Diferencia entre revisiones de «Relación 4»
De Razonamiento automático (2014-15)
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*} | *} | ||
− | -- "davoremar,javrodviv1, jeshorcob" | + | -- "davoremar,javrodviv1, jeshorcob, emimarriv" |
fun snoc :: "'a list ⇒ 'a ⇒ 'a list" where | fun snoc :: "'a list ⇒ 'a ⇒ 'a list" where | ||
Línea 31: | Línea 31: | ||
*} | *} | ||
− | -- "davoremar,javrodviv1, jeshorcob" | + | -- "davoremar,javrodviv1, jeshorcob, emimarriv" |
lemma "snoc xs a = xs @ [a]" | lemma "snoc xs a = xs @ [a]" | ||
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*} | *} | ||
− | -- "davoremar,javrodviv1, jeshorcob" | + | -- "davoremar,javrodviv1, jeshorcob, emimarriv" |
lemma snoc_append: "snoc xs a = xs @ [a]" | lemma snoc_append: "snoc xs a = xs @ [a]" |
Revisión del 22:57 16 nov 2014
header {* R4: Cons inverso *}
theory R4
imports Main
begin
text {*
---------------------------------------------------------------------
Ejercicio 1. Definir recursivamente la función
snoc :: "'a list ⇒ 'a ⇒ 'a list"
tal que (snoc xs a) es la lista obtenida al añadir el elemento a al
final de la lista xs. Por ejemplo,
value "snoc [2,5] (3::int)" == [2,5,3]
Nota: No usar @.
---------------------------------------------------------------------
*}
-- "davoremar,javrodviv1, jeshorcob, emimarriv"
fun snoc :: "'a list ⇒ 'a ⇒ 'a list" where
"snoc [] a = [a]"
| "snoc (x#xs) a = x # (snoc xs a)"
text {*
---------------------------------------------------------------------
Ejercicio 2. Demostrar automáticamente el siguiente teorema
snoc xs a = xs @ [a]
---------------------------------------------------------------------
*}
-- "davoremar,javrodviv1, jeshorcob, emimarriv"
lemma "snoc xs a = xs @ [a]"
by (induct xs) auto
(*Pedrosrei:no hace falta usar auto, sirve simp_all*)
text {*
---------------------------------------------------------------------
Ejercicio 3. Demostrar detalladamente el siguiente teorema
snoc xs a = xs @ [a]
---------------------------------------------------------------------
*}
-- "davoremar,javrodviv1, jeshorcob, emimarriv"
lemma snoc_append: "snoc xs a = xs @ [a]"
proof (induct xs)
show "snoc [] a = [] @ [a]" by simp
next
fix x xs
assume HI: "snoc xs a = xs @ [a]"
have "snoc (x#xs) a = x # (snoc xs a)" by simp
also have "... = x # (xs @ [a])" using HI by simp
finally show "snoc (x#xs) a = (x#xs) @ [a]" by simp
qed
text {*
---------------------------------------------------------------------
Ejercicio 4. Demostrar automáticamente el siguiente lema
rev (x # xs) = snoc (rev xs) x"
---------------------------------------------------------------------
*}
-- "davoremar,javrodviv1, jeshorcob"
lemma "rev (x # xs) = snoc (rev xs) x"
by (auto simp add: snoc_append)
(*Pedrosrei:no hace falta usar auto, sirve simp*)
text {*
---------------------------------------------------------------------
Ejercicio 5. Demostrar detalladamente el siguiente lema
rev (x # xs) = snoc (rev xs) x"
---------------------------------------------------------------------
*}
-- "davoremar,javrodviv1, jeshorcob"
lemma "rev (x # xs) = snoc (rev xs) x"
proof -
have "rev (x # xs) = (rev xs) @ [x]" by simp
also have "... = snoc (rev xs) x" by (simp add:snoc_append)
finally show "rev (x # xs) = snoc (rev xs) x" by simp
qed
end