En Lean la operación de concatenación de listas se representa por (++) y está caracterizada por los siguientes lemas

nil_append : [] ++ ys = ys cons_append : (x :: xs) ++ y = x :: (xs ++ ys) |

Demostrar que la concatenación es asociativa; es decir,

xs ++ (ys ++ zs) = (xs ++ ys) ++ zs |

Para ello, completar la siguiente teoría de Lean:

import data.list.basic import tactic open list variable {α : Type} variable (x : α) variables (xs ys zs : list α) example : xs ++ (ys ++ zs) = (xs ++ ys) ++ zs := sorry |

[expand title=»Soluciones con Lean»]

import data.list.basic import tactic open list variable {α : Type} variable (x : α) variables (xs ys zs : list α) -- 1ª demostración example : xs ++ (ys ++ zs) = (xs ++ ys) ++ zs := begin induction xs with a as HI, { calc [] ++ (ys ++ zs) = ys ++ zs : append.equations._eqn_1 (ys ++ zs) ... = ([] ++ ys) ++ zs : congr_arg2 (++) (append.equations._eqn_1 ys) rfl, }, { calc (a :: as) ++ (ys ++ zs) = a :: (as ++ (ys ++ zs)) : append.equations._eqn_2 a as (ys ++ zs) ... = a :: ((as ++ ys) ++ zs) : congr_arg2 (::) rfl HI ... = (a :: (as ++ ys)) ++ zs : (append.equations._eqn_2 a (as ++ ys) zs).symm ... = ((a :: as) ++ ys) ++ zs : congr_arg2 (++) (append.equations._eqn_2 a as ys).symm rfl, }, end -- 2ª demostración example : xs ++ (ys ++ zs) = (xs ++ ys) ++ zs := begin induction xs with a as HI, { calc [] ++ (ys ++ zs) = ys ++ zs : nil_append (ys ++ zs) ... = ([] ++ ys) ++ zs : congr_arg2 (++) (nil_append ys) rfl, }, { calc (a :: as) ++ (ys ++ zs) = a :: (as ++ (ys ++ zs)) : cons_append a as (ys ++ zs) ... = a :: ((as ++ ys) ++ zs) : congr_arg2 (::) rfl HI ... = (a :: (as ++ ys)) ++ zs : (cons_append a (as ++ ys) zs).symm ... = ((a :: as) ++ ys) ++ zs : congr_arg2 (++) (cons_append a as ys).symm rfl, }, end -- 3ª demostración example : xs ++ (ys ++ zs) = (xs ++ ys) ++ zs := begin induction xs with a as HI, { calc [] ++ (ys ++ zs) = ys ++ zs : by rw nil_append ... = ([] ++ ys) ++ zs : by rw nil_append, }, { calc (a :: as) ++ (ys ++ zs) = a :: (as ++ (ys ++ zs)) : by rw cons_append ... = a :: ((as ++ ys) ++ zs) : by rw HI ... = (a :: (as ++ ys)) ++ zs : by rw cons_append ... = ((a :: as) ++ ys) ++ zs : by rw ← cons_append, }, end -- 4ª demostración example : xs ++ (ys ++ zs) = (xs ++ ys) ++ zs := begin induction xs with a as HI, { calc [] ++ (ys ++ zs) = ys ++ zs : rfl ... = ([] ++ ys) ++ zs : rfl, }, { calc (a :: as) ++ (ys ++ zs) = a :: (as ++ (ys ++ zs)) : rfl ... = a :: ((as ++ ys) ++ zs) : by rw HI ... = (a :: (as ++ ys)) ++ zs : rfl ... = ((a :: as) ++ ys) ++ zs : rfl, }, end -- 5ª demostración example : xs ++ (ys ++ zs) = (xs ++ ys) ++ zs := begin induction xs with a as HI, { calc [] ++ (ys ++ zs) = ys ++ zs : by simp ... = ([] ++ ys) ++ zs : by simp, }, { calc (a :: as) ++ (ys ++ zs) = a :: (as ++ (ys ++ zs)) : by simp ... = a :: ((as ++ ys) ++ zs) : congr_arg (cons a) HI ... = (a :: (as ++ ys)) ++ zs : by simp ... = ((a :: as) ++ ys) ++ zs : by simp, }, end -- 6ª demostración example : xs ++ (ys ++ zs) = (xs ++ ys) ++ zs := begin induction xs with a as HI, { by simp, }, { by exact (cons_inj a).mpr HI, }, end -- 7ª demostración example : xs ++ (ys ++ zs) = (xs ++ ys) ++ zs := begin induction xs with a as HI, { rw nil_append, rw nil_append, }, { rw cons_append, rw HI, rw cons_append, rw cons_append, }, end -- 8ª demostración example : xs ++ (ys ++ zs) = (xs ++ ys) ++ zs := list.rec_on xs ( show [] ++ (ys ++ zs) = ([] ++ ys) ++ zs, from calc [] ++ (ys ++ zs) = ys ++ zs : by rw nil_append ... = ([] ++ ys) ++ zs : by rw nil_append ) ( assume a as, assume HI : as ++ (ys ++ zs) = (as ++ ys) ++ zs, show (a :: as) ++ (ys ++ zs) = ((a :: as) ++ ys) ++ zs, from calc (a :: as) ++ (ys ++ zs) = a :: (as ++ (ys ++ zs)) : by rw cons_append ... = a :: ((as ++ ys) ++ zs) : by rw HI ... = (a :: (as ++ ys)) ++ zs : by rw cons_append ... = ((a :: as) ++ ys) ++ zs : by rw ← cons_append) -- 9ª demostración example : xs ++ (ys ++ zs) = (xs ++ ys) ++ zs := list.rec_on xs (by simp) (by simp [*]) -- 10ª demostración lemma conc_asoc_1 : ∀ xs, xs ++ (ys ++ zs) = (xs ++ ys) ++ zs | [] := by calc [] ++ (ys ++ zs) = ys ++ zs : by rw nil_append ... = ([] ++ ys) ++ zs : by rw nil_append | (a :: as) := by calc (a :: as) ++ (ys ++ zs) = a :: (as ++ (ys ++ zs)) : by rw cons_append ... = a :: ((as ++ ys) ++ zs) : by rw conc_asoc_1 ... = (a :: (as ++ ys)) ++ zs : by rw cons_append ... = ((a :: as) ++ ys) ++ zs : by rw ← cons_append -- 11ª demostración example : (xs ++ ys) ++ zs = xs ++ (ys ++ zs) := -- by library_search append_assoc xs ys zs -- 12ª demostración example : (xs ++ ys) ++ zs = xs ++ (ys ++ zs) := by induction xs ; simp [*] -- 13ª demostración example : (xs ++ ys) ++ zs = xs ++ (ys ++ zs) := by simp |

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>

[/expand]

[expand title=»Soluciones con Isabelle/HOL»]

theory Asociatividad_de_la_concatenacion_de_listas imports Main begin (* 1ª demostración *) lemma "xs @ (ys @ zs) = (xs @ ys) @ zs" proof (induct xs) have "[] @ (ys @ zs) = ys @ zs" by (simp only: append_Nil) also have "… = ([] @ ys) @ zs" by (simp only: append_Nil) finally show "[] @ (ys @ zs) = ([] @ ys) @ zs" by this next fix x xs assume HI : "xs @ (ys @ zs) = (xs @ ys) @ zs" have "(x # xs) @ (ys @ zs) = x # (xs @ (ys @ zs))" by (simp only: append_Cons) also have "… = x # ((xs @ ys) @ zs)" by (simp only: HI) also have "… = (x # (xs @ ys)) @ zs" by (simp only: append_Cons) also have "… = ((x # xs) @ ys) @ zs" by (simp only: append_Cons) finally show "(x # xs) @ (ys @ zs) = ((x # xs) @ ys) @ zs" by this qed (* 2ª demostración *) lemma "xs @ (ys @ zs) = (xs @ ys) @ zs" proof (induct xs) have "[] @ (ys @ zs) = ys @ zs" by simp also have "… = ([] @ ys) @ zs" by simp finally show "[] @ (ys @ zs) = ([] @ ys) @ zs" . next fix x xs assume HI : "xs @ (ys @ zs) = (xs @ ys) @ zs" have "(x # xs) @ (ys @ zs) = x # (xs @ (ys @ zs))" by simp also have "… = x # ((xs @ ys) @ zs)" by simp also have "… = (x # (xs @ ys)) @ zs" by simp also have "… = ((x # xs) @ ys) @ zs" by simp finally show "(x # xs) @ (ys @ zs) = ((x # xs) @ ys) @ zs" . qed (* 3ª demostración *) lemma "xs @ (ys @ zs) = (xs @ ys) @ zs" proof (induct xs) show "[] @ (ys @ zs) = ([] @ ys) @ zs" by simp next fix x xs assume "xs @ (ys @ zs) = (xs @ ys) @ zs" then show "(x # xs) @ (ys @ zs) = ((x # xs) @ ys) @ zs" by simp qed (* 4ª demostración *) lemma "xs @ (ys @ zs) = (xs @ ys) @ zs" proof (induct xs) case Nil then show ?case by simp next case (Cons a xs) then show ?case by simp qed (* 5ª demostración *) lemma "xs @ (ys @ zs) = (xs @ ys) @ zs" by (rule append_assoc [symmetric]) (* 6ª demostración *) lemma "xs @ (ys @ zs) = (xs @ ys) @ zs" by (induct xs) simp_all (* 7ª demostración *) lemma "xs @ (ys @ zs) = (xs @ ys) @ zs" by simp end |

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

[/expand]