Mjoseh: Protegió «T5-1 sol» ([Editar=Solo administradores] (indefinido) [Trasladar=Solo administradores] (indefinido))

2020-06-17T15:36:57Z

Protegió «T5-1 sol» ([Editar=Solo administradores] (indefinido) [Trasladar=Solo administradores] (indefinido))

Mjoseh: Página creada con « theory ej_2_jun_sol imports Main begin lemma notnotI: "P ⟹ ¬¬ P" by auto lemma mt: "⟦F ⟶ G; ¬G⟧ ⟹ ¬F" by auto text ‹------…»

2020-06-17T15:36:45Z

Página creada con «<source lang = "isabelle"> theory ej_2_jun_sol imports Main begin lemma notnotI: "P ⟹ ¬¬ P" by auto lemma mt: "⟦F ⟶ G; ¬G⟧ ⟹ ¬F" by auto text ‹------…»

Página nueva

<source lang = "isabelle">

theory ej_2_jun_sol
imports Main
begin

lemma notnotI: "P ⟹ ¬¬ P"
by auto

lemma mt: "⟦F ⟶ G; ¬G⟧ ⟹ ¬F"
by auto

text ‹-----------------------------------------------------------------
Ejercicio 1. Consideremos el árbol binario definido por
datatype 'a arbol = H
| N "'a" "'a arbol" "'a arbol"

Por ejemplo, el árbol
e
/ \
/ \
c g
/ \ / \

se representa por "N e (N c H H) (N g H H)".

Se define las funciones
fun nHojas :: "'a arbol ⇒ nat" where
"nHojas H = 1"
| "nHojas (N x i d) = nHojas i + nHojas d"

fun nHojasAaux :: "'a arbol ⇒ nat ⇒ nat" where
"nHojasAaux H n = n + 1"
| "nHojasAaux (N x i d) n = nHojasAaux i (nHojasAaux d n)"

definition nHojasA :: "'a arbol ⇒ nat" where
"nHojasA a ≡ nHojasAaux a 0"
tales que
+ (nHojas a) es el número de hojas del árbol a. Por ejemplo,
nHojas (N e (N c H H) (N g H H)) = 4
+ (nHojasAA a) es el número de hojas del árbol a, calculado
con la función auxiliar nHojasAaux que el el segundo argumento va
acumulando el resultado. Por ejemplo,
nHojasA (N e (N c H H) (N g H H)) = 4

Demostrar detalladamente que las funciones nHojas y nHojasA son
equivalentes; es decir,
nHojasA a = nHojas a

Nota: Los únicos métodos que se pueden usar son induct y simp.
-------------------------------------------------------------------›

section ‹Definiciones›

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

fun nHojas :: "'a arbol ⇒ nat" where
"nHojas H = 1"
| "nHojas (N x i d) = nHojas i + nHojas d"

fun nHojasAaux :: "'a arbol ⇒ nat ⇒ nat" where
"nHojasAaux H n = n + 1"
| "nHojasAaux (N x i d) n = nHojasAaux i (nHojasAaux d n)"

definition nHojasA :: "'a arbol ⇒ nat" where
"nHojasA a ≡ nHojasAaux a 0"

value "nHojas (N e (N c H H) (N g H H)) = 4"
value "nHojasA (N e (N c H H) (N g H H)) = 4"

section ‹Demostraciones del lema auxiliar nHojasAux›

subsection ‹Demostración estructurada de nHojasAux›

lemma nHojasAuxE:
"nHojasAaux a n = nHojas a + n"
proof (induct a arbitrary: n)
case H
then show ?case by simp
next
case (N x1 a1 a2)
then show ?case by simp
qed

subsection ‹Demostración automática de nHojasAux›

lemma nHojasAuxA:
"nHojasAaux a n = nHojas a + n"
by (induct a arbitrary: n) simp_all

subsection ‹Demostración declarativa detallada de nHojasAux›

lemma nHojasAux:
"nHojasAaux a n = (nHojas a) + n"
proof (induct a arbitrary: n)
fix n
have "nHojasAaux H n = n + 1"
by (simp only: nHojasAaux.simps(1))
also have "… = nHojas H + n"
by (simp only:nHojas.simps(1))
finally show "nHojasAaux H n = nHojas H + n"
by this
next
fix x :: 'a and i d :: "'a arbol" and n :: nat
assume HI1: "⋀n. nHojasAaux i n = nHojas i + n"
and HI2: "⋀n. nHojasAaux d n = nHojas d + n"
have "nHojasAaux (N x i d) n = nHojasAaux i (nHojasAaux d n)"
by (simp only: nHojasAaux.simps(2))
also have "… = nHojasAaux i (nHojas d + n)"
by (simp only: HI2)
also have "… = nHojas i + (nHojas d + n)"
by (simp only: HI1)
also have "… = (nHojas i + nHojas d) + n"
by (simp only: add.commute)
also have "… = nHojas (N x i d) + n"
by (simp only: nHojas.simps(2))
finally show "nHojasAaux (N x i d) n = nHojas (N x i d) + n"
by this
qed

section ‹Demostraciones del teorema nHojasA›

subsection ‹Demostración automática de nHojasA›

theorem nHojasA_A:
"nHojasA a = nHojas a"
by (simp add: nHojasA_def nHojasAux)

subsection "Demostración declarativa detallada de nHojasA"

theorem nHojasA:
"nHojasA a = nHojas a"
proof -
have "nHojasA a = nHojasAaux a 0"
by (simp only: nHojasA_def)
also have "… = nHojas a + 0"
by (simp only: nHojasAux)
also have "... = nHojas a"
by (simp only: add_0_right)
finally show "nHojasA a = nHojas a"
by this
qed

text ‹------------------------------------------------------------------
Ejercicio 2. Sea G un grupo. Demostrar detalladamente lo siguiente:
x = y^ ⋅ z si y sólo si z = y ⋅ x

Nota: No usar ninguno de los métodos automáticos: auto, blast, force,
fast, arith o metis
------------------------------------------------------------------ ›

locale grupo =
fixes prod :: "['a, 'a] ⇒ 'a" (infixl "⋅" 70)
and neutro ("𝟭")
and inverso ("_^" [100] 100)
assumes asociativa: "(x ⋅ y) ⋅ z = x ⋅ (y ⋅ z)"
and neutro_i: "𝟭 ⋅ x = x"
and neutro_d: "x ⋅ 𝟭 = x"
and inverso_i: "x^ ⋅ x = 𝟭"

(* Notas sobre notación:
* El producto es ⋅ y se escribe con \ cdot (sin espacio entre ellos).
* El neutro es 𝟭 y se escribe con \ y one (sin espacio entre ellos).
* El inverso de x es x^ y se escribe pulsando 2 veces en ^. *)

context grupo
begin

section ‹Demostraciones de inverso_d›

subsection ‹Demostración automática de inverso_d›

lemma inverso_d_A:
"x ⋅ x^ = 𝟭"
by (metis asociativa neutro_i inverso_i)

subsection ‹Demostración declarativa detallada de inverso_d›

lemma inverso_d:
"x ⋅ x^ = 𝟭"
proof -
have "x ⋅ x^ = 𝟭 ⋅ (x ⋅ x^)"
by (simp only: neutro_i)
also have "… = (𝟭 ⋅ x) ⋅ x^"
by (simp only: asociativa)
also have "… = (((x^)^ ⋅ x^) ⋅ x) ⋅ x^"
by (simp only: inverso_i)
also have "… = ((x^)^ ⋅ (x^ ⋅ x)) ⋅ x^"
by (simp only: asociativa)
also have "… = ((x^)^ ⋅ 𝟭) ⋅ x^"
by (simp only: inverso_i)
also have "… = (x^)^ ⋅ (𝟭 ⋅ x^)"
by (simp only: asociativa)
also have "… = (x^)^ ⋅ x^"
by (simp only: neutro_i)
also have "… = 𝟭"
by (simp only: inverso_i)
finally show ?thesis
by this
qed

section ‹Demostraciones de condicion_necesaria›

subsection ‹Demostración automática de condicion_necesaria›

lemma condicion_necesaria_A:
assumes "x = y^ ⋅ z"
shows "z = y ⋅ x"
using assms
by (metis asociativa neutro_i inverso_d)

subsection ‹Demostración declarativa detallada de condicion_necesaria›

lemma condicion_necesaria:
assumes "x = y^ ⋅ z"
shows "z = y ⋅ x"
proof -
have "y ⋅ x = y ⋅ (y^ ⋅ z)"
using assms(1)
by (rule arg_cong)
also have "… = (y ⋅ y^) ⋅ z"
by (simp only: asociativa)
also have "… = 𝟭 ⋅ z"
by (simp only: inverso_d)
also have "… = z"
by (simp only: neutro_i)
finally have "y ⋅ x = z"
by this
then show "z = y ⋅ x"
by (rule sym)
qed

section ‹Demostraciones de condicion_suficiente›

subsection ‹Demostración automática de condicion_suficiente›

lemma condicion_suficiente_A:
assumes "z = y ⋅ x"
shows "x = y^ ⋅ z"
using assms(1)
by (metis asociativa neutro_i inverso_i)

subsection ‹Demostración declarativa detallada de condicion_suficiente›

lemma condicion_suficiente:
assumes "z = y ⋅ x"
shows "x = y^ ⋅ z"
proof -
have "y^ ⋅ z = y^ ⋅ (y ⋅ x)"
using assms(1)
by (rule arg_cong)
also have "… = (y^ ⋅ y) ⋅ x"
by (simp only: asociativa)
also have "… = 𝟭 ⋅ x"
by (simp only: inverso_i)
also have "… = x"
by (simp only: neutro_i)
finally have "y^ ⋅ z = x"
by this
then show "x = y^ ⋅ z"
by (rule sym)
qed

section ‹Demostraciones del teorema›

section ‹Demostraciones automática del teorema›

theorem condicion_necesaria_y_suficiente_A:
"x = y^ ⋅ z ⟷ z = y ⋅ x"
using condicion_necesaria
condicion_suficiente
by auto

section ‹Demostraciones declarativa detallada del teorema›

theorem condicion_necesaria_y_suficiente:
"x = y^ ⋅ z ⟷ z = y ⋅ x"
proof (rule iffI)
assume "x = y^ ⋅ z"
then show "z = y ⋅ x"
by (rule condicion_necesaria)
next
assume "z = y ⋅ x"
then show "x = y^ ⋅ z"
by (rule condicion_suficiente)
qed

end

end
</source>

← Revisión anterior	Revisión del 15:36 17 jun 2020
(Sin diferencias)

T5-1 sol - Historial de revisiones

Mjoseh: Protegió «T5-1 sol» ([Editar=Solo administradores] (indefinido) [Trasladar=Solo administradores] (indefinido))

Mjoseh: Página creada con « theory ej_2_jun_sol imports Main begin lemma notnotI: "P ⟹ ¬¬ P" by auto lemma mt: "⟦F ⟶ G; ¬G⟧ ⟹ ¬F" by auto text ‹------…»