Diferencia entre revisiones de «Relación 3»

Revisión actual del 17:46 16 jul 2018

header {* R3: Razonamiento sobre programas en Isabelle/HOL *}

theory R3
imports Main 
begin

text {* --------------------------------------------------------------- 
  Ejercicio 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
  ------------------------------------------------------------------ *}

-- "maresccas4 irealetei diecabmen1 domlloriv pabflomar marescpla"

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

value "sumaImpares 5" -- "= 25"

-- "julrobrel"
fun sumaImpares2 :: "nat ⇒ nat" where
  "sumaImpares2 0       = 0"
| "sumaImpares2 (Suc n) = (2 * (Suc n) - 1) + sumaImpares2 n"

value "sumaImpares2 5" -- "= 25"

-- "jaisalmen zoiruicha"
fun sumaImpares3 :: "nat ⇒ nat" where
  "sumaImpares3 0 = 0"
| "sumaImpares3 (n) = (2 * n) - 1 + sumaImpares3 (n - 1)"

text {* --------------------------------------------------------------- 
  Ejercicio 2. Demostrar que 
     sumaImpares n = n*n
  ------------------------------------------------------------------- *}

-- "maresccas4 pabflomar juaruipav marescpla jaisalmen zoiruicha"
-- "jaisalmen: con la definición sumaImpares3 me da algunos errores al tratar de probar por inducción"

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) = 2 * n + 1 + sumaImpares n" by (simp only: sumaImpares.simps(2))
  also have "... = 2 * n + 1 + n * n" using HI by simp
  finally show "sumaImpares (Suc n) = (Suc n) * (Suc n)" by simp
qed

-- "irealetei diecabmen1 domlloriv"
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) = (2*n + 1) + sumaImpares n  " by simp
  also have "... = 2*n + 1 + (n * n)" using HI  by simp
  also have "...=  (n+1)*(n+1)" by simp  
  finally show "sumaImpares (Suc n) =Suc n * Suc n" by simp
qed

-- "Sobrarían varios parentesis."

-- "julrobrel"
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) = (2 * (Suc n) - 1) + sumaImpares n" by simp
  also have "...= (n+1)*(n+1)" using HI by simp
  finally show "sumaImpares (Suc n) = (Suc n) * (Suc n)" by simp
qed

-- "ElyIvan"

lemma "sumaImpares n = n*n"
proof -
  by (induc n)
  show "sumaImpares 0 = 0*0" by simp
next
  assume HI: "sumaImpares n = n * n"
  have "sumaImpares n = 2 *  n - 1) + sumaImpares ( n- 1 )" by sumanImpares2
  also have "...= 2 * n - 1 + (n - 1)  * ( n - 1)" using HI by simp
  also have "...= 2 * n -1 + n * n - 2 * n + 1" by simp
  also have "...= n * n" by simp
  finally show "sumaImpares n = n * n" by simp
qed

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

-- "maresccas4 irealetei diecabmen1 domlloriv pabflomar julrobrel"

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

value "sumaPotenciasDeDosMasUno 3" -- "= 16"

-- "ElyIvan jaisalmen zoiruicha"

fun sumaPotenciasDeDosMasUno2 :: "nat ⇒ nat" where
  "sumaPotenciasDeDosMasUno2 0 = 2"
| "sumaPotenciasDeDosMasUno2 n = 2^n + sumaPotenciasDeDosMasUno2 (n-1)"

value "sumaPotenciasDeDosMasUno2 1" -- "= 16" 

text {* --------------------------------------------------------------- 
  Ejercicio 4. Demostrar que 
     sumaPotenciasDeDosMasUno n = 2^(n+1)
  ------------------------------------------------------------------- *}

-- "maresccas4 pabflomar juaruipav jaisalmen zoiruicha"

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) = 2^(Suc n) + sumaPotenciasDeDosMasUno n" by (simp only: sumaPotenciasDeDosMasUno.simps(2))
  also have "... = 2^(Suc n) + 2^(n+1)" using HI by simp
  also have "... = 2^(n+1) + 2^(n+1)" by simp
  also have "... = 2 * 2^(n+1)" by simp
  also have "... = 2 * 2^(Suc n)" by simp
  finally show "sumaPotenciasDeDosMasUno (Suc n) = 2^((Suc n)+1)" by simp
qed

--"irealetei"
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 (n+1) = 2^(n+1) + sumaPotenciasDeDosMasUno(n)" by simp
  also have "... = 2^(n+1) + 2^(n+1)" using HI by simp
  also have "...=2*2^(n+1)" by simp
  also have "...=2^(n+2)" by simp
  finally show "sumaPotenciasDeDosMasUno (Suc n) = 2^(Suc n +1)" by simp  
qed

--"diecabmen1, marescpla"
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) = 2^(Suc n)+sumaPotenciasDeDosMasUno n" by simp
  also have "… = (2^(Suc n)) + (2^(n+1))" using HI by simp
  also have "… = 2^(n+1) + 2^(n+1)" by simp
  also have "… = 2 * 2^(Suc n)" by simp
  finally show "sumaPotenciasDeDosMasUno (Suc n) =   2^(Suc n +1)" by simp
qed

--"en el primer have yo especifiqué la regla usada: by (simp only: sumaPotenciasDeDosMasUno.simps(2))  marescpla"

--"domlloriv"
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) = 2^(Suc n) + sumaPotenciasDeDosMasUno n" by simp
  also have "... = 2^(Suc n) +  2^(n+1)" using HI by simp
  finally show "sumaPotenciasDeDosMasUno (Suc n) = 2^((Suc n)+1)" by simp
qed

-- "julrobrel"
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) = 2^(Suc n) + sumaPotenciasDeDosMasUno n" by simp
  also have "...= 2^((n+1)+1)" using HI by simp
  finally show "sumaPotenciasDeDosMasUno (Suc n) = 2^((Suc n)+1)" by simp
qed

-- "ElyIvan"

lemma "sumaPotenciasDeDosMasUno2 n = 2^(n+1)"
proof (induct n)
show "sumaPotenciasDeDosMasUno2 0 = 2^(0+1)" by simp
next
 fix n
 assume HI "sumaPotenciasDeDosMasUno2 n = 2^(n+1)"
 have "sumaPotenciasDeDosMasUno2 n = 2^n + sumaPotenciasDeDosMasUno2 (n-1)" by sumaPotenciasDeDosMasUno2 2
 also have "...= 2^n + 2 ^((n-1)+1)" by HI
 also have "...= 2^n + 2^n" by simp
 finally show "sumaPotenciasDeDosMasUno2 n = 2^(n+1)" by simp
qed

text {* --------------------------------------------------------------- 
  Ejercicio 5. 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]
  ------------------------------------------------------------------ *}

-- "maresccas4 irealetei diecabmen1 domlloriv pabflomar julrobrel marescpla jaisalmen zoiruicha"

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

value "copia 3 x" -- "= [x,x,x]"

-- "ElyIvan"

fun copia2 :: "nat ⇒ 'a ⇒ 'a list" where
  "copia2 0 x = []"
| "copia2 n x = x # copia2 ( n - 1 ) x"

value "copia2 3 x" -- "= [x,x,x]"

text {* --------------------------------------------------------------- 
  Ejercicio 6. 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
  Nota: La conjunción se representa por ∧
  ----------------------------------------------------------------- *}

-- "maresccas4 irealetei diecabmen1 domlloriv pabflomar julrobrel marescpla"

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

value "todos (λx. x>(1::nat)) [2,6,4]" -- "= True"
value "todos (λx. x>(2::nat)) [2,6,4]" -- "= False"

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


text {* --------------------------------------------------------------- 
  Ejercicio 7. Demostrar que todos los elementos de (copia n x) son
  iguales a x. 
  ------------------------------------------------------------------- *}

-- "maresccas4"

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 only: copia.simps(2))
  also have "... = todos (λy. y=x) (copia n x)" by simp
  finally show "todos (λy. y=x) (copia (Suc n) x)" using HI by simp
qed

-- "irealetei diecabmen1 pabflomar jaisalmen zoiruicha"
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) (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

-- "domlloriv"
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) = ((λy. y=x) x ∧ todos (λy. y=x) (copia n x))" using HI by simp
  also have "... = True" using HI by simp
  finally show "todos (λy. y=x) (copia (Suc n) x)" by simp
qed

-- "julrobrel"
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)" using HI by simp
  finally show "todos (λy. y = x) (copia (Suc n) x)" by simp
qed

-- "juaruipav"
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]" using HI by simp
  finally show "todos (λy. y=x) (copia (Suc n) x)" by simp
qed


--"marescpla"

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

--"jajaja ya he visto que me he complicado demasiado. Es mejor resolverlo en el otro orden"



text {* --------------------------------------------------------------- 
  Ejercicio 8. Definir la función
    factR :: nat ⇒ nat
  tal que (factR n) es el factorial de n. Por ejemplo,
    factR 4 = 24
  ------------------------------------------------------------------ *}

-- "maresccas4 irealetei diecabmen1 domlloriv pabflomar julrobrel marescpla jaisalmen zoiruicha"

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

value "factR 4" -- "= 24"

text {* --------------------------------------------------------------- 
  Ejercicio 9. 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
  ------------------------------------------------------------------- *}

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

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

value "factI 4" -- "= 24"

-- "maresccas4"
     
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 (Suc n)*x" by (simp only: factI'.simps(2))
    also have "... = x * factI' n (Suc n)" by simp
    also have "... = x * ((Suc n) * factR n)" using HI by simp
    finally show "factI' (Suc n) x = x * factR (Suc n)" by simp
  qed
qed

-- "jaisalmen zoiruicha"
-- "jaisalmen" (*lo tengo similar sin embargo me sale un error q no entiendo*)
-- "irealetei" (*Éste ha sido la muerte a pellizcos, aún leyéndome todo el tema.*)
lemma fact2: "factI' n x = x * factR n"
proof (induct n arbitrary: x)
  show "⋀x. factI' 0 x = x * factR 0" by simp
next
  fix n x
  assume HI: "⋀x. factI' n x = x * factR n"
  have "factI' (Suc n) x = factI' n (Suc n)*x" by simp
  also have "... = x * factI' n (Suc n)" by simp
  also have "... = x * (Suc n * factR n)" using HI by simp
  also have "... = x * factR (Suc n)" by simp
  finally show "factI' (Suc n) x = x * factR (Suc n)" by simp
qed

-- "diecabmen1" (*Es el mismo que de irealetei solo que me parece que se puede eliminar el último also have y ponerlo directo en el finally.*)
lemma fact3: "factI' n x = x * factR n"
proof (induct n arbitrary: x)
  show "⋀x. factI' 0 x = x * factR 0" by simp
next
  fix n x
  assume HI: "⋀x. factI' n x = x * factR n"
  have "factI' (Suc n) x = factI' n (Suc n)*x" by simp
  also have "… = x * ((Suc n) * factR n)" using HI by simp
  finally show "factI' (Suc n) x = x * factR (Suc n)" by simp
qed

-- "domlloriv"
lemma fact4: "factI' n x = x * factR n"
proof (induct n arbitrary: x)
  show "⋀x. factI' 0 x = x * factR 0" by simp (*Para todo x se cumple*)
next
  fix n
  assume HI: "⋀x. factI' n x = x * factR n"
  have "factI' (Suc n) x = factI' n (Suc n)*x" using HI by simp
  also have "... = x * (Suc n * factR n)" using HI by simp
  finally show "⋀x. factI' (Suc n) x = x * factR (Suc n)" using HI by simp
qed

(* Comentario: "using HI" se necesita sólo una vez. *)

-- "julrobrel"
lemma fact5: "factI' n x = x * factR n"
proof (induct n arbitrary: x)
  show "⋀x. factI' 0 x = x * factR 0" by simp
next
  fix n x
  assume HI: "⋀x. factI' n x = x * factR n"
  have "factI' (Suc n) x = factI' n (Suc n)*x" by simp
  also have "...=x * ((Suc n) * factR n)" using HI by simp
  finally show "factI' (Suc n) x = x* factR (Suc n)" by simp
qed

-- "juaruipav"

lemma fact6: "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"
  have "factI'(Suc n) x= factI' n (Suc n)*x" by simp
  also have "...=x*((Suc n)*factR n)" using HI by simp
  finally show "⋀x. factI' (Suc n) x = x*factR (Suc n)" by simp
  qed

-- " juaruipav: No me termina de funcionar"

-- "pabflomar los entiendo casi todos pero yo no habría sacado ni uno de ellos"


--"marescpla (no me resuelve el paso de inducción"
lemma fact: "factI' n x = x * factR n"
proof (induct n)
show "factI' 0 x = x* factR 0" by (simp only: factI'.simps(1) factR.simps(1))
next
fix n
assume HI : "factI' n x = x * factR n"
have " factI' (Suc n) x= factI' n (Suc n)*x" by (simp only: factI'.simps(2))
also have "...= (Suc n)*x * factR n" using HI by simp
also have " ... = x * (Suc n)*factR n" by simp
also have " ... = x * factR (Suc n)" by (simp only: factR.simps(2))
finally show " factI' (Suc n) x = x * factR (Suc n)" by simp
qed


text {* --------------------------------------------------------------- 
  Ejercicio 10. Demostrar que
     factI n = factR n
  ------------------------------------------------------------------- *}

-- "maresccas4 julrobrel"

corollary "factI n = factR n"
proof - 
  show "factI n = factR n" by (simp add:fact)
qed

-- "irealetei diecabmen1 domlloriv pabflomar juaruipav jaisalmen zoiruicha"
corollary "factI n = factR n"
by (simp add: fact)


--"marescpla"
corollary "factI n = factR n"
proof-
have "factI n = factI' n 1" by (simp only: factI.simps(1))
also have " ...= 1*factR n" by (simp add: fact)
finally show "factI n = factR n" by simp
qed


text {* --------------------------------------------------------------- 
  Ejercicio 11. 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]
  ------------------------------------------------------------------ *}

-- "maresccas4 irealetei diecabmen1 domlloriv julrobrel marescpla jaisalmen zoiruicha"

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

value "amplia [d,a] t" -- "= [d,a,t]"

text {* --------------------------------------------------------------- 
  Ejercicio 12. Demostrar que 
     amplia xs y = xs @ [y]
  ------------------------------------------------------------------- *}

-- "maresccas4 marescpla"

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 only: amplia.simps(2))
  also have "... = x # xs @ [y]" using HI by simp
  finally show "amplia (x#xs) y = (x#xs) @ [y]" by simp
qed

--"irealetei diecabmen1 domlloriv pabflomar juaruipav jaisalmen zoiruicha"
lemma "amplia xs y = xs @ [y]"
proof (induct xs)
  show "amplia [] y = [] @ [y]" by simp
next
  fix a xs
  assume HI: "amplia xs y = xs @ [y]"
  have "amplia (a#xs) y = a # (amplia xs y)" by simp
  also have "... = (a # xs) @ [y]" using HI by simp
  finally show "amplia (a#xs) y =  (a # xs) @ [y]" by simp
qed

-- "julrobrel"
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
  finally show "amplia (x#xs) y = (x#xs) @ [y]" by simp
qed

end
De Razonamiento automático (2013-14)

Revisión actual del 17:46 16 jul 2018
