1 xs = inversa1 (tl xs)@ ((hd xs)#[])"

(*danrodcha, pablucoto*)
(* es igual que inversa sustituyendo x#[] por [x] *)
fun inversa2 :: "'a list ⇒ 'a list" where
"inversa2 [] = []"
| "inversa2 (x#xs) = (inversa2 xs)@[x] "

(*danrodcha*)
fun inversa3 :: "'a list ⇒ 'a list" where
"inversa3 [] = []"
| "inversa3 (x#xs) = concat [(inversa3 xs),[x]] "

(*crigomgom*)
fun inversa4 :: "'a list ⇒ 'a list" where
"inversa4 [] = []" |
"inversa4 xs = (last xs)#(inversa4(butlast xs)) "

value "inversa [a,d,c]" -- "= [c,d,a]"

text {* ---------------------------------------------------------------
Ejercicio 4. Definir la función
repite :: nat ⇒ 'a ⇒ 'a list
tal que (repite n x) es la lista formada por n copias del elemento
x. Por ejemplo,
repite 3 a = [a,a,a]
------------------------------------------------------------------ *}

(*wilmorort,marcarmor13, crigomgom, pablucoto, rubgonmar, manmorjim1*)
fun repite :: "nat ⇒ 'a ⇒ 'a list" where
"repite 0 x = [] " |
"repite n x = x # (repite(n-1) x) "

(*danrodcha*)
fun repite1 :: "nat ⇒ 'a ⇒ 'a list" where
"repite1 0 x = []"
| "repite1 (Suc n) x = x#(repite1 n x)"

value "repite 3 a" -- "= [a,a,a]"

text {* ---------------------------------------------------------------
Ejercicio 5. Definir la función
conc :: 'a list ⇒ 'a list ⇒ 'a list
tal que (conc xs ys) es la concatención de las listas xs e ys. Por
ejemplo,
conc [a,d] [b,d,a,c] = [a,d,b,d,a,c]
------------------------------------------------------------------ *}

(*marcarmor13*)
fun conc :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc xs ys = xs@ys"

(*danrodcha, crigomgom, rubgonmar*)
fun conc1 :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc1 [] ys = ys"
| "conc1 xs [] = xs" (*esta no hace falta*)
| "conc1 (x#xs) ys = x#(conc1 xs ys)"

(*pablucoto*)
fun conc2 :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc2 [] ys = ys" |
"conc2 (x#xs) ys = x # (conc2 xs ys)"

(* manmorjim1 *)
fun conc3 :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc3 [] ys = ys"
| "conc3 xs ys = (hd xs)#[] @ (conc3 (tl xs) ys)"

value "conc [a,d] [b,d,a,c]" -- "= [a,d,b,d,a,c]"

text {* ---------------------------------------------------------------
Ejercicio 6. Definir la función
coge :: nat ⇒ 'a list ⇒ 'a list
tal que (coge n xs) es la lista de los n primeros elementos de xs. Por
ejemplo,
coge 2 [a,c,d,b,e] = [a,c]
------------------------------------------------------------------ *}
(*marcarmor13, manmorjim1*)
fun coge :: "nat ⇒ 'a list ⇒ 'a list" where
"coge 0 xs = []"
| "coge n xs = (hd xs)#(coge (n-1) (tl xs)) "

(*danrodcha*)
fun coge1 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge1 0 _ = []"
| "coge1 _ [] = []"
| "coge1 (Suc n) (x#xs) = x#(coge1 n xs)"

(*crimgomgom*)
fun coge2 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge2 0 _ = []"|
"coge2 _ [] = []" |
"coge2 n (x#xs) = x#(coge2 (n-1) xs) "

(*pablucoto*)
fun coge3 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge3 0 xs = []" |
"coge3 n (x#xs) = (if n>length(x#xs) then (x#xs) else x # (coge3 (n-1) xs))"

value "coge 2 [a,c,d,b,e]" -- "= [a,c]"

text {* ---------------------------------------------------------------
Ejercicio 7. Definir la función
elimina :: nat ⇒ 'a list ⇒ 'a list
tal que (elimina n xs) es la lista obtenida eliminando los n primeros
elementos de xs. Por ejemplo,
elimina 2 [a,c,d,b,e] = [d,b,e]
------------------------------------------------------------------ *}
(*marcarmor13*)
fun elimina :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina 0 xs = xs"
| "elimina n xs = (elimina (n-1) (tl xs ))"

(*danrodcha, crigomgom*)
fun elimina1 :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina1 0 xs = xs"
| "elimina1 _ [] = []"
| "elimina1 (Suc n) (x#xs) = elimina1 n xs"

(*pablucoto*)
fun elimina2 :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina2 0 xs = xs" |
"elimina2 n (x#xs) = (if n>length(x#xs) then [] else (elimina2 (n-1) xs))"

value "elimina 2 [a,c,d,b,e]" -- "= [d,b,e]"

text {* ---------------------------------------------------------------
Ejercicio 8. Definir la función
esVacia :: 'a list ⇒ bool
tal que (esVacia xs) se verifica si xs es la lista vacía. Por ejemplo,
esVacia [] = True
esVacia [1] = False
------------------------------------------------------------------ *}

(*marcarmor13, rubgonmar, danrodcha, crigomgom*)
fun esVacia :: "'a list ⇒ bool" where
"esVacia [] = True "
| "esVacia xs = False"

(*danrodcha*)
fun esVacia1 :: "'a list ⇒ bool" where
"esVacia1 xs = (xs = [])"

(*pablucoto*)
fun esVacia2 :: "'a list ⇒ bool" where
"esVacia2 xs = (if xs=[] then True else False)"

value "esVacia []" -- "= True"
value "esVacia [1]" -- "= False"

text {* ---------------------------------------------------------------
Ejercicio 9. Definir la función
inversaAc :: 'a list ⇒ 'a list
tal que (inversaAc xs) es a inversa de xs calculada usando
acumuladores. Por ejemplo,
inversaAc [a,c,b,e] = [e,b,c,a]
------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13*)
fun inversaAcAux :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux [] ys = ys"
| "inversaAcAux xs ys = inversaAcAux (tl xs) (hd xs#ys) "

(*danrodcha, crigomgom*)
fun inversaAcAux1 :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux1 [] ys = ys"
| "inversaAcAux1 (x#xs) ys = inversaAcAux1 xs (x#ys)"

(*rubgonmar, danrodcha, crigomgom*)
fun inversaAc :: "'a list ⇒ 'a list" where
"inversaAc xs = inversaAcAux xs []"

(*pablucoto*)
fun inversaAcAux2 :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux2 [] [] = []"|
"inversaAcAux2 xs (y#ys) = (inversaAcAux2 [] ys) @ [y]"
fun inversaAc2 :: "'a list ⇒ 'a list" where
"inversaAc2 xs = inversaAcAux2 [] xs "

value "inversaAc [a,c,b,e]" -- "= [e,b,c,a]"

text {* ---------------------------------------------------------------
Ejercicio 10. Definir la función
sum :: nat list ⇒ nat
tal que (sum xs) es la suma de los elementos de xs. Por ejemplo,
sum [3,2,5] = 10
------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13*)
fun sum :: "nat list ⇒ nat" where
"sum [] = 0"
|"sum xs = hd xs + sum (tl xs)"

(*danrodcha, crigomgom, pablucoto*)
fun sum1 :: "nat list ⇒ nat" where
"sum1 [] = 0"
| "sum1 (x#xs) = x + sum1 xs"

(*danrodcha*)
fun sum2 :: "nat list ⇒ nat" where
"sum2 xs = fold (op +) xs 0"

value "sum [3,2,5]" -- "= 10"

text {* ---------------------------------------------------------------
Ejercicio 11. Definir la función
map :: ('a ⇒ 'b) ⇒ 'a list ⇒ 'b list
tal que (map f xs) es la lista obtenida aplicando la función f a los
elementos de xs. Por ejemplo,
map (λx. 2*x) [3,2,5] = [6,4,10]
------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13*)
fun map :: "('a ⇒ 'b) ⇒ 'a list ⇒ 'b list" where
"map f [] = []"
| "map f xs = f(hd xs)#map f (tl xs)"

(*wilmorort, danrodcha, crigomgom, pablucoto*)
fun map :: "('a ⇒ 'b) ⇒ 'a list ⇒ 'b list" where
"map f [] = []"
|"map f (x # xs) = f x # map f xs" (*yo pondría paréntesis, pero sin
ellos lo entiende*)

value "map (λx. 2*x) [3::nat,2,5]" -- "= [6,4,10]"

end
</source>

Relación 1

2016-10-30T13:25:36Z

Manmorjim1:

<source lang="isar">
chapter {* R1: Programación funcional en Isabelle *}

theory R1
imports Main
begin

text {* ----------------------------------------------------------------
Ejercicio 0. Definir, por recursión, la función
factorial :: nat ⇒ nat
tal que (factorial n) es el factorial de n. Por ejemplo,
factorial 4 = 24
------------------------------------------------------------------- *}

(* danrodcha, anaprarod, ivamenjim, serrodcal, manmorjim1 *)
fun factorial :: "nat ⇒ nat" where
"factorial 0 = 1"
| "factorial (Suc n) = (Suc n) * factorial n"

(*wilmorort, pablucoto,marcarmor13, crigomgom, rubgonmar*)
fun factorial1 :: "nat ⇒ nat" where
"factorial1 0 = 1 "
| "factorial1 n = n * factorial1(n-1)"

value "factorial 4" -- "24"

text {* ----------------------------------------------------------------
Ejercicio 1. Definir, por recursión, la función
longitud :: 'a list ⇒ nat
tal que (longitud xs) es la longitud de la listas xs. Por ejemplo,
longitud [4,2,5] = 3
------------------------------------------------------------------- *}

(*wilmorort*)
(* Para usar las lista en forma de [a,b,c] *)

translations
"[x, xs]" == "x#[xs]"
"[x]" == "x#[]"

(*wilmorort, serrodcal,crigomgom,rubgonmar*)
fun longitud :: "'a list ⇒ nat" where
"longitud [] = 0" |
"longitud (x # xs) = 1 + longitud xs "

(*pablucoto*)
fun longitud0 :: "'a list ⇒ nat " where
" longitud0 [] = 0"
|"longitud0 xs = 1 + longitud0 ((butlast xs)) "

fun longitud0_1 :: "'a list ⇒ nat " where
"longitud0_1 xs = (if xs =[] then 0 else 1 + longitud0_1((butlast xs))) "

(*marcarmor13, manmorjim1*)
fun longitud1 :: "'a list ⇒ nat" where
"longitud1 [] = 0 "
| "longitud1 xs = (1+ longitud2 (tl xs))"

(*danrodcha*)
fun longitud2 :: "'a list ⇒ nat" where
"longitud2 [] = 0"
| "longitud2 (x#xs) = Suc (longitud2 xs)"

(*serrodcal*)
fun longitud3 :: "'a list ⇒ nat" where
"longitud xs = length (xs)"

value "longitud [4,2,5] " -- "= 3"

text {* ---------------------------------------------------------------
Ejercicio 2. Definir la función
fun intercambia :: 'a × 'b ⇒ 'b × 'a
tal que (intercambia p) es el par obtenido intercambiando las
componentes del par p. Por ejemplo,
intercambia (u,v) = (v,u)
------------------------------------------------------------------ *}

(*wilmorort,marcarmor13,danrodcha,crigomgom,pablucoto, rubgonmar, manmorjim1*)
fun intercambia :: "'a × 'b ⇒ 'b × 'a" where
"intercambia (x,y) = (y,x) "

value "intercambia (u,v)"-- "= (v,u)"

text {* ---------------------------------------------------------------
Ejercicio 3. Definir, por recursión, la función
inversa :: 'a list ⇒ 'a list
tal que (inversa xs) es la lista obtenida invirtiendo el orden de los
elementos de xs. Por ejemplo,
inversa [a,d,c] = [c,d,a]
------------------------------------------------------------------ *}

(*wilmorort*)
(* @ :: "'a list => 'a list => 'a list", función agregación definida
en Theory Main, concatena dos listas: [a,b] @ [c,d] = [a,b,c,d] *)
fun inversa :: "'a list ⇒ 'a list" where
"inversa [] = []" |
"inversa (x # xs) = (inversa xs)@(x#[]) "

(*marcarmor13, manmorjim1*)
fun inversa1 :: "'a list ⇒ 'a list" where
"inversa1 [] = []"
| "inversa1 xs = inversa1 (tl xs)@ ((hd xs)#[])"

(*danrodcha, pablucoto*)
(* es igual que inversa sustituyendo x#[] por [x] *)
fun inversa2 :: "'a list ⇒ 'a list" where
"inversa2 [] = []"
| "inversa2 (x#xs) = (inversa2 xs)@[x] "

(*danrodcha*)
fun inversa3 :: "'a list ⇒ 'a list" where
"inversa3 [] = []"
| "inversa3 (x#xs) = concat [(inversa3 xs),[x]] "

(*crigomgom*)
fun inversa4 :: "'a list ⇒ 'a list" where
"inversa4 [] = []" |
"inversa4 xs = (last xs)#(inversa4(butlast xs)) "

value "inversa [a,d,c]" -- "= [c,d,a]"

text {* ---------------------------------------------------------------
Ejercicio 4. Definir la función
repite :: nat ⇒ 'a ⇒ 'a list
tal que (repite n x) es la lista formada por n copias del elemento
x. Por ejemplo,
repite 3 a = [a,a,a]
------------------------------------------------------------------ *}

(*wilmorort,marcarmor13, crigomgom, pablucoto, rubgonmar, manmorjim1*)
fun repite :: "nat ⇒ 'a ⇒ 'a list" where
"repite 0 x = [] " |
"repite n x = x # (repite(n-1) x) "

(*danrodcha*)
fun repite1 :: "nat ⇒ 'a ⇒ 'a list" where
"repite1 0 x = []"
| "repite1 (Suc n) x = x#(repite1 n x)"

value "repite 3 a" -- "= [a,a,a]"

text {* ---------------------------------------------------------------
Ejercicio 5. Definir la función
conc :: 'a list ⇒ 'a list ⇒ 'a list
tal que (conc xs ys) es la concatención de las listas xs e ys. Por
ejemplo,
conc [a,d] [b,d,a,c] = [a,d,b,d,a,c]
------------------------------------------------------------------ *}

(*marcarmor13*)
fun conc :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc xs ys = xs@ys"

(*danrodcha, crigomgom, rubgonmar*)
fun conc1 :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc1 [] ys = ys"
| "conc1 xs [] = xs" (*esta no hace falta*)
| "conc1 (x#xs) ys = x#(conc1 xs ys)"

(*pablucoto*)
fun conc2 :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc2 [] ys = ys" |
"conc2 (x#xs) ys = x # (conc2 xs ys)"

(* manmorjim1 *)
fun conc3 :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc3 [] ys = ys"
| "conc3 xs ys = (hd xs)#[] @ (conc3 (tl xs) ys)"

value "conc [a,d] [b,d,a,c]" -- "= [a,d,b,d,a,c]"

text {* ---------------------------------------------------------------
Ejercicio 6. Definir la función
coge :: nat ⇒ 'a list ⇒ 'a list
tal que (coge n xs) es la lista de los n primeros elementos de xs. Por
ejemplo,
coge 2 [a,c,d,b,e] = [a,c]
------------------------------------------------------------------ *}
(*marcarmor13*)
fun coge :: "nat ⇒ 'a list ⇒ 'a list" where
"coge 0 xs = []"
| "coge n xs = (hd xs)#(coge (n-1) (tl xs)) "

(*danrodcha*)
fun coge1 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge1 0 _ = []"
| "coge1 _ [] = []"
| "coge1 (Suc n) (x#xs) = x#(coge1 n xs)"

(*crimgomgom*)
fun coge2 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge2 0 _ = []"|
"coge2 _ [] = []" |
"coge2 n (x#xs) = x#(coge2 (n-1) xs) "

(*pablucoto*)
fun coge3 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge3 0 xs = []" |
"coge3 n (x#xs) = (if n>length(x#xs) then (x#xs) else x # (coge3 (n-1) xs))"

value "coge 2 [a,c,d,b,e]" -- "= [a,c]"

text {* ---------------------------------------------------------------
Ejercicio 7. Definir la función
elimina :: nat ⇒ 'a list ⇒ 'a list
tal que (elimina n xs) es la lista obtenida eliminando los n primeros
elementos de xs. Por ejemplo,
elimina 2 [a,c,d,b,e] = [d,b,e]
------------------------------------------------------------------ *}
(*marcarmor13*)
fun elimina :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina 0 xs = xs"
| "elimina n xs = (elimina (n-1) (tl xs ))"

(*danrodcha, crigomgom*)
fun elimina1 :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina1 0 xs = xs"
| "elimina1 _ [] = []"
| "elimina1 (Suc n) (x#xs) = elimina1 n xs"

(*pablucoto*)
fun elimina2 :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina2 0 xs = xs" |
"elimina2 n (x#xs) = (if n>length(x#xs) then [] else (elimina2 (n-1) xs))"

value "elimina 2 [a,c,d,b,e]" -- "= [d,b,e]"

text {* ---------------------------------------------------------------
Ejercicio 8. Definir la función
esVacia :: 'a list ⇒ bool
tal que (esVacia xs) se verifica si xs es la lista vacía. Por ejemplo,
esVacia [] = True
esVacia [1] = False
------------------------------------------------------------------ *}

(*marcarmor13, rubgonmar, danrodcha, crigomgom*)
fun esVacia :: "'a list ⇒ bool" where
"esVacia [] = True "
| "esVacia xs = False"

(*danrodcha*)
fun esVacia1 :: "'a list ⇒ bool" where
"esVacia1 xs = (xs = [])"

(*pablucoto*)
fun esVacia2 :: "'a list ⇒ bool" where
"esVacia2 xs = (if xs=[] then True else False)"

value "esVacia []" -- "= True"
value "esVacia [1]" -- "= False"

text {* ---------------------------------------------------------------
Ejercicio 9. Definir la función
inversaAc :: 'a list ⇒ 'a list
tal que (inversaAc xs) es a inversa de xs calculada usando
acumuladores. Por ejemplo,
inversaAc [a,c,b,e] = [e,b,c,a]
------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13*)
fun inversaAcAux :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux [] ys = ys"
| "inversaAcAux xs ys = inversaAcAux (tl xs) (hd xs#ys) "

(*danrodcha, crigomgom*)
fun inversaAcAux1 :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux1 [] ys = ys"
| "inversaAcAux1 (x#xs) ys = inversaAcAux1 xs (x#ys)"

(*rubgonmar, danrodcha, crigomgom*)
fun inversaAc :: "'a list ⇒ 'a list" where
"inversaAc xs = inversaAcAux xs []"

(*pablucoto*)
fun inversaAcAux2 :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux2 [] [] = []"|
"inversaAcAux2 xs (y#ys) = (inversaAcAux2 [] ys) @ [y]"
fun inversaAc2 :: "'a list ⇒ 'a list" where
"inversaAc2 xs = inversaAcAux2 [] xs "

value "inversaAc [a,c,b,e]" -- "= [e,b,c,a]"

text {* ---------------------------------------------------------------
Ejercicio 10. Definir la función
sum :: nat list ⇒ nat
tal que (sum xs) es la suma de los elementos de xs. Por ejemplo,
sum [3,2,5] = 10
------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13*)
fun sum :: "nat list ⇒ nat" where
"sum [] = 0"
|"sum xs = hd xs + sum (tl xs)"

(*danrodcha, crigomgom, pablucoto*)
fun sum1 :: "nat list ⇒ nat" where
"sum1 [] = 0"
| "sum1 (x#xs) = x + sum1 xs"

(*danrodcha*)
fun sum2 :: "nat list ⇒ nat" where
"sum2 xs = fold (op +) xs 0"

value "sum [3,2,5]" -- "= 10"

text {* ---------------------------------------------------------------
Ejercicio 11. Definir la función
map :: ('a ⇒ 'b) ⇒ 'a list ⇒ 'b list
tal que (map f xs) es la lista obtenida aplicando la función f a los
elementos de xs. Por ejemplo,
map (λx. 2*x) [3,2,5] = [6,4,10]
------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13*)
fun map :: "('a ⇒ 'b) ⇒ 'a list ⇒ 'b list" where
"map f [] = []"
| "map f xs = f(hd xs)#map f (tl xs)"

(*wilmorort, danrodcha, crigomgom, pablucoto*)
fun map :: "('a ⇒ 'b) ⇒ 'a list ⇒ 'b list" where
"map f [] = []"
|"map f (x # xs) = f x # map f xs" (*yo pondría paréntesis, pero sin
ellos lo entiende*)

value "map (λx. 2*x) [3::nat,2,5]" -- "= [6,4,10]"

end
</source>

Relación 1

2016-10-30T13:14:50Z

Manmorjim1:

<source lang="isar">
chapter {* R1: Programación funcional en Isabelle *}

theory R1
imports Main
begin

text {* ----------------------------------------------------------------
Ejercicio 0. Definir, por recursión, la función
factorial :: nat ⇒ nat
tal que (factorial n) es el factorial de n. Por ejemplo,
factorial 4 = 24
------------------------------------------------------------------- *}

(* danrodcha, anaprarod, ivamenjim, serrodcal, manmorjim1 *)
fun factorial :: "nat ⇒ nat" where
"factorial 0 = 1"
| "factorial (Suc n) = (Suc n) * factorial n"

(*wilmorort, pablucoto,marcarmor13, crigomgom, rubgonmar*)
fun factorial1 :: "nat ⇒ nat" where
"factorial1 0 = 1 "
| "factorial1 n = n * factorial1(n-1)"

value "factorial 4" -- "24"

text {* ----------------------------------------------------------------
Ejercicio 1. Definir, por recursión, la función
longitud :: 'a list ⇒ nat
tal que (longitud xs) es la longitud de la listas xs. Por ejemplo,
longitud [4,2,5] = 3
------------------------------------------------------------------- *}

(*wilmorort*)
(* Para usar las lista en forma de [a,b,c] *)

translations
"[x, xs]" == "x#[xs]"
"[x]" == "x#[]"

(*wilmorort, serrodcal,crigomgom,rubgonmar*)
fun longitud :: "'a list ⇒ nat" where
"longitud [] = 0" |
"longitud (x # xs) = 1 + longitud xs "

(*pablucoto*)
fun longitud0 :: "'a list ⇒ nat " where
" longitud0 [] = 0"
|"longitud0 xs = 1 + longitud0 ((butlast xs)) "

fun longitud0_1 :: "'a list ⇒ nat " where
"longitud0_1 xs = (if xs =[] then 0 else 1 + longitud0_1((butlast xs))) "

(*marcarmor13, manmorjim1*)
fun longitud1 :: "'a list ⇒ nat" where
"longitud1 [] = 0 "
| "longitud1 xs = (1+ longitud2 (tl xs))"

(*danrodcha*)
fun longitud2 :: "'a list ⇒ nat" where
"longitud2 [] = 0"
| "longitud2 (x#xs) = Suc (longitud2 xs)"

(*serrodcal*)
fun longitud3 :: "'a list ⇒ nat" where
"longitud xs = length (xs)"

value "longitud [4,2,5] " -- "= 3"

text {* ---------------------------------------------------------------
Ejercicio 2. Definir la función
fun intercambia :: 'a × 'b ⇒ 'b × 'a
tal que (intercambia p) es el par obtenido intercambiando las
componentes del par p. Por ejemplo,
intercambia (u,v) = (v,u)
------------------------------------------------------------------ *}

(*wilmorort,marcarmor13,danrodcha,crigomgom,pablucoto, rubgonmar, manmorjim1*)
fun intercambia :: "'a × 'b ⇒ 'b × 'a" where
"intercambia (x,y) = (y,x) "

value "intercambia (u,v)"-- "= (v,u)"

text {* ---------------------------------------------------------------
Ejercicio 3. Definir, por recursión, la función
inversa :: 'a list ⇒ 'a list
tal que (inversa xs) es la lista obtenida invirtiendo el orden de los
elementos de xs. Por ejemplo,
inversa [a,d,c] = [c,d,a]
------------------------------------------------------------------ *}

(*wilmorort*)
(* @ :: "'a list => 'a list => 'a list", función agregación definida
en Theory Main, concatena dos listas: [a,b] @ [c,d] = [a,b,c,d] *)
fun inversa :: "'a list ⇒ 'a list" where
"inversa [] = []" |
"inversa (x # xs) = (inversa xs)@(x#[]) "

(*marcarmor13, manmorjim1*)
fun inversa1 :: "'a list ⇒ 'a list" where
"inversa1 [] = []"
| "inversa1 xs = inversa1 (tl xs)@ ((hd xs)#[])"

(*danrodcha, pablucoto*)
(* es igual que inversa sustituyendo x#[] por [x] *)
fun inversa2 :: "'a list ⇒ 'a list" where
"inversa2 [] = []"
| "inversa2 (x#xs) = (inversa2 xs)@[x] "

(*danrodcha*)
fun inversa3 :: "'a list ⇒ 'a list" where
"inversa3 [] = []"
| "inversa3 (x#xs) = concat [(inversa3 xs),[x]] "

(*crigomgom*)
fun inversa4 :: "'a list ⇒ 'a list" where
"inversa4 [] = []" |
"inversa4 xs = (last xs)#(inversa4(butlast xs)) "

value "inversa [a,d,c]" -- "= [c,d,a]"

text {* ---------------------------------------------------------------
Ejercicio 4. Definir la función
repite :: nat ⇒ 'a ⇒ 'a list
tal que (repite n x) es la lista formada por n copias del elemento
x. Por ejemplo,
repite 3 a = [a,a,a]
------------------------------------------------------------------ *}

(*wilmorort,marcarmor13, crigomgom, pablucoto, rubgonmar, manmorjim1*)
fun repite :: "nat ⇒ 'a ⇒ 'a list" where
"repite 0 x = [] " |
"repite n x = x # (repite(n-1) x) "

(*danrodcha*)
fun repite1 :: "nat ⇒ 'a ⇒ 'a list" where
"repite1 0 x = []"
| "repite1 (Suc n) x = x#(repite1 n x)"

value "repite 3 a" -- "= [a,a,a]"

text {* ---------------------------------------------------------------
Ejercicio 5. Definir la función
conc :: 'a list ⇒ 'a list ⇒ 'a list
tal que (conc xs ys) es la concatención de las listas xs e ys. Por
ejemplo,
conc [a,d] [b,d,a,c] = [a,d,b,d,a,c]
------------------------------------------------------------------ *}

(*marcarmor13*)
fun conc :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc xs ys = xs@ys"

(*danrodcha, crigomgom, rubgonmar*)
fun conc1 :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc1 [] ys = ys"
| "conc1 xs [] = xs" (*esta no hace falta*)
| "conc1 (x#xs) ys = x#(conc1 xs ys)"

(*pablucoto*)
fun conc2 :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc2 [] ys = ys" |
"conc2 (x#xs) ys = x # (conc2 xs ys)"

value "conc [a,d] [b,d,a,c]" -- "= [a,d,b,d,a,c]"

text {* ---------------------------------------------------------------
Ejercicio 6. Definir la función
coge :: nat ⇒ 'a list ⇒ 'a list
tal que (coge n xs) es la lista de los n primeros elementos de xs. Por
ejemplo,
coge 2 [a,c,d,b,e] = [a,c]
------------------------------------------------------------------ *}
(*marcarmor13*)
fun coge :: "nat ⇒ 'a list ⇒ 'a list" where
"coge 0 xs = []"
| "coge n xs = (hd xs)#(coge (n-1) (tl xs)) "

(*danrodcha*)
fun coge1 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge1 0 _ = []"
| "coge1 _ [] = []"
| "coge1 (Suc n) (x#xs) = x#(coge1 n xs)"

(*crimgomgom*)
fun coge2 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge2 0 _ = []"|
"coge2 _ [] = []" |
"coge2 n (x#xs) = x#(coge2 (n-1) xs) "

(*pablucoto*)
fun coge3 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge3 0 xs = []" |
"coge3 n (x#xs) = (if n>length(x#xs) then (x#xs) else x # (coge3 (n-1) xs))"

value "coge 2 [a,c,d,b,e]" -- "= [a,c]"

text {* ---------------------------------------------------------------
Ejercicio 7. Definir la función
elimina :: nat ⇒ 'a list ⇒ 'a list
tal que (elimina n xs) es la lista obtenida eliminando los n primeros
elementos de xs. Por ejemplo,
elimina 2 [a,c,d,b,e] = [d,b,e]
------------------------------------------------------------------ *}
(*marcarmor13*)
fun elimina :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina 0 xs = xs"
| "elimina n xs = (elimina (n-1) (tl xs ))"

(*danrodcha, crigomgom*)
fun elimina1 :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina1 0 xs = xs"
| "elimina1 _ [] = []"
| "elimina1 (Suc n) (x#xs) = elimina1 n xs"

(*pablucoto*)
fun elimina2 :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina2 0 xs = xs" |
"elimina2 n (x#xs) = (if n>length(x#xs) then [] else (elimina2 (n-1) xs))"

value "elimina 2 [a,c,d,b,e]" -- "= [d,b,e]"

text {* ---------------------------------------------------------------
Ejercicio 8. Definir la función
esVacia :: 'a list ⇒ bool
tal que (esVacia xs) se verifica si xs es la lista vacía. Por ejemplo,
esVacia [] = True
esVacia [1] = False
------------------------------------------------------------------ *}

(*marcarmor13, rubgonmar, danrodcha, crigomgom*)
fun esVacia :: "'a list ⇒ bool" where
"esVacia [] = True "
| "esVacia xs = False"

(*danrodcha*)
fun esVacia1 :: "'a list ⇒ bool" where
"esVacia1 xs = (xs = [])"

(*pablucoto*)
fun esVacia2 :: "'a list ⇒ bool" where
"esVacia2 xs = (if xs=[] then True else False)"

value "esVacia []" -- "= True"
value "esVacia [1]" -- "= False"

text {* ---------------------------------------------------------------
Ejercicio 9. Definir la función
inversaAc :: 'a list ⇒ 'a list
tal que (inversaAc xs) es a inversa de xs calculada usando
acumuladores. Por ejemplo,
inversaAc [a,c,b,e] = [e,b,c,a]
------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13*)
fun inversaAcAux :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux [] ys = ys"
| "inversaAcAux xs ys = inversaAcAux (tl xs) (hd xs#ys) "

(*danrodcha, crigomgom*)
fun inversaAcAux1 :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux1 [] ys = ys"
| "inversaAcAux1 (x#xs) ys = inversaAcAux1 xs (x#ys)"

(*rubgonmar, danrodcha, crigomgom*)
fun inversaAc :: "'a list ⇒ 'a list" where
"inversaAc xs = inversaAcAux xs []"

(*pablucoto*)
fun inversaAcAux2 :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux2 [] [] = []"|
"inversaAcAux2 xs (y#ys) = (inversaAcAux2 [] ys) @ [y]"
fun inversaAc2 :: "'a list ⇒ 'a list" where
"inversaAc2 xs = inversaAcAux2 [] xs "

value "inversaAc [a,c,b,e]" -- "= [e,b,c,a]"

text {* ---------------------------------------------------------------
Ejercicio 10. Definir la función
sum :: nat list ⇒ nat
tal que (sum xs) es la suma de los elementos de xs. Por ejemplo,
sum [3,2,5] = 10
------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13*)
fun sum :: "nat list ⇒ nat" where
"sum [] = 0"
|"sum xs = hd xs + sum (tl xs)"

(*danrodcha, crigomgom, pablucoto*)
fun sum1 :: "nat list ⇒ nat" where
"sum1 [] = 0"
| "sum1 (x#xs) = x + sum1 xs"

(*danrodcha*)
fun sum2 :: "nat list ⇒ nat" where
"sum2 xs = fold (op +) xs 0"

value "sum [3,2,5]" -- "= 10"

text {* ---------------------------------------------------------------
Ejercicio 11. Definir la función
map :: ('a ⇒ 'b) ⇒ 'a list ⇒ 'b list
tal que (map f xs) es la lista obtenida aplicando la función f a los
elementos de xs. Por ejemplo,
map (λx. 2*x) [3,2,5] = [6,4,10]
------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13*)
fun map :: "('a ⇒ 'b) ⇒ 'a list ⇒ 'b list" where
"map f [] = []"
| "map f xs = f(hd xs)#map f (tl xs)"

(*wilmorort, danrodcha, crigomgom, pablucoto*)
fun map :: "('a ⇒ 'b) ⇒ 'a list ⇒ 'b list" where
"map f [] = []"
|"map f (x # xs) = f x # map f xs" (*yo pondría paréntesis, pero sin
ellos lo entiende*)

value "map (λx. 2*x) [3::nat,2,5]" -- "= [6,4,10]"

end
</source>

Relación 1

2016-10-30T13:10:23Z

Manmorjim1:

<source lang="isar">
chapter {* R1: Programación funcional en Isabelle *}

theory R1
imports Main
begin

text {* ----------------------------------------------------------------
Ejercicio 0. Definir, por recursión, la función
factorial :: nat ⇒ nat
tal que (factorial n) es el factorial de n. Por ejemplo,
factorial 4 = 24
------------------------------------------------------------------- *}

(* danrodcha, anaprarod, ivamenjim, serrodcal, manmorjim1 *)
fun factorial :: "nat ⇒ nat" where
"factorial 0 = 1"
| "factorial (Suc n) = (Suc n) * factorial n"

(*wilmorort, pablucoto,marcarmor13, crigomgom, rubgonmar*)
fun factorial1 :: "nat ⇒ nat" where
"factorial1 0 = 1 "
| "factorial1 n = n * factorial1(n-1)"

value "factorial 4" -- "24"

text {* ----------------------------------------------------------------
Ejercicio 1. Definir, por recursión, la función
longitud :: 'a list ⇒ nat
tal que (longitud xs) es la longitud de la listas xs. Por ejemplo,
longitud [4,2,5] = 3
------------------------------------------------------------------- *}

(*wilmorort*)
(* Para usar las lista en forma de [a,b,c] *)

translations
"[x, xs]" == "x#[xs]"
"[x]" == "x#[]"

(*wilmorort, serrodcal,crigomgom,rubgonmar*)
fun longitud :: "'a list ⇒ nat" where
"longitud [] = 0" |
"longitud (x # xs) = 1 + longitud xs "

(*pablucoto*)
fun longitud0 :: "'a list ⇒ nat " where
" longitud0 [] = 0"
|"longitud0 xs = 1 + longitud0 ((butlast xs)) "

fun longitud0_1 :: "'a list ⇒ nat " where
"longitud0_1 xs = (if xs =[] then 0 else 1 + longitud0_1((butlast xs))) "

(*marcarmor13, manmorjim1*)
fun longitud1 :: "'a list ⇒ nat" where
"longitud1 [] = 0 "
| "longitud1 xs = (1+ longitud2 (tl xs))"

(*danrodcha*)
fun longitud2 :: "'a list ⇒ nat" where
"longitud2 [] = 0"
| "longitud2 (x#xs) = Suc (longitud2 xs)"

(*serrodcal*)
fun longitud3 :: "'a list ⇒ nat" where
"longitud xs = length (xs)"

value "longitud [4,2,5] " -- "= 3"

text {* ---------------------------------------------------------------
Ejercicio 2. Definir la función
fun intercambia :: 'a × 'b ⇒ 'b × 'a
tal que (intercambia p) es el par obtenido intercambiando las
componentes del par p. Por ejemplo,
intercambia (u,v) = (v,u)
------------------------------------------------------------------ *}

(*wilmorort,marcarmor13,danrodcha,crigomgom,pablucoto, rubgonmar, manmorjim1*)
fun intercambia :: "'a × 'b ⇒ 'b × 'a" where
"intercambia (x,y) = (y,x) "

value "intercambia (u,v)"-- "= (v,u)"

text {* ---------------------------------------------------------------
Ejercicio 3. Definir, por recursión, la función
inversa :: 'a list ⇒ 'a list
tal que (inversa xs) es la lista obtenida invirtiendo el orden de los
elementos de xs. Por ejemplo,
inversa [a,d,c] = [c,d,a]
------------------------------------------------------------------ *}

(*wilmorort*)
(* @ :: "'a list => 'a list => 'a list", función agregación definida
en Theory Main, concatena dos listas: [a,b] @ [c,d] = [a,b,c,d] *)
fun inversa :: "'a list ⇒ 'a list" where
"inversa [] = []" |
"inversa (x # xs) = (inversa xs)@(x#[]) "

(*marcarmor13, manmorjim1*)
fun inversa1 :: "'a list ⇒ 'a list" where
"inversa1 [] = []"
| "inversa1 xs = inversa1 (tl xs)@ ((hd xs)#[])"

(*danrodcha, pablucoto*)
(* es igual que inversa sustituyendo x#[] por [x] *)
fun inversa2 :: "'a list ⇒ 'a list" where
"inversa2 [] = []"
| "inversa2 (x#xs) = (inversa2 xs)@[x] "

(*danrodcha*)
fun inversa3 :: "'a list ⇒ 'a list" where
"inversa3 [] = []"
| "inversa3 (x#xs) = concat [(inversa3 xs),[x]] "

(*crigomgom*)
fun inversa4 :: "'a list ⇒ 'a list" where
"inversa4 [] = []" |
"inversa4 xs = (last xs)#(inversa4(butlast xs)) "

value "inversa [a,d,c]" -- "= [c,d,a]"

text {* ---------------------------------------------------------------
Ejercicio 4. Definir la función
repite :: nat ⇒ 'a ⇒ 'a list
tal que (repite n x) es la lista formada por n copias del elemento
x. Por ejemplo,
repite 3 a = [a,a,a]
------------------------------------------------------------------ *}

(*wilmorort,marcarmor13, crigomgom, pablucoto, rubgonmar*)
fun repite :: "nat ⇒ 'a ⇒ 'a list" where
"repite 0 x = [] " |
"repite n x = x # (repite(n-1) x) "

(*danrodcha*)
fun repite1 :: "nat ⇒ 'a ⇒ 'a list" where
"repite1 0 x = []"
| "repite1 (Suc n) x = x#(repite1 n x)"

value "repite 3 a" -- "= [a,a,a]"

text {* ---------------------------------------------------------------
Ejercicio 5. Definir la función
conc :: 'a list ⇒ 'a list ⇒ 'a list
tal que (conc xs ys) es la concatención de las listas xs e ys. Por
ejemplo,
conc [a,d] [b,d,a,c] = [a,d,b,d,a,c]
------------------------------------------------------------------ *}

(*marcarmor13*)
fun conc :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc xs ys = xs@ys"

(*danrodcha, crigomgom, rubgonmar*)
fun conc1 :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc1 [] ys = ys"
| "conc1 xs [] = xs" (*esta no hace falta*)
| "conc1 (x#xs) ys = x#(conc1 xs ys)"

(*pablucoto*)
fun conc2 :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc2 [] ys = ys" |
"conc2 (x#xs) ys = x # (conc2 xs ys)"

value "conc [a,d] [b,d,a,c]" -- "= [a,d,b,d,a,c]"

text {* ---------------------------------------------------------------
Ejercicio 6. Definir la función
coge :: nat ⇒ 'a list ⇒ 'a list
tal que (coge n xs) es la lista de los n primeros elementos de xs. Por
ejemplo,
coge 2 [a,c,d,b,e] = [a,c]
------------------------------------------------------------------ *}
(*marcarmor13*)
fun coge :: "nat ⇒ 'a list ⇒ 'a list" where
"coge 0 xs = []"
| "coge n xs = (hd xs)#(coge (n-1) (tl xs)) "

(*danrodcha*)
fun coge1 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge1 0 _ = []"
| "coge1 _ [] = []"
| "coge1 (Suc n) (x#xs) = x#(coge1 n xs)"

(*crimgomgom*)
fun coge2 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge2 0 _ = []"|
"coge2 _ [] = []" |
"coge2 n (x#xs) = x#(coge2 (n-1) xs) "

(*pablucoto*)
fun coge3 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge3 0 xs = []" |
"coge3 n (x#xs) = (if n>length(x#xs) then (x#xs) else x # (coge3 (n-1) xs))"

value "coge 2 [a,c,d,b,e]" -- "= [a,c]"

text {* ---------------------------------------------------------------
Ejercicio 7. Definir la función
elimina :: nat ⇒ 'a list ⇒ 'a list
tal que (elimina n xs) es la lista obtenida eliminando los n primeros
elementos de xs. Por ejemplo,
elimina 2 [a,c,d,b,e] = [d,b,e]
------------------------------------------------------------------ *}
(*marcarmor13*)
fun elimina :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina 0 xs = xs"
| "elimina n xs = (elimina (n-1) (tl xs ))"

(*danrodcha, crigomgom*)
fun elimina1 :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina1 0 xs = xs"
| "elimina1 _ [] = []"
| "elimina1 (Suc n) (x#xs) = elimina1 n xs"

(*pablucoto*)
fun elimina2 :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina2 0 xs = xs" |
"elimina2 n (x#xs) = (if n>length(x#xs) then [] else (elimina2 (n-1) xs))"

value "elimina 2 [a,c,d,b,e]" -- "= [d,b,e]"

text {* ---------------------------------------------------------------
Ejercicio 8. Definir la función
esVacia :: 'a list ⇒ bool
tal que (esVacia xs) se verifica si xs es la lista vacía. Por ejemplo,
esVacia [] = True
esVacia [1] = False
------------------------------------------------------------------ *}

(*marcarmor13, rubgonmar, danrodcha, crigomgom*)
fun esVacia :: "'a list ⇒ bool" where
"esVacia [] = True "
| "esVacia xs = False"

(*danrodcha*)
fun esVacia1 :: "'a list ⇒ bool" where
"esVacia1 xs = (xs = [])"

(*pablucoto*)
fun esVacia2 :: "'a list ⇒ bool" where
"esVacia2 xs = (if xs=[] then True else False)"

value "esVacia []" -- "= True"
value "esVacia [1]" -- "= False"

text {* ---------------------------------------------------------------
Ejercicio 9. Definir la función
inversaAc :: 'a list ⇒ 'a list
tal que (inversaAc xs) es a inversa de xs calculada usando
acumuladores. Por ejemplo,
inversaAc [a,c,b,e] = [e,b,c,a]
------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13*)
fun inversaAcAux :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux [] ys = ys"
| "inversaAcAux xs ys = inversaAcAux (tl xs) (hd xs#ys) "

(*danrodcha, crigomgom*)
fun inversaAcAux1 :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux1 [] ys = ys"
| "inversaAcAux1 (x#xs) ys = inversaAcAux1 xs (x#ys)"

(*rubgonmar, danrodcha, crigomgom*)
fun inversaAc :: "'a list ⇒ 'a list" where
"inversaAc xs = inversaAcAux xs []"

(*pablucoto*)
fun inversaAcAux2 :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux2 [] [] = []"|
"inversaAcAux2 xs (y#ys) = (inversaAcAux2 [] ys) @ [y]"
fun inversaAc2 :: "'a list ⇒ 'a list" where
"inversaAc2 xs = inversaAcAux2 [] xs "

value "inversaAc [a,c,b,e]" -- "= [e,b,c,a]"

text {* ---------------------------------------------------------------
Ejercicio 10. Definir la función
sum :: nat list ⇒ nat
tal que (sum xs) es la suma de los elementos de xs. Por ejemplo,
sum [3,2,5] = 10
------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13*)
fun sum :: "nat list ⇒ nat" where
"sum [] = 0"
|"sum xs = hd xs + sum (tl xs)"

(*danrodcha, crigomgom, pablucoto*)
fun sum1 :: "nat list ⇒ nat" where
"sum1 [] = 0"
| "sum1 (x#xs) = x + sum1 xs"

(*danrodcha*)
fun sum2 :: "nat list ⇒ nat" where
"sum2 xs = fold (op +) xs 0"

value "sum [3,2,5]" -- "= 10"

text {* ---------------------------------------------------------------
Ejercicio 11. Definir la función
map :: ('a ⇒ 'b) ⇒ 'a list ⇒ 'b list
tal que (map f xs) es la lista obtenida aplicando la función f a los
elementos de xs. Por ejemplo,
map (λx. 2*x) [3,2,5] = [6,4,10]
------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13*)
fun map :: "('a ⇒ 'b) ⇒ 'a list ⇒ 'b list" where
"map f [] = []"
| "map f xs = f(hd xs)#map f (tl xs)"

(*wilmorort, danrodcha, crigomgom, pablucoto*)
fun map :: "('a ⇒ 'b) ⇒ 'a list ⇒ 'b list" where
"map f [] = []"
|"map f (x # xs) = f x # map f xs" (*yo pondría paréntesis, pero sin
ellos lo entiende*)

value "map (λx. 2*x) [3::nat,2,5]" -- "= [6,4,10]"

end
</source>

Relación 1

2016-10-30T13:09:23Z

Manmorjim1:

<source lang="isar">
chapter {* R1: Programación funcional en Isabelle *}

theory R1
imports Main
begin

text {* ----------------------------------------------------------------
Ejercicio 0. Definir, por recursión, la función
factorial :: nat ⇒ nat
tal que (factorial n) es el factorial de n. Por ejemplo,
factorial 4 = 24
------------------------------------------------------------------- *}

(* danrodcha, anaprarod, ivamenjim, serrodcal, manmorjim1 *)
fun factorial :: "nat ⇒ nat" where
"factorial 0 = 1"
| "factorial (Suc n) = (Suc n) * factorial n"

(*wilmorort, pablucoto,marcarmor13, crigomgom, rubgonmar*)
fun factorial1 :: "nat ⇒ nat" where
"factorial1 0 = 1 "
| "factorial1 n = n * factorial1(n-1)"

value "factorial 4" -- "24"

text {* ----------------------------------------------------------------
Ejercicio 1. Definir, por recursión, la función
longitud :: 'a list ⇒ nat
tal que (longitud xs) es la longitud de la listas xs. Por ejemplo,
longitud [4,2,5] = 3
------------------------------------------------------------------- *}

(*wilmorort*)
(* Para usar las lista en forma de [a,b,c] *)

translations
"[x, xs]" == "x#[xs]"
"[x]" == "x#[]"

(*wilmorort, serrodcal,crigomgom,rubgonmar*)
fun longitud :: "'a list ⇒ nat" where
"longitud [] = 0" |
"longitud (x # xs) = 1 + longitud xs "

(*pablucoto*)
fun longitud0 :: "'a list ⇒ nat " where
" longitud0 [] = 0"
|"longitud0 xs = 1 + longitud0 ((butlast xs)) "

fun longitud0_1 :: "'a list ⇒ nat " where
"longitud0_1 xs = (if xs =[] then 0 else 1 + longitud0_1((butlast xs))) "

(*marcarmor13, manmorjim1*)
fun longitud1 :: "'a list ⇒ nat" where
"longitud1 [] = 0 "
| "longitud1 xs = (1+ longitud2 (tl xs))"

(*danrodcha*)
fun longitud2 :: "'a list ⇒ nat" where
"longitud2 [] = 0"
| "longitud2 (x#xs) = Suc (longitud2 xs)"

(*serrodcal*)
fun longitud3 :: "'a list ⇒ nat" where
"longitud xs = length (xs)"

value "longitud [4,2,5] " -- "= 3"

text {* ---------------------------------------------------------------
Ejercicio 2. Definir la función
fun intercambia :: 'a × 'b ⇒ 'b × 'a
tal que (intercambia p) es el par obtenido intercambiando las
componentes del par p. Por ejemplo,
intercambia (u,v) = (v,u)
------------------------------------------------------------------ *}

(*wilmorort,marcarmor13,danrodcha,crigomgom,pablucoto, rubgonmar, manmorjim1*)
fun intercambia :: "'a × 'b ⇒ 'b × 'a" where
"intercambia (x,y) = (y,x) "

value "intercambia (u,v)"-- "= (v,u)"

text {* ---------------------------------------------------------------
Ejercicio 3. Definir, por recursión, la función
inversa :: 'a list ⇒ 'a list
tal que (inversa xs) es la lista obtenida invirtiendo el orden de los
elementos de xs. Por ejemplo,
inversa [a,d,c] = [c,d,a]
------------------------------------------------------------------ *}

(*wilmorort*)
(* @ :: "'a list => 'a list => 'a list", función agregación definida
en Theory Main, concatena dos listas: [a,b] @ [c,d] = [a,b,c,d] *)
fun inversa :: "'a list ⇒ 'a list" where
"inversa [] = []" |
"inversa (x # xs) = (inversa xs)@(x#[]) "

(*marcarmor13*)
fun inversa1 :: "'a list ⇒ 'a list" where
"inversa1 [] = []"
| "inversa1 xs = inversa1 (tl xs)@ ((hd xs)#[])"

(*danrodcha, pablucoto*)
(* es igual que inversa sustituyendo x#[] por [x] *)
fun inversa2 :: "'a list ⇒ 'a list" where
"inversa2 [] = []"
| "inversa2 (x#xs) = (inversa2 xs)@[x] "

(*danrodcha*)
fun inversa3 :: "'a list ⇒ 'a list" where
"inversa3 [] = []"
| "inversa3 (x#xs) = concat [(inversa3 xs),[x]] "

(*crigomgom*)
fun inversa4 :: "'a list ⇒ 'a list" where
"inversa4 [] = []" |
"inversa4 xs = (last xs)#(inversa4(butlast xs)) "

value "inversa [a,d,c]" -- "= [c,d,a]"

(* manmorjim1 *)
fun inversa5 :: "'a list ⇒ 'a list" where
"inversa5 [] = []"
| "inversa5 xs = (inversa5 (tl xs)) @ (hd xs)#[]"

text {* ---------------------------------------------------------------
Ejercicio 4. Definir la función
repite :: nat ⇒ 'a ⇒ 'a list
tal que (repite n x) es la lista formada por n copias del elemento
x. Por ejemplo,
repite 3 a = [a,a,a]
------------------------------------------------------------------ *}

(*wilmorort,marcarmor13, crigomgom, pablucoto, rubgonmar*)
fun repite :: "nat ⇒ 'a ⇒ 'a list" where
"repite 0 x = [] " |
"repite n x = x # (repite(n-1) x) "

(*danrodcha*)
fun repite1 :: "nat ⇒ 'a ⇒ 'a list" where
"repite1 0 x = []"
| "repite1 (Suc n) x = x#(repite1 n x)"

value "repite 3 a" -- "= [a,a,a]"

text {* ---------------------------------------------------------------
Ejercicio 5. Definir la función
conc :: 'a list ⇒ 'a list ⇒ 'a list
tal que (conc xs ys) es la concatención de las listas xs e ys. Por
ejemplo,
conc [a,d] [b,d,a,c] = [a,d,b,d,a,c]
------------------------------------------------------------------ *}

(*marcarmor13*)
fun conc :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc xs ys = xs@ys"

(*danrodcha, crigomgom, rubgonmar*)
fun conc1 :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc1 [] ys = ys"
| "conc1 xs [] = xs" (*esta no hace falta*)
| "conc1 (x#xs) ys = x#(conc1 xs ys)"

(*pablucoto*)
fun conc2 :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc2 [] ys = ys" |
"conc2 (x#xs) ys = x # (conc2 xs ys)"

value "conc [a,d] [b,d,a,c]" -- "= [a,d,b,d,a,c]"

text {* ---------------------------------------------------------------
Ejercicio 6. Definir la función
coge :: nat ⇒ 'a list ⇒ 'a list
tal que (coge n xs) es la lista de los n primeros elementos de xs. Por
ejemplo,
coge 2 [a,c,d,b,e] = [a,c]
------------------------------------------------------------------ *}
(*marcarmor13*)
fun coge :: "nat ⇒ 'a list ⇒ 'a list" where
"coge 0 xs = []"
| "coge n xs = (hd xs)#(coge (n-1) (tl xs)) "

(*danrodcha*)
fun coge1 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge1 0 _ = []"
| "coge1 _ [] = []"
| "coge1 (Suc n) (x#xs) = x#(coge1 n xs)"

(*crimgomgom*)
fun coge2 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge2 0 _ = []"|
"coge2 _ [] = []" |
"coge2 n (x#xs) = x#(coge2 (n-1) xs) "

(*pablucoto*)
fun coge3 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge3 0 xs = []" |
"coge3 n (x#xs) = (if n>length(x#xs) then (x#xs) else x # (coge3 (n-1) xs))"

value "coge 2 [a,c,d,b,e]" -- "= [a,c]"

text {* ---------------------------------------------------------------
Ejercicio 7. Definir la función
elimina :: nat ⇒ 'a list ⇒ 'a list
tal que (elimina n xs) es la lista obtenida eliminando los n primeros
elementos de xs. Por ejemplo,
elimina 2 [a,c,d,b,e] = [d,b,e]
------------------------------------------------------------------ *}
(*marcarmor13*)
fun elimina :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina 0 xs = xs"
| "elimina n xs = (elimina (n-1) (tl xs ))"

(*danrodcha, crigomgom*)
fun elimina1 :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina1 0 xs = xs"
| "elimina1 _ [] = []"
| "elimina1 (Suc n) (x#xs) = elimina1 n xs"

(*pablucoto*)
fun elimina2 :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina2 0 xs = xs" |
"elimina2 n (x#xs) = (if n>length(x#xs) then [] else (elimina2 (n-1) xs))"

value "elimina 2 [a,c,d,b,e]" -- "= [d,b,e]"

text {* ---------------------------------------------------------------
Ejercicio 8. Definir la función
esVacia :: 'a list ⇒ bool
tal que (esVacia xs) se verifica si xs es la lista vacía. Por ejemplo,
esVacia [] = True
esVacia [1] = False
------------------------------------------------------------------ *}

(*marcarmor13, rubgonmar, danrodcha, crigomgom*)
fun esVacia :: "'a list ⇒ bool" where
"esVacia [] = True "
| "esVacia xs = False"

(*danrodcha*)
fun esVacia1 :: "'a list ⇒ bool" where
"esVacia1 xs = (xs = [])"

(*pablucoto*)
fun esVacia2 :: "'a list ⇒ bool" where
"esVacia2 xs = (if xs=[] then True else False)"

value "esVacia []" -- "= True"
value "esVacia [1]" -- "= False"

text {* ---------------------------------------------------------------
Ejercicio 9. Definir la función
inversaAc :: 'a list ⇒ 'a list
tal que (inversaAc xs) es a inversa de xs calculada usando
acumuladores. Por ejemplo,
inversaAc [a,c,b,e] = [e,b,c,a]
------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13*)
fun inversaAcAux :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux [] ys = ys"
| "inversaAcAux xs ys = inversaAcAux (tl xs) (hd xs#ys) "

(*danrodcha, crigomgom*)
fun inversaAcAux1 :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux1 [] ys = ys"
| "inversaAcAux1 (x#xs) ys = inversaAcAux1 xs (x#ys)"

(*rubgonmar, danrodcha, crigomgom*)
fun inversaAc :: "'a list ⇒ 'a list" where
"inversaAc xs = inversaAcAux xs []"

(*pablucoto*)
fun inversaAcAux2 :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux2 [] [] = []"|
"inversaAcAux2 xs (y#ys) = (inversaAcAux2 [] ys) @ [y]"
fun inversaAc2 :: "'a list ⇒ 'a list" where
"inversaAc2 xs = inversaAcAux2 [] xs "

value "inversaAc [a,c,b,e]" -- "= [e,b,c,a]"

text {* ---------------------------------------------------------------
Ejercicio 10. Definir la función
sum :: nat list ⇒ nat
tal que (sum xs) es la suma de los elementos de xs. Por ejemplo,
sum [3,2,5] = 10
------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13*)
fun sum :: "nat list ⇒ nat" where
"sum [] = 0"
|"sum xs = hd xs + sum (tl xs)"

(*danrodcha, crigomgom, pablucoto*)
fun sum1 :: "nat list ⇒ nat" where
"sum1 [] = 0"
| "sum1 (x#xs) = x + sum1 xs"

(*danrodcha*)
fun sum2 :: "nat list ⇒ nat" where
"sum2 xs = fold (op +) xs 0"

value "sum [3,2,5]" -- "= 10"

text {* ---------------------------------------------------------------
Ejercicio 11. Definir la función
map :: ('a ⇒ 'b) ⇒ 'a list ⇒ 'b list
tal que (map f xs) es la lista obtenida aplicando la función f a los
elementos de xs. Por ejemplo,
map (λx. 2*x) [3,2,5] = [6,4,10]
------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13*)
fun map :: "('a ⇒ 'b) ⇒ 'a list ⇒ 'b list" where
"map f [] = []"
| "map f xs = f(hd xs)#map f (tl xs)"

(*wilmorort, danrodcha, crigomgom, pablucoto*)
fun map :: "('a ⇒ 'b) ⇒ 'a list ⇒ 'b list" where
"map f [] = []"
|"map f (x # xs) = f x # map f xs" (*yo pondría paréntesis, pero sin
ellos lo entiende*)

value "map (λx. 2*x) [3::nat,2,5]" -- "= [6,4,10]"

end
</source>

Relación 1

2016-10-30T12:38:58Z

Manmorjim1:

<source lang="isar">
chapter {* R1: Programación funcional en Isabelle *}

theory R1
imports Main
begin

text {* ----------------------------------------------------------------
Ejercicio 0. Definir, por recursión, la función
factorial :: nat ⇒ nat
tal que (factorial n) es el factorial de n. Por ejemplo,
factorial 4 = 24
------------------------------------------------------------------- *}

(* danrodcha, anaprarod, ivamenjim, serrodcal, manmorjim1 *)
fun factorial :: "nat ⇒ nat" where
"factorial 0 = 1"
| "factorial (Suc n) = (Suc n) * factorial n"

(*wilmorort, pablucoto,marcarmor13, crigomgom, rubgonmar*)
fun factorial1 :: "nat ⇒ nat" where
"factorial1 0 = 1 "
| "factorial1 n = n * factorial1(n-1)"

value "factorial 4" -- "24"

text {* ----------------------------------------------------------------
Ejercicio 1. Definir, por recursión, la función
longitud :: 'a list ⇒ nat
tal que (longitud xs) es la longitud de la listas xs. Por ejemplo,
longitud [4,2,5] = 3
------------------------------------------------------------------- *}

(*wilmorort*)
(* Para usar las lista en forma de [a,b,c] *)

translations
"[x, xs]" == "x#[xs]"
"[x]" == "x#[]"

(*wilmorort, serrodcal,crigomgom,rubgonmar*)
fun longitud :: "'a list ⇒ nat" where
"longitud [] = 0" |
"longitud (x # xs) = 1 + longitud xs "

(*pablucoto*)
fun longitud0 :: "'a list ⇒ nat " where
" longitud0 [] = 0"
|"longitud0 xs = 1 + longitud0 ((butlast xs)) "

fun longitud0_1 :: "'a list ⇒ nat " where
"longitud0_1 xs = (if xs =[] then 0 else 1 + longitud0_1((butlast xs))) "

(*marcarmor13, manmorjim1*)
fun longitud1 :: "'a list ⇒ nat" where
"longitud1 [] = 0 "
| "longitud1 xs = (1+ longitud2 (tl xs))"

(*danrodcha*)
fun longitud2 :: "'a list ⇒ nat" where
"longitud2 [] = 0"
| "longitud2 (x#xs) = Suc (longitud2 xs)"

(*serrodcal*)
fun longitud3 :: "'a list ⇒ nat" where
"longitud xs = length (xs)"

value "longitud [4,2,5] " -- "= 3"

text {* ---------------------------------------------------------------
Ejercicio 2. Definir la función
fun intercambia :: 'a × 'b ⇒ 'b × 'a
tal que (intercambia p) es el par obtenido intercambiando las
componentes del par p. Por ejemplo,
intercambia (u,v) = (v,u)
------------------------------------------------------------------ *}

(*wilmorort,marcarmor13,danrodcha,crigomgom,pablucoto, rubgonmar, manmorjim1*)
fun intercambia :: "'a × 'b ⇒ 'b × 'a" where
"intercambia (x,y) = (y,x) "

value "intercambia (u,v)"-- "= (v,u)"

text {* ---------------------------------------------------------------
Ejercicio 3. Definir, por recursión, la función
inversa :: 'a list ⇒ 'a list
tal que (inversa xs) es la lista obtenida invirtiendo el orden de los
elementos de xs. Por ejemplo,
inversa [a,d,c] = [c,d,a]
------------------------------------------------------------------ *}

(*wilmorort*)
(* @ :: "'a list => 'a list => 'a list", función agregación definida
en Theory Main, concatena dos listas: [a,b] @ [c,d] = [a,b,c,d] *)
fun inversa :: "'a list ⇒ 'a list" where
"inversa [] = []" |
"inversa (x # xs) = (inversa xs)@(x#[]) "

(*marcarmor13*)
fun inversa1 :: "'a list ⇒ 'a list" where
"inversa1 [] = []"
| "inversa1 xs = inversa1 (tl xs)@ ((hd xs)#[])"

(*danrodcha, pablucoto*)
(* es igual que inversa sustituyendo x#[] por [x] *)
fun inversa2 :: "'a list ⇒ 'a list" where
"inversa2 [] = []"
| "inversa2 (x#xs) = (inversa2 xs)@[x] "

(*danrodcha*)
fun inversa3 :: "'a list ⇒ 'a list" where
"inversa3 [] = []"
| "inversa3 (x#xs) = concat [(inversa3 xs),[x]] "

(*crigomgom*)
fun inversa4 :: "'a list ⇒ 'a list" where
"inversa4 [] = []" |
"inversa4 xs = (last xs)#(inversa4(butlast xs)) "

value "inversa [a,d,c]" -- "= [c,d,a]"

text {* ---------------------------------------------------------------
Ejercicio 4. Definir la función
repite :: nat ⇒ 'a ⇒ 'a list
tal que (repite n x) es la lista formada por n copias del elemento
x. Por ejemplo,
repite 3 a = [a,a,a]
------------------------------------------------------------------ *}

(*wilmorort,marcarmor13, crigomgom, pablucoto, rubgonmar*)
fun repite :: "nat ⇒ 'a ⇒ 'a list" where
"repite 0 x = [] " |
"repite n x = x # (repite(n-1) x) "

(*danrodcha*)
fun repite1 :: "nat ⇒ 'a ⇒ 'a list" where
"repite1 0 x = []"
| "repite1 (Suc n) x = x#(repite1 n x)"

value "repite 3 a" -- "= [a,a,a]"

text {* ---------------------------------------------------------------
Ejercicio 5. Definir la función
conc :: 'a list ⇒ 'a list ⇒ 'a list
tal que (conc xs ys) es la concatención de las listas xs e ys. Por
ejemplo,
conc [a,d] [b,d,a,c] = [a,d,b,d,a,c]
------------------------------------------------------------------ *}

(*marcarmor13*)
fun conc :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc xs ys = xs@ys"

(*danrodcha, crigomgom, rubgonmar*)
fun conc1 :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc1 [] ys = ys"
| "conc1 xs [] = xs" (*esta no hace falta*)
| "conc1 (x#xs) ys = x#(conc1 xs ys)"

(*pablucoto*)
fun conc2 :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc2 [] ys = ys" |
"conc2 (x#xs) ys = x # (conc2 xs ys)"

value "conc [a,d] [b,d,a,c]" -- "= [a,d,b,d,a,c]"

text {* ---------------------------------------------------------------
Ejercicio 6. Definir la función
coge :: nat ⇒ 'a list ⇒ 'a list
tal que (coge n xs) es la lista de los n primeros elementos de xs. Por
ejemplo,
coge 2 [a,c,d,b,e] = [a,c]
------------------------------------------------------------------ *}
(*marcarmor13*)
fun coge :: "nat ⇒ 'a list ⇒ 'a list" where
"coge 0 xs = []"
| "coge n xs = (hd xs)#(coge (n-1) (tl xs)) "

(*danrodcha*)
fun coge1 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge1 0 _ = []"
| "coge1 _ [] = []"
| "coge1 (Suc n) (x#xs) = x#(coge1 n xs)"

(*crimgomgom*)
fun coge2 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge2 0 _ = []"|
"coge2 _ [] = []" |
"coge2 n (x#xs) = x#(coge2 (n-1) xs) "

(*pablucoto*)
fun coge3 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge3 0 xs = []" |
"coge3 n (x#xs) = (if n>length(x#xs) then (x#xs) else x # (coge3 (n-1) xs))"

value "coge 2 [a,c,d,b,e]" -- "= [a,c]"

text {* ---------------------------------------------------------------
Ejercicio 7. Definir la función
elimina :: nat ⇒ 'a list ⇒ 'a list
tal que (elimina n xs) es la lista obtenida eliminando los n primeros
elementos de xs. Por ejemplo,
elimina 2 [a,c,d,b,e] = [d,b,e]
------------------------------------------------------------------ *}
(*marcarmor13*)
fun elimina :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina 0 xs = xs"
| "elimina n xs = (elimina (n-1) (tl xs ))"

(*danrodcha, crigomgom*)
fun elimina1 :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina1 0 xs = xs"
| "elimina1 _ [] = []"
| "elimina1 (Suc n) (x#xs) = elimina1 n xs"

(*pablucoto*)
fun elimina2 :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina2 0 xs = xs" |
"elimina2 n (x#xs) = (if n>length(x#xs) then [] else (elimina2 (n-1) xs))"

value "elimina 2 [a,c,d,b,e]" -- "= [d,b,e]"

text {* ---------------------------------------------------------------
Ejercicio 8. Definir la función
esVacia :: 'a list ⇒ bool
tal que (esVacia xs) se verifica si xs es la lista vacía. Por ejemplo,
esVacia [] = True
esVacia [1] = False
------------------------------------------------------------------ *}

(*marcarmor13, rubgonmar, danrodcha, crigomgom*)
fun esVacia :: "'a list ⇒ bool" where
"esVacia [] = True "
| "esVacia xs = False"

(*danrodcha*)
fun esVacia1 :: "'a list ⇒ bool" where
"esVacia1 xs = (xs = [])"

(*pablucoto*)
fun esVacia2 :: "'a list ⇒ bool" where
"esVacia2 xs = (if xs=[] then True else False)"

value "esVacia []" -- "= True"
value "esVacia [1]" -- "= False"

text {* ---------------------------------------------------------------
Ejercicio 9. Definir la función
inversaAc :: 'a list ⇒ 'a list
tal que (inversaAc xs) es a inversa de xs calculada usando
acumuladores. Por ejemplo,
inversaAc [a,c,b,e] = [e,b,c,a]
------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13*)
fun inversaAcAux :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux [] ys = ys"
| "inversaAcAux xs ys = inversaAcAux (tl xs) (hd xs#ys) "

(*danrodcha, crigomgom*)
fun inversaAcAux1 :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux1 [] ys = ys"
| "inversaAcAux1 (x#xs) ys = inversaAcAux1 xs (x#ys)"

(*rubgonmar, danrodcha, crigomgom*)
fun inversaAc :: "'a list ⇒ 'a list" where
"inversaAc xs = inversaAcAux xs []"

(*pablucoto*)
fun inversaAcAux2 :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux2 [] [] = []"|
"inversaAcAux2 xs (y#ys) = (inversaAcAux2 [] ys) @ [y]"
fun inversaAc2 :: "'a list ⇒ 'a list" where
"inversaAc2 xs = inversaAcAux2 [] xs "

value "inversaAc [a,c,b,e]" -- "= [e,b,c,a]"

text {* ---------------------------------------------------------------
Ejercicio 10. Definir la función
sum :: nat list ⇒ nat
tal que (sum xs) es la suma de los elementos de xs. Por ejemplo,
sum [3,2,5] = 10
------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13*)
fun sum :: "nat list ⇒ nat" where
"sum [] = 0"
|"sum xs = hd xs + sum (tl xs)"

(*danrodcha, crigomgom, pablucoto*)
fun sum1 :: "nat list ⇒ nat" where
"sum1 [] = 0"
| "sum1 (x#xs) = x + sum1 xs"

(*danrodcha*)
fun sum2 :: "nat list ⇒ nat" where
"sum2 xs = fold (op +) xs 0"

value "sum [3,2,5]" -- "= 10"

text {* ---------------------------------------------------------------
Ejercicio 11. Definir la función
map :: ('a ⇒ 'b) ⇒ 'a list ⇒ 'b list
tal que (map f xs) es la lista obtenida aplicando la función f a los
elementos de xs. Por ejemplo,
map (λx. 2*x) [3,2,5] = [6,4,10]
------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13*)
fun map :: "('a ⇒ 'b) ⇒ 'a list ⇒ 'b list" where
"map f [] = []"
| "map f xs = f(hd xs)#map f (tl xs)"

(*wilmorort, danrodcha, crigomgom, pablucoto*)
fun map :: "('a ⇒ 'b) ⇒ 'a list ⇒ 'b list" where
"map f [] = []"
|"map f (x # xs) = f x # map f xs" (*yo pondría paréntesis, pero sin
ellos lo entiende*)

value "map (λx. 2*x) [3::nat,2,5]" -- "= [6,4,10]"

end
</source>

Relación 1

2016-10-30T12:36:19Z

Manmorjim1:

<source lang="isar">
chapter {* R1: Programación funcional en Isabelle *}

theory R1
imports Main
begin

text {* ----------------------------------------------------------------
Ejercicio 0. Definir, por recursión, la función
factorial :: nat ⇒ nat
tal que (factorial n) es el factorial de n. Por ejemplo,
factorial 4 = 24
------------------------------------------------------------------- *}

(* danrodcha, anaprarod, ivamenjim, serrodcal, manmorjim1 *)
fun factorial :: "nat ⇒ nat" where
"factorial 0 = 1"
| "factorial (Suc n) = (Suc n) * factorial n"

(*wilmorort, pablucoto,marcarmor13, crigomgom, rubgonmar*)
fun factorial1 :: "nat ⇒ nat" where
"factorial1 0 = 1 "
| "factorial1 n = n * factorial1(n-1)"

value "factorial 4" -- "24"

text {* ----------------------------------------------------------------
Ejercicio 1. Definir, por recursión, la función
longitud :: 'a list ⇒ nat
tal que (longitud xs) es la longitud de la listas xs. Por ejemplo,
longitud [4,2,5] = 3
------------------------------------------------------------------- *}

(*wilmorort*)
(* Para usar las lista en forma de [a,b,c] *)

translations
"[x, xs]" == "x#[xs]"
"[x]" == "x#[]"

(*wilmorort, serrodcal,crigomgom,rubgonmar*)
fun longitud :: "'a list ⇒ nat" where
"longitud [] = 0" |
"longitud (x # xs) = 1 + longitud xs "

(*pablucoto*)
fun longitud0 :: "'a list ⇒ nat " where
" longitud0 [] = 0"
|"longitud0 xs = 1 + longitud0 ((butlast xs)) "

fun longitud0_1 :: "'a list ⇒ nat " where
"longitud0_1 xs = (if xs =[] then 0 else 1 + longitud0_1((butlast xs))) "

(*marcarmor13, manmorjim1*)
fun longitud1 :: "'a list ⇒ nat" where
"longitud1 [] = 0 "
| "longitud1 xs = (1+ longitud2 (tl xs))"

(*danrodcha*)
fun longitud2 :: "'a list ⇒ nat" where
"longitud2 [] = 0"
| "longitud2 (x#xs) = Suc (longitud2 xs)"

(*serrodcal*)
fun longitud3 :: "'a list ⇒ nat" where
"longitud xs = length (xs)"

value "longitud [4,2,5] " -- "= 3"

text {* ---------------------------------------------------------------
Ejercicio 2. Definir la función
fun intercambia :: 'a × 'b ⇒ 'b × 'a
tal que (intercambia p) es el par obtenido intercambiando las
componentes del par p. Por ejemplo,
intercambia (u,v) = (v,u)
------------------------------------------------------------------ *}

(*wilmorort,marcarmor13,danrodcha,crigomgom,pablucoto, rubgonmar*)
fun intercambia :: "'a × 'b ⇒ 'b × 'a" where
"intercambia (x,y) = (y,x) "

value "intercambia (u,v)"-- "= (v,u)"

text {* ---------------------------------------------------------------
Ejercicio 3. Definir, por recursión, la función
inversa :: 'a list ⇒ 'a list
tal que (inversa xs) es la lista obtenida invirtiendo el orden de los
elementos de xs. Por ejemplo,
inversa [a,d,c] = [c,d,a]
------------------------------------------------------------------ *}

(*wilmorort*)
(* @ :: "'a list => 'a list => 'a list", función agregación definida
en Theory Main, concatena dos listas: [a,b] @ [c,d] = [a,b,c,d] *)
fun inversa :: "'a list ⇒ 'a list" where
"inversa [] = []" |
"inversa (x # xs) = (inversa xs)@(x#[]) "

(*marcarmor13*)
fun inversa1 :: "'a list ⇒ 'a list" where
"inversa1 [] = []"
| "inversa1 xs = inversa1 (tl xs)@ ((hd xs)#[])"

(*danrodcha, pablucoto*)
(* es igual que inversa sustituyendo x#[] por [x] *)
fun inversa2 :: "'a list ⇒ 'a list" where
"inversa2 [] = []"
| "inversa2 (x#xs) = (inversa2 xs)@[x] "

(*danrodcha*)
fun inversa3 :: "'a list ⇒ 'a list" where
"inversa3 [] = []"
| "inversa3 (x#xs) = concat [(inversa3 xs),[x]] "

(*crigomgom*)
fun inversa4 :: "'a list ⇒ 'a list" where
"inversa4 [] = []" |
"inversa4 xs = (last xs)#(inversa4(butlast xs)) "

value "inversa [a,d,c]" -- "= [c,d,a]"

text {* ---------------------------------------------------------------
Ejercicio 4. Definir la función
repite :: nat ⇒ 'a ⇒ 'a list
tal que (repite n x) es la lista formada por n copias del elemento
x. Por ejemplo,
repite 3 a = [a,a,a]
------------------------------------------------------------------ *}

(*wilmorort,marcarmor13, crigomgom, pablucoto, rubgonmar*)
fun repite :: "nat ⇒ 'a ⇒ 'a list" where
"repite 0 x = [] " |
"repite n x = x # (repite(n-1) x) "

(*danrodcha*)
fun repite1 :: "nat ⇒ 'a ⇒ 'a list" where
"repite1 0 x = []"
| "repite1 (Suc n) x = x#(repite1 n x)"

value "repite 3 a" -- "= [a,a,a]"

text {* ---------------------------------------------------------------
Ejercicio 5. Definir la función
conc :: 'a list ⇒ 'a list ⇒ 'a list
tal que (conc xs ys) es la concatención de las listas xs e ys. Por
ejemplo,
conc [a,d] [b,d,a,c] = [a,d,b,d,a,c]
------------------------------------------------------------------ *}

(*marcarmor13*)
fun conc :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc xs ys = xs@ys"

(*danrodcha, crigomgom, rubgonmar*)
fun conc1 :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc1 [] ys = ys"
| "conc1 xs [] = xs" (*esta no hace falta*)
| "conc1 (x#xs) ys = x#(conc1 xs ys)"

(*pablucoto*)
fun conc2 :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc2 [] ys = ys" |
"conc2 (x#xs) ys = x # (conc2 xs ys)"

value "conc [a,d] [b,d,a,c]" -- "= [a,d,b,d,a,c]"

text {* ---------------------------------------------------------------
Ejercicio 6. Definir la función
coge :: nat ⇒ 'a list ⇒ 'a list
tal que (coge n xs) es la lista de los n primeros elementos de xs. Por
ejemplo,
coge 2 [a,c,d,b,e] = [a,c]
------------------------------------------------------------------ *}
(*marcarmor13*)
fun coge :: "nat ⇒ 'a list ⇒ 'a list" where
"coge 0 xs = []"
| "coge n xs = (hd xs)#(coge (n-1) (tl xs)) "

(*danrodcha*)
fun coge1 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge1 0 _ = []"
| "coge1 _ [] = []"
| "coge1 (Suc n) (x#xs) = x#(coge1 n xs)"

(*crimgomgom*)
fun coge2 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge2 0 _ = []"|
"coge2 _ [] = []" |
"coge2 n (x#xs) = x#(coge2 (n-1) xs) "

(*pablucoto*)
fun coge3 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge3 0 xs = []" |
"coge3 n (x#xs) = (if n>length(x#xs) then (x#xs) else x # (coge3 (n-1) xs))"

value "coge 2 [a,c,d,b,e]" -- "= [a,c]"

text {* ---------------------------------------------------------------
Ejercicio 7. Definir la función
elimina :: nat ⇒ 'a list ⇒ 'a list
tal que (elimina n xs) es la lista obtenida eliminando los n primeros
elementos de xs. Por ejemplo,
elimina 2 [a,c,d,b,e] = [d,b,e]
------------------------------------------------------------------ *}
(*marcarmor13*)
fun elimina :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina 0 xs = xs"
| "elimina n xs = (elimina (n-1) (tl xs ))"

(*danrodcha, crigomgom*)
fun elimina1 :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina1 0 xs = xs"
| "elimina1 _ [] = []"
| "elimina1 (Suc n) (x#xs) = elimina1 n xs"

(*pablucoto*)
fun elimina2 :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina2 0 xs = xs" |
"elimina2 n (x#xs) = (if n>length(x#xs) then [] else (elimina2 (n-1) xs))"

value "elimina 2 [a,c,d,b,e]" -- "= [d,b,e]"

text {* ---------------------------------------------------------------
Ejercicio 8. Definir la función
esVacia :: 'a list ⇒ bool
tal que (esVacia xs) se verifica si xs es la lista vacía. Por ejemplo,
esVacia [] = True
esVacia [1] = False
------------------------------------------------------------------ *}

(*marcarmor13, rubgonmar, danrodcha, crigomgom*)
fun esVacia :: "'a list ⇒ bool" where
"esVacia [] = True "
| "esVacia xs = False"

(*danrodcha*)
fun esVacia1 :: "'a list ⇒ bool" where
"esVacia1 xs = (xs = [])"

(*pablucoto*)
fun esVacia2 :: "'a list ⇒ bool" where
"esVacia2 xs = (if xs=[] then True else False)"

value "esVacia []" -- "= True"
value "esVacia [1]" -- "= False"

text {* ---------------------------------------------------------------
Ejercicio 9. Definir la función
inversaAc :: 'a list ⇒ 'a list
tal que (inversaAc xs) es a inversa de xs calculada usando
acumuladores. Por ejemplo,
inversaAc [a,c,b,e] = [e,b,c,a]
------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13*)
fun inversaAcAux :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux [] ys = ys"
| "inversaAcAux xs ys = inversaAcAux (tl xs) (hd xs#ys) "

(*danrodcha, crigomgom*)
fun inversaAcAux1 :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux1 [] ys = ys"
| "inversaAcAux1 (x#xs) ys = inversaAcAux1 xs (x#ys)"

(*rubgonmar, danrodcha, crigomgom*)
fun inversaAc :: "'a list ⇒ 'a list" where
"inversaAc xs = inversaAcAux xs []"

(*pablucoto*)
fun inversaAcAux2 :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux2 [] [] = []"|
"inversaAcAux2 xs (y#ys) = (inversaAcAux2 [] ys) @ [y]"
fun inversaAc2 :: "'a list ⇒ 'a list" where
"inversaAc2 xs = inversaAcAux2 [] xs "

value "inversaAc [a,c,b,e]" -- "= [e,b,c,a]"

text {* ---------------------------------------------------------------
Ejercicio 10. Definir la función
sum :: nat list ⇒ nat
tal que (sum xs) es la suma de los elementos de xs. Por ejemplo,
sum [3,2,5] = 10
------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13*)
fun sum :: "nat list ⇒ nat" where
"sum [] = 0"
|"sum xs = hd xs + sum (tl xs)"

(*danrodcha, crigomgom, pablucoto*)
fun sum1 :: "nat list ⇒ nat" where
"sum1 [] = 0"
| "sum1 (x#xs) = x + sum1 xs"

(*danrodcha*)
fun sum2 :: "nat list ⇒ nat" where
"sum2 xs = fold (op +) xs 0"

value "sum [3,2,5]" -- "= 10"

text {* ---------------------------------------------------------------
Ejercicio 11. Definir la función
map :: ('a ⇒ 'b) ⇒ 'a list ⇒ 'b list
tal que (map f xs) es la lista obtenida aplicando la función f a los
elementos de xs. Por ejemplo,
map (λx. 2*x) [3,2,5] = [6,4,10]
------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13*)
fun map :: "('a ⇒ 'b) ⇒ 'a list ⇒ 'b list" where
"map f [] = []"
| "map f xs = f(hd xs)#map f (tl xs)"

(*wilmorort, danrodcha, crigomgom, pablucoto*)
fun map :: "('a ⇒ 'b) ⇒ 'a list ⇒ 'b list" where
"map f [] = []"
|"map f (x # xs) = f x # map f xs" (*yo pondría paréntesis, pero sin
ellos lo entiende*)

value "map (λx. 2*x) [3::nat,2,5]" -- "= [6,4,10]"

end
</source>

Relación 1

2016-10-30T12:33:38Z

Manmorjim1:

<source lang="isar">
chapter {* R1: Programación funcional en Isabelle *}

theory R1
imports Main
begin

text {* ----------------------------------------------------------------
Ejercicio 0. Definir, por recursión, la función
factorial :: nat ⇒ nat
tal que (factorial n) es el factorial de n. Por ejemplo,
factorial 4 = 24
------------------------------------------------------------------- *}

(* danrodcha, anaprarod, ivamenjim, serrodcal, manmorjim1 *)
fun factorial :: "nat ⇒ nat" where
"factorial 0 = 1"
| "factorial (Suc n) = (Suc n) * factorial n"

(*wilmorort, pablucoto,marcarmor13, crigomgom, rubgonmar*)
fun factorial1 :: "nat ⇒ nat" where
"factorial1 0 = 1 "
| "factorial1 n = n * factorial1(n-1)"

value "factorial 4" -- "24"

text {* ----------------------------------------------------------------
Ejercicio 1. Definir, por recursión, la función
longitud :: 'a list ⇒ nat
tal que (longitud xs) es la longitud de la listas xs. Por ejemplo,
longitud [4,2,5] = 3
------------------------------------------------------------------- *}

(*wilmorort*)
(* Para usar las lista en forma de [a,b,c] *)

translations
"[x, xs]" == "x#[xs]"
"[x]" == "x#[]"

(*wilmorort, serrodcal,crigomgom,rubgonmar*)
fun longitud :: "'a list ⇒ nat" where
"longitud [] = 0" |
"longitud (x # xs) = 1 + longitud xs "

(*pablucoto*)
fun longitud0 :: "'a list ⇒ nat " where
" longitud0 [] = 0"
|"longitud0 xs = 1 + longitud0 ((butlast xs)) "

fun longitud0_1 :: "'a list ⇒ nat " where
"longitud0_1 xs = (if xs =[] then 0 else 1 + longitud0_1((butlast xs))) "

(*marcarmor13*)
fun longitud1 :: "'a list ⇒ nat" where
"longitud1 [] = 0 "
| "longitud1 xs = (1+ longitud2 (tl xs))"

(*danrodcha*)
fun longitud2 :: "'a list ⇒ nat" where
"longitud2 [] = 0"
| "longitud2 (x#xs) = Suc (longitud2 xs)"

(*serrodcal*)
fun longitud3 :: "'a list ⇒ nat" where
"longitud xs = length (xs)"

value "longitud [4,2,5] " -- "= 3"

text {* ---------------------------------------------------------------
Ejercicio 2. Definir la función
fun intercambia :: 'a × 'b ⇒ 'b × 'a
tal que (intercambia p) es el par obtenido intercambiando las
componentes del par p. Por ejemplo,
intercambia (u,v) = (v,u)
------------------------------------------------------------------ *}

(*wilmorort,marcarmor13,danrodcha,crigomgom,pablucoto, rubgonmar*)
fun intercambia :: "'a × 'b ⇒ 'b × 'a" where
"intercambia (x,y) = (y,x) "

value "intercambia (u,v)"-- "= (v,u)"

text {* ---------------------------------------------------------------
Ejercicio 3. Definir, por recursión, la función
inversa :: 'a list ⇒ 'a list
tal que (inversa xs) es la lista obtenida invirtiendo el orden de los
elementos de xs. Por ejemplo,
inversa [a,d,c] = [c,d,a]
------------------------------------------------------------------ *}

(*wilmorort*)
(* @ :: "'a list => 'a list => 'a list", función agregación definida
en Theory Main, concatena dos listas: [a,b] @ [c,d] = [a,b,c,d] *)
fun inversa :: "'a list ⇒ 'a list" where
"inversa [] = []" |
"inversa (x # xs) = (inversa xs)@(x#[]) "

(*marcarmor13*)
fun inversa1 :: "'a list ⇒ 'a list" where
"inversa1 [] = []"
| "inversa1 xs = inversa1 (tl xs)@ ((hd xs)#[])"

(*danrodcha, pablucoto*)
(* es igual que inversa sustituyendo x#[] por [x] *)
fun inversa2 :: "'a list ⇒ 'a list" where
"inversa2 [] = []"
| "inversa2 (x#xs) = (inversa2 xs)@[x] "

(*danrodcha*)
fun inversa3 :: "'a list ⇒ 'a list" where
"inversa3 [] = []"
| "inversa3 (x#xs) = concat [(inversa3 xs),[x]] "

(*crigomgom*)
fun inversa4 :: "'a list ⇒ 'a list" where
"inversa4 [] = []" |
"inversa4 xs = (last xs)#(inversa4(butlast xs)) "

value "inversa [a,d,c]" -- "= [c,d,a]"

text {* ---------------------------------------------------------------
Ejercicio 4. Definir la función
repite :: nat ⇒ 'a ⇒ 'a list
tal que (repite n x) es la lista formada por n copias del elemento
x. Por ejemplo,
repite 3 a = [a,a,a]
------------------------------------------------------------------ *}

(*wilmorort,marcarmor13, crigomgom, pablucoto, rubgonmar*)
fun repite :: "nat ⇒ 'a ⇒ 'a list" where
"repite 0 x = [] " |
"repite n x = x # (repite(n-1) x) "

(*danrodcha*)
fun repite1 :: "nat ⇒ 'a ⇒ 'a list" where
"repite1 0 x = []"
| "repite1 (Suc n) x = x#(repite1 n x)"

value "repite 3 a" -- "= [a,a,a]"

text {* ---------------------------------------------------------------
Ejercicio 5. Definir la función
conc :: 'a list ⇒ 'a list ⇒ 'a list
tal que (conc xs ys) es la concatención de las listas xs e ys. Por
ejemplo,
conc [a,d] [b,d,a,c] = [a,d,b,d,a,c]
------------------------------------------------------------------ *}

(*marcarmor13*)
fun conc :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc xs ys = xs@ys"

(*danrodcha, crigomgom, rubgonmar*)
fun conc1 :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc1 [] ys = ys"
| "conc1 xs [] = xs" (*esta no hace falta*)
| "conc1 (x#xs) ys = x#(conc1 xs ys)"

(*pablucoto*)
fun conc2 :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc2 [] ys = ys" |
"conc2 (x#xs) ys = x # (conc2 xs ys)"

value "conc [a,d] [b,d,a,c]" -- "= [a,d,b,d,a,c]"

text {* ---------------------------------------------------------------
Ejercicio 6. Definir la función
coge :: nat ⇒ 'a list ⇒ 'a list
tal que (coge n xs) es la lista de los n primeros elementos de xs. Por
ejemplo,
coge 2 [a,c,d,b,e] = [a,c]
------------------------------------------------------------------ *}
(*marcarmor13*)
fun coge :: "nat ⇒ 'a list ⇒ 'a list" where
"coge 0 xs = []"
| "coge n xs = (hd xs)#(coge (n-1) (tl xs)) "

(*danrodcha*)
fun coge1 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge1 0 _ = []"
| "coge1 _ [] = []"
| "coge1 (Suc n) (x#xs) = x#(coge1 n xs)"

(*crimgomgom*)
fun coge2 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge2 0 _ = []"|
"coge2 _ [] = []" |
"coge2 n (x#xs) = x#(coge2 (n-1) xs) "

(*pablucoto*)
fun coge3 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge3 0 xs = []" |
"coge3 n (x#xs) = (if n>length(x#xs) then (x#xs) else x # (coge3 (n-1) xs))"

value "coge 2 [a,c,d,b,e]" -- "= [a,c]"

text {* ---------------------------------------------------------------
Ejercicio 7. Definir la función
elimina :: nat ⇒ 'a list ⇒ 'a list
tal que (elimina n xs) es la lista obtenida eliminando los n primeros
elementos de xs. Por ejemplo,
elimina 2 [a,c,d,b,e] = [d,b,e]
------------------------------------------------------------------ *}
(*marcarmor13*)
fun elimina :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina 0 xs = xs"
| "elimina n xs = (elimina (n-1) (tl xs ))"

(*danrodcha, crigomgom*)
fun elimina1 :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina1 0 xs = xs"
| "elimina1 _ [] = []"
| "elimina1 (Suc n) (x#xs) = elimina1 n xs"

(*pablucoto*)
fun elimina2 :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina2 0 xs = xs" |
"elimina2 n (x#xs) = (if n>length(x#xs) then [] else (elimina2 (n-1) xs))"

value "elimina 2 [a,c,d,b,e]" -- "= [d,b,e]"

text {* ---------------------------------------------------------------
Ejercicio 8. Definir la función
esVacia :: 'a list ⇒ bool
tal que (esVacia xs) se verifica si xs es la lista vacía. Por ejemplo,
esVacia [] = True
esVacia [1] = False
------------------------------------------------------------------ *}

(*marcarmor13, rubgonmar, danrodcha, crigomgom*)
fun esVacia :: "'a list ⇒ bool" where
"esVacia [] = True "
| "esVacia xs = False"

(*danrodcha*)
fun esVacia1 :: "'a list ⇒ bool" where
"esVacia1 xs = (xs = [])"

(*pablucoto*)
fun esVacia2 :: "'a list ⇒ bool" where
"esVacia2 xs = (if xs=[] then True else False)"

value "esVacia []" -- "= True"
value "esVacia [1]" -- "= False"

text {* ---------------------------------------------------------------
Ejercicio 9. Definir la función
inversaAc :: 'a list ⇒ 'a list
tal que (inversaAc xs) es a inversa de xs calculada usando
acumuladores. Por ejemplo,
inversaAc [a,c,b,e] = [e,b,c,a]
------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13*)
fun inversaAcAux :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux [] ys = ys"
| "inversaAcAux xs ys = inversaAcAux (tl xs) (hd xs#ys) "

(*danrodcha, crigomgom*)
fun inversaAcAux1 :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux1 [] ys = ys"
| "inversaAcAux1 (x#xs) ys = inversaAcAux1 xs (x#ys)"

(*rubgonmar, danrodcha, crigomgom*)
fun inversaAc :: "'a list ⇒ 'a list" where
"inversaAc xs = inversaAcAux xs []"

(*pablucoto*)
fun inversaAcAux2 :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux2 [] [] = []"|
"inversaAcAux2 xs (y#ys) = (inversaAcAux2 [] ys) @ [y]"
fun inversaAc2 :: "'a list ⇒ 'a list" where
"inversaAc2 xs = inversaAcAux2 [] xs "

value "inversaAc [a,c,b,e]" -- "= [e,b,c,a]"

text {* ---------------------------------------------------------------
Ejercicio 10. Definir la función
sum :: nat list ⇒ nat
tal que (sum xs) es la suma de los elementos de xs. Por ejemplo,
sum [3,2,5] = 10
------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13*)
fun sum :: "nat list ⇒ nat" where
"sum [] = 0"
|"sum xs = hd xs + sum (tl xs)"

(*danrodcha, crigomgom, pablucoto*)
fun sum1 :: "nat list ⇒ nat" where
"sum1 [] = 0"
| "sum1 (x#xs) = x + sum1 xs"

(*danrodcha*)
fun sum2 :: "nat list ⇒ nat" where
"sum2 xs = fold (op +) xs 0"

value "sum [3,2,5]" -- "= 10"

text {* ---------------------------------------------------------------
Ejercicio 11. Definir la función
map :: ('a ⇒ 'b) ⇒ 'a list ⇒ 'b list
tal que (map f xs) es la lista obtenida aplicando la función f a los
elementos de xs. Por ejemplo,
map (λx. 2*x) [3,2,5] = [6,4,10]
------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13*)
fun map :: "('a ⇒ 'b) ⇒ 'a list ⇒ 'b list" where
"map f [] = []"
| "map f xs = f(hd xs)#map f (tl xs)"

(*wilmorort, danrodcha, crigomgom, pablucoto*)
fun map :: "('a ⇒ 'b) ⇒ 'a list ⇒ 'b list" where
"map f [] = []"
|"map f (x # xs) = f x # map f xs" (*yo pondría paréntesis, pero sin
ellos lo entiende*)

value "map (λx. 2*x) [3::nat,2,5]" -- "= [6,4,10]"

end
</source>