Relación 4
De Razonamiento automático (2017-18)
Revisión del 03:49 14 dic 2017 de Rafferrod (discusión | contribuciones)
chapter {* R4: Cuantificadores sobre listas *}
theory R4_Cuantificadores_sobre_listas
imports Main
begin
text {*
---------------------------------------------------------------------
Ejercicio 1. Definir la función
todos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool
tal que (todos p xs) se verifica si todos los elementos de la lista
xs cumplen la propiedad p. Por ejemplo, se verifica
todos (λx. 1<length x) [[2,1,4],[1,3]]
¬todos (λx. 1<length x) [[2,1,4],[3]]
Nota: La función todos es equivalente a la predefinida list_all.
---------------------------------------------------------------------
*}
(*anddonram edupalhid luicedval cesgongut jescudero rafcabgon diwu2 macmerflo rafferrod*)
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos p [] = True"
| "todos p (x#xs) = (p x ∧ todos p xs)"
value "todos (λx. 1<length x) [[2,1,4],[1,3::nat]] = True"
value " ¬todos (λx. 1<length x) [[2,1,4],[3::nat]] = True"
text {*
---------------------------------------------------------------------
Ejercicio 2. Definir la función
algunos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool
tal que (algunos p xs) se verifica si algunos elementos de la lista
xs cumplen la propiedad p. Por ejemplo, se verifica
algunos (λx. 1<length x) [[2,1,4],[3]]
¬algunos (λx. 1<length x) [[],[3]]"
Nota: La función algunos es equivalente a la predefinida list_ex.
---------------------------------------------------------------------
*}
(*anddonram edupalhid luicedval cesgongut jescudero rafcabgon diwu2 macmerflo rafferrod*)
fun algunos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"algunos p [] = False"
| "algunos p (x#xs) = (p x ∨ algunos p xs)"
value " algunos (λx. 1<length x) [[2::nat,1,4],[3]] = True"
value " ¬algunos (λx. 1<length x) [[],[3::nat]] = True"
text {*
---------------------------------------------------------------------
Ejercicio 3.1. Demostrar o refutar automáticamente
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)
---------------------------------------------------------------------
*}
(*anddonram edupalhid luicedval cesgongut jescudero rafcabgon diwu2 macmerflo rafferrod*)
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
by (induct xs) auto
text {*
---------------------------------------------------------------------
Ejercicio 3.2. Demostrar o refutar detalladamente
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)
---------------------------------------------------------------------
*}
(*anddonram[conmutatividad and porque no sé hacerlo de otra forma] *)
lemma and_comm: "(a ∧ b) = (b ∧ a)"
by (cases a) auto
(*anddonram diwu2*)
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix a xs
assume HI:"todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (a # xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp
also have "... = (P a ∧ Q a ∧ todos P xs ∧ todos Q xs)" using HI by simp
also have "... = (P a ∧ Q a ∧ todos Q xs ∧ todos P xs)" by (simp add:and_comm)
also have "... = (P a ∧ todos Q (a # xs) ∧ todos P xs)" by simp
also have "... = (P a ∧ todos P xs ∧ Q a ∧ todos Q xs)" by (simp add:and_comm)
finally show "todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))"
by simp
qed
(* edupalhid *)
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix n xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (n # xs) =
((P n ∧ Q n) ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "... = ((P n ∧ todos P xs) ∧ (Q n ∧ todos Q xs))" by arith
also have "... = ((todos P(n#xs)) ∧ (todos Q(n#xs)))" by simp
finally show "todos (λx. P x ∧ Q x) (n#xs) =
(todos P (n#xs) ∧ todos Q (n#xs))" by simp
qed
(*rafferrod*)
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix a xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (a#xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp
also have "... = (P a ∧ Q a ∧ todos P xs ∧ todos Q xs)" using HI by simp
also have "... = (P a ∧ todos P (xs) ∧ Q a ∧ todos Q (xs))" by auto
finally show "todos (λx. P x ∧ Q x) (a#xs) = (todos P (a#xs) ∧ todos Q (a#xs))" by simp
qed
text {*
---------------------------------------------------------------------
Ejercicio 4.1. Demostrar o refutar automáticamente
todos P (x @ y) = (todos P x ∧ todos P y)
---------------------------------------------------------------------
*}
(*anddonram edupalhid cesgongut luicedval jescudero rafcabgon diwu2 macmerflo cesgongut rafferrod*)
lemma "todos P (x @ y) = (todos P x ∧ todos P y)"
by (induct x) auto
text {*
---------------------------------------------------------------------
Ejercicio 4.2. Demostrar o refutar detalladamente
todos P (x @ y) = (todos P x ∧ todos P y)
---------------------------------------------------------------------
*}
(*anddonram edupalhid luicedval rafcabgon diwu2 jescudero macmerflo cesgongut rafferrod*)
lemma todos_append:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x)
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix a x
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"
have "todos P ((a # x) @ y) = (P a ∧ todos P (x @ y))" by simp
also have "... = (P a ∧ todos P x ∧ todos P y)" using HI by simp
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp
qed
text {*
---------------------------------------------------------------------
Ejercicio 5.1. Demostrar o refutar automáticamente
todos P (rev xs) = todos P xs
---------------------------------------------------------------------
*}
(*anddonram*)
lemma "todos P (rev xs) = todos P xs"
by (induct xs) (simp_all add: todos_append and_comm)
(* edupalhid diwu2 macmerflo rafferrod*)
lemma "todos P (rev xs) = todos P xs"
by (induct xs) (auto simp add: todos_append)
text {*
---------------------------------------------------------------------
Ejercicio 5.2. Demostrar o refutar detalladamente
todos P (rev xs) = todos P xs
---------------------------------------------------------------------
*}
(*anddonram diwu2*)
lemma "todos P (rev xs) = todos P xs"
proof (induct xs)
show "todos P (rev []) = todos P []" by simp
next
fix a xs
assume HI: "todos P (rev xs) = todos P xs"
have " todos P (rev (a # xs)) = todos P (rev xs @ [a])" by simp
also have "... = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)
also have "... = (todos P xs ∧ todos P [a])" using HI by simp
finally show "todos P (rev (a # xs)) = todos P (a # xs)" by (simp add:and_comm)
qed
(* edupalhid *)
lemma "todos P (rev xs) = todos P xs"
proof (induct xs)
show "todos P (rev []) = todos P []" by simp
next
fix a xs
assume HI: "todos P (rev xs) = todos P xs"
have "todos P (rev (a # xs)) = (todos P ((rev xs)@[a]))" by simp
also have "... = (todos P (rev xs) ∧ todos P [a])"
by (simp add:todos_append)
also have "... = (todos P xs ∧ P a)" using HI by simp
also have "... = (P a ∧ todos P xs)" by arith
also have "... = (todos P (a#xs))" by simp
finally show "todos P (rev (a # xs)) = (todos P (a#xs))" by simp
qed
(*rafferrod*)
lemma "todos P (rev xs) = todos P xs"
proof (induct xs)
show "todos P (rev []) = todos P []" by simp
next
fix a xs
assume HI: "todos P (rev xs) = todos P xs"
have "todos P (rev (a#xs)) = todos P (rev xs @ [a])" by simp
also have "... = (todos P (rev xs) ∧ todos P [a])" by (simp add: todos_append)
also have "... = (todos P xs ∧ todos P [a])" using HI by simp
also have "... = (todos P [a] ∧ todos P xs)" by auto
finally show "todos P (rev (a#xs)) = todos P (a#xs)" by simp
qed
text {*
---------------------------------------------------------------------
Ejercicio 6. Demostrar o refutar:
algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)
---------------------------------------------------------------------
*}
(*anddonram Contraejemplo*)
value "let xs=[True,False]
in (algunos (λx. (λx. (x=False)) x ∧ (λx. x) x) xs =
(algunos (λx. (x=False)) xs ∧ algunos (λx. x) xs))"
(* edupalhid diwu2 rafferrod*)
lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)"
quickcheck
oops
text {*
---------------------------------------------------------------------
Ejercicio 7.1. Demostrar o refutar automáticamente
algunos P (map f xs) = algunos (P ∘ f) xs
---------------------------------------------------------------------
*}
(*anddonram edupalhid jescudero luicedval rafcabgon diwu2 macmerflo cesgongut rafferrod*)
lemma "algunos P (map f xs) = algunos (P o f) xs"
by (induct xs) auto
text {*
---------------------------------------------------------------------
Ejercicio 7.2. Demostrar o refutar datalladamente
algunos P (map f xs) = algunos (P ∘ f) xs
---------------------------------------------------------------------
*}
(*anddonram edupalhid luicedval rafcabgon diwu2 jescudero macmerflo rafferrod*)
lemma "algunos P (map f xs) = algunos (P o f) xs"
proof(induct xs)
show "algunos P (map f []) = algunos (P o f) []" by simp
next
fix a xs
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"
have "algunos P (map f (a # xs)) = algunos P (f a#map f xs)" by simp
also have "... = ((P ∘ f) a ∨ algunos P (map f xs))" by simp
also have "... = ((P ∘ f) a ∨ algunos (P ∘ f) xs) " using HI by simp
finally show "algunos P (map f (a # xs)) = algunos (P ∘ f) (a # xs)" by simp
qed
(* cesgongut *)
lemma "algunos P (map f xs) = algunos (P o f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P o f) []" by simp
next
fix x xs
assume HI: "algunos P (map f xs) = algunos (P o f) xs"
have "algunos P (map f (x#xs)) = (P (f x) ∨ algunos P (map f xs))" by simp
also have "... = (P (f x) ∨ algunos (P o f) xs)" using HI by simp
finally show "algunos P (map f (x#xs)) = algunos (P o f) (x#xs)" by simp
qed
text {*
---------------------------------------------------------------------
Ejercicio 8.1. Demostrar o refutar automáticamente
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)
---------------------------------------------------------------------
*}
(*anddonram edupalhid luicedval rafcabgon diwu2 jescudero macmerflo cesgongut rafferrod*)
lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
by (induct xs) auto
text {*
---------------------------------------------------------------------
Ejercicio 8.2. Demostrar o refutar detalladamente
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)
---------------------------------------------------------------------
*}
(*anddonram edupalhid luicedval rafcabgon diwu2 jescudero macmerflo cesgongut rafferrod*)
lemma algunos_append:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
proof (induct xs)
show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp
next
fix a xs
assume HI: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
have "algunos P ((a # xs) @ ys) = (P a ∨ algunos P (xs@ys) )" by simp
also have "... = (P a ∨ (algunos P xs ∨ algunos P ys))" using HI by simp
finally show "algunos P ((a # xs) @ ys) = (algunos P (a # xs) ∨ algunos P ys)" by simp
qed
text {*
---------------------------------------------------------------------
Ejercicio 9.1. Demostrar o refutar automáticamente
algunos P (rev xs) = algunos P xs
---------------------------------------------------------------------
*}
(*anddonram[conmutatividad or porque no sé hacerlo de otra forma] *)
lemma or_comm: "(a ∨ b) = (b ∨ a)"
by (cases a) auto
(*anddonram*)
lemma "algunos P (rev xs) = algunos P xs"
by (induct xs) (simp_all add: algunos_append or_comm)
(* edupalhid diwu2 rafferrod*)
lemma "algunos P (rev xs) = algunos P xs"
by (induct xs) (auto simp add: algunos_append)
text {*
---------------------------------------------------------------------
Ejercicio 9.2. Demostrar o refutar detalladamente
algunos P (rev xs) = algunos P xs
---------------------------------------------------------------------
*}
(*anddonram diwu2*)
lemma "algunos P (rev xs) = algunos P xs"
proof (induct xs)
show "algunos P (rev []) = algunos P []" by simp
next
fix a xs
assume HI:" algunos P (rev xs) = algunos P xs"
have " algunos P (rev (a # xs)) = algunos P (rev xs @[a]) " by simp
also have "... = (algunos P (rev xs) ∨ algunos P [a])" by (simp add: algunos_append)
also have "... = (algunos P xs ∨ P a)" using HI by simp
also have "... = (P a ∨ algunos P xs)" by (simp add:or_comm)
finally show " algunos P (rev (a # xs)) = algunos P (a # xs) " by simp
qed
(* edupalhid *)
lemma "algunos P (rev xs) = algunos P xs"
proof (induct xs)
show "algunos P (rev []) = algunos P []" by simp
next
fix a xs
assume HI:"algunos P (rev xs) = algunos P xs"
have "algunos P (rev (a # xs)) = algunos P ((rev xs) @ [a])" by simp
also have "... = ((algunos P (rev xs)) ∨ (algunos P [a]))"
by (simp add: algunos_append)
also have "... = ((algunos P xs) ∨ (algunos P [a]))" using HI by simp
also have "... = ((algunos P [a]) ∨ (algunos P xs))" by arith
finally show "algunos P (rev (a # xs)) = algunos P (a # xs)" by simp
qed
text {*
---------------------------------------------------------------------
Ejercicio 10. Encontrar un término no trivial Z tal que sea cierta la
siguiente ecuación:
algunos (λx. P x ∨ Q x) xs = Z
y demostrar la equivalencia de forma automática y detallada.
---------------------------------------------------------------------
*}
(*anddonram edupalhid diwu2 jescudero macmerflo*)
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
by (induct xs) auto
(*anddonram*)
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
proof (induct xs)
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q [])" by simp
next
fix a xs
assume HI: " algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
have "algunos (λx. P x ∨ Q x) (a # xs) = (P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)"
by simp
also have "... = (P a ∨ Q a ∨ algunos P xs ∨ algunos Q xs)" using HI by simp
also have "... = (P a ∨ Q a ∨ algunos Q xs ∨ algunos P xs)" by (simp add: or_comm)
also have "... = (P a ∨ algunos Q (a#xs) ∨ algunos P xs)" by simp
also have "... = (P a ∨ algunos P xs ∨ algunos Q (a#xs))" by (simp add: or_comm)
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (algunos P (a # xs) ∨ algunos Q (a # xs))"
by simp
qed
(* edupalhid *)
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
(is "?T xs")
proof (induct xs)
show "?T []" by simp
next
fix a xs assume HI: "?T xs"
have p1:" (Q a ∨ algunos P xs) = (algunos P xs ∨ Q a)"
by (simp add: HOL.disj_commute)
have "algunos (λx. P x ∨ Q x) (a # xs) =
(algunos P [a] ∨ algunos Q [a] ∨ algunos (λx. P x ∨ Q x) xs)"
by simp
also have "… = (P a ∨ (Q a ∨ algunos P xs) ∨ algunos Q xs)"
using HI by simp
also have "… = (P a ∨ algunos P xs ∨ Q a ∨ algunos Q xs)"
using p1 by simp
also have "… = (algunos P (a # xs) ∨ algunos Q (a # xs))" by simp
finally show "?T (a # xs)" by simp
qed
text {*
---------------------------------------------------------------------
Ejercicio 11.1. Demostrar o refutar automáticamente
algunos P xs = (¬ todos (λx. (¬ P x)) xs)
---------------------------------------------------------------------
*}
(*anddonram edupalhid diwu2 cesgongut*)
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
by (induct xs) auto
text {*
---------------------------------------------------------------------
Ejercicio 11.2. Demostrar o refutar datalladamente
algunos P xs = (¬ todos (λx. (¬ P x)) xs)
---------------------------------------------------------------------
*}
(*anddonram edupalhid diwu2*)
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
proof (induct xs)
show "algunos P [] = (¬ todos (λx. (¬ P x)) [])" by simp
next
fix a xs
assume HI:"algunos P xs = (¬ todos (λx. ¬ P x) xs)"
have " algunos P (a # xs) =(P a ∨ algunos P xs)" by simp
also have "... = (P a ∨ (¬ todos (λx. ¬ P x) xs))" using HI by simp
also have "... = (¬ (¬ P a ∧ todos (λx. ¬ P x) xs))" by simp
finally show " algunos P (a # xs) = (¬ todos (λx. ¬ P x) (a # xs))" by simp
qed
text {*
---------------------------------------------------------------------
Ejercicio 12. 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
---------------------------------------------------------------------
*}
(*anddonram edupalhd luicedval rafcabgon diwu2 jescudero macmerflo cesgongut*)
fun estaEn :: "'a ⇒ 'a list ⇒ bool" where
"estaEn x [] = 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"
text {*
---------------------------------------------------------------------
Ejercicio 13. Expresar la relación existente entre estaEn y algunos.
Demostrar dicha relación de forma automática y detallada.
---------------------------------------------------------------------
*}
(*anddonram edupalhid luicedval rafcabgon diwu2 jescudero macmerflo cesgongut*)
lemma "estaEn x xs=algunos (λy.(y=x)) xs"
by (induct xs) auto
(*anddonram edupalhid luicedval rafcabgon jescudero macmerflo cesgongut*)
lemma "estaEn x xs=algunos (λy.(y=x)) xs"
proof (induct xs)
show "estaEn x []=algunos (λy.(y=x)) []" by simp
next
fix a xs
assume HI:"estaEn x xs = algunos (λy. y = x) xs"
have "estaEn x (a # xs) =((a=x) ∨ estaEn x xs)" by simp
also have "... = ((a=x) ∨ algunos (λy. y = x) xs)" using HI by simp
finally show "estaEn x (a # xs) = algunos (λy. y = x) (a # xs)" by simp
qed
end