LMF2019: Razonamiento por casos y por inducción en Isabelle/HOL
En la clase de hoy del curso de Razonamiento automático hemos profundizado en el estudio de las demostraciones por casos y por inducción. En concreto, se ha estudiado
- el razonamiento por casos booleanos,
- el razonamiento por casos booleanos sobre una variable,
- el razonamiento por casos sobre listas,
- el razonamiento por inducción sobre números naturales con patrones,
- el razonamiento sobre definiciones con existenciales,
- el uso de librerías auxiliares (como Parity) y
- el uso de otros métodos de demostración (como presburg).
La clase se ha dado mediante videoconferencia y el vídeo correspondiente es:
La teoría con los ejemplos presentados en la clase es la siguiente:
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chapter ‹Tema 6: Razonamiento sobre programas› theory T6_Razonamiento_sobre_programas imports Main begin chapter ‹Tema 6: Razonamiento sobre programas› theory T6_Razonamiento_sobre_programas imports Main begin text ‹--------------------------------------------------------------- Ejemplo 13. (p. 30) Demostrar que longitud (conc xs ys) = longitud xs + longitud ys ------------------------------------------------------------------- › (* La demostración aplicativa es *) lemma "longitud (conc xs ys) = longitud xs + longitud ys" apply (induct xs) (* 1. longitud (conc [] ys) = longitud [] + longitud ys 2. ⋀a xs. longitud (conc xs ys) = longitud xs + longitud ys ⟹ longitud (conc (a # xs) ys) = longitud (a # xs) + longitud ys *) apply simp_all (* No subgoals! *) done (* La demostración automática es *) lemma "longitud (conc xs ys) = longitud xs + longitud ys" by (induct xs) simp_all (* La demostración declarativa es *) lemma "longitud (conc xs ys) = longitud xs + longitud ys" proof (induct xs) show "longitud (conc [] ys) = longitud [] + longitud ys" by simp next fix x xs assume HI: "longitud (conc xs ys) = longitud xs + longitud ys" have "longitud (conc (x # xs) ys) = longitud (x # (conc xs ys))" by simp also have "… = 1 + longitud (conc xs ys)" by simp also have "… = 1 + longitud xs + longitud ys" using HI by simp also have "… = longitud (x # xs) + longitud ys" by simp finally show "longitud (conc (x # xs) ys) = longitud (x # xs) + longitud ys" by simp qed (* La demostración declarativa detallada es *) lemma "longitud (conc xs ys) = longitud xs + longitud ys" proof (induct xs) show "longitud (conc [] ys) = longitud [] + longitud ys" by (simp only: conc.simps(1) longitud.simps(1)) next fix x xs assume HI: "longitud (conc xs ys) = longitud xs + longitud ys" have "longitud (conc (x # xs) ys) = longitud (x # (conc xs ys))" by (simp only: conc.simps(2)) also have "… = 1 + longitud (conc xs ys)" by (simp only: longitud.simps(2)) also have "… = 1 + longitud xs + longitud ys" using HI by (simp only:) also have "… = longitud (x # xs) + longitud ys" by (simp only: longitud.simps(2)) finally show "longitud (conc (x # xs) ys) = longitud (x # xs) + longitud ys" by this qed section ‹Inducción correspondiente a la definición recursiva › text ‹--------------------------------------------------------------- Ejemplo 14. 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] ------------------------------------------------------------------ › fun coge :: "nat ⇒ 'a list ⇒ 'a list" where "coge n [] = []" | "coge 0 xs = []" | "coge (Suc n) (x#xs) = x # (coge n xs)" value "coge 2 [a,c,d,b,e] = [a,c]" text ‹--------------------------------------------------------------- Ejemplo 15. 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] ------------------------------------------------------------------ › fun elimina :: "nat ⇒ 'a list ⇒ 'a list" where "elimina n [] = []" | "elimina 0 xs = xs" | "elimina (Suc n) (x#xs) = elimina n xs" value "elimina 2 [a,c,d,b,e] = [d,b,e]" text ‹La definición coge genera el esquema de inducción coge.induct: ⟦⋀n. P n []; ⋀x xs. P 0 (x#xs); ⋀n x xs. P n xs ⟹ P (Suc n) (x#xs)⟧ ⟹ P n xs Puede verse usando "thm coge.induct". › thm elimina.induct thm coge.induct text ‹--------------------------------------------------------------- Ejemplo 16. (p. 35) Demostrar que conc (coge n xs) (elimina n xs) = xs ------------------------------------------------------------------- › (* La demostración aplicativa es *) lemma "conc (coge n xs) (elimina n xs) = xs" apply (induct rule: coge.induct) (* 1. ⋀n. conc (coge n []) (elimina n []) = [] 2. ⋀v va. conc (coge 0 (v # va)) (elimina 0 (v # va)) = v # va 3. ⋀n x xs. conc (coge n xs) (elimina n xs) = xs ⟹ conc (coge (Suc n) (x # xs)) (elimina (Suc n) (x # xs)) = x # xs *) apply simp_all (* No subgoals! *) done (* La demostración automática es *) lemma "conc (coge n xs) (elimina n xs) = xs" by (induct rule: coge.induct) simp_all (* La demostración declarativa es *) lemma "conc (coge n xs) (elimina n xs) = xs" proof (induct rule: coge.induct) fix n show "conc (coge n []) (elimina n []) = []" by simp next fix x :: "'a" and xs :: "'a list" show "conc (coge 0 (x#xs)) (elimina 0 (x#xs)) = x#xs" by simp next fix n and x :: "'a" and xs :: "'a list" assume HI: "conc (coge n xs) (elimina n xs) = xs" have "conc (coge (Suc n) (x#xs)) (elimina (Suc n) (x#xs)) = conc (x#(coge n xs)) (elimina n xs)" by simp also have "… = x#(conc (coge n xs) (elimina n xs))" by simp also have "… = x#xs" using HI by simp finally show "conc (coge (Suc n) (x#xs)) (elimina (Suc n) (x#xs)) = x#xs" by simp qed text ‹Comentario sobre la demostración anterior: · (induct rule: coge.induct) indica que el método de demostración es por el esquema de inducción correspondiente a la definición de la función coge. · Se generan 3 subobjetivos: · 1. ⋀n. conc (coge n []) (elimina n []) = [] · 2. ⋀x xs. conc (coge 0 (x#xs)) (elimina 0 (x#xs)) = x#xs · 3. ⋀n x xs. conc (coge n xs) (elimina n xs) = xs ⟹ conc (coge (Suc n) (x#xs)) (elimina (Suc n) (x#xs)) = x#xs › (* La demostración declarativa detallada es *) lemma "conc (coge n xs) (elimina n xs) = xs" proof (induct rule: coge.induct) fix n show "conc (coge n []) (elimina n []) = []" by (simp only: coge.simps(1) elimina.simps(1) conc.simps(1)) next fix x :: "'a" and xs :: "'a list" show "conc (coge 0 (x#xs)) (elimina 0 (x#xs)) = x#xs" by (simp only: coge.simps(2) elimina.simps(2) conc.simps(1)) next fix n and x :: "'a" and xs :: "'a list" assume HI: "conc (coge n xs) (elimina n xs) = xs" have "conc (coge (Suc n) (x#xs)) (elimina (Suc n) (x#xs)) = conc (x#(coge n xs)) (elimina n xs)" by (simp only: coge.simps(3) elimina.simps(3)) also have "… = x#(conc (coge n xs) (elimina n xs))" by (simp only: conc.simps(2)) also have "… = x#xs" using HI by (simp only:) finally show "conc (coge (Suc n) (x#xs)) (elimina (Suc n) (x#xs)) = x#xs" by this qed section ‹Razonamiento por casos › text ‹ --------------------------------------------------------------- Ejemplo 17. 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 ------------------------------------------------------------------ › fun esVacia :: "'a list ⇒ bool" where "esVacia [] = True" | "esVacia (x#xs) = False" value "esVacia [] = True" value "esVacia [a] = False" text ‹--------------------------------------------------------------- Ejemplo 18 (p. 39) . Demostrar que esVacia xs = esVacia (conc xs xs) ------------------------------------------------------------------- › (* La demostración aplicativa es *) lemma "esVacia xs = esVacia (conc xs xs)" apply (cases xs) (* 1. xs = [] ⟹ esVacia xs = esVacia (conc xs xs) 2. ⋀a list. xs = a # list ⟹ esVacia xs = esVacia (conc xs xs) *) apply simp_all (* No subgoals! *) done (* La demostración automática es *) lemma "esVacia xs = esVacia (conc xs xs)" by (cases xs) simp_all (* La demostración declarativa es *) lemma "esVacia xs = esVacia (conc xs xs)" proof (cases xs) assume "xs = []" then show "esVacia xs = esVacia (conc xs xs)" by simp next fix y ys assume "xs = y#ys" then show "esVacia xs = esVacia (conc xs xs)" by simp qed text ‹Comentarios sobre la demostración anterior: · "(cases xs)" es el método de demostración por casos según xs. · Se generan dos subobjetivos correspondientes a los dos constructores de listas: · 1. xs = [] ⟹ esVacia xs = esVacia (conc xs xs) · 2. ⋀y ys. xs = y#ys ⟹ esVacia xs = esVacia (conc xs xs) · "then" indica "usando la propiedad anterior" › (* La demostración declarativa detallada es *) lemma "esVacia xs = esVacia (conc xs xs)" proof (cases xs) assume "xs = []" then show "esVacia xs = esVacia (conc xs xs)" by (simp only: conc.simps(1)) next fix y ys assume "xs = y#ys" then show "esVacia xs = esVacia (conc xs xs)" by (simp only: esVacia.simps(2) conc.simps(2)) qed (* La demostración declarativa simplificada es *) lemma "esVacia xs = esVacia (conc xs xs)" proof (cases xs) case Nil then show "esVacia xs = esVacia (conc xs xs)" by simp next case Cons then show "esVacia xs = esVacia (conc xs xs)" by simp qed text ‹Comentarios sobre la demostración anterior: · "case Nil" es una abreviatura de "assume xs = []" · "case Cons" es una abreviatura de "fix y ys assume xs = y#ys"› lemma "esVacia xs = esVacia (conc xs xs)" proof (cases xs) case Nil then show ?thesis by simp next case (Cons a list) then show ?thesis by simp qed (* La demostración con el patrón sugerido es *) lemma "esVacia xs = esVacia (conc xs xs)" proof (cases xs) case Nil then show ?thesis by simp next case (Cons x xs) then show ?thesis by simp qed section ‹Heurística de generalización› text ‹--------------------------------------------------------------- Ejemplo 19. 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] ------------------------------------------------------------------ › fun inversaAcAux :: "'a list ⇒ 'a list ⇒ 'a list" where "inversaAcAux [] ys = ys" | "inversaAcAux (x#xs) ys = inversaAcAux xs (x#ys)" definition inversaAc :: "'a list ⇒ 'a list" where "inversaAc xs = inversaAcAux xs []" value "inversaAc [a,c,b,e] = [e,b,c,a]" text ‹Lema. [Ejemplo de equivalencia entre las definiciones] La inversa de [a,b,c] es lo mismo calculada con la primera definición que con la segunda.› lemma "inversaAc [a,b,c] = inversa [a,b,c]" by (simp add: inversaAc_def) text ‹Nota. [Ejemplo fallido de demostración por inducción] El siguiente intento de demostrar que para cualquier lista xs, se tiene que "inversaAc xs = inversa xs" falla.› lemma "inversaAc xs = inversa xs" proof (induct xs) show "inversaAc [] = inversa []" by (simp add: inversaAc_def) next fix a :: "'b" and xs :: "'b list" assume HI: "inversaAc xs = inversa xs" have "inversaAc (a#xs) = inversaAcAux (a#xs) []" by (simp add: inversaAc_def) also have "… = inversaAcAux xs [a]" by simp also have "… = inversa (a#xs)" (* Problema: la hipótesis de inducción no es aplicable. *) oops text ‹Nota. [Heurística de generalización] Cuando se use demostración estructural, cuantificar universalmente las variables libres (o, equivalentemente, considerar las variables libres como variables arbitrarias). Lema. [Lema con generalización] Para toda lista ys se tiene inversaAcAux xs ys = (inversa xs) @ ys › text ‹--------------------------------------------------------------- Ejemplo 20. (p. 44) Demostrar que inversaAcAux xs ys = (inversa xs) @ ys ------------------------------------------------------------------- › (* La demostración aplicativa es *) lemma "inversaAcAux xs ys = (inversa xs) @ ys" apply (induct xs arbitrary: ys) (* 1. ⋀ys. inversaAcAux [] ys = inversa [] @ ys 2. ⋀a xs ys. (⋀ys. inversaAcAux xs ys = inversa xs @ ys) ⟹ inversaAcAux (a # xs) ys = inversa (a # xs) @ ys *) apply simp_all (* No subgoals! *) done (* La demostración automática es *) lemma "inversaAcAux xs ys = (inversa xs) @ ys" by (induct xs arbitrary: ys) simp_all (* La demostración declarativa es *) lemma "inversaAcAux xs ys = (inversa xs) @ ys" proof (induct xs arbitrary: ys) show "⋀ys. inversaAcAux [] ys = (inversa []) @ ys" by simp next fix a :: "'b" and xs :: "'b list" assume HI: "⋀ys. inversaAcAux xs ys = inversa xs @ ys" show "⋀ys. inversaAcAux (a#xs) ys = inversa (a#xs) @ ys" proof - fix ys have "inversaAcAux (a#xs) ys = inversaAcAux xs (a#ys)" by simp also have "… = inversa xs @ (a#ys)" using HI by simp also have "… = inversa (a#xs) @ ys" (* using [[simp_trace]] *) by simp finally show "inversaAcAux (a#xs) ys = inversa (a#xs) @ ys" by simp qed qed text ‹Comentarios sobre la demostración anterior: · "(induct xs arbitrary: ys)" es el método de demostración por inducción sobre xs usando ys como variable arbitraria. · Se generan dos subobjetivos: · 1. ⋀ys. inversaAcAux [] ys = inversa [] @ ys · 2. ⋀a xs ys. (⋀ys. inversaAcAux xs ys = inversa xs @ ys) ⟹ inversaAcAux (a # xs) ys = inversa (a # xs) @ ys · Dentro de una demostración se pueden incluir otras demostraciones. · Para demostrar la propiedad universal "⋀ys. P(ys)" se elige una lista arbitraria (con "fix ys") y se demuestra "P(ys)". › text ‹Nota. En el paso "inversa xs@(a#ys) = inversa (a#xs)@ys" se usan lemas de la teoría List. Se puede observar, insertano using [[simp_trace]] entre la igualdad y by simp, que los lemas usados son · append_simps_1: []@ys = ys · append_simps_2: (x#xs)@ys = x#(xs@ys) · append_assoc: (xs @ ys) @ zs = xs @ (ys @ zs) Las dos primeras son las ecuaciones de la definición de append. En la siguiente demostración se detallan los lemas utilizados.› ― ‹Demostración aplicativa detallada› lemma "(inversa xs) @ (a#ys) = (inversa (a#xs)) @ ys" apply (simp only: inversa.simps(2)) (* inversa xs @ (a # ys) = (inversa xs @ [a]) @ ys*) apply (simp only: append_assoc) (* inversa xs @ (a # ys) = inversa xs @ ([a] @ ys) *) apply (simp only: append.simps(2)) (* inversa xs @ (a # ys) = inversa xs @ (a # ([] @ ys)) *) apply (simp only: append.simps(1)) (* *) done ― ‹Demostración declarativa detallada del lema auxiliar› lemma auxiliar: "(inversa xs) @ (a#ys) = (inversa (a#xs)) @ ys" proof - have "(inversa xs) @ (a#ys) = (inversa xs) @ (a # ([] @ ys))" by (simp only: append.simps(1)) also have "… = (inversa xs) @ ([a] @ ys)" by (simp only: append.simps(2)) also have "… = ((inversa xs) @ [a]) @ ys" by (simp only: append_assoc) also have "… = (inversa (a#xs)) @ ys" by (simp only: inversa.simps(2)) finally show ?thesis by this qed (* La demostración declarativa detallada es *) lemma inversaAcAux_es_inversa: "inversaAcAux xs ys = (inversa xs) @ ys" proof (induct xs arbitrary: ys) fix ys :: "'b list" show "inversaAcAux [] ys = (inversa []) @ ys" by (simp only: inversaAcAux.simps(1) inversa.simps(1) append.simps(1)) next fix a :: "'b" and xs :: "'b list" assume HI: "⋀ys. inversaAcAux xs ys = inversa xs @ ys" show "⋀ys. inversaAcAux (a#xs) ys = inversa (a#xs) @ ys" proof - fix ys have "inversaAcAux (a#xs) ys = inversaAcAux xs (a#ys)" by (simp only: inversaAcAux.simps(2)) also have "… = inversa xs @ (a#ys)" using HI by (simp only:) also have "… = inversa (a#xs) @ ys" by (rule auxiliar) finally show "inversaAcAux (a#xs) ys = inversa (a#xs) @ ys" by this qed qed text ‹--------------------------------------------------------------- Ejemplo 21. (p. 43) Demostrar que inversaAc xs = inversa xs ------------------------------------------------------------------- › (* La demostración aplicativa es *) corollary "inversaAc xs = inversa xs" apply (simp add: inversaAcAux_es_inversa inversaAc_def) done (* La demostración automática es *) corollary "inversaAc xs = inversa xs" by (simp add: inversaAcAux_es_inversa inversaAc_def) text ‹Comentario de la demostración anterior: · "(simp add: inversaAcAux_es_inversa inversaAc_def)" es el método de demostración por simplificación usando como regla de simplificación las propiedades inversaAcAux_es_inversa e inversaAc_def. › section ‹Demostración por inducción para funciones de orden superior › text ‹ --------------------------------------------------------------- Ejemplo 22. 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 ------------------------------------------------------------------ › fun sum :: "nat list ⇒ nat" where "sum [] = 0" | "sum (x#xs) = x + sum xs" value "sum [3,2,5] = 10" text ‹--------------------------------------------------------------- Ejemplo 23. 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::nat,2,5] = [6,4,10] map ((*) 2) [3::nat,2,5] = [6,4,10] map ((+) 2) [3::nat,2,5] = [5,4,7] ------------------------------------------------------------------ › fun map :: "('a ⇒ 'b) ⇒ 'a list ⇒ 'b list" where "map f [] = []" | "map f (x#xs) = (f x) # map f xs" value "map (λx. 2*x) [3::nat,2,5] = [6,4,10]" value "map ((*) 2) [3::nat,2,5] = [6,4,10]" value "map ((+) 2) [3::nat,2,5] = [5,4,7]" text ‹--------------------------------------------------------------- Ejemplo 24. (p. 45) Demostrar que sum (map ((*) 2) xs) = 2 * (sum xs) ------------------------------------------------------------------- › declare [[names_short]] (* La demostración aplicativa es *) lemma "sum (map ((*) 2) xs) = 2 * (sum xs)" apply (induct xs) (* 1. sum (map (( * ) 2) []) = 2 * sum [] 2. ⋀a xs. sum (map (( * ) 2) xs) = 2 * sum xs ⟹ sum (map (( * ) 2) (a # xs)) = 2 * sum (a # xs) *) apply simp_all (* No subgoals! *) done (* La demostración automática es *) lemma "sum (map ((*) 2) xs) = 2 * (sum xs)" by (induct xs) simp_all (* La demostración declarativa es *) lemma "sum (map ((*) 2) xs) = 2 * (sum xs)" proof (induct xs) show "sum (map ((*) 2) []) = 2 * (sum [])" by simp next fix a xs assume HI: "sum (map ((*) 2) xs) = 2 * (sum xs)" have "sum (map ((*) 2) (a#xs)) = sum ((2*a) # (map ((*) 2) xs))" by simp also have "… = 2*a + sum (map ((*) 2) xs)" by simp also have "… = 2*a + 2*(sum xs)" using HI by simp also have "… = 2*(a + sum xs)" by simp also have "… = 2*(sum (a#xs))" by simp finally show "sum (map ((*) 2) (a#xs)) = 2*(sum (a#xs))" by simp qed (* La demostración declarativa detallada es *) lemma "sum (map ((*) 2) xs) = 2 * (sum xs)" proof (induct xs) show "sum (map ((*) 2) []) = 2 * (sum [])" by (simp only: map.simps(1) sum.simps(1)) next fix a xs assume HI: "sum (map ((*) 2) xs) = 2 * (sum xs)" have "sum (map ((*) 2) (a#xs)) = sum ((2*a)#(map ((*) 2) xs))" by (simp only: map.simps(2)) also have "… = 2*a + sum (map ((*) 2) xs)" by (simp only: sum.simps(2)) also have "… = 2*a + 2*(sum xs)" using HI by (simp only:) also have "… = 2*(a + sum xs)" (* find_theorems "_ * (_ + _)" *) by (simp only: add_mult_distrib2) also have "… = 2*(sum (a#xs))" by (simp only: sum.simps(2)) finally show "sum (map ((*) 2) (a#xs)) = 2*(sum (a#xs))" by this qed text ‹ --------------------------------------------------------------- Ejemplo 25. (p. 48) Demostrar que longitud (map f xs) = longitud xs ------------------------------------------------------------------- › (* La demostración aplicativa es *) lemma "longitud (map f xs) = longitud xs" apply (induct xs) (* 1. longitud (map f []) = longitud [] 2. ⋀a xs. longitud (map f xs) = longitud xs ⟹ longitud (map f (a # xs)) = longitud (a # xs) *) apply simp_all (* No subgoals! *) done (* La demostración automática es *) lemma "longitud (map f xs) = longitud xs" by (induct xs) simp_all (* La demostración declarativa es *) lemma "longitud (map f xs) = longitud xs" proof (induct xs) show "longitud (map f []) = longitud []" by simp next fix a xs assume HI: "longitud (map f xs) = longitud xs" have "longitud (map f (a#xs)) = longitud (f a # (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 (a#xs)" by simp finally show "longitud (map f (a#xs)) = longitud (a#xs)" by simp qed (* La demostración declarativa detallada es *) lemma "longitud (map f xs) = longitud xs" proof (induct xs) show "longitud (map f []) = longitud []" by (simp only: map.simps(1) longitud.simps(1)) next fix a xs assume HI: "longitud (map f xs) = longitud xs" have "longitud (map f (a#xs)) = longitud (f a # (map f xs))" by (simp only: map.simps(2)) also have "… = 1 + longitud (map f xs)" by (simp only: longitud.simps(2)) also have "… = 1 + longitud xs" using HI by (simp only:) also have "… = longitud (a#xs)" by (simp only: longitud.simps(2)) finally show "longitud (map f (a#xs)) = longitud (a#xs)" by this qed section ‹Referencias› text ‹ · J.A. Alonso. "Razonamiento sobre programas" http://goo.gl/R06O3 · G. Hutton. "Programming in Haskell". Cap. 13 "Reasoning about programms". · S. Thompson. "Haskell: the Craft of Functional Programming, 3rd Edition. Cap. 8 "Reasoning about programms". · L. Paulson. "ML for the Working Programmer, 2nd Edition". Cap. 6. "Reasoning about functional programs". › end |
Como tarea se ha propuesto la resolución de los ejercicios de la relación 12.