En Lean, está definida la función

reverse : list α → list α |

tal que (reverse xs) es la lista obtenida invirtiendo el orden de los elementos de xs. Por ejemplo,

reverse [3,2,5,1] = [1,5,2,3] |

Su definición es

def reverse_core : list α → list α → list α | [] r := r | (a::l) r := reverse_core l (a::r) def reverse : list α → list α := λ l, reverse_core l [] |

Una definición alternativa es

def inversa : list α → list α | [] := [] | (x :: xs) := inversa xs ++ [x] |

Demostrar que las dos definiciones son equivalentes; es decir,

reverse xs = inversa xs |

Para ello, completar la siguiente teoría de Lean:

import data.list.basic open list variable {α : Type*} variable (x : α) variables (xs ys : list α) def inversa : list α → list α | [] := [] | (x :: xs) := inversa xs ++ [x] example : reverse xs = inversa xs := sorry |

import data.list.basic open list variable {α : Type*} variable (x : α) variables (xs ys : list α) -- Definición y reglas de simplificación de inversa -- ================================================ def inversa : list α → list α | [] := [] | (x :: xs) := inversa xs ++ [x] @[simp] lemma inversa_nil : inversa ([] : list α) = [] := rfl @[simp] lemma inversa_cons : inversa (x :: xs) = inversa xs ++ [x] := rfl -- Reglas de simplificación de reverse_core -- ======================================== @[simp] lemma reverse_core_nil : reverse_core [] ys = ys := rfl @[simp] lemma reverse_core_cons : reverse_core (x :: xs) ys = reverse_core xs (x :: ys) := rfl -- Lema auxiliar: reverse_core xs ys = (inversa xs) ++ ys -- ====================================================== -- 1ª demostración del lema auxiliar example : reverse_core xs ys = (inversa xs) ++ ys := begin induction xs with a as HI generalizing ys, { calc reverse_core [] ys = ys : reverse_core_nil ys ... = [] ++ ys : (nil_append ys).symm ... = inversa [] ++ ys : congr_arg2 (++) inversa_nil.symm rfl, }, { calc reverse_core (a :: as) ys = reverse_core as (a :: ys) : reverse_core_cons a as ys ... = inversa as ++ (a :: ys) : (HI (a :: ys)) ... = inversa as ++ ([a] ++ ys) : congr_arg2 (++) rfl singleton_append ... = (inversa as ++ [a]) ++ ys : (append_assoc (inversa as) [a] ys).symm ... = inversa (a :: as) ++ ys : congr_arg2 (++) (inversa_cons a as).symm rfl}, end -- 2ª demostración del lema auxiliar example : reverse_core xs ys = (inversa xs) ++ ys := begin induction xs with a as HI generalizing ys, { calc reverse_core [] ys = ys : by rw reverse_core_nil ... = [] ++ ys : by rw nil_append ... = inversa [] ++ ys : by rw inversa_nil }, { calc reverse_core (a :: as) ys = reverse_core as (a :: ys) : by rw reverse_core_cons ... = inversa as ++ (a :: ys) : by rw (HI (a :: ys)) ... = inversa as ++ ([a] ++ ys) : by rw singleton_append ... = (inversa as ++ [a]) ++ ys : by rw append_assoc ... = inversa (a :: as) ++ ys : by rw inversa_cons }, end -- 3ª demostración del lema auxiliar example : reverse_core xs ys = (inversa xs) ++ ys := begin induction xs with a as HI generalizing ys, { calc reverse_core [] ys = ys : rfl ... = [] ++ ys : rfl ... = inversa [] ++ ys : rfl }, { calc reverse_core (a :: as) ys = reverse_core as (a :: ys) : rfl ... = inversa as ++ (a :: ys) : (HI (a :: ys)) ... = inversa as ++ ([a] ++ ys) : rfl ... = (inversa as ++ [a]) ++ ys : by rw append_assoc ... = inversa (a :: as) ++ ys : rfl }, end -- 3ª demostración del lema auxiliar example : reverse_core xs ys = (inversa xs) ++ ys := begin induction xs with a as HI generalizing ys, { calc reverse_core [] ys = ys : by simp ... = [] ++ ys : by simp ... = inversa [] ++ ys : by simp }, { calc reverse_core (a :: as) ys = reverse_core as (a :: ys) : by simp ... = inversa as ++ (a :: ys) : (HI (a :: ys)) ... = inversa as ++ ([a] ++ ys) : by simp ... = (inversa as ++ [a]) ++ ys : by simp ... = inversa (a :: as) ++ ys : by simp }, end -- 4ª demostración del lema auxiliar example : reverse_core xs ys = (inversa xs) ++ ys := begin induction xs with a as HI generalizing ys, { by simp, }, { calc reverse_core (a :: as) ys = reverse_core as (a :: ys) : by simp ... = inversa as ++ (a :: ys) : (HI (a :: ys)) ... = inversa (a :: as) ++ ys : by simp }, end -- 5ª demostración del lema auxiliar example : reverse_core xs ys = (inversa xs) ++ ys := begin induction xs with a as HI generalizing ys, { simp, }, { simp [HI (a :: ys)], }, end -- 6ª demostración del lema auxiliar example : reverse_core xs ys = (inversa xs) ++ ys := by induction xs generalizing ys ; simp [*] -- 7ª demostración del lema auxiliar example : reverse_core xs ys = (inversa xs) ++ ys := begin induction xs with a as HI generalizing ys, { rw reverse_core_nil, rw inversa_nil, rw nil_append, }, { rw reverse_core_cons, rw (HI (a :: ys)), rw inversa_cons, rw append_assoc, rw singleton_append, }, end -- 8ª demostración del lema auxiliar @[simp] lemma inversa_equiv : ∀ xs : list α, ∀ ys, reverse_core xs ys = (inversa xs) ++ ys | [] := by simp | (a :: as) := by simp [inversa_equiv as] -- Demostraciones del lema principal -- ================================= -- 1ª demostración example : reverse xs = inversa xs := calc reverse xs = reverse_core xs [] : rfl ... = inversa xs ++ [] : by rw inversa_equiv ... = inversa xs : by rw append_nil -- 2ª demostración example : reverse xs = inversa xs := by simp [inversa_equiv, reverse] -- 3ª demostración example : reverse xs = inversa xs := by simp [reverse] |

Se puede interactuar con la prueba anterior en esta sesión con Lean.

En los comentarios se pueden escribir otras soluciones, escribiendo el código entre una línea con <pre lang="lean"> y otra con </pre>

theory Pruebas_de_equivalencia_de_definiciones_de_inversa imports Main begin (* Definición alternativa *) (* ====================== *) fun inversa_aux :: "'a list ⇒ 'a list ⇒ 'a list" where "inversa_aux [] ys = ys" | "inversa_aux (x#xs) ys = inversa_aux xs (x#ys)" fun inversa :: "'a list ⇒ 'a list" where "inversa xs = inversa_aux xs []" (* Lema auxiliar: inversa_aux xs ys = (rev xs) @ ys *) (* ================================================ *) (* 1ª demostración del lema auxiliar *) lemma "inversa_aux xs ys = (rev xs) @ ys" proof (induct xs arbitrary: ys) fix ys :: "'a list" have "inversa_aux [] ys = ys" by (simp only: inversa_aux.simps(1)) also have "… = [] @ ys" by (simp only: append.simps(1)) also have "… = rev [] @ ys" by (simp only: rev.simps(1)) finally show "inversa_aux [] ys = rev [] @ ys" by this next fix a ::'a and xs :: "'a list" assume HI: "⋀ys. inversa_aux xs ys = rev xs@ys" show "⋀ys. inversa_aux (a#xs) ys = rev (a#xs)@ys" proof - fix ys have "inversa_aux (a#xs) ys = inversa_aux xs (a#ys)" by (simp only: inversa_aux.simps(2)) also have "… = rev xs@(a#ys)" by (simp only: HI) also have "… = rev xs @ ([a] @ ys)" by (simp only: append.simps) also have "… = (rev xs @ [a]) @ ys" by (simp only: append_assoc) also have "… = rev (a # xs) @ ys" by (simp only: rev.simps(2)) finally show "inversa_aux (a#xs) ys = rev (a#xs)@ys" by this qed qed (* 2ª demostración del lema auxiliar *) lemma "inversa_aux xs ys = (rev xs) @ ys" proof (induct xs arbitrary: ys) fix ys :: "'a list" have "inversa_aux [] ys = ys" by simp also have "… = [] @ ys" by simp also have "… = rev [] @ ys" by simp finally show "inversa_aux [] ys = rev [] @ ys" . next fix a ::'a and xs :: "'a list" assume HI: "⋀ys. inversa_aux xs ys = rev xs@ys" show "⋀ys. inversa_aux (a#xs) ys = rev (a#xs)@ys" proof - fix ys have "inversa_aux (a#xs) ys = inversa_aux xs (a#ys)" by simp also have "… = rev xs@(a#ys)" using HI by simp also have "… = rev xs @ ([a] @ ys)" by simp also have "… = (rev xs @ [a]) @ ys" by simp also have "… = rev (a # xs) @ ys" by simp finally show "inversa_aux (a#xs) ys = rev (a#xs)@ys" . qed qed (* 3ª demostración del lema auxiliar *) lemma "inversa_aux xs ys = (rev xs) @ ys" proof (induct xs arbitrary: ys) fix ys :: "'a list" show "inversa_aux [] ys = rev [] @ ys" by simp next fix a ::'a and xs :: "'a list" assume HI: "⋀ys. inversa_aux xs ys = rev xs@ys" show "⋀ys. inversa_aux (a#xs) ys = rev (a#xs)@ys" proof - fix ys have "inversa_aux (a#xs) ys = rev xs@(a#ys)" using HI by simp also have "… = rev (a # xs) @ ys" by simp finally show "inversa_aux (a#xs) ys = rev (a#xs)@ys" . qed qed (* 4ª demostración del lema auxiliar *) lemma "inversa_aux xs ys = (rev xs) @ ys" proof (induct xs arbitrary: ys) show "⋀ys. inversa_aux [] ys = rev [] @ ys" by simp next fix a ::'a and xs :: "'a list" assume "⋀ys. inversa_aux xs ys = rev xs@ys" then show "⋀ys. inversa_aux (a#xs) ys = rev (a#xs)@ys" by simp qed (* 5ª demostración del lema auxiliar *) lemma "inversa_aux xs ys = (rev xs) @ ys" proof (induct xs arbitrary: ys) case Nil then show ?case by simp next case (Cons a xs) then show ?case by simp qed (* 6ª demostración del lema auxiliar *) lemma inversa_equiv: "inversa_aux xs ys = (rev xs) @ ys" by (induct xs arbitrary: ys) simp_all (* Demostraciones del lema principal *) (* ================================= *) (* 1ª demostración *) lemma "inversa xs = rev xs" proof - have "inversa xs = inversa_aux xs []" by (rule inversa.simps) also have "… = (rev xs) @ []" by (rule inversa_equiv) also have "… = rev xs" by (rule append.right_neutral) finally show "inversa xs = rev xs" by this qed (* 2ª demostración *) lemma "inversa xs = rev xs" proof - have "inversa xs = inversa_aux xs []" by simp also have "… = (rev xs) @ []" by (rule inversa_equiv) also have "… = rev xs" by simp finally show "inversa xs = rev xs" . qed (* 3ª demostración *) lemma "inversa xs = rev xs" by (simp add: inversa_equiv) end |

En los comentarios se pueden escribir otras soluciones, escribiendo el código entre una línea con <pre lang="isar"> y otra con </pre>