<source lang="isar">
chapter {* R3: Razonamiento sobre programas *}

theory R3_Razonamiento_sobre_programas
imports Main
begin

text {* ---------------------------------------------------------------
Ejercicio 1.1. Definir la función
sumaImpares :: nat ⇒ nat
tal que (sumaImpares n) es la suma de los n primeros números
impares. Por ejemplo,
sumaImpares 5 = 25
------------------------------------------------------------------ *}

fun sumaImpares :: "nat ⇒ nat" where
"sumaImpares 0 = 0"
| "sumaImpares (Suc n) = sumaImpares n + (2*n+1)"

text {* ---------------------------------------------------------------
Ejercicio 1.2. Escribir la demostración detallada de
sumaImpares n = n*n
------------------------------------------------------------------- *}

-- "La demostración detallada es"
(* crigomgom fraortmoy marpoldia1 ivamenjim serrodcal rubgonmar ferrenseg juacabsou wilmorort josgarsan lucnovdos paupeddeg bowma pabrodmac*)
lemma "sumaImpares n = n*n"
proof (induct n)
show "sumaImpares 0 = 0 * 0" by simp
next
fix n
assume HI: "sumaImpares n = n * n"
have "sumaImpares (Suc n) = sumaImpares n + 2*n + 1" by simp
also have "... = n*n + 2*n + 1" using HI by simp
also have "... = Suc n * Suc n" by simp
finally show "sumaImpares (Suc n) = Suc n * Suc n" by simp
qed

(*danrodcha ; es la misma demostración que la anterior pero uso ?P para sustituir la propiedad.*)
lemma "sumaImpares n = n*n" (is "?P n")
proof (induct n)
show "?P 0" by simp
next
fix n
assume HI: "?P n"
have "sumaImpares (Suc n) = sumaImpares n + 2*n + 1" by simp
also have "... = n*n + 2*n + 1" using HI by simp
also have "... = Suc n * Suc n" by simp
finally show "?P (Suc n)" by simp
qed

(* migtermor *)
lemma "sumaImpares n = n*n"
proof (induct n)
show "sumaImpares 0 = 0*0" by simp
next
fix n
assume HI: "sumaImpares n = n*n"
have "sumaImpares (Suc n) = sumaImpares n + 2*n+1"
by (simp only: sumaImpares.simps(2))
also have "sumaImpares (Suc n) = n*n + 2*n+1" using HI by simp
finally show "sumaImpares (Suc n) = (Suc n)*(Suc n)" by simp
qed

(* fracorjim1 - Especifico la regla de simplificación y el paso del desarrollo al cuadrado para hacerlo más legible*)
lemma "sumaImpares n = n*n"
proof (induct n)
show "sumaImpares 0 = 0 * 0" by simp
next
fix n
assume HI : "sumaImpares n = n * n"
have "sumaImpares (Suc n) = sumaImpares n + (2 * n + 1)" by (simp only: sumaImpares.simps(2))
also have "... = n * n + 2 * n + 1" using HI by simp
also have "... = Suc n * Suc n" by simp
finally show "sumaImpares (Suc n) = Suc n * Suc n" by simp
qed

(*palucoto anaprarod*)
lemma "sumaImpares n = n*n"
proof (induct n)
show "sumaImpares 0 = 0 * 0" by simp
next
fix n
assume HI : "sumaImpares n = n * n"
have " sumaImpares (Suc n) = sumaImpares n + 2*n+1" by simp
also have "... = n*n + 2*n+1" using HI by simp
finally show "sumaImpares (Suc n) = Suc n * Suc n" by simp
qed

text {* ---------------------------------------------------------------
Ejercicio 2.1. Definir la función
sumaPotenciasDeDosMasUno :: nat ⇒ nat
tal que
(sumaPotenciasDeDosMasUno n) = 1 + 2^0 + 2^1 + 2^2 + ... + 2^n.
Por ejemplo,
sumaPotenciasDeDosMasUno 3 = 16
------------------------------------------------------------------ *}

fun sumaPotenciasDeDosMasUno :: "nat ⇒ nat" where
"sumaPotenciasDeDosMasUno 0 = 2"
| "sumaPotenciasDeDosMasUno (Suc n) =
sumaPotenciasDeDosMasUno n + 2^(n+1)"

text {* ---------------------------------------------------------------
Ejercicio 2.2. Escribir la demostración detallada de
sumaPotenciasDeDosMasUno n = 2^(n+1)
------------------------------------------------------------------- *}

(*crigomgom ivamenjim danrodcha serrodcal rubgonmar ferrenseg juacabsou wilmorort anaprarod marpoldia1 josgarsan paupeddeg pabrodmac bowma*)
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"
proof (induct n)
show "sumaPotenciasDeDosMasUno 0 = 2 ^ (0 + 1)" by simp
next
fix n
assume HI: "sumaPotenciasDeDosMasUno n = 2 ^ (n + 1)"
have "sumaPotenciasDeDosMasUno (Suc n) = sumaPotenciasDeDosMasUno n + 2^(n+1)" by simp
also have "... = 2 ^ (n + 1) + 2 ^ (n + 1)" using HI by simp
also have "... = 2 ^ ((Suc n) + 1)" by simp
finally show "sumaPotenciasDeDosMasUno (Suc n) = 2 ^ (Suc n + 1)" by simp
qed

(* fraortmoy *)
(* es la misma demostración, pero quise probar a delimitar lo que se usa en el "by simp" *)

lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"
proof (induct n)
show "sumaPotenciasDeDosMasUno 0 = 2 ^ (0 + 1)" by simp
next
fix n
assume H1:" sumaPotenciasDeDosMasUno n = 2 ^ (n + 1)"
have "sumaPotenciasDeDosMasUno (Suc n) = sumaPotenciasDeDosMasUno n + 2^(n + 1)" by (simp only : sumaPotenciasDeDosMasUno.simps(2))
also have "... = 2 ^ (n + 1) + 2 ^ (n + 1)" using H1 by simp
also have "... = 2 ^ (Suc n + 1)" by simp
finally show "sumaPotenciasDeDosMasUno (Suc n) = 2 ^ (Suc n + 1)" by simp
qed

(* migtermor *)
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"
proof (induct n)
show "sumaPotenciasDeDosMasUno 0 = 2^(0+1)" by simp
next
fix n
assume HI: "sumaPotenciasDeDosMasUno n = 2^(n+1)"
have "sumaPotenciasDeDosMasUno (Suc n) = sumaPotenciasDeDosMasUno n + 2^(n+1)"
by (simp only: sumaPotenciasDeDosMasUno.simps(2))
also have "sumaPotenciasDeDosMasUno (Suc n) = 2^(n+1)+2^(n+1)" using HI by simp
finally show "sumaPotenciasDeDosMasUno (Suc n) = 2^((Suc n)+1)" by simp
qed

(*pablucoto*)
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"
proof (induct n)
show "sumaPotenciasDeDosMasUno 0 = 2 ^ (0 + 1)" by simp
next
fix n
assume HI : "sumaPotenciasDeDosMasUno n = 2 ^ (n + 1)"
have "sumaPotenciasDeDosMasUno (Suc n) = sumaPotenciasDeDosMasUno n + 2^(n+1) " by simp
also have "... = 2^(n + 1) + 2^( n + 1) " using HI by simp
finally show " sumaPotenciasDeDosMasUno (Suc n) = 2 ^ (Suc n + 1)" by simp
qed

(* fracorjim1 - Hago explícita toda la manipulación algebraica. Quizás excesivo.*)
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"
proof (induction n)
show "sumaPotenciasDeDosMasUno 0 = 2 ^ (0 + 1)" by simp
next
fix n
assume HI : "sumaPotenciasDeDosMasUno n = 2 ^ (n + 1)"
have "sumaPotenciasDeDosMasUno (Suc n) = sumaPotenciasDeDosMasUno n + 2 ^ (n + 1)"
by (simp only:sumaPotenciasDeDosMasUno.simps(2))
also have "... = 2 ^ (n + 1) + 2 ^ (n + 1)" using HI by simp
also have "... = 2 ^ (n + 1) * 2" by simp
also have "... = 2 ^ ((n + 1) + 1)" by simp
also have "... = 2 ^ (Suc n + 1)" by simp
finally show "sumaPotenciasDeDosMasUno (Suc n) = 2 ^ (Suc n + 1)" by simp
qed

(*lucnovdos*)
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"
proof (induct n)
show "sumaPotenciasDeDosMasUno 0 = 2 ^ (0 + 1)" by simp
next
fix n
assume HI: "sumaPotenciasDeDosMasUno n = 2 ^ (n + 1)"
have "sumaPotenciasDeDosMasUno (Suc n) = sumaPotenciasDeDosMasUno n + 2 ^ (n + 1)" by simp
also have "... = 2 ^ (n + 1) + 2 ^ (n + 1)" using HI by simp
finally show "sumaPotenciasDeDosMasUno (Suc n) = 2 ^ (Suc n + 1)" by simp
qed

text {* ---------------------------------------------------------------
Ejercicio 3.1. Definir la función
copia :: nat ⇒ 'a ⇒ 'a list
tal que (copia n x) es la lista formado por n copias del elemento
x. Por ejemplo,
copia 3 x = [x,x,x]
------------------------------------------------------------------ *}

fun copia :: "nat ⇒ 'a ⇒ 'a list" where
"copia 0 x = []"
| "copia (Suc n) x = x # copia n x"

text {* ---------------------------------------------------------------
Ejercicio 3.2. Definir la función
todos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool
tal que (todos p xs) se verifica si todos los elementos de xs cumplen
la propiedad p. Por ejemplo,
todos (λx. x>(1::nat)) [2,6,4] = True
todos (λx. x>(2::nat)) [2,6,4] = False
------------------------------------------------------------------ *}

fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos p [] = True"
| "todos p (x#xs) = (p x ∧ todos p xs)"

text {* ---------------------------------------------------------------
Ejercicio 3.2. Demostrar detalladamente que todos los elementos de
(copia n x) son iguales a x.
------------------------------------------------------------------- *}

(* crigomgom ivamenjim serrodcal ferrenseg wilmorort juacabsou josgarsan*)
lemma "todos (λy. y=x) (copia n x)"
proof (induct n)
show "todos (λy. y = x) (copia 0 x)" by simp
next
fix n
assume HI: "todos (λy. y = x) (copia n x)"
have "todos (λy. y = x) (copia (Suc n) x) = todos (λy. y = x) (x # (copia n x))" by simp
also have "... = ((x = x) ∧ (todos (λy. y = x) (copia n x)))" by simp
also have "... = True" using HI by simp
finally show "todos (λy. y = x) (copia (Suc n) x)" by simp
qed

(* fraortmoy *)
lemma "todos (λy. y=x) (copia n x)"
proof (induct n)
have "todos (λy. y = x) (copia 0 x) = todos (λy. y = x) []" by (simp only: copia.simps(1))
also have "... = True" by (simp only: todos.simps(1))
show "todos (λy. y = x) (copia 0 x)" by simp
next
fix n
assume H1 : " todos (λy. y = x) (copia n x) "
have "todos (λy. y = x) (copia (Suc n) x) = ((todos (λy. y = x) (x#[])) ∧ (todos (λy. y = x) (copia n x) )) " by simp
also have " ... = (todos (λy. y = x) (x#[]))" using H1 by simp
also have " ... = ((λy. y = x) x ∧ todos (λy. y = x) [])" by simp
also have " ... = True" by simp
finally show " todos (λy. y = x) (copia (Suc n) x)" by simp
qed

(*danrodcha rubgonmar*)
lemma "todos (λy. y=x) (copia n x)" (is "?P n")
proof (induct n)
show "?P 0" by simp
next
fix n
assume HI: "?P n"
have "todos (λy. y=x) (copia (Suc n) x) = todos (λy. y=x) (x # copia n x)" by simp
also have "... = ((λy. y=x) x ∧ todos (λy. y=x) (copia n x))" by simp
also have "... = todos (λy. y=x) (copia n x)" by simp
finally show "?P (Suc n)" using HI by simp
qed

(* migtermor *)
lemma "todos (λy. y=x) (copia n x)"
proof (induct n)
have "todos (λy. y=x) (copia 0 x) = todos (λy. y=x) []" by (simp only: copia.simps(1))
also have "todos (λy. y=x) []" by (simp only: todos.simps(1))
show "todos (λy. y=x) (copia 0 x)" by simp
next
fix n
assume HI: "todos (λy. y=x) (copia n x)"
have "todos (λy. y=x) (copia (Suc n) x) = todos (λy. y=x) (x # (copia n x))"
by (simp only: copia.simps(2))
also have "todos (λy. y=x) (x#(copia n x)) = ((λy. y=x) x ∧ todos (λy. y=x) (copia n x))"
by (simp only: todos.simps(2))
also have "todos (λy. y=x) (copia (Suc n) x) = ((λy. y=x) x)" using HI by simp
also have "((λy. y=x) x) = True" by simp
finally show "(todos (λy. y=x) (copia (Suc n) x))" by simp
qed

(*pablucoto pabrodmac*)

lemma "todos (λy. y=x) (copia n x)"
proof (induct n)
show "todos (λy. y = x) (copia 0 x)" by simp
next
fix n
assume HI: "todos (λy. y=x) (copia n x)"
have "todos (λy. y = x) (copia (Suc n) x) = todos (λy. y=x) (x # copia n x)" by simp
also have "... = ( (λy. y=x) x ∧ todos (λy. y=x) (copia (Suc n) x) ) " by simp
also have "... = True" using HI by simp
finally show " todos (λy. y = x) (copia (Suc n) x) " by simp
qed

(* anaprarod marpoldia1 *)
lemma "todos (λy. y=x) (copia n x)"
proof (induct n)
show "todos (λy. y = x) (copia 0 x)" by simp
next
fix n
assume HI: "todos (λy. y = x) (copia n x)"
have "todos (λy. y = x) (copia (Suc n) x) =todos (λy. y = x) (x # copia n x)" by simp
also have "... = ((λy. y = x) x ∧ todos (λy. y = x) ( copia n x))" by simp
finally show "todos (λy. y = x) (copia (Suc n) x)" using HI by simp
qed

(*paupeddeg*)
lemma "todos (λy. y=x) (copia n x)"
proof (induct n)
show "todos (λy. y = x) (copia 0 x)" by simp
next
fix n
assume HI: "todos (λy. y = x) (copia n x)"
have "todos (λy. y = x) (copia (Suc n) x) = (todos (λy. y = x) (x # copia n x))" by simp
also have "... = ((todos (λy. y = x) [x]) ∧ (todos (λy. y = x) (copia n x)))" by simp
also have "... = True" using HI by simp
finally show "todos(λy. y = x) (copia (Suc n) x)" by simp
qed

text {* ---------------------------------------------------------------
Ejercicio 4.1. Definir la función
factR :: nat ⇒ nat
tal que (factR n) es el factorial de n. Por ejemplo,
factR 4 = 24
------------------------------------------------------------------ *}

fun factR :: "nat ⇒ nat" where
"factR 0 = 1"
| "factR (Suc n) = Suc n * factR n"

text {* ---------------------------------------------------------------
Ejercicio 4.2. Se considera la siguiente definición iterativa de la
función factorial
factI :: "nat ⇒ nat" where
factI n = factI' n 1

factI' :: nat ⇒ nat ⇒ nat" where
factI' 0 x = x
factI' (Suc n) x = factI' n (Suc n)*x
Demostrar que, para todo n y todo x, se tiene
factI' n x = x * factR n
Indicación: La propiedad mult_Suc es
(Suc m) * n = n + m * n
Puede que se necesite desactivarla en un paso con
(simp del: mult_Suc)
------------------------------------------------------------------- *}

fun factI' :: "nat ⇒ nat ⇒ nat" where
"factI' 0 x = x"
| "factI' (Suc n) x = factI' n (x * Suc n)"

fun factI :: "nat ⇒ nat" where
"factI n = factI' n 1"

(* fraortmoy , danrodcha serrodcal pablucoto wilmorort anaprarod marpoldia1 juacabsou josgarsan rubgonmar paupeddeg pabrodmac*)
lemma fact: "factI' n x = x * factR n"
proof (induct n)
show "factI' 0 x = x * factR 0" by simp
next
fix n
assume H1 : "factI' n x = x * factR n"
have "factI' (Suc n) x = factI' n (x * Suc n)" by (simp only:factI'.simps(2))
also have "... = (x * Suc n) * factR n" using H1 by simp (* no entiendo por qué no hace esto bien y luego todo funciona *) (*Creo que es porque no le estas dando la hipótesis de inducción sino otra que se le parece*)
also have "... = x * factR (Suc n)" by (simp del: mult_Suc)
finally show "factI' (Suc n) x = x * factR (Suc n)" by simp
qed

(* crigomgom ivamenjim ferrenseg *)
lemma fact: "factI' n x = x * factR n"
proof (induct n arbitrary: x)
show "⋀x. factI' 0 x = x * factR 0" by simp
next
fix n
assume HI: "⋀x. factI' n x = x * factR n"
show "⋀x. factI' (Suc n) x = x * factR (Suc n)"
proof -
fix x
have "factI' (Suc n) x = factI' n (x * Suc n)" by simp
also have "... = (x * Suc n) * factR n" using HI by simp
also have "... = x * (Suc n * factR n)" by (simp del: mult_Suc)
also have "... = x * factR (Suc n)" by simp
finally show "factI' (Suc n) x = x * factR (Suc n)" by simp
qed
qed

(* migtermor *)

lemma fact: "factI' n x = x * factR n"
proof (induct n)
have "x * factR 0 = x" by (simp only: factR.simps(1))
also have "factI' 0 x = x" by (simp only: factI'.simps(1))
show "factI' 0 x = x * factR 0" by simp
next
fix n
assume HI: "∀x. factI' n x = x * factR n"
fix x
have "factI' (Suc n) x = factI' n (x * Suc n)" by (simp only: factI'.simps(2))
also have "factI' (Suc n) x = factI' n (x * Suc n)" by simp
have "factI' (Suc n) x = x * Suc n * factR n" using HI by simp
also have "... = x * factR (Suc n)" by (simp del: mult_Suc)
finally show "factI' (Suc n) x = x * factR (Suc n)" by simp (* No entiendo por qué no acepta esto como demostrado *)
qed

text {* ---------------------------------------------------------------
Ejercicio 4.3. Escribir la demostración detallada de
factI n = factR n
------------------------------------------------------------------- *}

(* fraortmoy danrodcha crigomgom ivamenjim serrodcal ferrenseg pablucoto wilmorort rubgonmar marpoldia1 juacabsou josgarsan paupeddeg*)
corollary "factI n = factR n"
proof -
have "factI n = factI' n 1" by simp
also have "... = 1 * factR n" by (simp add: fact)
finally show "factI n = factR n" by simp
qed

(* migtermor *)
corollary "factI n = factR n"
proof -
have "factI n = factI' n 1" by simp
also have "... = 1 * factI' n 1" by simp
also have "... = 1 * factR n" using fact by simp
finally show "factI n = factR n" by simp
qed

(*pabrodmac)
corollary "factI n = factR n"
proof -
have "factI n = factI' n 1" by simp
also have "... = 1 * factR n" using fact by simp
finally show "factI n = factR n" by simp
qed

text {* ---------------------------------------------------------------
Ejercicio 5.1. Definir, recursivamente y sin usar (@), la función
amplia :: 'a list ⇒ 'a ⇒ 'a list
tal que (amplia xs y) es la lista obtenida añadiendo el elemento y al
final de la lista xs. Por ejemplo,
amplia [d,a] t = [d,a,t]
------------------------------------------------------------------ *}

fun amplia :: "'a list ⇒ 'a ⇒ 'a list" where
"amplia [] y = [y]"
| "amplia (x#xs) y = x # amplia xs y"

text {* ---------------------------------------------------------------
Ejercicio 5.2. Escribir la demostración detallada de
amplia xs y = xs @ [y]
------------------------------------------------------------------- *}

(* crigomgom fraortmoy rubgonmar ivamenjim serrodcal pablucoto wilmorort anaprarod marpoldia1 juacabsou paupeddeg *)
lemma "amplia xs y = xs @ [y]"
proof (induct xs)
show "amplia [] y = [] @ [y]" by simp
next
fix x xs
assume HI: "amplia xs y = xs @ [y]"
have "amplia (x # xs) y = x # (amplia xs y)" by simp
also have "... = x # (xs @ [y])" using HI by simp
also have "... = (x # xs) @ [y]" by simp
finally show "amplia (x # xs) y = (x # xs) @ [y]" by simp
qed

(*danrodcha ferrenseg *)
lemma "amplia xs y = xs @ [y]" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix x xs
assume HI: "?P xs"
have "amplia (x # xs) y = x # amplia xs y" by simp
also have "... = (x # xs) @ [y]" using HI by simp
finally show "?P (x#xs)" by simp
qed

(* migtermor *)
lemma "amplia xs y = xs @ [y]"
proof (induct xs)
have "amplia [] y = [y]" by (simp only: amplia.simps(1))
have "[] @ [y] = [y]" by simp
show "amplia [] y = [] @ [y]" by simp
next
fix x xs
assume HI: "amplia xs y = xs @ [y]"
have "amplia (x # xs) y = x # amplia xs y" by (simp only: amplia.simps(2))
also have "... = x # (xs @ [y])" using HI by simp
also have "... = (x # xs) @ [y]" by simp
finally show "amplia (x # xs) y = (x # xs) @ [y]" by simp
qed

end
</source>

Relación 3

2016-11-16T10:32:04Z

Pabrodmac:

<source lang="isar">
chapter {* R3: Razonamiento sobre programas *}

theory R3_Razonamiento_sobre_programas
imports Main
begin

text {* ---------------------------------------------------------------
Ejercicio 1.1. Definir la función
sumaImpares :: nat ⇒ nat
tal que (sumaImpares n) es la suma de los n primeros números
impares. Por ejemplo,
sumaImpares 5 = 25
------------------------------------------------------------------ *}

fun sumaImpares :: "nat ⇒ nat" where
"sumaImpares 0 = 0"
| "sumaImpares (Suc n) = sumaImpares n + (2*n+1)"

text {* ---------------------------------------------------------------
Ejercicio 1.2. Escribir la demostración detallada de
sumaImpares n = n*n
------------------------------------------------------------------- *}

-- "La demostración detallada es"
(* crigomgom fraortmoy marpoldia1 ivamenjim serrodcal rubgonmar ferrenseg juacabsou wilmorort josgarsan lucnovdos paupeddeg bowma pabrodmac*)
lemma "sumaImpares n = n*n"
proof (induct n)
show "sumaImpares 0 = 0 * 0" by simp
next
fix n
assume HI: "sumaImpares n = n * n"
have "sumaImpares (Suc n) = sumaImpares n + 2*n + 1" by simp
also have "... = n*n + 2*n + 1" using HI by simp
also have "... = Suc n * Suc n" by simp
finally show "sumaImpares (Suc n) = Suc n * Suc n" by simp
qed

(*danrodcha ; es la misma demostración que la anterior pero uso ?P para sustituir la propiedad.*)
lemma "sumaImpares n = n*n" (is "?P n")
proof (induct n)
show "?P 0" by simp
next
fix n
assume HI: "?P n"
have "sumaImpares (Suc n) = sumaImpares n + 2*n + 1" by simp
also have "... = n*n + 2*n + 1" using HI by simp
also have "... = Suc n * Suc n" by simp
finally show "?P (Suc n)" by simp
qed

(* migtermor *)
lemma "sumaImpares n = n*n"
proof (induct n)
show "sumaImpares 0 = 0*0" by simp
next
fix n
assume HI: "sumaImpares n = n*n"
have "sumaImpares (Suc n) = sumaImpares n + 2*n+1"
by (simp only: sumaImpares.simps(2))
also have "sumaImpares (Suc n) = n*n + 2*n+1" using HI by simp
finally show "sumaImpares (Suc n) = (Suc n)*(Suc n)" by simp
qed

(* fracorjim1 - Especifico la regla de simplificación y el paso del desarrollo al cuadrado para hacerlo más legible*)
lemma "sumaImpares n = n*n"
proof (induct n)
show "sumaImpares 0 = 0 * 0" by simp
next
fix n
assume HI : "sumaImpares n = n * n"
have "sumaImpares (Suc n) = sumaImpares n + (2 * n + 1)" by (simp only: sumaImpares.simps(2))
also have "... = n * n + 2 * n + 1" using HI by simp
also have "... = Suc n * Suc n" by simp
finally show "sumaImpares (Suc n) = Suc n * Suc n" by simp
qed

(*palucoto anaprarod*)
lemma "sumaImpares n = n*n"
proof (induct n)
show "sumaImpares 0 = 0 * 0" by simp
next
fix n
assume HI : "sumaImpares n = n * n"
have " sumaImpares (Suc n) = sumaImpares n + 2*n+1" by simp
also have "... = n*n + 2*n+1" using HI by simp
finally show "sumaImpares (Suc n) = Suc n * Suc n" by simp
qed

text {* ---------------------------------------------------------------
Ejercicio 2.1. Definir la función
sumaPotenciasDeDosMasUno :: nat ⇒ nat
tal que
(sumaPotenciasDeDosMasUno n) = 1 + 2^0 + 2^1 + 2^2 + ... + 2^n.
Por ejemplo,
sumaPotenciasDeDosMasUno 3 = 16
------------------------------------------------------------------ *}

fun sumaPotenciasDeDosMasUno :: "nat ⇒ nat" where
"sumaPotenciasDeDosMasUno 0 = 2"
| "sumaPotenciasDeDosMasUno (Suc n) =
sumaPotenciasDeDosMasUno n + 2^(n+1)"

text {* ---------------------------------------------------------------
Ejercicio 2.2. Escribir la demostración detallada de
sumaPotenciasDeDosMasUno n = 2^(n+1)
------------------------------------------------------------------- *}

(*crigomgom ivamenjim danrodcha serrodcal rubgonmar ferrenseg juacabsou wilmorort anaprarod marpoldia1 josgarsan paupeddeg pabrodmac bowma*)
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"
proof (induct n)
show "sumaPotenciasDeDosMasUno 0 = 2 ^ (0 + 1)" by simp
next
fix n
assume HI: "sumaPotenciasDeDosMasUno n = 2 ^ (n + 1)"
have "sumaPotenciasDeDosMasUno (Suc n) = sumaPotenciasDeDosMasUno n + 2^(n+1)" by simp
also have "... = 2 ^ (n + 1) + 2 ^ (n + 1)" using HI by simp
also have "... = 2 ^ ((Suc n) + 1)" by simp
finally show "sumaPotenciasDeDosMasUno (Suc n) = 2 ^ (Suc n + 1)" by simp
qed

(* fraortmoy *)
(* es la misma demostración, pero quise probar a delimitar lo que se usa en el "by simp" *)

lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"
proof (induct n)
show "sumaPotenciasDeDosMasUno 0 = 2 ^ (0 + 1)" by simp
next
fix n
assume H1:" sumaPotenciasDeDosMasUno n = 2 ^ (n + 1)"
have "sumaPotenciasDeDosMasUno (Suc n) = sumaPotenciasDeDosMasUno n + 2^(n + 1)" by (simp only : sumaPotenciasDeDosMasUno.simps(2))
also have "... = 2 ^ (n + 1) + 2 ^ (n + 1)" using H1 by simp
also have "... = 2 ^ (Suc n + 1)" by simp
finally show "sumaPotenciasDeDosMasUno (Suc n) = 2 ^ (Suc n + 1)" by simp
qed

(* migtermor *)
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"
proof (induct n)
show "sumaPotenciasDeDosMasUno 0 = 2^(0+1)" by simp
next
fix n
assume HI: "sumaPotenciasDeDosMasUno n = 2^(n+1)"
have "sumaPotenciasDeDosMasUno (Suc n) = sumaPotenciasDeDosMasUno n + 2^(n+1)"
by (simp only: sumaPotenciasDeDosMasUno.simps(2))
also have "sumaPotenciasDeDosMasUno (Suc n) = 2^(n+1)+2^(n+1)" using HI by simp
finally show "sumaPotenciasDeDosMasUno (Suc n) = 2^((Suc n)+1)" by simp
qed

(*pablucoto*)
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"
proof (induct n)
show "sumaPotenciasDeDosMasUno 0 = 2 ^ (0 + 1)" by simp
next
fix n
assume HI : "sumaPotenciasDeDosMasUno n = 2 ^ (n + 1)"
have "sumaPotenciasDeDosMasUno (Suc n) = sumaPotenciasDeDosMasUno n + 2^(n+1) " by simp
also have "... = 2^(n + 1) + 2^( n + 1) " using HI by simp
finally show " sumaPotenciasDeDosMasUno (Suc n) = 2 ^ (Suc n + 1)" by simp
qed

(* fracorjim1 - Hago explícita toda la manipulación algebraica. Quizás excesivo.*)
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"
proof (induction n)
show "sumaPotenciasDeDosMasUno 0 = 2 ^ (0 + 1)" by simp
next
fix n
assume HI : "sumaPotenciasDeDosMasUno n = 2 ^ (n + 1)"
have "sumaPotenciasDeDosMasUno (Suc n) = sumaPotenciasDeDosMasUno n + 2 ^ (n + 1)"
by (simp only:sumaPotenciasDeDosMasUno.simps(2))
also have "... = 2 ^ (n + 1) + 2 ^ (n + 1)" using HI by simp
also have "... = 2 ^ (n + 1) * 2" by simp
also have "... = 2 ^ ((n + 1) + 1)" by simp
also have "... = 2 ^ (Suc n + 1)" by simp
finally show "sumaPotenciasDeDosMasUno (Suc n) = 2 ^ (Suc n + 1)" by simp
qed

(*lucnovdos*)
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"
proof (induct n)
show "sumaPotenciasDeDosMasUno 0 = 2 ^ (0 + 1)" by simp
next
fix n
assume HI: "sumaPotenciasDeDosMasUno n = 2 ^ (n + 1)"
have "sumaPotenciasDeDosMasUno (Suc n) = sumaPotenciasDeDosMasUno n + 2 ^ (n + 1)" by simp
also have "... = 2 ^ (n + 1) + 2 ^ (n + 1)" using HI by simp
finally show "sumaPotenciasDeDosMasUno (Suc n) = 2 ^ (Suc n + 1)" by simp
qed

text {* ---------------------------------------------------------------
Ejercicio 3.1. Definir la función
copia :: nat ⇒ 'a ⇒ 'a list
tal que (copia n x) es la lista formado por n copias del elemento
x. Por ejemplo,
copia 3 x = [x,x,x]
------------------------------------------------------------------ *}

fun copia :: "nat ⇒ 'a ⇒ 'a list" where
"copia 0 x = []"
| "copia (Suc n) x = x # copia n x"

text {* ---------------------------------------------------------------
Ejercicio 3.2. Definir la función
todos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool
tal que (todos p xs) se verifica si todos los elementos de xs cumplen
la propiedad p. Por ejemplo,
todos (λx. x>(1::nat)) [2,6,4] = True
todos (λx. x>(2::nat)) [2,6,4] = False
------------------------------------------------------------------ *}

fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos p [] = True"
| "todos p (x#xs) = (p x ∧ todos p xs)"

text {* ---------------------------------------------------------------
Ejercicio 3.2. Demostrar detalladamente que todos los elementos de
(copia n x) son iguales a x.
------------------------------------------------------------------- *}

(* crigomgom ivamenjim serrodcal ferrenseg wilmorort juacabsou josgarsan*)
lemma "todos (λy. y=x) (copia n x)"
proof (induct n)
show "todos (λy. y = x) (copia 0 x)" by simp
next
fix n
assume HI: "todos (λy. y = x) (copia n x)"
have "todos (λy. y = x) (copia (Suc n) x) = todos (λy. y = x) (x # (copia n x))" by simp
also have "... = ((x = x) ∧ (todos (λy. y = x) (copia n x)))" by simp
also have "... = True" using HI by simp
finally show "todos (λy. y = x) (copia (Suc n) x)" by simp
qed

(* fraortmoy *)
lemma "todos (λy. y=x) (copia n x)"
proof (induct n)
have "todos (λy. y = x) (copia 0 x) = todos (λy. y = x) []" by (simp only: copia.simps(1))
also have "... = True" by (simp only: todos.simps(1))
show "todos (λy. y = x) (copia 0 x)" by simp
next
fix n
assume H1 : " todos (λy. y = x) (copia n x) "
have "todos (λy. y = x) (copia (Suc n) x) = ((todos (λy. y = x) (x#[])) ∧ (todos (λy. y = x) (copia n x) )) " by simp
also have " ... = (todos (λy. y = x) (x#[]))" using H1 by simp
also have " ... = ((λy. y = x) x ∧ todos (λy. y = x) [])" by simp
also have " ... = True" by simp
finally show " todos (λy. y = x) (copia (Suc n) x)" by simp
qed

(*danrodcha rubgonmar*)
lemma "todos (λy. y=x) (copia n x)" (is "?P n")
proof (induct n)
show "?P 0" by simp
next
fix n
assume HI: "?P n"
have "todos (λy. y=x) (copia (Suc n) x) = todos (λy. y=x) (x # copia n x)" by simp
also have "... = ((λy. y=x) x ∧ todos (λy. y=x) (copia n x))" by simp
also have "... = todos (λy. y=x) (copia n x)" by simp
finally show "?P (Suc n)" using HI by simp
qed

(* migtermor *)
lemma "todos (λy. y=x) (copia n x)"
proof (induct n)
have "todos (λy. y=x) (copia 0 x) = todos (λy. y=x) []" by (simp only: copia.simps(1))
also have "todos (λy. y=x) []" by (simp only: todos.simps(1))
show "todos (λy. y=x) (copia 0 x)" by simp
next
fix n
assume HI: "todos (λy. y=x) (copia n x)"
have "todos (λy. y=x) (copia (Suc n) x) = todos (λy. y=x) (x # (copia n x))"
by (simp only: copia.simps(2))
also have "todos (λy. y=x) (x#(copia n x)) = ((λy. y=x) x ∧ todos (λy. y=x) (copia n x))"
by (simp only: todos.simps(2))
also have "todos (λy. y=x) (copia (Suc n) x) = ((λy. y=x) x)" using HI by simp
also have "((λy. y=x) x) = True" by simp
finally show "(todos (λy. y=x) (copia (Suc n) x))" by simp
qed

(*pablucoto pabrodmac*)

lemma "todos (λy. y=x) (copia n x)"
proof (induct n)
show "todos (λy. y = x) (copia 0 x)" by simp
next
fix n
assume HI: "todos (λy. y=x) (copia n x)"
have "todos (λy. y = x) (copia (Suc n) x) = todos (λy. y=x) (x # copia n x)" by simp
also have "... = ( (λy. y=x) x ∧ todos (λy. y=x) (copia (Suc n) x) ) " by simp
also have "... = True" using HI by simp
finally show " todos (λy. y = x) (copia (Suc n) x) " by simp
qed

(* anaprarod marpoldia1 *)
lemma "todos (λy. y=x) (copia n x)"
proof (induct n)
show "todos (λy. y = x) (copia 0 x)" by simp
next
fix n
assume HI: "todos (λy. y = x) (copia n x)"
have "todos (λy. y = x) (copia (Suc n) x) =todos (λy. y = x) (x # copia n x)" by simp
also have "... = ((λy. y = x) x ∧ todos (λy. y = x) ( copia n x))" by simp
finally show "todos (λy. y = x) (copia (Suc n) x)" using HI by simp
qed

(*paupeddeg*)
lemma "todos (λy. y=x) (copia n x)"
proof (induct n)
show "todos (λy. y = x) (copia 0 x)" by simp
next
fix n
assume HI: "todos (λy. y = x) (copia n x)"
have "todos (λy. y = x) (copia (Suc n) x) = (todos (λy. y = x) (x # copia n x))" by simp
also have "... = ((todos (λy. y = x) [x]) ∧ (todos (λy. y = x) (copia n x)))" by simp
also have "... = True" using HI by simp
finally show "todos(λy. y = x) (copia (Suc n) x)" by simp
qed

text {* ---------------------------------------------------------------
Ejercicio 4.1. Definir la función
factR :: nat ⇒ nat
tal que (factR n) es el factorial de n. Por ejemplo,
factR 4 = 24
------------------------------------------------------------------ *}

fun factR :: "nat ⇒ nat" where
"factR 0 = 1"
| "factR (Suc n) = Suc n * factR n"

text {* ---------------------------------------------------------------
Ejercicio 4.2. Se considera la siguiente definición iterativa de la
función factorial
factI :: "nat ⇒ nat" where
factI n = factI' n 1

factI' :: nat ⇒ nat ⇒ nat" where
factI' 0 x = x
factI' (Suc n) x = factI' n (Suc n)*x
Demostrar que, para todo n y todo x, se tiene
factI' n x = x * factR n
Indicación: La propiedad mult_Suc es
(Suc m) * n = n + m * n
Puede que se necesite desactivarla en un paso con
(simp del: mult_Suc)
------------------------------------------------------------------- *}

fun factI' :: "nat ⇒ nat ⇒ nat" where
"factI' 0 x = x"
| "factI' (Suc n) x = factI' n (x * Suc n)"

fun factI :: "nat ⇒ nat" where
"factI n = factI' n 1"

(* fraortmoy , danrodcha serrodcal pablucoto wilmorort anaprarod marpoldia1 juacabsou josgarsan rubgonmar paupeddeg pabrodmac*)
lemma fact: "factI' n x = x * factR n"
proof (induct n)
show "factI' 0 x = x * factR 0" by simp
next
fix n
assume H1 : "factI' n x = x * factR n"
have "factI' (Suc n) x = factI' n (x * Suc n)" by (simp only:factI'.simps(2))
also have "... = (x * Suc n) * factR n" using H1 by simp (* no entiendo por qué no hace esto bien y luego todo funciona *) (*Creo que es porque no le estas dando la hipótesis de inducción sino otra que se le parece*)
also have "... = x * factR (Suc n)" by (simp del: mult_Suc)
finally show "factI' (Suc n) x = x * factR (Suc n)" by simp
qed

(* crigomgom ivamenjim ferrenseg *)
lemma fact: "factI' n x = x * factR n"
proof (induct n arbitrary: x)
show "⋀x. factI' 0 x = x * factR 0" by simp
next
fix n
assume HI: "⋀x. factI' n x = x * factR n"
show "⋀x. factI' (Suc n) x = x * factR (Suc n)"
proof -
fix x
have "factI' (Suc n) x = factI' n (x * Suc n)" by simp
also have "... = (x * Suc n) * factR n" using HI by simp
also have "... = x * (Suc n * factR n)" by (simp del: mult_Suc)
also have "... = x * factR (Suc n)" by simp
finally show "factI' (Suc n) x = x * factR (Suc n)" by simp
qed
qed

(* migtermor *)

lemma fact: "factI' n x = x * factR n"
proof (induct n)
have "x * factR 0 = x" by (simp only: factR.simps(1))
also have "factI' 0 x = x" by (simp only: factI'.simps(1))
show "factI' 0 x = x * factR 0" by simp
next
fix n
assume HI: "∀x. factI' n x = x * factR n"
fix x
have "factI' (Suc n) x = factI' n (x * Suc n)" by (simp only: factI'.simps(2))
also have "factI' (Suc n) x = factI' n (x * Suc n)" by simp
have "factI' (Suc n) x = x * Suc n * factR n" using HI by simp
also have "... = x * factR (Suc n)" by (simp del: mult_Suc)
finally show "factI' (Suc n) x = x * factR (Suc n)" by simp (* No entiendo por qué no acepta esto como demostrado *)
qed

text {* ---------------------------------------------------------------
Ejercicio 4.3. Escribir la demostración detallada de
factI n = factR n
------------------------------------------------------------------- *}

(* fraortmoy danrodcha crigomgom ivamenjim serrodcal ferrenseg pablucoto wilmorort rubgonmar marpoldia1 juacabsou josgarsan paupeddeg*)
corollary "factI n = factR n"
proof -
have "factI n = factI' n 1" by simp
also have "... = 1 * factR n" by (simp add: fact)
finally show "factI n = factR n" by simp
qed

(* migtermor *)
corollary "factI n = factR n"
proof -
have "factI n = factI' n 1" by simp
also have "... = 1 * factI' n 1" by simp
also have "... = 1 * factR n" using fact by simp
finally show "factI n = factR n" by simp
qed

text {* ---------------------------------------------------------------
Ejercicio 5.1. Definir, recursivamente y sin usar (@), la función
amplia :: 'a list ⇒ 'a ⇒ 'a list
tal que (amplia xs y) es la lista obtenida añadiendo el elemento y al
final de la lista xs. Por ejemplo,
amplia [d,a] t = [d,a,t]
------------------------------------------------------------------ *}

fun amplia :: "'a list ⇒ 'a ⇒ 'a list" where
"amplia [] y = [y]"
| "amplia (x#xs) y = x # amplia xs y"

text {* ---------------------------------------------------------------
Ejercicio 5.2. Escribir la demostración detallada de
amplia xs y = xs @ [y]
------------------------------------------------------------------- *}

(* crigomgom fraortmoy rubgonmar ivamenjim serrodcal pablucoto wilmorort anaprarod marpoldia1 juacabsou paupeddeg *)
lemma "amplia xs y = xs @ [y]"
proof (induct xs)
show "amplia [] y = [] @ [y]" by simp
next
fix x xs
assume HI: "amplia xs y = xs @ [y]"
have "amplia (x # xs) y = x # (amplia xs y)" by simp
also have "... = x # (xs @ [y])" using HI by simp
also have "... = (x # xs) @ [y]" by simp
finally show "amplia (x # xs) y = (x # xs) @ [y]" by simp
qed

(*danrodcha ferrenseg *)
lemma "amplia xs y = xs @ [y]" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix x xs
assume HI: "?P xs"
have "amplia (x # xs) y = x # amplia xs y" by simp
also have "... = (x # xs) @ [y]" using HI by simp
finally show "?P (x#xs)" by simp
qed

(* migtermor *)
lemma "amplia xs y = xs @ [y]"
proof (induct xs)
have "amplia [] y = [y]" by (simp only: amplia.simps(1))
have "[] @ [y] = [y]" by simp
show "amplia [] y = [] @ [y]" by simp
next
fix x xs
assume HI: "amplia xs y = xs @ [y]"
have "amplia (x # xs) y = x # amplia xs y" by (simp only: amplia.simps(2))
also have "... = x # (xs @ [y])" using HI by simp
also have "... = (x # xs) @ [y]" by simp
finally show "amplia (x # xs) y = (x # xs) @ [y]" by simp
qed

end
</source>