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| − | <source lang = "Isar">
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| − | header {* Tema 15: Razonamiento sobre programas en Isabelle *}
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| − |
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| − | theory T15
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| − | imports Main
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| − | begin
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| − |
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| − | text {*
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| − | En este tema se demuestra con Isabelle las propiedades de los
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| − | programas funcionales como se expone en el tema 8 del curso
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| − | "Informática" que puede leerse en
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| − | http://www.cs.us.es/~jalonso/cursos/i1m/temas/tema-8t.pdf
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| − | *}
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| − |
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| − | section {* Razonamiento ecuacional *}
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| − |
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| − | text {* ----------------------------------------------------------------
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| − | Ejercicio 1. Definir, por recursión, la función
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| − | longitud :: "'a list ⇒ nat" where
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| − | tal que (longitud xs) es la longitud de la listas xs. Por ejemplo,
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| − | longitud [4,2,5] = 3
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| − | ------------------------------------------------------------------- *}
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| − |
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| − | fun longitud :: "'a list ⇒ nat" where
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| − | "longitud [] = 0"
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| − | | "longitud (x#xs) = 1 + longitud xs"
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| − |
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| − | value "longitud [4,2,5]" -- "= 3"
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| − |
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| − | text {* ---------------------------------------------------------------
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| − | Ejercicio 2. Demostrar que
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| − | longitud [4,2,5] = 3
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| − | ------------------------------------------------------------------- *}
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| − |
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| − | lemma "longitud [4,2,5] = 3"
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| − | by simp
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| − |
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| − | text {* ---------------------------------------------------------------
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| − | Ejercicio 3. Definir la función
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| − | fun intercambia :: "'a × 'b ⇒ 'b × 'a"
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| − | tal que (intercambia p) es el par obtenido intercambiando las
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| − | componentes del par p. Por ejemplo,
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| − | "intercambia (2,3) = (3,2)
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| − | ------------------------------------------------------------------ *}
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| − |
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| − | fun intercambia :: "'a × 'b ⇒ 'b × 'a" where
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| − | "intercambia (x,y) = (y,x)"
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| − |
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| − | value "intercambia (2,3)" -- "= (3,2)"
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| − |
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| − | text {* ---------------------------------------------------------------
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| − | Ejercicio 4. Demostrar que
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| − | intercambia (intercambia (x,y)) = (x,y)
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| − | ------------------------------------------------------------------- *}
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| − |
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| − | lemma "intercambia (intercambia (x,y)) = (x,y)"
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| − | by simp
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| − |
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| − | text {* ---------------------------------------------------------------
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| − | Ejercicio 5. Definir, por recursión, la función
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| − | inversa :: "'a list ⇒ 'a list"
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| − | tal que (inversa xs) es la lista obtenida invirtiendo el orden de los
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| − | elementos de xs. Por ejemplo,
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| − | inversa [3,2,5] = [5,2,3]
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| − | ------------------------------------------------------------------ *}
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| − |
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| − | fun inversa :: "'a list ⇒ 'a list" where
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| − | "inversa [] = []"
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| − | | "inversa (x#xs) = inversa xs @ [x]"
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| − |
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| − | value "inversa [3,2,5]" -- "= [5,2,3]"
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| − |
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| − | text {* ---------------------------------------------------------------
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| − | Ejercicio 6. Demostrar que
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| − | inversa [x] = [x]
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| − | ------------------------------------------------------------------- *}
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| − |
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| − | lemma "inversa [x] = [x]"
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| − | by simp
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| − |
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| − | text {* ---------------------------------------------------------------
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| − | Ejercicio 7. Definir la función
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| − | repite :: "nat ⇒ 'a ⇒ 'a list" where
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| − | tal que . Por ejemplo,
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| − | repite 3 5 = [5,5,5]
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| − | ------------------------------------------------------------------ *}
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| − |
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| − | fun repite :: "nat ⇒ 'a ⇒ 'a list" where
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| − | "repite 0 x = []"
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| − | | "repite (Suc n) x = x # (repite n x)"
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| − |
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| − | value "repite 3 5" -- "= [5,5,5]"
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| − |
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| − | text {* ---------------------------------------------------------------
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| − | Ejercicio 8. Demostrar que
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| − | longitud (repite n x) = n
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| − | ------------------------------------------------------------------- *}
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| − |
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| − | lemma "longitud (repite n x) = n"
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| − | apply (induct_tac n)
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| − | apply (auto)
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| − | done
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| − |
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| − |
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| − | lemma "longitud (repite n x) = n"
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| − | by (induct n) auto
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| − |
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| − | lemma longitud_repite:
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| − | "longitud (repite n x) = n"
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| − | proof (induct n)
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| − | show "longitud (repite 0 x) = 0" by simp
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| − | next
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| − | fix n
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| − | assume hi: "longitud (repite n x) = n"
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| − | show "longitud (repite (Suc n) x) = Suc n"
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| − | proof -
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| − | have "longitud (repite (Suc n) x) = longitud (x # (repite n x))" by simp
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| − | also have "... = 1 + longitud (repite n x)" by simp
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| − | also have "... = 1 + n" using hi by simp
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| − | also have "... = Suc n" by simp
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| − | finally show ?thesis .
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| − | qed
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| − | qed
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| − |
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| − | text {* ---------------------------------------------------------------
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| − | Ejercicio 9. Definir la función
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| − | fun conc :: "'a list ⇒ 'a list ⇒ 'a list"
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| − | tal que . Por ejemplo,
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| − | conc [2,3] [4,3,5] = [2,3,4,3,5]
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| − | ------------------------------------------------------------------ *}
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| − |
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| − | fun conc :: "'a list ⇒ 'a list ⇒ 'a list" where
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| − | "conc [] ys = ys"
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| − | | "conc (x#xs) ys = x # (conc xs ys)"
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| − |
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| − | value "conc [2,3] [4,3,5]" -- "= [2,3,4,3,5]"
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| − |
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| − | text {* ---------------------------------------------------------------
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| − | Ejercicio 10. Demostrar que
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| − | conc xs (conc ys zs) = (conc xs ys) zs
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| − | ------------------------------------------------------------------- *}
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| − |
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| − | lemma "conc xs (conc ys zs) = conc (conc xs ys) zs"
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| − | by (induct xs) auto
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| − |
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| − | lemma conc_asociativa:
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| − | "conc xs (conc ys zs) = conc (conc xs ys) zs"
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| − | proof (induct xs)
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| − | show "conc [] (conc ys zs) = conc (conc [] ys) zs" by simp
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| − | next
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| − | fix x xs
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| − | assume hi: "conc xs (conc ys zs) = conc (conc xs ys) zs"
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| − | show "conc (x # xs) (conc ys zs) = conc (conc (x # xs) ys) zs"
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| − | proof -
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| − | have "conc (x # xs) (conc ys zs) = x # (conc xs (conc ys zs))" by simp
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| − | also have "... = x # (conc (conc xs ys) zs)" using hi by simp
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| − | also have "... = conc (x#(conc xs ys)) zs" by simp
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| − | also have "... = conc (conc (x # xs) ys) zs" by simp
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| − | finally show ?thesis .
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| − | qed
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| − | qed
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| − |
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| − |
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| − | text {* ---------------------------------------------------------------
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| − | Ejercicio 11. Refutar que
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| − | conc xs ys = conc ys xs
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| − | ------------------------------------------------------------------- *}
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| − |
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| − | lemma "conc xs ys = conc ys xs"
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| − | quickcheck
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| − | oops
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| − |
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| − | text {* Encuentra el contraejemplo,
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| − | xs = [a\<^isub>2]
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| − | ys = [a\<^isub>1] *}
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| − |
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| − | text {* ---------------------------------------------------------------
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| − | Ejercicio 12. Demostrar que
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| − | conc xs [] = xs
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| − | ------------------------------------------------------------------- *}
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| − |
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| − | lemma "conc xs [] = xs"
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| − | by (induct xs) auto
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| − |
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| − | text {* ---------------------------------------------------------------
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| − | Ejercicio 13. Demostrar que
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| − | longitud (conc xs ys) = longitud xs + longitud ys
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| − | ------------------------------------------------------------------- *}
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| − |
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| − | lemma "longitud (conc xs ys) = longitud xs + longitud ys"
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| − | by (induct xs) auto
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| − |
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| − | lemma long_conc:
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| − | "longitud (conc xs ys) = longitud xs + longitud ys"
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| − | proof (induct xs)
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| − | show "longitud (conc [] ys) = longitud [] + longitud ys" by simp
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| − | next
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| − | fix x xs
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| − | assume hi: "longitud (conc xs ys) = longitud xs + longitud ys"
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| − | show "longitud (conc (x # xs) ys) = longitud (x # xs) + longitud ys"
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| − | proof -
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| − | have "longitud (conc (x # xs) ys) = longitud (x # (conc xs ys))" by simp
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| − | also have "... = 1 + longitud (conc xs ys)" by simp
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| − | also have "... = 1 + (longitud xs + longitud ys)" using hi by simp
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| − | also have "... = (1+ longitud xs) + longitud ys" by simp
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| − | also have "... = longitud (x # xs) + longitud ys" by simp
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| − | finally show ?thesis .
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| − | qed
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| − | qed
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| − |
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| − |
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| − | text {* ---------------------------------------------------------------
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| − | Ejercicio 14. Definir la función
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| − | coge :: "nat ⇒ 'a list ⇒ 'a list"
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| − | tal que (coge n xs) es la lista de los n primeros elementos de xs. Por
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| − | ejemplo,
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| − | coge 2 [3,7,5,4] = [3,7]
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| − | ------------------------------------------------------------------ *}
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| − |
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| − | fun coge :: "nat ⇒ 'a list ⇒ 'a list" where
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| − | "coge n [] = []"
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| − | | "coge 0 xs = []"
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| − | | "coge (Suc n) (x#xs) = x # (coge n xs)"
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| − |
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| − | value "coge 2 [3,7,5,4]" -- "= [3,7]"
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| − |
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| − | text {* ---------------------------------------------------------------
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| − | Ejercicio 15. Definir la función
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| − | elimina :: "nat ⇒ 'a list ⇒ 'a list"
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| − | tal que (coge n xs) es la lista de los n primeros elementos de xs. Por
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| − | ejemplo,
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| − | elimina 2 [3,7,5,4] = [5,4]
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| − | ------------------------------------------------------------------ *}
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| − |
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| − | fun elimina :: "nat ⇒ 'a list ⇒ 'a list" where
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| − | "elimina n [] = []"
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| − | | "elimina 0 xs = xs"
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| − | | "elimina (Suc n) (x#xs) = elimina n xs"
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| − |
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| − | value "elimina 2 [3,7,5,4]" -- "= [5,4]"
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| − |
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| − | text {* ---------------------------------------------------------------
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| − | Ejercicio 16. Demostrar que
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| − | conc (coge n xs) (elimina n xs) = xs
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| − | ------------------------------------------------------------------- *}
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| − |
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| − | lemma "conc (coge n xs) (elimina n xs) = xs"
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| − | by (induct rule: coge.induct) auto
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| − |
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| − | text {* coge.induct es el esquema de inducción asociado a la definición
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| − | de la función coge. Puede verse como sigue: *}
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| − |
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| − | thm coge.induct
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| − |
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| − | lemma "conc (coge n xs) (elimina n xs) = xs"
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| − | by (induct rule: elimina.induct) auto
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| − |
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| − | thm elimina.induct
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| − |
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| − | lemma conc_coge_elimina:
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| − | "conc (coge n xs) (elimina n xs) = xs"
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| − | proof (induct rule: coge.induct)
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| − | show "⋀n. conc (coge n []) (elimina n []) = []" by simp
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| − | next
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| − | show "⋀v va. conc (coge 0 (v # va)) (elimina 0 (v # va)) = v # va" by simp
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| − | next
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| − | fix x xs n
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| − | assume hi: "conc (coge n xs) (elimina n xs) = xs"
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| − | show "conc (coge (Suc n) (x # xs)) (elimina (Suc n) (x # xs)) = x # xs"
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| − | proof -
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| − | have "conc (coge (Suc n) (x # xs)) (elimina (Suc n) (x # xs)) = conc (coge (Suc n) (x # xs)) (elimina n xs)" by simp
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| − | also have "... = conc (x# (coge n xs)) (elimina n xs)" by simp
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| − | also have "... = x#(conc (coge n xs) (elimina n xs))" by simp
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| − | also have "... = (x#xs)" using hi by simp
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| − | finally show ?thesis .
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| − | qed
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| − | qed
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| − |
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| − |
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| − | text {* ---------------------------------------------------------------
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| − | Ejercicio 17. Definir la función
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| − | esVacia :: "'a list ⇒ bool"
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| − | tal que (esVacia xs) se verifica si xs es la lista vacía. Por ejemplo,
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| − | esVacia [] = True
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| − | esVacia [1] = False
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| − | ------------------------------------------------------------------ *}
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| − |
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| − | fun esVacia :: "'a list ⇒ bool" where
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| − | "esVacia [] = True"
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| − | | "esVacia (x#xs) = False"
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| − |
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| − | value "esVacia []" -- "= True"
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| − | value "esVacia [1]" -- "= False"
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| − |
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| − | text {* ---------------------------------------------------------------
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| − | Ejercicio 18. Demostrar que
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| − | esVacia xs = esVacia (conc xs xs)
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| − | ------------------------------------------------------------------- *}
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| − |
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| − | lemma "esVacia xs = esVacia (conc xs xs)"
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| − | by (induct xs) auto
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| − |
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| − | lemma vacia_conc:
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| − | "esVacia xs = esVacia (conc xs xs)"
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| − | proof (induct xs)
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| − | show "esVacia [] = esVacia (conc [] [])" by simp
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| − | next
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| − | fix x xs
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| − | assume hi: "esVacia xs = esVacia (conc xs xs)"
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| − | show "esVacia (x # xs) = esVacia (conc (x # xs) (x # xs))"
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| − | proof -
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| − | have "esVacia (conc (x # xs) (x # xs)) = esVacia (x# (conc xs (x#xs)))" by simp
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| − | also have "... = esVacia (x # xs)" by simp
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| − | finally show ?thesis by simp
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| − | qed
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| − | qed
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| − |
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| − | lemma vacia_conc':
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| − | "esVacia xs = esVacia (conc xs xs)"
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| − | proof (cases xs)
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| − | case Nil thus ?thesis by simp
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| − | next
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| − | case Cons thus ?thesis by simp
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| − | qed
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| − |
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| − |
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| − | lemma vacia_conc'':
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| − | "esVacia xs = esVacia (conc xs xs)"
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| − | by (cases xs) simp_all
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| − |
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| − |
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| − | text {* ---------------------------------------------------------------
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| − | Ejercicio 19. Definir la función
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| − | inversaAc :: "'a list ⇒ 'a list"
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| − | tal que (inversaAc xs) es a inversa de xs calculada usando
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| − | acumuladores. Por ejemplo,
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| − | inversaAc [3,2,5] = [5,2,3]
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| − | ------------------------------------------------------------------ *}
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| − |
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| − | fun inversaAcAux :: "'a list ⇒ 'a list ⇒ 'a list" where
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| − | "inversaAcAux [] ys = ys"
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| − | | "inversaAcAux (x#xs) ys = inversaAcAux xs (x#ys)"
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| − |
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| − | fun inversaAc :: "'a list ⇒ 'a list" where
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| − | "inversaAc xs = inversaAcAux xs []"
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| − |
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| − | value "inversaAc [3,2,5]" -- "= [5,2,3]"
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| − |
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| − | text {* ---------------------------------------------------------------
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| − | Ejercicio 20. Demostrar que
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| − | inversaAcAux xs ys = (inversa xs)@ys
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| − | ------------------------------------------------------------------- *}
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| − |
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| − | lemma inversaAcAux_es_inversa:
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| − | "inversaAcAux xs ys = (inversa xs)@ys"
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| − | by (induct xs arbitrary: ys) auto
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| − |
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| − |
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| − | lemma inversaAcAux_es_inversa_b:
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| − | "inversaAcAux xs ys = (inversa xs)@ys"
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| − | proof (induct xs arbitrary: ys)
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| − | show "⋀ys. inversaAcAux [] ys = inversa [] @ ys" by simp
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| − | next
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| − | fix x xs zs
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| − | assume hi: "⋀ys. inversaAcAux xs ys = inversa xs @ ys"
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| − | show "inversaAcAux (x#xs) zs = inversa (x#xs) @ zs"
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| − | proof -
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| − | have "inversaAcAux (x#xs) zs = inversaAcAux xs (x#zs)" by simp
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| − | also have "... = inversa xs @ (x#zs)" using hi by simp
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| − | also have "... = inversa xs @ [x] @ zs" by simp
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| − | also have "... = inversa (x#xs) @ zs" by simp
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| − | finally show ?thesis .
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| − | qed
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| − | qed
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| − |
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| − |
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| − | text {* ---------------------------------------------------------------
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| − | Ejercicio 21. Demostrar que
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| − | inversaAc xs = inversa xs
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| − | ------------------------------------------------------------------- *}
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| − |
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| − | corollary "inversaAc xs = inversa xs"
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| − | by (simp add: inversaAcAux_es_inversa)
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| − |
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| − | text {* ---------------------------------------------------------------
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| − | Ejercicio 22. Definir la función
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| − | sum :: "int list ⇒ int"
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| − | tal que (sum xs) es la suma de los elementos de xs. Por ejemplo,
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| − | sum [3,2,5] = 10
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| − | ------------------------------------------------------------------ *}
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| − |
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| − | fun sum :: "int list ⇒ int" where
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| − | "sum [] = 0"
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| − | | "sum (x#xs) = x + sum xs"
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| − |
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| − | value "sum [3,2,5]" -- "= 10"
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| − |
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| − | text {* ---------------------------------------------------------------
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| − | Ejercicio 23. Definir la función
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| − | map :: ('a ⇒ 'b) ⇒ 'a list ⇒ 'b list
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| − | tal que (map f xs) es la lista obtenida aplicando la función f a los
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| − | elementos de xs. Por ejemplo,
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| − | map (λx. 2*x) [3,2,5] = [6,4,10]
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| − | ------------------------------------------------------------------ *}
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| − |
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| − | fun map :: "('a ⇒ 'b) ⇒ 'a list ⇒ 'b list" where
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| − | "map f [] = []"
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| − | | "map f (x#xs) = (f x) # map f xs"
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| − |
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| − | value "map (λx. 2*x) [3::int,2,5]" -- "= [6,4,10]"
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| − |
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| − | value "map (λx. 6*x) [3::int,2,5]" -- "= [18,12,30]"
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| − |
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| − | text {* ---------------------------------------------------------------
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| − | Ejercicio 24. Demostrar que
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| − | sum (map (λx. 2*x) xs) = 2 * (sum xs)
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| − | ------------------------------------------------------------------- *}
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| − |
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| − | lemma "sum (map (λx. 2*x) xs) = 2 * (sum xs)"
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| − | by (induct xs) auto
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| − |
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| − |
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| − | lemma sum_map:
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| − | "sum (map (λx. 2*x) xs) = 2 * (sum xs)"
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| − | proof (induct xs)
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| − | show "sum (map (λx. 2*x) []) = 2 * (sum [])" by simp
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| − | next
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| − | fix a xs
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| − | assume hi: "sum (map (λx. 2*x) xs) = 2 * (sum xs)"
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| − | show "sum (map (λx. 2*x) (a#xs)) = 2 * (sum (a#xs))"
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| − | proof -
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| − | have "sum (map (λx. 2*x) (a#xs)) = sum ((2*a)#(map (λx. 2*x) xs))" by simp
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| − | also have "... = 2*a + sum (map (λx. 2*x) xs)" by simp
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| − | also have "... = 2*a + 2 * (sum xs)" using hi by simp
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| − | also have "... = 2*(a + sum xs)" by simp
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| − | also have "... = 2 * (sum (a#xs))"by simp
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| − | finally show ?thesis .
| |
| − | qed
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| − | qed
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| − |
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| − |
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| − | text {* ---------------------------------------------------------------
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| − | Ejercicio 25. Demostrar que
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| − | longitud (map f xs) = longitud xs
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| − | ------------------------------------------------------------------- *}
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| − |
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| − | lemma "longitud (map f xs) = longitud xs"
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| − | by (induct xs) auto
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| − |
| |
| − |
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| − | lemma long_map:
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| − | "longitud (map f xs) = longitud xs"
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| − | proof (induct xs)
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| − | show "longitud (map f []) = longitud []" by simp
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| − | next
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| − | fix x xs
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| − | assume hi: "longitud (map f xs) = longitud xs"
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| − | show "longitud (map f (x#xs)) = longitud (x#xs)"
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| − | proof -
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| − | have "longitud (map f (x#xs)) = longitud ((f x)#(map f xs))" by simp
| |
| − | also have "... = 1 + longitud (map f xs)" by simp
| |
| − | also have "... = 1 + longitud xs" using hi by simp
| |
| − | also have "... = longitud (x#xs)" by simp
| |
| − | finally show ?thesis .
| |
| − | qed
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| − | qed
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| − |
| |
| − |
| |
| − | end
| |
| − | </source>
| |
