Lógica Matemática y fundamentos (2015-16) - Contribuciones del usuario [es]

Relación 7

2016-06-17T11:03:47Z

Albromgon:

<source lang = "haskell">

-- TablerosSemánticos.hs
-- Tableros semánticos proposicionales.
-- José A. Alonso Jiménez <jalonso@us.es>
-- ---------------------------------------------------------------------

module TablerosSemanticos where

-- ---------------------------------------------------------------------
-- Librerías auxiliares --
-- ---------------------------------------------------------------------

import SintaxisSemantica
import Data.List

-- ---------------------------------------------------------------------
-- Literales --
-- ---------------------------------------------------------------------

-- ---------------------------------------------------------------------
-- Ejercicio 6: Definir la función
-- literal :: Prop -> Bool
-- tal que (literal f) se verifica si la fórmula F es un literal. Por
-- ejemplo,
-- literal p ==> True
-- literal (no p) ==> True
-- literal (no (p --> q)) ==> False
-- ---------------------------------------------------------------------

literal :: Prop -> Bool
literal (Atom f) = True
literal (Neg (Atom f)) = True
literal _ = False

-- ---------------------------------------------------------------------
-- Notación uniforme --
-- ---------------------------------------------------------------------

-- ---------------------------------------------------------------------
-- Ejercicio 1: Definir la función
-- dobleNegacion :: Prop -> Bool
-- tal que (dobleNegacion f) se verifica si f es una doble negación. Por
-- ejemplo,
-- dobleNegacion (no (no p)) ==> True
-- dobleNegacion (no (p --> q)) ==> False
-- ---------------------------------------------------------------------

dobleNegacion :: Prop -> Bool
dobleNegacion (Neg (Neg _)) = True
dobleNegación _ = False

-- ---------------------------------------------------------------------
-- Ejercicio 2: Definir la función
-- alfa :: Prop -> Bool
-- tal que (alfa f) se verifica si f es una fórmula alfa.
-- ---------------------------------------------------------------------

alfa :: Prop -> Bool
alfa (Conj _ _) = True
alfa (Neg (Impl _ _)) = True
alfa (Neg (Disj _ _)) = True
alfa _ = False

-- ---------------------------------------------------------------------
-- Ejercicio 3: Definir la función
-- beta :: Prop -> Bool
-- tal que (beta d) se verifica si f es una fórmula beta.
-- ---------------------------------------------------------------------

beta :: Prop -> Bool
beta (Disj _ _) = True
beta (Impl _ _) = True
beta (Neg (Conj _ _)) = True
beta (Equi _ _) = True
beta (Neg (Equi _ _)) = True
beta _ = False

-- ---------------------------------------------------------------------
-- Ejercicio 4: Definir la función
-- componentes :: Prop -> [Prop]
-- tal que (componentes ) es la lista de las componentes de la fórmula
-- f. Por ejemplo,
-- componentes (p /\ q --> r) ==> [no (p /\ q),r]
-- componentes (no (p /\ q --> r)) ==> [(p /\ q),no r]
-- ---------------------------------------------------------------------

componentes :: Prop -> [Prop]
componentes (Neg (Neg f)) = [f]
componentes (Conj f g) = [f, g]
componentes (Neg (Impl f g)) = [f, Neg g]
componentes (Neg (Disj f g)) = [Neg f, Neg g]
componentes (Disj f g) = [f, g]
componentes (Impl f g) = [Neg f, g]
componentes (Neg (Conj f g)) = [Neg f, Neg g]
componentes (Equi f g) = [Conj f g, Conj (Neg f) (Neg g)]
componentes (Neg (Equi f g)) = [Conj f (Neg g), Conj (Neg f) g]
-- ---------------------------------------------------------------------
-- Modelos mediante tableros --
-- ---------------------------------------------------------------------

-- ---------------------------------------------------------------------
-- Ejercicio 5: Definir la función
-- conjuntoDeLiterales :: [Prop] -> Bool
-- tal que (conjuntoDeLiterales fs) se verifica si fs es un conjunto de
-- literales. Por ejemplo,
-- conjuntoDeLiterales [p --> q, no r, r /\ s, p] ==> False
-- conjuntoDeLiterales [p, no q, r] ==> True
-- ---------------------------------------------------------------------

conjuntoDeLiterales :: [Prop] -> Bool
conjuntoDeLiterales fs = and [literal f | f <- fs]

-- ---------------------------------------------------------------------
-- Ejercicio 6: Definir la función
-- tieneContradiccion :: [Prop] -> Bool
-- tal que (tieneContradiccion fs) se verifica si fs contiene una
-- fórmula y su negación. Por ejemplo,
-- tieneContradiccion [r, p /\ q, s, no(p /\ q)] ==> True
-- ---------------------------------------------------------------------

tieneContradiccion :: [Prop] -> Bool
tieneContradiccion fs = [f | f <- fs, elem (Neg f) fs] /= []

-- Otra forma usando pliegues
tieneContradiccion2 :: [Prop] -> Bool
tieneContradiccion2 fs = foldl (\acc x -> if (elem (Neg x) fs) then True else acc) False fs
-- ---------------------------------------------------------------------
-- Ejercicio 7: Definir la función
-- expansionDN :: [Prop] -> Prop -> [[Prop]]
-- tal que (expansionDN fs f) es la expansión de fs mediante la doble
-- negación f. Por ejemplo,
-- expansionDN [p, no(no q), r] (no(no q)) ==> [[q,p,r]]
-- ---------------------------------------------------------------------

expansionDN :: [Prop] -> Prop -> [[Prop]]
expansionDN fs f = [(componentes f) `union` (delete f fs)]

-- ---------------------------------------------------------------------
-- Ejercicio 8: Definir la función
-- expansionAlfa :: [Prop] -> Prop -> [[Prop]]
-- tal que (expansionAlfa fs f) es la expansión de fs mediante la
-- fórmula alfa f. Por ejemplo,
-- expansionAlfa [q, (p1 /\ p2) , r] (p1 /\ p2) ==> [[p1,p2,q,r]]
-- ---------------------------------------------------------------------

expansionAlfa :: [Prop] -> Prop -> [[Prop]]
expansionAlfa fs f = [(componentes f) `union` (delete f fs)]

-- ---------------------------------------------------------------------
-- Ejercicio 9: Definir la función
-- expansionBeta :: [Prop] -> Prop -> [[Prop]]
-- tal que (expansionBeta fs f) es la expansión de fs mediante la
-- fórmula beta f. Por ejemplo,
-- expansionBeta [q, (p1 \/ p2) , r] (p1 \/ p2) ==> [[p1,q,r],[p2,q,r]]
-- ---------------------------------------------------------------------

expansionBeta :: [Prop] -> Prop -> [[Prop]]
expansionBeta fs f = [[g] `union` (delete f fs) | g <- componentes f]

-- ---------------------------------------------------------------------
-- Ejercicio 10: Definir la función
-- sucesores :: [Prop] -> [[Prop]]
-- tal que (sucesores fs) es la lista de sucesores de fs. Por ejemplo,
-- sucesores [q \/ s, no(no r), p1 /\ p2] => [[r,(q \/ s),(p1 /\ p2)]]
-- sucesores [r,(q \/ s),(p1 /\ p2)] => [[p1,p2,r,(q \/ s)]]
-- sucesores [p1,p2,r,(q \/ s)] => [[q,p1,p2,r],[s,p1,p2,r]]
-- ---------------------------------------------------------------------

sucesores :: [Prop] -> [[Prop]]
sucesores fs
| doblesNegación /= [] = expansiónDN fs (head doblesNegación)
| alfas /= [] = expansiónAlfa fs (head alfas)
| betas /= [] = expansiónBeta fs (head betas)
where doblesNegación = [f | f <- fs, dobleNegación f]
alfas = [f | f <- fs, alfa f]
betas = [f | f <- fs, beta f]
-- ---------------------------------------------------------------------
-- Ejercicio 11: Definir la función
-- modelosTab :: [Prop] -> [[Prop]]
-- tal que (modelosTab fs) es el conjunto de los modelos de fs
-- calculados mediante el método de tableros semánticos. Por ejemplo,
-- modelosTab [p --> q, no(q --> p)]
-- ==> [[no p,q],[q,no p]]
-- modelosTab [p --> q, no q --> no p]
-- ==> [[q,no p],[no p],[q],[no p,q]]
-- ---------------------------------------------------------------------

modelosTab :: [Prop] -> [[Prop]]
modelosTab fs
| tieneContradicción fs = []
| conjuntoDeLiterales fs = [fs]
| otherwise = uniónGeneral [modelosTab gs
| gs <- sucesores fs]

-- ---------------------------------------------------------------------
-- Ejercicio 12: Definir la función
-- subconjunto :: Eq a => [a] -> [a] -> Bool
-- tal que (subconjunto x y) se verifica si x es subconjunto de y. Por
-- ejemplo,
-- subconjunto [1,3] [3,2,1] ==> True
-- subconjunto [1,3,5] [3,2,1] ==> False
-- ---------------------------------------------------------------------

subconjunto :: Eq a => [a] -> [a] -> Bool
subconjunto xs ys = and [elem x ys | x <- xs]

-- ---------------------------------------------------------------------
-- Ejercicio 13: Definir la función
-- modelosGenerales :: [Prop] -> [[Prop]]
-- tal que (modelosGenerales fs) es el conjunto de los modelos generales
-- de fs calculados mediante el método de tableros semánticos. Por
-- ejemplo,
-- modelosGenerales [p --> q, no q --> no p] ==> [[no p],[q]]
-- ---------------------------------------------------------------------

modelosGenerales :: [Prop] -> [[Prop]]
modelosGenerales fs = [gs | gs <- modelos, [hs | hs <- delete gs modelos, subconjunto hs gs] == []]
where modelos = modelosTab fs

-- ---------------------------------------------------------------------
-- Teoremas por tableros --
-- ---------------------------------------------------------------------

-- ---------------------------------------------------------------------
-- Ejercicio 14: Definir la función
-- esTeoremaPorTableros :: Prop -> Bool
-- tal que (esTeoremaPorTableros f) se verifica si la fórmula f es
-- teorema (mediante tableros semánticos). Por ejemplo,
-- esTeoremaPorTableros (p --> p) ==> True
-- esTeoremaPorTableros (p --> q) ==> False
-- ---------------------------------------------------------------------

esTeoremaPorTableros :: Prop -> Bool
esTeoremaPorTableros f = modelosTab [Neg f] == []
-- ---------------------------------------------------------------------
-- Consecuencia por tableros --
-- ---------------------------------------------------------------------

-- ---------------------------------------------------------------------
-- Ejercicio 15: Definir la función
-- esDeduciblePorTableros :: [Prop] -> Prop -> Bool
-- tal que (esDeduciblePorTableros fs f) se verifica si la fórmula f es
-- consecuencia (mediante tableros) del conjunto de fórmulas fs. Por
-- ejemplo,
-- esDeduciblePorTableros [p --> q, q --> r] (p --> r) ==> True
-- esDeduciblePorTableros [p --> q, q --> r] (p <--> r) ==> False
-- ---------------------------------------------------------------------

esDeduciblePorTableros :: [Prop] -> Prop -> Bool
esDeduciblePorTableros fs f = modelosTab ((Neg f):fs) == []

</source>

Relación 7

2016-06-17T10:57:34Z

Albromgon: Otra opción para la función tieneContradiccion

<source lang = "haskell">

-- TablerosSemánticos.hs
-- Tableros semánticos proposicionales.
-- José A. Alonso Jiménez <jalonso@us.es>
-- ---------------------------------------------------------------------

module TablerosSemanticos where

-- ---------------------------------------------------------------------
-- Librerías auxiliares --
-- ---------------------------------------------------------------------

import SintaxisSemantica
import Data.List

-- ---------------------------------------------------------------------
-- Literales --
-- ---------------------------------------------------------------------

-- ---------------------------------------------------------------------
-- Ejercicio 6: Definir la función
-- literal :: Prop -> Bool
-- tal que (literal f) se verifica si la fórmula F es un literal. Por
-- ejemplo,
-- literal p ==> True
-- literal (no p) ==> True
-- literal (no (p --> q)) ==> False
-- ---------------------------------------------------------------------

literal :: Prop -> Bool
literal (Atom f) = True
literal (Neg (Atom f)) = True
literal _ = False

-- ---------------------------------------------------------------------
-- Notación uniforme --
-- ---------------------------------------------------------------------

-- ---------------------------------------------------------------------
-- Ejercicio 1: Definir la función
-- dobleNegacion :: Prop -> Bool
-- tal que (dobleNegacion f) se verifica si f es una doble negación. Por
-- ejemplo,
-- dobleNegacion (no (no p)) ==> True
-- dobleNegacion (no (p --> q)) ==> False
-- ---------------------------------------------------------------------

dobleNegacion :: Prop -> Bool
dobleNegacion (Neg (Neg _)) = True
dobleNegación _ = False

-- ---------------------------------------------------------------------
-- Ejercicio 2: Definir la función
-- alfa :: Prop -> Bool
-- tal que (alfa f) se verifica si f es una fórmula alfa.
-- ---------------------------------------------------------------------

alfa :: Prop -> Bool
alfa (Conj _ _) = True
alfa (Neg (Impl _ _)) = True
alfa (Neg (Disj _ _)) = True
alfa _ = False

-- ---------------------------------------------------------------------
-- Ejercicio 3: Definir la función
-- beta :: Prop -> Bool
-- tal que (beta d) se verifica si f es una fórmula beta.
-- ---------------------------------------------------------------------

beta :: Prop -> Bool
beta (Disj _ _) = True
beta (Impl _ _) = True
beta (Neg (Conj _ _)) = True
beta (Equi _ _) = True
beta (Neg (Equi _ _)) = True
beta _ = False

-- ---------------------------------------------------------------------
-- Ejercicio 4: Definir la función
-- componentes :: Prop -> [Prop]
-- tal que (componentes ) es la lista de las componentes de la fórmula
-- f. Por ejemplo,
-- componentes (p /\ q --> r) ==> [no (p /\ q),r]
-- componentes (no (p /\ q --> r)) ==> [(p /\ q),no r]
-- ---------------------------------------------------------------------

componentes :: Prop -> [Prop]
componentes (Neg (Neg f)) = [f]
componentes (Conj f g) = [f, g]
componentes (Neg (Impl f g)) = [f, Neg g]
componentes (Neg (Disj f g)) = [Neg f, Neg g]
componentes (Disj f g) = [f, g]
componentes (Impl f g) = [Neg f, g]
componentes (Neg (Conj f g)) = [Neg f, Neg g]
componentes (Equi f g) = [Conj f g, Conj (Neg f) (Neg g)]
componentes (Neg (Equi f g)) = [Conj f (Neg g), Conj (Neg f) g]
-- ---------------------------------------------------------------------
-- Modelos mediante tableros --
-- ---------------------------------------------------------------------

-- ---------------------------------------------------------------------
-- Ejercicio 5: Definir la función
-- conjuntoDeLiterales :: [Prop] -> Bool
-- tal que (conjuntoDeLiterales fs) se verifica si fs es un conjunto de
-- literales. Por ejemplo,
-- conjuntoDeLiterales [p --> q, no r, r /\ s, p] ==> False
-- conjuntoDeLiterales [p, no q, r] ==> True
-- ---------------------------------------------------------------------

conjuntoDeLiterales :: [Prop] -> Bool
conjuntoDeLiterales fs = and [literal f | f <- fs]

-- ---------------------------------------------------------------------
-- Ejercicio 6: Definir la función
-- tieneContradiccion :: [Prop] -> Bool
-- tal que (tieneContradiccion fs) se verifica si fs contiene una
-- fórmula y su negación. Por ejemplo,
-- tieneContradiccion [r, p /\ q, s, no(p /\ q)] ==> True
-- ---------------------------------------------------------------------

tieneContradiccion :: [Prop] -> Bool
tieneContradiccion fs = [f | f <- fs, elem (Neg f) fs] /= []

-- Otra forma usando pliegues
tieneContradiccion :: [Prop] -> Bool
tieneContradiccion fs = foldl (\acc x -> if (elem (Neg x) fs) then True else acc) False fs
-- ---------------------------------------------------------------------
-- Ejercicio 7: Definir la función
-- expansionDN :: [Prop] -> Prop -> [[Prop]]
-- tal que (expansionDN fs f) es la expansión de fs mediante la doble
-- negación f. Por ejemplo,
-- expansionDN [p, no(no q), r] (no(no q)) ==> [[q,p,r]]
-- ---------------------------------------------------------------------

expansionDN :: [Prop] -> Prop -> [[Prop]]
expansionDN fs f = [(componentes f) `union` (delete f fs)]

-- ---------------------------------------------------------------------
-- Ejercicio 8: Definir la función
-- expansionAlfa :: [Prop] -> Prop -> [[Prop]]
-- tal que (expansionAlfa fs f) es la expansión de fs mediante la
-- fórmula alfa f. Por ejemplo,
-- expansionAlfa [q, (p1 /\ p2) , r] (p1 /\ p2) ==> [[p1,p2,q,r]]
-- ---------------------------------------------------------------------

expansionAlfa :: [Prop] -> Prop -> [[Prop]]
expansionAlfa fs f = [(componentes f) `union` (delete f fs)]

-- ---------------------------------------------------------------------
-- Ejercicio 9: Definir la función
-- expansionBeta :: [Prop] -> Prop -> [[Prop]]
-- tal que (expansionBeta fs f) es la expansión de fs mediante la
-- fórmula beta f. Por ejemplo,
-- expansionBeta [q, (p1 \/ p2) , r] (p1 \/ p2) ==> [[p1,q,r],[p2,q,r]]
-- ---------------------------------------------------------------------

expansionBeta :: [Prop] -> Prop -> [[Prop]]
expansionBeta fs f = [[g] `union` (delete f fs) | g <- componentes f]

-- ---------------------------------------------------------------------
-- Ejercicio 10: Definir la función
-- sucesores :: [Prop] -> [[Prop]]
-- tal que (sucesores fs) es la lista de sucesores de fs. Por ejemplo,
-- sucesores [q \/ s, no(no r), p1 /\ p2] => [[r,(q \/ s),(p1 /\ p2)]]
-- sucesores [r,(q \/ s),(p1 /\ p2)] => [[p1,p2,r,(q \/ s)]]
-- sucesores [p1,p2,r,(q \/ s)] => [[q,p1,p2,r],[s,p1,p2,r]]
-- ---------------------------------------------------------------------

sucesores :: [Prop] -> [[Prop]]
sucesores fs
| doblesNegación /= [] = expansiónDN fs (head doblesNegación)
| alfas /= [] = expansiónAlfa fs (head alfas)
| betas /= [] = expansiónBeta fs (head betas)
where doblesNegación = [f | f <- fs, dobleNegación f]
alfas = [f | f <- fs, alfa f]
betas = [f | f <- fs, beta f]
-- ---------------------------------------------------------------------
-- Ejercicio 11: Definir la función
-- modelosTab :: [Prop] -> [[Prop]]
-- tal que (modelosTab fs) es el conjunto de los modelos de fs
-- calculados mediante el método de tableros semánticos. Por ejemplo,
-- modelosTab [p --> q, no(q --> p)]
-- ==> [[no p,q],[q,no p]]
-- modelosTab [p --> q, no q --> no p]
-- ==> [[q,no p],[no p],[q],[no p,q]]
-- ---------------------------------------------------------------------

modelosTab :: [Prop] -> [[Prop]]
modelosTab fs
| tieneContradicción fs = []
| conjuntoDeLiterales fs = [fs]
| otherwise = uniónGeneral [modelosTab gs
| gs <- sucesores fs]

-- ---------------------------------------------------------------------
-- Ejercicio 12: Definir la función
-- subconjunto :: Eq a => [a] -> [a] -> Bool
-- tal que (subconjunto x y) se verifica si x es subconjunto de y. Por
-- ejemplo,
-- subconjunto [1,3] [3,2,1] ==> True
-- subconjunto [1,3,5] [3,2,1] ==> False
-- ---------------------------------------------------------------------

subconjunto :: Eq a => [a] -> [a] -> Bool
subconjunto xs ys = and [elem x ys | x <- xs]

-- ---------------------------------------------------------------------
-- Ejercicio 13: Definir la función
-- modelosGenerales :: [Prop] -> [[Prop]]
-- tal que (modelosGenerales fs) es el conjunto de los modelos generales
-- de fs calculados mediante el método de tableros semánticos. Por
-- ejemplo,
-- modelosGenerales [p --> q, no q --> no p] ==> [[no p],[q]]
-- ---------------------------------------------------------------------

modelosGenerales :: [Prop] -> [[Prop]]
modelosGenerales fs = [gs | gs <- modelos, [hs | hs <- delete gs modelos, subconjunto hs gs] == []]
where modelos = modelosTab fs

-- ---------------------------------------------------------------------
-- Teoremas por tableros --
-- ---------------------------------------------------------------------

-- ---------------------------------------------------------------------
-- Ejercicio 14: Definir la función
-- esTeoremaPorTableros :: Prop -> Bool
-- tal que (esTeoremaPorTableros f) se verifica si la fórmula f es
-- teorema (mediante tableros semánticos). Por ejemplo,
-- esTeoremaPorTableros (p --> p) ==> True
-- esTeoremaPorTableros (p --> q) ==> False
-- ---------------------------------------------------------------------

esTeoremaPorTableros :: Prop -> Bool
esTeoremaPorTableros f = modelosTab [Neg f] == []
-- ---------------------------------------------------------------------
-- Consecuencia por tableros --
-- ---------------------------------------------------------------------

-- ---------------------------------------------------------------------
-- Ejercicio 15: Definir la función
-- esDeduciblePorTableros :: [Prop] -> Prop -> Bool
-- tal que (esDeduciblePorTableros fs f) se verifica si la fórmula f es
-- consecuencia (mediante tableros) del conjunto de fórmulas fs. Por
-- ejemplo,
-- esDeduciblePorTableros [p --> q, q --> r] (p --> r) ==> True
-- esDeduciblePorTableros [p --> q, q --> r] (p <--> r) ==> False
-- ---------------------------------------------------------------------

esDeduciblePorTableros :: [Prop] -> Prop -> Bool
esDeduciblePorTableros fs f = modelosTab ((Neg f):fs) == []

</source>

Relación 2a

2016-06-16T19:06:34Z

Albromgon: Otra opción para la función esConsecuencia

<source lang = "haskell">

-- SintaxisSemanticaProp.hs
-- Lógica proposicional: Sintaxis y semántica
-- José A. Alonso Jiménez <jalonso@us.es>
-- ---------------------------------------------------------------------

module SintaxisSemantica where

-- ---------------------------------------------------------------------
-- Librerías auxiliares --
-- ---------------------------------------------------------------------

import Data.List

-- ---------------------------------------------------------------------
-- Gramática de fórmulas prosicionales --
-- ---------------------------------------------------------------------

-- ---------------------------------------------------------------------
-- Ejercicio 1: Definir los siguientes tipos de datos:
-- * SímboloProposicional para representar los símbolos de proposiciones
-- * Prop para representar las fórmulas proposicionales usando los
-- constructores Atom, Neg, Conj, Disj, Impl y Equi para las fórmulas
-- atómicas, negaciones, conjunciones, implicaciones y equivalencias,
-- respectivamente.
-- ---------------------------------------------------------------------

type SimboloProposicional = String

data Prop = Atom SimboloProposicional
| Neg Prop
| Conj Prop Prop
| Disj Prop Prop
| Impl Prop Prop
| Equi Prop Prop
deriving (Eq,Ord)

instance Show Prop where
show (Atom p) = p
show (Neg p) = "no " ++ show p
show (Conj p q) = "(" ++ show p ++ " /\\ " ++ show q ++ ")"
show (Disj p q) = "(" ++ show p ++ " \\/ " ++ show q ++ ")"
show (Impl p q) = "(" ++ show p ++ " --> " ++ show q ++ ")"
show (Equi p q) = "(" ++ show p ++ " <--> " ++ show q ++ ")"

-- ---------------------------------------------------------------------
-- Ejercicio 2: Definir las siguientes fórmulas proposicionales
-- atómicas: p, p1, p2, q, r, s, t y u.
-- ---------------------------------------------------------------------

p, p1, p2, q, r, s, t, u :: Prop
p = Atom "p"
p1 = Atom "p1"
p2 = Atom "p2"
q = Atom "q"
r = Atom "r"
s = Atom "s"
t = Atom "t"
u = Atom "u"

-- ---------------------------------------------------------------------
-- Ejercicio 3: Definir la función
-- no :: Prop -> Prop
-- tal que (no f) es la negación de f.
-- ---------------------------------------------------------------------

no :: Prop -> Prop
no = Neg

-- ---------------------------------------------------------------------
-- Ejercicio 4: Definir los siguientes operadores
-- (/\), (\/), (-->), (<-->) :: Prop -> Prop -> Prop
-- tales que
-- f /\ g es la conjunción de f y g
-- f \/ g es la disyunción de f y g
-- f --> g es la implicación de f a g
-- f <--> g es la equivalencia entre f y g
-- ---------------------------------------------------------------------

infixr 5 \/
infixr 4 /\
infixr 3 -->
infixr 2 <-->
(/\), (\/), (-->), (<-->) :: Prop -> Prop -> Prop
(/\) = Conj
(\/) = Disj
(-->) = Impl
(<-->) = Equi

-- ---------------------------------------------------------------------
-- Símbolos proposicionales de una fórmula --
-- ---------------------------------------------------------------------

-- ---------------------------------------------------------------------
-- Ejercicio 5: Definir la función
-- símbolosPropFórm :: Prop -> [Prop]
-- tal que (símbolosPropFórm f) es el conjunto formado por todos los
-- símbolos proposicionales que aparecen en f. Por ejemplo,
-- símbolosPropFórm (p /\ q --> p) ==> [p,q]
-- ---------------------------------------------------------------------

simbolosPropForm :: Prop -> [Prop]
simbolosPropForm (Atom f) = [(Atom f)]
simbolosPropForm (Neg f) = simbolosPropForm f
simbolosPropForm (Conj f g) = union (simbolosPropForm f) (simbolosPropForm g)
simbolosPropForm (Disj f g) = union (simbolosPropForm f) (simbolosPropForm g)
simbolosPropForm (Impl f g) = union (simbolosPropForm f) (simbolosPropForm g)
simbolosPropForm (Equi f g) = union (simbolosPropForm f) (simbolosPropForm g)

--------------------------TIENE UN FALLO, pues devuelve una lista con simbolos repetidos... una posible solución:
-->

eliminaRepetidos :: [Prop] -> [Prop]
eliminaRepetidos [(Atom x)] = [(Atom x)]
eliminaRepetidos xs = if (head xs) `elem`
(tail xs) then
eliminaRepetidos (tail
xs)
else [head xs] ++ eliminaRepetidos (tail xs)

simbolosPropForm :: Prop -> [Prop]
simbolosPropForm f = reverse (eliminaRepetidos (simbolosPropFormAUX f))

simbolosPropFormAUX (Atom f) = [(Atom f)]
simbolosPropFormAUX (Neg f) = simbolosPropFormAUX f
simbolosPropFormAUX (Conj f g)= (simbolosPropFormAUX f) ++ (simbolosPropFormAUX g)
simbolosPropFormAUX (Disj f g)= (simbolosPropFormAUX f) ++ (simbolosPropFormAUX g)
simbolosPropFormAUX (Impl f g)= (simbolosPropFormAUX f) ++ (simbolosPropFormAUX g)
simbolosPropFormAUX (Equi f g)= (simbolosPropFormAUX f) ++ (simbolosPropFormAUX g)

-------------------------------------------------
-- nub xs : elimina los elementos repetidos de la lista xs.

simbolosPropForm1 :: Prop -> [Prop]
simbolosPropForm1 (Atom f) = [(Atom f)]
simbolosPropForm1 (Neg f) = nub (simbolosPropForm f)
simbolosPropForm1 (Conj f g) = nub (simbolosPropForm f ++ simbolosPropForm g)
simbolosPropForm1 (Disj f g) = nub (simbolosPropForm f ++ simbolosPropForm g)
simbolosPropForm1 (Impl f g) = nub (simbolosPropForm f ++ simbolosPropForm g)
simbolosPropForm1 (Equi f g) = nub (simbolosPropForm f ++ simbolosPropForm g)

-- ---------------------------------------------------------------------
-- Interpretaciones --
-- ---------------------------------------------------------------------

-- ---------------------------------------------------------------------
-- Ejercicio 6: Definir el tipo de datos Interpretación para
-- representar las interpretaciones como listas de fórmulas atómicas.
-- ---------------------------------------------------------------------

type Interpretacion = [Prop]

-- ---------------------------------------------------------------------
-- Significado de una fórmula en una interpretación --
-- ---------------------------------------------------------------------

-- ---------------------------------------------------------------------
-- Ejercicio 7: Definir la función
-- significado :: Prop -> Interpretación -> Bool
-- tal que (significado f i) es el significado de f en i. Por ejemplo,
-- significado ((p \/ q) /\ ((no q) \/ r)) [r] ==> False
-- significado ((p \/ q) /\ ((no q) \/ r)) [p,r] ==> True
-- ---------------------------------------------------------------------

significado :: Prop -> Interpretacion -> Bool
significado (Atom f) i = (elem (Atom f) i)
significado (Neg f) i = not (significado f i)
significado (Conj f g) i = (significado f i)&&(significado g i)
significado (Disj f g) i = (significado f i)||(significado g i)
significado (Equi f g) i = (significado f i)==(significado g i)
significado (Impl f g) i
|((significado f i)==True)&&((significado g i))==False=False
|otherwise = True

{-Otra opción:
esAtomica :: Prop -> Bool
esAtomica (Atom _) = True
esAtomica f = False
significado f y | esAtomica f = elem f y
significado (Neg x) xs = not (significado x xs)
significado (Disj x y) xs = not (all (==False) [significado x xs,
significado
y xs])
significado (Conj x y) xs = all(==True) [significado x xs, significado y xs]
significado (Impl x y) xs |significado x xs == True && significado y xs
== False = False
|otherwise = True
significado (Equi x y) xs = significado x xs == significado y xs
-}

--otra forma mas rapida es:
significado2 :: Prop -> Interpretacion -> Bool
significado2 (Atom p) xs = elem (Atom p) xs
significado2 (Neg p) xs = not (significado2 p xs )
significado2 (Conj p q) xs = (significado2 p xs) && (significado2 q xs)
significado2 (Disj p q) xs = (significado2 p xs) ||(significado2 q xs)
significado2 (Impl p q) xs |(significado2 p xs) = (significado2 q xs)
|otherwise = True
significado2 (Equi p q) xs = (significado2 p xs) == (significado2 q xs)

-- ---------------------------------------------------------------------
-- Interpretaciones de una fórmula --
-- ---------------------------------------------------------------------

-- ---------------------------------------------------------------------
-- Ejercicio 8: Definir la función
-- subconjuntos :: [a] -> [[a]]
-- tal que (subconjuntos x) es la lista de los subconjuntos de x. Por
-- ejmplo,
-- subconjuntos "abc" ==> ["abc","ab","ac","a","bc","b","c",""]
-- ---------------------------------------------------------------------

subconjuntos :: [a] -> [[a]]
subconjuntos [] = [[]]
subconjuntos (x:xs) = (map (x:) (subconjuntos xs))++(subconjuntos xs)

--otra manera:
subconjuntos1 :: [a] -> [[a]]
subconjuntos1 [] = [[]]
subconjuntos1 (x:xs)= [x:ys|ys<- subconjuntos1 xs] ++ subconjuntos1 xs

-- ---------------------------------------------------------------------
-- Ejercicio 9: Definir la función
-- interpretacionesForm :: Prop -> [Interpretación]
-- tal que (interpretacionesForm f) es la lista de todas las
-- interpretaciones de f. Por ejemplo,
-- interpretacionesForm (p /\ q --> p) ==> [[p,q],[p],[q],[]]
-- ---------------------------------------------------------------------

interpretacionesForm :: Prop -> [Interpretacion]
interpretacionesForm f = subconjuntos (simbolosPropForm f)

{- Otra opción:
interpretacionesForm = subconjuntos.simbolosPropForm
-}

-- ---------------------------------------------------------------------
-- Modelos de fórmulas --
-- ---------------------------------------------------------------------

-- ---------------------------------------------------------------------
-- Ejercicio 10: Definir la función
-- esModeloFormula :: Interpretacion -> Prop -> Bool
-- tal que (esModeloFormula i f) se verifica si i es un modelo de f. Por
-- ejemplo,
-- esModeloFormula [r] ((p \/ q) /\ ((no q) \/ r)) ==> False
-- esModeloFormula [p,r] ((p \/ q) /\ ((no q) \/ r)) ==> True
-- ---------------------------------------------------------------------

esModeloFormula :: Interpretacion -> Prop -> Bool
esModeloFormula i f = significado f i

-- ---------------------------------------------------------------------
-- Ejercicio 11: Definir la función
-- modelosFormula :: Prop -> [Interpretacion]
-- tal que (modelosFormula f) es la lista de todas las interpretaciones
-- de f que son modelo de F. Por ejemplo,
-- modelosFormula ((p \/ q) /\ ((no q) \/ r))
-- ==> [[p,q,r],[p,r],[p],[q,r]]
-- ---------------------------------------------------------------------

modelosFormula :: Prop -> [Interpretacion]
modelosFormula f = [i|i <-(interpretacionesForm f), esModeloFormula i f]

-- ---------------------------------------------------------------------
-- Fórmulas válidas, satisfacibles e insatisfacibles --
-- ---------------------------------------------------------------------

-- ---------------------------------------------------------------------
-- Ejercicio 12: Definir la función
-- esValida :: Prop -> Bool
-- tal que (esValida f) se verifica si f es válida. Por ejemplo,
-- esValida (p --> p) ==> True
-- esValida (p --> q) ==> False
-- esValida ((p --> q) \/ (q --> p)) ==> True
-- ---------------------------------------------------------------------

esValida :: Prop -> Bool
esValida f = null((interpretacionesForm f)\\( modelosFormula f))

{- Otra opción:
esValida f = length(modelosFormula f)==length(interpretacionesForm f)
-}

--mas corta que esta anterior tenemos :
esValida1 :: Prop -> Bool
esValida1 f = modelosFormula f == interpretacionesForm f
-- ---------------------------------------------------------------------
-- Ejercicio 13: Definir la función
-- esInsatisfacible :: Prop -> Bool
-- tal que (esInsatisfacible f) se verifica si f es insatisfacible. Por
-- ejemplo,
-- esInsatisfacible (p /\ (no p)) ==> True
-- esInsatisfacible ((p --> q) /\ (q --> r)) ==> False
-- ---------------------------------------------------------------------

esInsatisfacible :: Prop -> Bool
esInsatisfacible f = null (modelosFormula f)

{- Otra opción:
esInsatisfacible = null. modelosFormula
-}

-- ---------------------------------------------------------------------
-- Ejercicio 14: Definir la función
-- esSatisfacible :: Prop -> Bool
-- tal que (esSatisfacible f) se verifica si f es satisfacible. Por
-- ejemplo,
-- esSatisfacible (p /\ (no p)) ==> False
-- esSatisfacible ((p --> q) /\ (q --> r)) ==> True
-- ---------------------------------------------------------------------

esSatisfacible :: Prop -> Bool
esSatisfacible f = not( null (modelosFormula f))

{- Otra opción:
esSatisfacible = not.esInsatisfacible
-}

-- ---------------------------------------------------------------------
-- Símbolos proposicionales de un conjunto de fórmulas --
-- ---------------------------------------------------------------------

-- ---------------------------------------------------------------------
-- Ejercicio 15: Definir la función
-- unionGeneral :: Eq a => [[a]] -> [a]
-- tal que (unionGeneral x) es la unión de los conjuntos de la lista de
-- conjuntos x. Por ejemplo,
-- unionGeneral [] ==> []
-- unionGeneral [[1]] ==> [1]
-- unionGeneral [[1],[1,2],[2,3]] ==> [1,2,3]
-- ---------------------------------------------------------------------

unionGeneral :: Eq a => [[a]] -> [a]
unionGeneral [] = []
unionGeneral (xs:xss) = union xs (unionGeneral xss)

-- Otra opción
unionG :: Eq a => [[a]] -> [a]
unionG = nub . concat
-- ---------------------------------------------------------------------
-- Ejercicio 16: Definir la función
-- simbolosPropConj :: [Prop] -> [Prop]
-- tal que (simbolosPropConj s) es el conjunto de los símbolos
-- proposiciones de s. Por ejemplo,
-- simbolosPropConj [p /\ q --> r, p --> s] ==> [p,q,r,s]
-- ---------------------------------------------------------------------

simbolosPropConj :: [Prop] -> [Prop]
simbolosPropConj s = unionGeneral [simbolosPropForm f|f<-s]

{- Otra Opción:
simbolosPropConj =nub.concat.map(simbolosPropForm)
-}
{- Otra forma combinando las anteriores:
simbolosPropConj s = unionGeneral (map simbolosPropForm s)
-}

-- ---------------------------------------------------------------------
-- Interpretaciones de un conjunto de fórmulas --
-- ---------------------------------------------------------------------

-- ---------------------------------------------------------------------
-- Ejercicio 17: Definir la función
-- interpretacionesConjunto :: [Prop] -> [Interpretacion]
-- tal que (interpretacionesConjunto s) es la lista de las
-- interpretaciones de s. Por ejemplo,
-- interpretacionesConjunto [p --> q, q --> r]
-- ==> [[p,q,r],[p,q],[p,r],[p],[q,r],[q],[r],[]]
-- ---------------------------------------------------------------------

interpretacionesConjunto :: [Prop] -> [Interpretacion]
interpretacionesConjunto s = subconjuntos.simbolosPropConj

-- ---------------------------------------------------------------------
-- Modelos de conjuntos de fórmulas --
-- ---------------------------------------------------------------------

-- ---------------------------------------------------------------------
-- Ejercicio 18: Definir la función
-- esModeloConjunto :: Interpretacion -> [Prop] -> Bool
-- tal que (esModeloConjunto i s) se verifica si i es modelo de s. Por
-- ejemplo,
-- esModeloConjunto [p,r] [(p \/ q) /\ ((no q) \/ r), q --> r]
-- ==> True
-- esModeloConjunto [p,r] [(p \/ q) /\ ((no q) \/ r), r --> q]
-- ==> False
-- ---------------------------------------------------------------------

esModeloConjunto :: Interpretacion -> [Prop] -> Bool
esModeloConjunto i s = and [esModeloFormula i f|f<-s]

{- Otra forma (usando map):

esModeloConjunto i s = and(map f s)
where f p = esModeloFormula i p

-}
-- ---------------------------------------------------------------------
-- Ejercicio 19: Definir la función
-- modelosConjunto :: [Prop] -> [Interpretacion]
-- tal que (modelosConjunto s) es la lista de modelos del conjunto
-- s. Por ejemplo,
-- modelosConjunto [(p \/ q) /\ ((no q) \/ r), q --> r]
-- ==> [[p,q,r],[p,r],[p],[q,r]]
-- modelosConjunto [(p \/ q) /\ ((no q) \/ r), r --> q]
-- ==> [[p,q,r],[p],[q,r]]
-- ---------------------------------------------------------------------

modelosConjunto :: [Prop] -> [Interpretacion]
modelosConjunto s = [i|i<-interpretacionesConjunto s, esModeloConjunto i s]

{- Otra forma (usando filter):

modelosConjunto s = filter p (interpretacionesConjunto s)
where p i = esModeloConjunto i s

-}
-- ---------------------------------------------------------------------
-- Conjuntos consistentes e inconsistentes de fórmulas --
-- ---------------------------------------------------------------------

-- ---------------------------------------------------------------------
-- Ejercicio 20: Definir la función
-- esConsistente :: [Prop] -> Bool
-- tal que (esConsistente s) se verifica si s es consistente. Por
-- ejemplo,
-- esConsistente [(p \/ q) /\ ((no q) \/ r), p --> r]
-- ==> True
-- esConsistente [(p \/ q) /\ ((no q) \/ r), p --> r, no r]
-- ==> False
-- ---------------------------------------------------------------------

esConsistente :: [Prop] -> Bool
esConsistente s = not.null.modelosConjunto

--otra seria:
esConsistente1 :: [Prop] -> Bool
esConsistente1 s = modelosConjunto s /= []

-- ---------------------------------------------------------------------
-- Ejercicio 21: Definir la función
-- esInconsistente :: [Prop] -> Bool
-- tal que (esInconsistente s) se verifica si s es inconsistente. Por
-- ejemplo,
-- esInconsistente [(p \/ q) /\ ((no q) \/ r), p --> r]
-- ==> False
-- esInconsistente [(p \/ q) /\ ((no q) \/ r), p --> r, no r]
-- ==> True
-- ---------------------------------------------------------------------

esInconsistente :: [Prop] -> Bool
esInconsistente s = not.esConsistente
-- usando el comando null tambien lo tendriamos:
esInconsistente1 :: [Prop] -> Bool
esInconsistente1 s = null (modelosConjunto s)
-- ---------------------------------------------------------------------
-- Consecuencia lógica --
-- ---------------------------------------------------------------------

-- ---------------------------------------------------------------------
-- Ejercicio 22: Definir la función
-- esConsecuencia :: [Prop] -> Prop -> Bool
-- tal que (esConsecuencia s f) se verifica si f es consecuencia de
-- s. Por ejemplo,
-- esConsecuencia [p --> q, q --> r] (p --> r) ==> True
-- esConsecuencia [p] (p /\ q) ==> False
-- ---------------------------------------------------------------------

esConsecuencia :: [Prop] -> Prop -> Bool
esConsecuencia s f = and[esModeloFormula i f|i<- modelosConjunto s]

{- Otra forma (usando map):

esConsecuencia s f = and (map fun (modelosConjunto s))
where fun i = esModeloFormula i f

-}
-- Una forma legible
esConsecuencia2 :: [Prop] -> Prop -> Bool
esConsecuencia2 s f = all (`esModeloFormula` f) (modelosConjunto s)

</source>

Relación 2a

2016-06-16T18:05:09Z

Albromgon: Función unionGeneral usando composición

Relación 1

2016-06-16T10:54:38Z

Albromgon: Alternativa a la función profundidad

<source lang = "haskell">
-- LMF 2015-16: Rel_1.hs
-- Introducción a la programación con Haskell.
-- Departamento de Ciencias de la Computación e I.A.
-- Universidad de Sevilla
-- =====================================================================

-- ---------------------------------------------------------------------
-- Introducción --
-- ---------------------------------------------------------------------

-- En esta relación de ejercicios hacemos una introducción a Haskell, en
-- la que se recuerdan:
-- * las definiciones elementales de funciones,
-- * las definiciones de funciones por comprensión,
-- * las definiciones de funciones por recursión y
-- * los tipos de datos.

-- ---------------------------------------------------------------------
-- Importación de librerías auxiliares
-- ---------------------------------------------------------------------

import Test.QuickCheck
import Data.Char
import Control.Monad

-- ---------------------------------------------------------------------
-- Ejercicio 1. Definir la función media3 tal que (media3 x y z) es
-- la media aritmética de los números x, y y z. Por ejemplo,
-- media3 1 3 8 == 4.0
-- media3 (-1) 0 7 == 2.0
-- media3 (-3) 0 3 == 0.0
-- ---------------------------------------------------------------------

media3 :: Float -> Float -> Float -> Float
media3 x y z = (x+y+z)/3

-- ---------------------------------------------------------------------
-- Ejercicio 2. Definir la función ultimaCifra tal que (ultimaCifra x)
-- es la última cifra del nímero x. Por ejemplo,
-- ultimaCifra 325 == 5
-- ---------------------------------------------------------------------

ultimaCifra :: Integer -> Integer
ultimaCifra x = abs(rem x 10)

-- ---------------------------------------------------------------------
-- Ejercicio 3. Definir la función rota tal que (rota n xs) es la lista
-- obtenida poniendo los n primeros elementos de xs al final de la
-- lista. Por ejemplo,
-- rota 1 [3,2,5,7] == [2,5,7,3]
-- rota 2 [3,2,5,7] == [5,7,3,2]
-- rota 3 [3,2,5,7] == [7,3,2,5]
-- ---------------------------------------------------------------------

rota :: Int -> [a] -> [a]
rota n xs | n < length xs = (drop n xs)++(take n xs)
|otherwise = rota ( mod n (length xs)) xs

-- ---------------------------------------------------------------------
-- Ejercicio 4.1. La disyunción excluyente xor de dos fórmulas se
-- verifica si una es verdadera y la otra es falsa.
--
-- Definir la función xor_1 que calcule la disyunción excluyente a
-- partir de la tabla de verdad. Usar 4 ecuaciones, una por cada línea
-- de la tabla.
-- ---------------------------------------------------------------------

xor_1 :: Bool -> Bool -> Bool
xor_1 True True = False
xor_1 True False = True
xor_1 False True = True
xor_1 False False = False

-- ---------------------------------------------------------------------
-- Ejercicio 4.2. Definir la función xor_2 que calcule la disyunción
-- excluyente a partir de la tabla de verdad y patrones. Usar 2
-- ecuaciones, una por cada valor del primer argumento.
-- ---------------------------------------------------------------------

xor_2 :: Bool -> Bool -> Bool
xor_2 True q = not q
xor_2 False q = q
-- ---------------------------------------------------------------------
-- Ejercicio 4.3. Definir la función xor_3 que calcule la disyunción
-- excluyente a partir de la disyunción (||), conjunción (&&) y negación
-- (not). Usar 1 ecuación.
-- ---------------------------------------------------------------------

xor_3 :: Bool -> Bool -> Bool
xor_3 p q = (p && (not q)) || ((not p) && q)

-- ---------------------------------------------------------------------
-- Ejercicio 4.4. Definir la función xor_4 que calcule la disyunción
-- excluyente a partir de desigualdad (/=). Usar 1 ecuación.
-- ---------------------------------------------------------------------

xor_4 :: Bool -> Bool -> Bool
xor_4 p q = p /= q

-- ---------------------------------------------------------------------
-- Ejercicio 5. Definir, por comprensión, la función
-- sumaDeCuadrados :: Integer -> Integer
-- tal que (sumaDeCuadrados n) es la suma de los cuadrados de los
-- primeros n números; es decir, 1^2 + 2^2 + ... + 100^2. Por ejemplo,
-- sumaDeCuadrados 3 == 14
-- sumaDeCuadrados 100 == 338350
-- ---------------------------------------------------------------------

sumaDeCuadrados :: Integer -> Integer
sumaDeCuadrados n = sum (map (^2) [1..n])

--otra forma:
sumaDeCuadrados2 :: Integer -> Integer
sumaDeCuadrados2 n = sum [x^2 | x <- [1..n]]

-- ---------------------------------------------------------------------
-- Ejercicio 6. Una terna (x,y,z) de enteros positivos es pitagórica si
-- x^2 + y^2 = z^2. Usando una lista por comprensión, definir la función
-- pitagoricas :: Int -> [(Int, Int, Int)]
-- tal que (pitagoricas n) es la lista de todas las ternas pitagóricas
-- cuyas componentes están entre 1 y n. Por ejemplo,
-- *Main> pitagoricas 10
-- [(3,4,5),(4,3,5),(6,8,10),(8,6,10)]
-- ---------------------------------------------------------------------
pitagoricas :: Int -> [(Int, Int, Int)]
pitagoricas n = [(a,b,c)|a<-[1..n],b<-[1..n], c<-[1..n], a^2 + b^2==c^2]

-- ---------------------------------------------------------------------
-- Ejercicio 7. Un entero positivo es perfecto si es igual a la suma de
-- sus factores, excluyendo el propio número. Usando una lista por
-- comprensión y la función factores (del tema), definir la función
-- perfectos :: Int -> [Int]
-- tal que (perfectos n) es la lista de todos los números perfectos
-- menores que n. Por ejemplo:
-- *Main> perfectos 500
-- [6,28,496]
-- ---------------------------------------------------------------------
factores :: Int -> [Int]
factores n = [x|x<-[1..(div n 2)], mod n x ==0]
perfectos :: Int -> [Int]
perfectos n = [x|x<-[1..n], sum (factores x)== x]

-- Otra forma
factores :: Int -> [Int]
factores n = [x|x<-[1..n],mod n x == 0]

perfectos :: Int -> [Int]
perfectos n = [i|i<-[1..(n-1)],i==(sum((factores i))-i)]

-- ---------------------------------------------------------------------
-- Ejercicio 8.1. Definir, por comprensión, la función
-- cuadradosC :: [Integer] -> [Integer]
-- tal que (cuadradosC xs) es la lista de los cuadrados de xs. Por
-- ejemplo,
-- cuadradosC [1,2,3] == [1,4,9]
-- ---------------------------------------------------------------------

cuadradosC :: [Integer] -> [Integer]
cuadradosC xs= [y^2| y<-xs]

-- ---------------------------------------------------------------------
-- Ejercicio 8.2. Definir, por recursión, la función
-- cuadradosR :: [Integer] -> [Integer]
-- tal que (cuadradosR xs) es la lista de los cuadrados de xs. Por
-- ejemplo,
-- cuadradosR [1,2,3] == [1,4,9]
-- ---------------------------------------------------------------------

cuadradosR :: [Integer] -> [Integer]
cuadradosR [] = []
cuadradoR (x:xs) = [x^2] ++ cuadradosR xs

-- otra forma
cuadradosR :: [Integer] -> [Integer]
cuadradosR []=[]
cuadradosR (x:xs) = (x^2):cuadradosR xs

-- ---------------------------------------------------------------------
-- Ejercicio 9.1. Definir, por comprensión, la función
-- imparesC :: [Integer] -> [Integer]
-- tal que (imparesC xs) es la lista de los números impares de xs. Por
-- ejemplo,
-- imparesC [1,2,3] == [1,3]
-- ---------------------------------------------------------------------

imparesC :: [Integer] -> [Integer]
imparesC xs = [x|x<-xs, odd x]
-- ---------------------------------------------------------------------
-- Ejercicio 9.2. Definir, por recursión, la función
-- imparesR :: [Integer] -> [Integer]
-- tal que (imparesR xs) es la lista de los números impares de xs. Por
-- ejemplo,
-- imparesR [1,2,3] == [1,3]
-- ---------------------------------------------------------------------

imparesR :: [Integer] -> [Integer]
imparesR []= []
imparesR (x:xs) | odd x= x: (imparesR xs)
| otherwise = imparesR xs

-- ---------------------------------------------------------------------
-- Ejercicio 10. Definir, por comprensión, la función
-- sumaConsecutivos :: [Int] -> [Int]
-- tal que (sumaConsecutivos xs) es la suma de los pares de elementos
-- consecutivos de la lista xs. Por ejemplo,
-- sumaConsecutivos [3,1,5,2] == [4,6,7]
-- sumaConsecutivos [3] == []
-- ---------------------------------------------------------------------

sumaConsecutivos :: [Int] -> [Int]
sumaConsecutivos xs = zipWith (+) xs (tail xs)

-- ---------------------------------------------------------------------
-- Ejercicio 11. La distancia de Hamming entre dos listas es el número
-- de posiciones en que los correspondientes elementos son
-- distintos. Por ejemplo, la distancia de Hamming entre "roma" y "loba"
-- es 2 (porque hay 2 posiciones en las que los elementos
-- correspondientes son distintos: la 1ª y la 3ª).
--
-- Definir la función distancia tal que (distancia xs ys) es la
-- distancia de Hamming entre xs e ys. Por ejemplo,
-- distancia "romano" "comino" == 2
-- distancia "romano" "camino" == 3
-- distancia "roma" "comino" == 2
-- distancia "roma" "camino" == 3
-- distancia "romano" "ron" == 1
-- distancia "romano" "cama" == 2
-- distancia "romano" "rama" == 1
-- ---------------------------------------------------------------------

distancia :: Eq a => [a] -> [a] -> Int
distancia [] _ = 0
distancia _ [] = 0
distancia (x:xs) (y:ys) | x==y= distancia xs ys
|otherwise = 1 + distancia xs ys

-- ---------------------------------------------------------------------
-- Ejercicio 12. Definir la función
-- factoriales :: [Integer]
-- tal que factoriales es la lista de los factoriales. Por ejemplo,
-- take 10 factoriales == [1,1,2,6,24,120,720,5040,40320,362880]
-- ---------------------------------------------------------------------

-- Por comprensión:

factoriales1 :: [Integer]
factoriales1 = 1:[factorial x|x<-[1..]]
factorial x = product[1..x]

--otra forma sin necesidad de añadir el 1 a la lista

factoriales1 :: [Integer]
factoriales1 = [product [1..n]|n<-[0..]]

-- Usando zipWith:

factoriales2 :: [Integer]
factoriales2 = 1: zipWith (*) [1..] factoriales2

-- Por recursión:

factoriales3 :: [Integer]
factoriales3 = 1: (aux 1 [1..])
where aux n (x:xs) = (n*x):(aux (n*x) xs)

--otra forma
factoriales3 :: [Integer]
factoriales3 = aux (1,1)
where aux (x,y) = x: aux (x*y,y+1)

-- Usando scanl1:

factoriales4 :: [Integer]
factoriales4 = 1:scanl1 (*) [1..]

-- Usando iterate:

factoriales5 :: [Integer]
factoriales5 = map snd aux
where aux = iterate aux2 (1,1) where aux2 (x,y) = (x+1,x*y)

--otra forma
factoriales5 :: [Integer]
factoriales5 = map (fst) (iterate (\(x,y)->(x*y,y+1)) (1,1))

-- Comparación de los tiempos de evaluación:
-- *Main> :set +s
-- take 100 factoriales1 -> (0.20 secs, 0 bytes)
-- take 100 factoriales2 -> (0.05 secs, 0 bytes)
-- take 100 factoriales3 -> (0.02 secs, 0 bytes)
-- take 100 factoriales4 -> (0.00 secs, 20,605,760 bytes)
-- take 100 factoriales5 -> (0.02 secs, 0 bytes)

-- ---------------------------------------------------------------------
-- Ejercicio 13.0. En los siguientes ejercicios se demostrarán
-- propiedades de los árboles binarios definidos como sigue
-- data Arbol a = Hoja
-- | Nodo a (Arbol a) (Arbol a)
-- deriving (Show, Eq)
-- Como ejemplos se usarán los árboles
-- ---------------------------------------------------------------------

data Arbol a = Hoja
| Nodo a (Arbol a) (Arbol a)
deriving (Show, Eq)

arbol_1 = Nodo 9
(Nodo 3
(Nodo 2 Hoja Hoja)
(Nodo 4 Hoja Hoja))
(Nodo 7 Hoja Hoja)

-- ---------------------------------------------------------------------
-- Ejercicio 13.1. Definir la función
-- espejo :: Arbol a -> Arbol a
-- tal que (espejo x) es la imagen especular del árbol x. Por ejemplo,
-- *Main> espejo arbol_1
-- Nodo 9
-- (Nodo 7 Hoja Hoja)
-- (Nodo 3
-- (Nodo 4 Hoja Hoja)
-- (Nodo 2 Hoja Hoja))
-- ---------------------------------------------------------------------

espejo :: Arbol a -> Arbol a
espejo Hoja = Hoja
espejo (Nodo a i d) = Nodo a (espejo d) (espejo i)

-- ---------------------------------------------------------------------
-- Ejercicio 13.2. Comprobar con QuickCheck que para todo árbol x,
-- espejo (espejo x) = x
-- ---------------------------------------------------------------------

prop_espejo :: Arbol Int -> Bool
prop_espejo x = espejo (espejo x) == x

-- ---------------------------------------------------------------------
-- Ejercicio 13.3. Demostrar por inducción que para todo árbol x,
-- espejo (espejo x) = x
-- ---------------------------------------------------------------------
{-
-- Demostración por inducción en x
-- Caso base: El árbol es una Hoja. Por definición de espejo, espejo Hoja = Hoja, luego espejo (espejo Hoja) = Hoja.
-- Caso inductivo: El árbol tiene profundidad n. Se supone cierto que para este árbol se verifica que espejo (espejo x) = x
-- Demostración para un árbol de profundidad (n+1). Este árbol es de la forma (Nodo a i d) donde i, d son árboles de profundidad igual o menor
-- que n.
-- espejo (espejo (Nodo a i d)) = espejo (Nodo a (espejo d) (espejo i)) = Nodo a (espejo(espejo i)) (espejo(espejo d)) por definición de la
-- función espejo. Por hipótesis de inducción espejo(espejo i) = i y espejo(espejo d) = d, y queda demostrado.
-}

-- ---------------------------------------------------------------------
-- Ejercicio 13.4. Definir la función
-- preorden :: Arbol a -> [a]
-- tal que (preorden x) es la lista correspondiente al recorrido
-- preorden del árbol x; es decir, primero visita la raíz del árbol, a
-- continuación recorre el subárbol izquierdo y, finalmente, recorre el
-- subárbol derecho. Por ejemplo,
-- *Main> arbol_1
-- Nodo 9 (Nodo 3 (Nodo 2 Hoja Hoja) (Nodo 4 Hoja Hoja)) (Nodo 7 Hoja Hoja)
-- *Main> preorden arbol_1
-- [9,3,2,4,7]
-- ---------------------------------------------------------------------

preorden :: Arbol a -> [a]
preorden Hoja = []
preorden (Nodo a i d)= a:(preorden i)++(preorden d)

-- ---------------------------------------------------------------------
-- Ejercicio 13.5. Definir la función
-- postorden :: Arbol a -> [a]
-- tal que (postorden x) es la lista correspondiente al recorrido
-- postorden del árbol x; es decir, primero recorre el subárbol
-- izquierdo, a continuación el subárbol derecho y, finalmente, la raíz
-- del árbol. Por ejemplo,
-- *Main> arbol_1
-- Nodo 9 (Nodo 3 (Nodo 2 Hoja Hoja) (Nodo 4 Hoja Hoja)) (Nodo 7 Hoja Hoja)
-- *Main> postorden arbol_1
-- [2,4,3,7,9]
-- ---------------------------------------------------------------------

postorden :: Arbol a -> [a]
postorden Hoja = []
postorden (Nodo a i d) = (postorden i)++(postorden d)++[a]

-- ---------------------------------------------------------------------
-- Ejercicio 13.6. Comprobar con QuickCheck que para todo árbol x,
-- postorden (espejo x) = reverse (preorden x)
-- ---------------------------------------------------------------------

-- La propiedad es
prop_recorrido :: Arbol Int -> Bool
prop_recorrido x = postorden (espejo x) == reverse (preorden x)

-- La comprobación es

-- ---------------------------------------------------------------------
-- Ejercicio 13.7. Demstrar por inducción que para todo árbol x,
-- postorden (espejo x) = reverse (preorden x)
-- ---------------------------------------------------------------------

{-
Demostración por inducción en x.

-}

-- ---------------------------------------------------------------------
-- Ejercicio 13.8. Comprobar con QuickCheck que para todo árbol binario
-- x, se tiene que
-- reverse (preorden (espejo x)) = postorden x
-- ---------------------------------------------------------------------

-- La propiedad es
prop_reverse_preorden_espejo :: Arbol Int -> Bool
prop_reverse_preorden_espejo x =
reverse (preorden (espejo x)) == postorden x

-- La comprobación es
-- *Main> quickCheck prop_reverse_preorden_espejo
-- OK, passed 100 tests.

-- ---------------------------------------------------------------------
-- Ejercicio 13.9. Demostrar que para todo árbol binario x, se tiene que
-- reverse (preorden (espejo x)) = preorden x
-- ---------------------------------------------------------------------

{-
Demostración:
reverse (preorden (espejo x))
= postorden (espejo (espejo x)) [por ejercicio 13.7]
= postorden x [por ejercicio 13.3]
-}

-- ---------------------------------------------------------------------
-- Ejercicio 13.10. Definir la función
-- nNodos :: Arbol a -> Int
-- tal que (nNodos x) es el número de nodos del árbol x. Por ejemplo,
-- *Main> arbol_1
-- Nodo 9 (Nodo 3 (Nodo 2 Hoja Hoja) (Nodo 4 Hoja Hoja)) (Nodo 7 Hoja Hoja)
-- *Main> nNodos arbol_1
-- 5
-- ---------------------------------------------------------------------

nNodos :: Arbol a -> Int
nNodos Hoja = 0
nNodos (Nodo n i d) = 1 + nNodos i + nNodos d

-- ---------------------------------------------------------------------
-- Ejercicio 13.11. Comprobar con QuickCheck que el número de nodos de la
-- imagen especular de un árbol es el mismo que el número de nodos del
-- árbol.
-- ---------------------------------------------------------------------

-- La propiedad es
prop_nNodos_espejo :: Arbol Int -> Bool
prop_nNodos_espejo x =
nNodos (espejo x) == nNodos x

-- La comprobación es
-- *Main> quickCheck prop_nNodos_espejo
-- +++ OK, passed 100 tests.

-- ---------------------------------------------------------------------
-- Ejercicio 13.12. Demostrar por inducción que el número de nodos de la
-- imagen especular de un árbol es el mismo que el número de nodos del
-- árbol.
-- ---------------------------------------------------------------------

{-
Demostración:

-}

-- ---------------------------------------------------------------------
-- Ejercicio 13.13. Comprobar con QuickCheck que la longitud de la lista
-- obtenida recorriendo un árbol en sentido preorden es igual al número
-- de nodos del árbol.
-- ---------------------------------------------------------------------

-- La propiedad es
prop_length_preorden :: Arbol Int -> Bool
prop_length_preorden x =
length (preorden x) == length (postorden x)

--
prop_length_preorden :: Arbol Int -> Bool
prop_length_preorden x = length(preorden x) == nNodos x

-- La comprobación es
-- *Main> quickCheck prop_length_preorden
-- +++ OK, passed 100 tests.

-- ---------------------------------------------------------------------
-- Ejercicio 13.14. Demostrar por inducción que la longitud de la lista
-- obtenida recorriendo un árbol en sentido preorden es igual al número
-- de nodos del árbol.
-- ---------------------------------------------------------------------

{-
Demostración:

-}

-- ---------------------------------------------------------------------
-- Ejercicio 13.15. Definir la función
-- profundidad :: Arbol a -> Int
-- tal que (profundidad x) es la profundidad del árbol x. Por ejemplo,
-- *Main> arbol_1
-- Nodo 9 (Nodo 3 (Nodo 2 Hoja Hoja) (Nodo 4 Hoja Hoja)) (Nodo 7 Hoja Hoja)
-- *Main> profundidad arbol_1
-- 3
-- ---------------------------------------------------------------------

profundidad :: Arbol a -> Int
profundidad Hoja = 0
profundidad (Nodo n i d)
| length (preorden i) > length (preorden d) = 1 + profundidad i
| otherwise = 1 + profundidad d

{- Otra forma:
profundidad :: Arbol a -> Int
profundidad Hoja = 0
profundidad (Nodo x c b) = 1 + (max (profundidad c) (profundidad b))
-}

-- ---------------------------------------------------------------------
-- Ejercicio 13.16. Comprobar con QuickCheck que para todo árbol binario
-- x, se tiene que
-- nNodos x <= 2^(profundidad x) - 1
-- ---------------------------------------------------------------------

-- La propiedad es
prop_nNodosProfundidad :: Arbol Int -> Bool
prop_nNodosProfundidad x =
nNodos x <= 2^(profundidad x) - 1

-- La comprobación es
-- quickCheck prop_nNodosProfundidad
-- +++ OK, passed 100 tests.

-- ---------------------------------------------------------------------
-- Nota. Para comprobar propiedades de árboles con QuickCheck se
-- utilizará el siguiente generador.
-- ---------------------------------------------------------------------

instance Arbitrary a => Arbitrary (Arbol a) where
arbitrary = sized arbol
where
arbol 0 = return Hoja
arbol n | n>0 = oneof [return Hoja,
liftM3 Nodo arbitrary subarbol subarbol]
where subarbol = arbol (div n 2)

</source>

Relación 3

2016-03-25T18:37:00Z

Albromgon: Otra opción ejercicio 55

<source lang = "isar">header {* R3: Deducción natural proposicional *}

theory R3
imports Main
begin

text {*
---------------------------------------------------------------------
El objetivo de esta relación es demostrar cada uno de los ejercicios
usando sólo las reglas básicas de deducción natural de la lógica
proposicional (sin usar el método auto).

Las reglas básicas de la deducción natural son las siguientes:
· conjI: ⟦P; Q⟧ ⟹ P ∧ Q
· conjunct1: P ∧ Q ⟹ P
· conjunct2: P ∧ Q ⟹ Q
· notnotD: ¬¬ P ⟹ P
· notnotI: P ⟹ ¬¬ P
· mp: ⟦P ⟶ Q; P⟧ ⟹ Q
· mt: ⟦F ⟶ G; ¬G⟧ ⟹ ¬F
· impI: (P ⟹ Q) ⟹ P ⟶ Q
· disjI1: P ⟹ P ∨ Q
· disjI2: Q ⟹ P ∨ Q
· disjE: ⟦P ∨ Q; P ⟹ R; Q ⟹ R⟧ ⟹ R
· FalseE: False ⟹ P
· notE: ⟦¬P; P⟧ ⟹ R
· notI: (P ⟹ False) ⟹ ¬P
· iffI: ⟦P ⟹ Q; Q ⟹ P⟧ ⟹ P = Q
· iffD1: ⟦Q = P; Q⟧ ⟹ P
· iffD2: ⟦P = Q; Q⟧ ⟹ P
· ccontr: (¬P ⟹ False) ⟹ P
---------------------------------------------------------------------
*}

text {*
Se usarán las reglas notnotI y mt que demostramos a continuación. *}

lemma notnotI: "P ⟹ ¬¬ P"
by auto

lemma mt: "⟦F ⟶ G; ¬G⟧ ⟹ ¬F"
by auto

section {* Implicaciones *}

text {* ---------------------------------------------------------------
Ejercicio 1. Demostrar
p ⟶ q, p ⊢ q
------------------------------------------------------------------ *}

lemma ejercicio_1:
assumes "p ⟶ q"
"p"
shows "q"

proof -
show "q" using assms(1,2) by (rule mp)
qed

lemma ejercicio_1_b:
assumes "p ⟶ q"
"p"
shows "q"
proof-
show "q" using assms(1) assms(2) by (rule mp)
qed

text {* ---------------------------------------------------------------
Ejercicio 2. Demostrar
p ⟶ q, q ⟶ r, p ⊢ r
------------------------------------------------------------------ *}

lemma ejercicio_2:
assumes "p ⟶ q"
"q ⟶ r"
"p"
shows "r"

proof-
have 1: "q" using assms(1,3) by (rule mp)
show "r" using assms(2) 1 by (rule mp)
qed

text {* ---------------------------------------------------------------
Ejercicio 3. Demostrar
p ⟶ (q ⟶ r), p ⟶ q, p ⊢ r
------------------------------------------------------------------ *}

lemma ejercicio_3:
assumes "p ⟶ (q ⟶ r)"
"p ⟶ q"
"p"
shows "r"

proof-
have 1: "q" using assms(2,3) by (rule mp)
have 2: "q⟶r" using assms(1,3) by (rule mp)
show "r" using 2 1 by (rule mp)
qed

text {* ---------------------------------------------------------------
Ejercicio 4. Demostrar
p ⟶ q, q ⟶ r ⊢ p ⟶ r
------------------------------------------------------------------ *}

lemma ejercicio_4:
assumes "p ⟶ q"
"q ⟶ r"
shows "p ⟶ r"

proof-
{assume 1: "p"
have 2: "q" using assms(1) 1 ..
have 3: "r" using assms(2) 2 ..}
thus "p⟶r" by (rule impI)
qed

lemma ejercicio_4:
assumes "p ⟶ q"
"q ⟶ r"
shows "p ⟶ r"
proof-
{assume 1 : "p"
have 2 : "q" using assms(1) 1 by (rule mp)
have 3 : "r" using assms(2) 2 by (rule mp)}
thus "p⟶r" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 5. Demostrar
p ⟶ (q ⟶ r) ⊢ q ⟶ (p ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_5:
assumes "p ⟶ (q ⟶ r)"
shows "q ⟶ (p ⟶ r)"

{assume 1: "q"
{assume 2: "p"
have 3: "q⟶r" using assms 2 by (rule mp)
have "r" using 3 1 by (rule mp)}
hence "p⟶r" by (rule impI)}
thus "(q⟶(p⟶r))" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 6. Demostrar
p ⟶ (q ⟶ r) ⊢ (p ⟶ q) ⟶ (p ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_6:
assumes "p ⟶ (q ⟶ r)"
shows "(p ⟶ q) ⟶ (p ⟶ r)"

proof-
{assume 1:"p⟶q"
{assume 2: "p"
have 3: "q⟶r" using assms 2 by (rule mp)
have 4: "q" using 1 2 by (rule mp)
have "r" using 3 4 by (rule mp)}
hence "p⟶r" by (rule impI)}
thus "(p⟶q)⟶(p⟶r)" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 7. Demostrar
p ⊢ q ⟶ p
------------------------------------------------------------------ *}

lemma ejercicio_7:
assumes "p"
shows "q ⟶ p"

proof-
{assume 1: "q"
have "p" using assms .}
thus "q⟶p" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 8. Demostrar
⊢ p ⟶ (q ⟶ p)
------------------------------------------------------------------ *}

lemma ejercicio_8:
"p ⟶ (q ⟶ p)"

proof-
{assume 1:"p"
{assume 2:"q"
have 3: "p" using 1 .}
hence "q⟶p" ..}
thus "p⟶(q⟶p)" ..
qed

lemma ejercicio_8:
"p ⟶ (q ⟶ p)"
proof-
{assume 1 : "p"
{assume 2 : "q"
have 3 : "p" using 1 .}
hence "q⟶p" by (rule impI)}
thus "p⟶(q⟶p)" by (rule impI)
qed

lemma ejercicio_8:
"p ⟶ (q ⟶ p)"
proof (rule impI)
assume 1: "p"
show "q ⟶ p"
proof (rule impI)
assume "q"
show "p" using 1 by this
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 9. Demostrar
p ⟶ q ⊢ (q ⟶ r) ⟶ (p ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_9:
assumes "p ⟶ q"
shows "(q ⟶ r) ⟶ (p ⟶ r)"

proof-
{assume 1:"q⟶r"
{assume 2:"p"
have 3:"q" using assms 2 ..
have "r" using 1 3 ..}
hence "p⟶r" ..}
thus "(q⟶r)⟶(p⟶r) " ..
qed

lemma ejercicio_9:
assumes "p ⟶ q"
shows "(q ⟶ r) ⟶ (p ⟶ r)"
proof-
{assume 1 : "q⟶r"
{assume 2 : "p"
have 3 : "q" using assms(1) 2 by (rule mp)
have 4 : "r" using 1 3 by (rule mp)}
hence "p⟶r" by (rule impI)}
thus "(q⟶r)⟶(p⟶r)" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 10. Demostrar
p ⟶ (q ⟶ (r ⟶ s)) ⊢ r ⟶ (q ⟶ (p ⟶ s))
------------------------------------------------------------------ *}

lemma ejercicio_10:
assumes "p ⟶ (q ⟶ (r ⟶ s))"
shows "r ⟶ (q ⟶ (p ⟶ s))"

proof-
{assume 1:"r"
{assume 2:"q"
{assume 3:"p"
have 4:"q⟶(r⟶s)" using assms 3 ..
have 5: "r⟶s" using 4 2 ..
have 6: "s" using 5 1 ..}
hence "p⟶s" ..}
hence "q⟶(p⟶s)" ..}
thus "r⟶(q⟶(p⟶s))" ..
qed

lemma ejercicio_10:
assumes "p ⟶ (q ⟶ (r ⟶ s))"
shows "r ⟶ (q ⟶ (p ⟶ s))"
proof-
{assume 1 : "r"
{assume 2 : "q"
{assume 3 : "p"
have 4 : "q ⟶ (r ⟶ s)" using assms(1) 3 by (rule mp)
have 5 : "r⟶s" using 4 2 by (rule mp)
have 6 : "s" using 5 1 by (rule mp)}
hence "p⟶s" by (rule impI)}
hence "q⟶(p⟶s)" by (rule impI)}
thus "r⟶(q⟶(p⟶s))" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 11. Demostrar
⊢ (p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))
------------------------------------------------------------------ *}

lemma ejercicio_11:
"(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))"

proof (rule impI)
assume 3: "p⟶(q⟶r)"
show "(p⟶q)⟶(p⟶r)"
proof(rule impI)
assume 2:"p⟶q"
show "p⟶r"
proof (rule impI)
assume 1: "p"
have 4:"q" using 2 1 ..
have 5:"q⟶r" using 3 1 ..
show "r" using 5 4 ..
qed
qed
qed

lemma ejercicio_11:
"(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))"
proof-
{assume 1 : "p ⟶ (q ⟶ r)"
{assume 2 : "p⟶q"
{assume 3 : "p"
have 4 : "q⟶r" using 1 3 by (rule mp)
have 5 : "q" using 2 3 by (rule mp)
have 6 : "r" using 4 5 by (rule mp)}
hence "p⟶r" by (rule impI)}
hence "(p ⟶ q) ⟶ (p ⟶ r)" by (rule impI)}
thus "(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 12. Demostrar
(p ⟶ q) ⟶ r ⊢ p ⟶ (q ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_12:
assumes "(p ⟶ q) ⟶ r"
shows "p ⟶ (q ⟶ r)"

proof-
{assume 1:"p"
{assume 3:"q"
have 4:"p" using 1 .
have 5:"q" using 3 .
then have 6:"p⟶q" ..
have "r" using assms 6 ..}
hence "q⟶r" ..}
thus "p⟶(q⟶r)" ..
qed

lemma ejercicio_12:
assumes "(p ⟶ q) ⟶ r"
shows "p ⟶ (q ⟶ r)"
proof-
{assume 1 : "p"
{assume 2 : "q"
{assume 3 : "p"
have 4 : "q" using 2 .}
hence 5 : "p⟶q" by (rule impI)
have 6 : "r" using assms(1) 5 by (rule mp)}
hence "q⟶r" by (rule impI)}
thus "p⟶(q⟶r)" by (rule impI)
qed

section {* Conjunciones *}

text {* ---------------------------------------------------------------
Ejercicio 13. Demostrar
p, q ⊢ p ∧ q
------------------------------------------------------------------ *}

lemma ejercicio_13:
assumes "p"
"q"
shows "p ∧ q"

proof-
show "p∧q" using assms(1) assms(2) by (rule conjI)
qed

lemma ejercicio_13:
assumes "p"
"q"
shows "p ∧ q"
proof-
show "p ∧ q" using assms(1,2) by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 14. Demostrar
p ∧ q ⊢ p
------------------------------------------------------------------ *}

lemma ejercicio_14:
assumes "p ∧ q"
shows "p"

proof-
show "p" using assms by (rule conjunct1)
qed

text {* ---------------------------------------------------------------
Ejercicio 15. Demostrar
p ∧ q ⊢ q
------------------------------------------------------------------ *}

lemma ejercicio_15:
assumes "p ∧ q"
shows "q"

proof-
show "q" using assms by (rule conjunct2)
qed

text {* ---------------------------------------------------------------
Ejercicio 16. Demostrar
p ∧ (q ∧ r) ⊢ (p ∧ q) ∧ r
------------------------------------------------------------------ *}

lemma ejercicio_16:
assumes "p ∧ (q ∧ r)"
shows "(p ∧ q) ∧ r"

proof-
have 1:"p" using assms by (rule conjunct1)
have 2:"q∧r" using assms by (rule conjunct2)
then have 3:"q" by (rule conjunct1)
have 4:"r" using 2 by (rule conjunct2)
have "p∧q" using 1 3 by (rule conjI)
then show "(p∧q)∧r" using 4 by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 17. Demostrar
(p ∧ q) ∧ r ⊢ p ∧ (q ∧ r)
------------------------------------------------------------------ *}

lemma ejercicio_17:
assumes "(p ∧ q) ∧ r"
shows "p ∧ (q ∧ r)"

proof-
have 1:"p∧q" using assms by (rule conjunct1)
have 2:"r" using assms by (rule conjunct2)
have 3:"p" using 1 by (rule conjunct1)
have 4:"q" using 1 by (rule conjunct2)
have 5:"q∧r" using 4 2 by (rule conjI)
show "p∧(q∧r)" using 3 5 by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 18. Demostrar
p ∧ q ⊢ p ⟶ q
------------------------------------------------------------------ *}

lemma ejercicio_18:
assumes "p ∧ q"
shows "p ⟶ q"
proof-
have 1:"p" using assms ..
have 2:"q" using assms ..
then show "p⟶q" ..
qed

lemma ejercicio_18:
assumes "p ∧ q"
shows "p ⟶ q"
proof-
{assume "p"
have "q" using assms by (rule conjunct2)}
thus "p⟶q" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 19. Demostrar
(p ⟶ q) ∧ (p ⟶ r) ⊢ p ⟶ q ∧ r
------------------------------------------------------------------ *}

lemma ejercicio_19:
assumes "(p ⟶ q) ∧ (p ⟶ r)"
shows "p ⟶ q ∧ r"
proof-
have 1:"p⟶q" using assms ..
have 2:"p⟶r" using assms ..
{assume 3:"p"
have 4:"q" using 1 3 ..
have 5:"r" using 2 3 ..
have 6:"q∧r" using 4 5 ..}
thus "p ⟶ q ∧ r" ..
qed

lemma ejercicio_19:
assumes "(p ⟶ q) ∧ (p ⟶ r)"
shows "p ⟶ q ∧ r"
proof-
{assume 1 : "p"
have 2 : "p⟶q" using assms by (rule conjunct1)
have 3 : "p⟶r" using assms by (rule conjunct2)
have 4 : "q" using 2 1 by (rule mp)
have 5 : "r" using 3 1 by (rule mp)
have 6 : "q ∧ r" using 4 5 by (rule conjI)}
thus "p⟶(q ∧ r)" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 20. Demostrar
p ⟶ q ∧ r ⊢ (p ⟶ q) ∧ (p ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_20:
assumes "p ⟶ q ∧ r"
shows "(p ⟶ q) ∧ (p ⟶ r)"

proof-
{assume 1: p
have 2:"q∧r" using assms 1 ..
have "r" using 2 ..}
hence 1:"p⟶r" ..
{assume 1: p
have 2:"q∧r" using assms 1 ..
have "q" using 2 ..}
hence 2:"p⟶q" ..
then show "(p ⟶ q) ∧ (p ⟶ r)" using 1 ..
qed

lemma ejercicio_20:
assumes "p ⟶ q ∧ r"
shows "(p ⟶ q) ∧ (p ⟶ r)"
proof-
{assume 1 : "p"
have 2 : "q ∧ r" using assms 1 by (rule mp)
have 3 : "q" using 2 by (rule conjunct1)}
hence 4 : "p⟶q" by (rule impI)
{assume 5 : "p"
have 6 : "q ∧ r" using assms 5 by (rule mp)
have 7 : "r" using 6 by (rule conjunct2)}
hence 8 : "p⟶r" by (rule impI)
show "(p ⟶ q) ∧ (p ⟶ r)" using 4 8 by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 21. Demostrar
p ⟶ (q ⟶ r) ⊢ p ∧ q ⟶ r
------------------------------------------------------------------ *}

lemma ejercicio_21:
assumes "p ⟶ (q ⟶ r)"
shows "p ∧ q ⟶ r"

proof-
{assume 1:"p∧q"
have 2:"p" using 1 ..
have 3:"q⟶r" using assms 2 ..
have 4: "q" using 1 ..
have "r" using 3 4 ..}
thus "p∧q ⟶r" ..
qed

lemma ejercicio_21:
assumes "p ⟶ (q ⟶ r)"
shows "p ∧ q ⟶ r"
proof-
{assume 1 : "p ∧ q"
have 2 : "p" using 1 by (rule conjunct1)
have 3 : "q" using 1 by (rule conjunct2)
have 4 : "q⟶r" using assms 2 by (rule mp)
have 5 : "r" using 4 3 by (rule mp)}
thus "p ∧ q ⟶ r" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 22. Demostrar
p ∧ q ⟶ r ⊢ p ⟶ (q ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_22:
assumes "p ∧ q ⟶ r"
shows "p ⟶ (q ⟶ r)"

proof-
{assume 1:"p"
{assume 2:"q"
have 3:"p∧q" using 1 2 ..
have "r" using assms 3 ..}
hence "q⟶r" ..}
thus "p⟶(q⟶r)" ..
qed

lemma ejercicio_22:
assumes "p ∧ q ⟶ r"
shows "p ⟶ (q ⟶ r)"
proof-
{assume 1 : "p"
{assume 2 : "q"
have 3 : "p ∧ q" using 1 2 by (rule conjI)
have 4 : "r" using assms 3 by (rule mp)}
hence "q⟶r" by (rule impI)}
thus "p⟶(q⟶r)" by (rule impI)
qed

lemma ejercicio_22:
assumes 1:"p ∧ q ⟶ r"
shows "p ⟶ (q ⟶ r)"
proof (rule impI)
assume 2:"p"
show "q ⟶ r"
proof (rule impI)
assume 3:"q"
have 4:"p ∧ q" using 2 3 by (rule conjI)
show "r" using 1 4 by (rule mp)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 23. Demostrar
(p ⟶ q) ⟶ r ⊢ p ∧ q ⟶ r
------------------------------------------------------------------ *}

lemma ejercicio_23:
assumes "(p ⟶ q) ⟶ r"
shows "p ∧ q ⟶ r"

proof-
{assume 1:"p∧q"
have 2:"p" using 1 ..
{assume "p"
have 3: "q" using 1 ..}
hence 4:"p⟶q" ..
have "r" using assms 4 ..}
thus "p∧q ⟶r" ..
qed

lemma ejercicio_23:
assumes "(p ⟶ q) ⟶ r"
shows "p ∧ q ⟶ r"
proof-
{assume 1 : "p ∧ q"
{assume 2 : "p"
have 3 : "q" using 1 by (rule conjunct2)}
hence 4 : "p⟶q" by (rule impI)
have 5 : "r" using assms 4 by (rule mp)}
thus "p ∧ q ⟶ r" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 24. Demostrar
p ∧ (q ⟶ r) ⊢ (p ⟶ q) ⟶ r
------------------------------------------------------------------ *}

lemma ejercicio_24:
assumes "p ∧ (q ⟶ r)"
shows "(p ⟶ q) ⟶ r"

proof-
{assume 1: "p⟶q"
have 2:"p" using assms ..
have 3:"q⟶r" using assms..
have 4:"q" using 1 2 ..
have "r" using 3 4 ..}
thus "(p⟶q)⟶r" ..
qed

lemma ejercicio_24:
assumes "p ∧ (q ⟶ r)"
shows "(p ⟶ q) ⟶ r"
proof-
{assume 1 : "p⟶q"
have 2 : "p" using assms by (rule conjunct1)
have 3 : "q⟶r" using assms by (rule conjunct2)
have 4 : "q" using 1 2 by (rule mp)
have 5 : "r" using 3 4 by (rule mp)}
thus "(p⟶q)⟶r" by (rule impI)
qed

lemma ejercicio_24:
assumes 1:"p ∧ (q ⟶ r)"
shows "(p ⟶ q) ⟶ r"
proof (rule impI)
assume 2:"p ⟶ q"
have 3:"p" using 1 by (rule conjunct1)
have 4:"q" using 2 3 by (rule mp)
have 5:"q ⟶ r" using 1 by (rule conjunct2)
show "r" using 5 4 by (rule mp)
qed

section {* Disyunciones *}

text {* ---------------------------------------------------------------
Ejercicio 25. Demostrar
p ⊢ p ∨ q
------------------------------------------------------------------ *}

lemma ejercicio_25:
assumes "p"
shows "p ∨ q"

proof-
show "p∨q" using assms by (rule disjI1)
qed

text {* ---------------------------------------------------------------
Ejercicio 26. Demostrar
q ⊢ p ∨ q
------------------------------------------------------------------ *}

lemma ejercicio_26:
assumes "q"
shows "p ∨ q"

proof-
show "p∨q" using assms by (rule disjI2)
qed

text {* ---------------------------------------------------------------
Ejercicio 27. Demostrar
p ∨ q ⊢ q ∨ p
------------------------------------------------------------------ *}

lemma ejercicio_27:
assumes "p ∨ q"
shows "q ∨ p"

proof -
have "p ∨ q" using assms by this
moreover
{ assume 2: "p"
have "q ∨ p" using 2 by (rule disjI2) }
moreover
{ assume 3: "q"
have "q ∨ p" using 3 by (rule disjI1) }
ultimately show "q ∨ p" by (rule disjE)
qed

lemma ejercicio_27:
assumes 1:"p ∨ q"
shows "q ∨ p"
using 1
proof (rule disjE)
{assume 2:"p"
show "q ∨ p" using 2 by (rule disjI2)}
next
{assume 4:"q"
show "q ∨ p" using 4 by (rule disjI1)}
qed

text {* ---------------------------------------------------------------
Ejercicio 28. Demostrar
q ⟶ r ⊢ p ∨ q ⟶ p ∨ r
------------------------------------------------------------------ *}

lemma ejercicio_28:
assumes "q ⟶ r"
shows "p ∨ q ⟶ p ∨ r"
proof-
{assume "p ∨ q"
moreover
{ assume 1:"p"
have "p ∨ r" using 1 by (rule disjI1)}
moreover
{ assume 2:"q"
have 3:"r" using assms 2 by (rule mp)
have "p ∨ r" using 3 by (rule disjI2) }
ultimately have " p ∨ r" by (rule disjE)}
thus "p ∨ q ⟶ p ∨ r" by (rule impI)
qed

lemma ejercicio_28:
assumes 1:"q ⟶ r"
shows "p ∨ q ⟶ p ∨ r"
proof (rule impI)
assume 2:"p ∨ q"
thus "p ∨ r"
proof (rule disjE)
{assume 3:"p"
show "p ∨ r" using 3 by (rule disjI1)}
next
{assume 4:"q"
have 5:"r" using 1 4 by (rule mp)
show "p ∨ r" using 5 by (rule disjI2)}
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 29. Demostrar
p ∨ p ⊢ p
------------------------------------------------------------------ *}

lemma ejercicio_29:
assumes "p ∨ p"
shows "p"

proof-
have 1: "p ∨ p" using assms by this
moreover
{assume 1:"p"
have "p" using 1 by this}
moreover
{assume 2:"p"
have "p" using 2 by this}
ultimately show "p" by (rule disjE)

qed
text{* Otra forma:
using assms
proof
assume "p"
thus "p" .
qed}

lemma ejercicio_29:
assumes "p ∨ p"
shows "p"
proof-
have "p ∨ p" using assms by this
moreover
{assume 1 : "p"
have "p" using 1 .}
moreover
{assume 2 : "p"
have "p" using 2 .}
ultimately show "p" by (rule disjE)
qed*}

lemma ejercicio_29:
assumes 1:"p ∨ p"
shows "p"
using 1
proof (rule disjE)
{assume 2:"p"
show "p" using 2 by this}
next
{assume 3:"p"
show "p" using 3 by this}
qed

text {* ---------------------------------------------------------------
Ejercicio 30. Demostrar
p ⊢ p ∨ p
------------------------------------------------------------------ *}

lemma ejercicio_30:
assumes "p"
shows "p ∨ p"
proof-
show "p ∨ p" using assms by (rule disjI1)
qed

text{* Otra forma:
using assms
proof
qed}

text {* ---------------------------------------------------------------
Ejercicio 31. Demostrar
p ∨ (q ∨ r) ⊢ (p ∨ q) ∨ r
------------------------------------------------------------------ *}

lemma ejercicio_31:
assumes "p ∨ (q ∨ r)"
shows "(p ∨ q) ∨ r"
proof -
have "p ∨ (q ∨ r)" using assms(1) by this
moreover
{ assume 1: "p"
have 2: "p ∨ q" using 1 by (rule disjI1)
have 3: "(p ∨ q) ∨ r" using 2 by (rule disjI1)}
moreover
{ assume 4: "q ∨ r"
moreover
{ assume 5: "q"
have 6: "p ∨ q" using 5 by (rule disjI2)
have 7: "(p ∨ q) ∨ r" using 6 by (rule disjI1)}
moreover
{ assume 8: "r"
have 9: "(p ∨ q) ∨ r" using 8 by (rule disjI2)}
ultimately have "(p ∨ q) ∨ r" by (rule disjE)}
ultimately show "(p ∨ q) ∨ r" by (rule disjE)
qed

text{* Otra forma:
using assms
proof
assume "p"
then have "p∨q" ..
thus "(p∨q)∨r" ..
next
assume 1:"q∨r"
thus "(p ∨ q) ∨ r"
proof
assume "q"
then have "p∨q" ..
thus 1: "(p ∨ q) ∨ r" ..
next
assume "r"
then have 2:"(p∨q)∨r" ..
then have "q∨r⟶(p ∨ q) ∨ r" ..
thus "(p ∨ q) ∨ r" using 1 ..
qed}*}

lemma ejercicio_31:
assumes 1:"p ∨ (q ∨ r)"
shows "(p ∨ q) ∨ r"
using 1
proof (rule disjE)
{assume 2:"p"
have 3:"p ∨ q" using 2 by (rule disjI1)
show "(p ∨ q) ∨ r" using 3 by (rule disjI1)}
next
{assume 4:"q ∨ r"
thus "(p ∨ q) ∨ r"
proof (rule disjE)
{assume 5:"q"
have 6:"p ∨ q" using 5 by (rule disjI2)
show "(p ∨ q) ∨ r" using 6 by (rule disjI1)}
next
{assume 7:"r"
show "(p ∨ q) ∨ r" using 7 by (rule disjI2)}
qed}
qed

text {* ---------------------------------------------------------------
Ejercicio 32. Demostrar
(p ∨ q) ∨ r ⊢ p ∨ (q ∨ r)
------------------------------------------------------------------ *}

lemma ejercicio_32:
assumes "(p ∨ q) ∨ r"
shows "p ∨ (q ∨ r)"
proof -
have "(p ∨ q) ∨ r" using assms(1) by this
moreover
{ assume 1: "p ∨ q"
moreover
{ assume 2: "p"
have 3: "p ∨ (q ∨ r)" using 2 by (rule disjI1)}
moreover
{ assume 4: "q"
have 5: "q ∨ r" using 4 by (rule disjI1)
have 6: "p ∨ (q ∨ r)" using 5 by (rule disjI2)}
ultimately have "p ∨ (q ∨ r)" by (rule disjE)}
moreover
{ assume 7: "r"
have 8: "q ∨ r" using 7 by (rule disjI2)
have 9: "p ∨ (q ∨ r)" using 8 by (rule disjI2)}
ultimately show "p ∨ (q ∨ r)" by (rule disjE)
qed

lemma ejercicio_32:
assumes 1:"(p ∨ q) ∨ r"
shows "p ∨ (q ∨ r)"
using 1
proof (rule disjE)
{assume 2:"p ∨ q"
thus "p ∨ (q ∨ r)"
proof (rule disjE)
{assume 3:"p"
show "p ∨ (q ∨ r)" using 3 by (rule disjI1)}
next
{assume 4:"q"
have 5:"q ∨ r" using 4 by (rule disjI1)
show "p ∨ (q ∨ r)" using 5 by (rule disjI2)}
qed}
next
{assume 6:"r"
have 7:"q ∨ r" using 6 by (rule disjI2)
show "p ∨ (q ∨ r)" using 7 by (rule disjI2)}
qed

text {* ---------------------------------------------------------------
Ejercicio 33. Demostrar
p ∧ (q ∨ r) ⊢ (p ∧ q) ∨ (p ∧ r)
------------------------------------------------------------------ *}

lemma ejercicio_33:
assumes "p ∧ (q ∨ r)"
shows "(p ∧ q) ∨ (p ∧ r)"
proof -
have 1: "p" using assms(1) by (rule conjunct1)
have 2: "q ∨ r" using assms(1) by (rule conjunct2)
moreover
{ assume 3: "q"
have 4: "p ∧ q" using 1 3 by (rule conjI)
have 5: "(p ∧ q) ∨ (p ∧ r)" using 4 by (rule disjI1)}
moreover
{ assume 6: "r"
have 7: "p ∧ r" using 1 6 by (rule conjI)
have 8: "(p ∧ q) ∨ (p ∧ r)" using 7 by (rule disjI2)}
ultimately show "(p ∧ q) ∨ (p ∧ r)" by (rule disjE)
qed

lemma ejercicio_33:
assumes 1:"p ∧ (q ∨ r)"
shows "(p ∧ q) ∨ (p ∧ r)"
proof -
have 2:"p" using 1 by (rule conjunct1)
have 3:"q ∨ r" using 1 by (rule conjunct2)
thus "(p ∧ q) ∨ (p ∧ r)"
proof (rule disjE)
{assume 4:"q"
have 5:"p ∧ q" using 2 4 by (rule conjI)
show "(p ∧ q) ∨ (p ∧ r)" using 5 by (rule disjI1)}
next
{assume 6:"r"
have 7:"p ∧ r" using 2 6 by (rule conjI)
show "(p ∧ q) ∨ (p ∧ r)" using 7 by (rule disjI2)}
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 34. Demostrar
(p ∧ q) ∨ (p ∧ r) ⊢ p ∧ (q ∨ r)
------------------------------------------------------------------ *}

lemma ejercicio_34:
assumes "(p ∧ q) ∨ (p ∧ r)"
shows "p ∧ (q ∨ r)"
proof -
have "(p ∧ q) ∨ (p ∧ r)" using assms(1) by this
moreover
{ assume 1: "p ∧ q"
have 2: "q" using 1 by (rule conjunct2)
have 3: "q ∨ r" using 2 by (rule disjI1)
have 4: "p" using 1 by (rule conjunct1)
have 5: "p ∧ (q ∨ r)" using 4 3 by (rule conjI)}
moreover
{ assume 6: "p ∧ r"
have 7: "r" using 6 by (rule conjunct2)
have 8: "q ∨ r" using 7 by (rule disjI2)
have 9: "p" using 6 by (rule conjunct1)
have 10: "p ∧ (q ∨ r)" using 9 8 by (rule conjI)}
ultimately show "p ∧ (q ∨ r)" by (rule disjE)
qed

lemma ejercicio_34:
assumes 1:"(p ∧ q) ∨ (p ∧ r)"
shows "p ∧ (q ∨ r)"
using 1
proof (rule disjE)
{assume 2:"p ∧ q"
have 3:"p" using 2 by (rule conjunct1)
have 4:"q" using 2 by (rule conjunct2)
have 5:"q ∨ r" using 4 by (rule disjI1)
show "p ∧ (q ∨ r)" using 3 5 by (rule conjI)}
next
{assume 6:"p ∧ r"
have 7:"p" using 6 by (rule conjunct1)
have 8:"r" using 6 by (rule conjunct2)
have 9:"q ∨ r" using 8 by (rule disjI2)
show "p ∧ (q ∨ r)" using 7 9 by (rule conjI)}
qed

text {* ---------------------------------------------------------------
Ejercicio 35. Demostrar
p ∨ (q ∧ r) ⊢ (p ∨ q) ∧ (p ∨ r)
------------------------------------------------------------------ *}

lemma ejercicio_35:
assumes "p ∨ (q ∧ r)"
shows "(p ∨ q) ∧ (p ∨ r)"
proof -
have "p ∨ (q ∧ r)" using assms(1) by this
moreover
{ assume 1: "p"
have 2: "p ∨ q" using 1 by (rule disjI1)
have 3: "p ∨ r" using 1 by (rule disjI1)
have 4: "(p ∨ q) ∧ (p ∨ r)" using 2 3 by (rule conjI)}
moreover
{ assume 5: "q ∧ r"
have 6: "q" using 5 by (rule conjunct1)
have 7: "p ∨ q" using 6 by (rule disjI2)
have 8: "r" using 5 by (rule conjunct2)
have 9: "p ∨ r" using 8 by (rule disjI2)
have 10: "(p ∨ q) ∧ (p ∨ r)" using 7 9 by (rule conjI)}
ultimately show "(p ∨ q) ∧ (p ∨ r)" by (rule disjE)
qed

lemma ejercicio_35:
assumes 1:"p ∨ (q ∧ r)"
shows "(p ∨ q) ∧ (p ∨ r)"
using 1
proof (rule disjE)
{assume 2:"p"
have 3:"p ∨ q" using 2 by (rule disjI1)
have 4:"p ∨ r" using 2 by (rule disjI1)
show "(p ∨ q) ∧ (p ∨ r)" using 3 4 by (rule conjI)}
next
{assume 5:"q ∧ r"
have 6:"q" using 5 by (rule conjunct1)
have 7:"r" using 5 by (rule conjunct2)
have 8:"p ∨ q" using 6 by (rule disjI2)
have 9:"p ∨ r" using 7 by (rule disjI2)
show "(p ∨ q) ∧ (p ∨ r)" using 8 9 by (rule conjI)}
qed

text {* ---------------------------------------------------------------
Ejercicio 36. Demostrar
(p ∨ q) ∧ (p ∨ r) ⊢ p ∨ (q ∧ r)
------------------------------------------------------------------ *}

lemma ejercicio_36:
assumes "(p ∨ q) ∧ (p ∨ r)"
shows "p ∨ (q ∧ r)"
proof-
have "(p ∨ q)" using assms by (rule conjunct1)
moreover
{assume 1:"p"
have "p ∨ (q ∧ r)" using 1 by (rule disjI1) }
moreover
{assume 1:"q"
have "(p ∨ r)" using assms by (rule conjunct2)
moreover
{assume 2:"p"
have "p ∨ (q ∧ r)" using 2 by (rule disjI1)}
moreover
{assume 2:"r"
have 3:"(q ∧ r)" using 1 2 by (rule conjI)
have "p ∨ (q ∧ r)" using 3 by (rule disjI2)}
ultimately have "p ∨ (q ∧ r)" by (rule disjE)}
ultimately show "p ∨ (q ∧ r)" by (rule disjE)
qed

------------------- ----------

proof-
have 1: "p∨q" using assms by (rule conjunct1)
thus "p∨(q∧r)"
proof(rule disjE)
{assume 3: "p"
show "p∨(q∧r)" using 3 by ( rule disjI1)}
next
{assume 4: "q"
have 5:"p∨r" using assms by (rule conjunct2)
thus "p∨(q∧r)"
proof(rule disjE)
{assume 6: "p"
show "p∨(q∧r)" using 6 by (rule disjI1)}
next
{assume 7: "r"
have 8: "q∧r" using 4 7 by (rule conjI)
show "p∨(q∧r)" using 8 by (rule disjI2)}
qed}
qed
qed
text {* ---------------------------------------------------------------
Ejercicio 37. Demostrar
(p ⟶ r) ∧ (q ⟶ r) ⊢ p ∨ q ⟶ r
------------------------------------------------------------------ *}

lemma ejercicio_37:
assumes "(p ⟶ r) ∧ (q ⟶ r)"
shows "p ∨ q ⟶ r"
proof-
{assume "p ∨ q"
moreover
{assume 1:"p"
have 2:"(p ⟶ r)" using assms by (rule conjunct1)
have "r" using 2 1 by (rule mp)}
moreover
{assume 1:"q"
have 2:"(q ⟶ r)" using assms by (rule conjunct2)
have "r" using 2 1 by (rule mp)}
ultimately have "r" by (rule disjE)}
thus "p ∨ q ⟶ r" by (rule impI)
qed

lemma ejercicio_37:
assumes 1:"(p ⟶ r) ∧ (q ⟶ r)"
shows "p ∨ q ⟶ r"
proof (rule impI)
assume 2:"p ∨ q"
thus "r"
proof (rule disjE)
{assume 3:"p"
have 4:"p ⟶ r" using 1 by (rule conjunct1)
show "r" using 4 3 by (rule mp)}
next
{assume 5:"q"
have 6:"q ⟶ r" using 1 by (rule conjunct2)
show "r" using 6 5 by (rule mp)}
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 38. Demostrar
p ∨ q ⟶ r ⊢ (p ⟶ r) ∧ (q ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_38:
assumes "p ∨ q ⟶ r"
shows "(p ⟶ r) ∧ (q ⟶ r)"
proof-
{assume 1 : "p"
have 2 : "p ∨ q" using 1 by (rule disjI1)
have 3 : "r" using assms 2 by (rule mp)}
hence 4 : "p⟶r" by (rule impI)
{assume 5 : "q"
have 6 : "p ∨ q" using 5 by (rule disjI2)
have 7 : "r" using assms 6 by (rule mp)}
hence 8 : "q⟶r" by (rule impI)
show "(p ⟶ r) ∧ (q ⟶ r)" using 4 8 by (rule conjI)
qed

section {* Negaciones *}

text {* ---------------------------------------------------------------
Ejercicio 39. Demostrar
p ⊢ ¬¬p
------------------------------------------------------------------ *}

lemma ejercicio_39:
assumes "p"
shows "¬¬p"
proof-
show "¬¬p" using assms by (rule notnotI)
qed

text {* ---------------------------------------------------------------
Ejercicio 40. Demostrar
¬p ⊢ p ⟶ q
------------------------------------------------------------------ *}

lemma ejercicio_40:
assumes "¬p"
shows "p ⟶ q"

proof-
{assume 1: "p"
have "q" using assms 1 ..}
thus "p⟶q" ..
qed
-------------------------------
lemma ejercicio_40:
assumes "¬p"
shows "p ⟶ q"
proof-
{assume 1 : "p"
have 2 : "False" using assms 1 by (rule notE)
have 3 : "q" using 2 by (rule FalseE)}
thus "p⟶q" by (rule impI)
qed

------------------------------
lemma ejercicio_40:
assumes "¬p"
shows "p ⟶ q"

proof-
{assume 1: "p"
have 2: "q" using assms 1 by (rule notE)}
thus "p ⟶ q" by ( rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 41. Demostrar
p ⟶ q ⊢ ¬q ⟶ ¬p
------------------------------------------------------------------ *}

lemma ejercicio_41:
assumes "p ⟶ q"
shows "¬q ⟶ ¬p"
proof-
{assume 1:"¬q"
have "¬p" using assms 1 by (rule mt)}
thus "¬q ⟶ ¬p" ..
qed

text {* ---------------------------------------------------------------
Ejercicio 42. Demostrar
p∨q, ¬q ⊢ p
------------------------------------------------------------------ *}

lemma ejercicio_42:
assumes "p∨q"
"¬q"
shows "p"

using assms(1)
proof
assume "p"
thus "p" .
next
assume 1:"q"
show "p" using assms(2) 1 ..
qed

lemma ejercicio_42:
assumes "p∨q"
"¬q"
shows "p"
proof-
have "p ∨ q" using assms(1) by this
moreover
{assume 1 : "p"
have 2 : "p" using 1 .}
moreover
{assume 3 : "q"
have 4 : "¬q" using assms(2) .
have 5 : "False" using 4 3 by (rule notE)
have 6 : "p" using 5 by (rule FalseE)}
ultimately show "p" by (rule disjE)
qed

--------**---
using assms(1)
proof(rule disjE)
{assume 1: "p"
have "p" using 1 by this}
next
{assume 2: "q"
show "p" using assms(2) 2 by (rule notE)}
qed

text {* ---------------------------------------------------------------
Ejercicio 43. Demostrar
p ∨ q, ¬p ⊢ q
------------------------------------------------------------------ *}

lemma ejercicio_43:
assumes "p ∨ q"
"¬p"
shows "q"

using assms(1)
proof
assume "q"
thus "q" .
next
assume 1:"p"
show "q" using assms(2) 1 ..
qed

lemma ejercicio_43:
assumes "p ∨ q"
"¬p"
shows "q"
proof-
have "p ∨ q" using assms(1) by this
moreover
{assume 1 : "p"
have 2 : "¬p" using assms(2) .
have 3 : "False" using 2 1 by (rule notE)
have 4 : "q" using 3 by (rule FalseE)}
moreover
{assume 5 : "q"
have 6 : "q" using 5 .}
ultimately show "q" by (rule disjE)
qed

---**--
using assms(1)
proof(rule disjE)
assume 1: "p"
show "q" using assms(2) 1 by (rule notE)
next
assume 2:"p"
have "p" using 2 by this
qed

text {* ---------------------------------------------------------------
Ejercicio 44. Demostrar
p ∨ q ⊢ ¬(¬p ∧ ¬q)
------------------------------------------------------------------ *}

lemma ejercicio_44:
assumes "p ∨ q"
shows "¬(¬p ∧ ¬q)"
proof-
{assume 1 : "¬p ∧ ¬q"
have "p ∨ q" using assms by this
moreover
{assume 2 : "p"
have 3 : "¬p" using 1 by (rule conjunct1)
have 4 : "False" using 3 2 by (rule notE)}
moreover
{assume 5 : "q"
have 6 : "¬q" using 1 by (rule conjunct2)
have 7 : "False" using 6 5 by (rule notE)}
ultimately have 8 : "False" by (rule disjE)}
thus "¬(¬p ∧ ¬q)" by (rule notI)
qed

---**
proof(rule notI)
assume 2 :"¬p∧¬q"
show False
using assms
proof(rule disjE)
{assume 1: "p"
have 3: "¬p" using 2 by (rule conjunct1)
show False using 3 1 by (rule notE)}
{assume 4: "q"
have 5: "¬q" using 2 by (rule conjunct2)
show False using 5 4 by (rule notE)}
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 45. Demostrar
p ∧ q ⊢ ¬(¬p ∨ ¬q)
------------------------------------------------------------------ *}

lemma ejercicio_45:
assumes "p ∧ q"
shows "¬(¬p ∨ ¬q)"
proof-
{assume 1 : "¬p ∨ ¬q"
have "¬p ∨ ¬q" using 1 by this
moreover
{assume 2 : "¬p"
have 3 : "p" using assms by (rule conjunct1)
have 4 : "False" using 2 3 by (rule notE)}
moreover
{assume 5 : "¬q"
have 6 : "q" using assms by (rule conjunct2)
have 7 : "False" using 5 6 by (rule notE)}
ultimately have "False" by (rule disjE)}
thus "¬(¬p ∨ ¬q)" by (rule notI)
qed

---**

assumes "p ∧ q"
shows "¬(¬p ∨ ¬q)"

proof (rule notI)
assume 1: "¬p∨¬q"
show False
using 1
proof(rule disjE)
assume 2: "¬p"
have 3: "p" using assms by (rule conjunct1)
show False using 2 3 by (rule notE)
next
assume 4: "¬q"
have 5: "q" using assms by (rule conjunct2)
show False using 4 5 by (rule notE)
qed
qed
text {* ---------------------------------------------------------------
Ejercicio 46. Demostrar
¬(p ∨ q) ⊢ ¬p ∧ ¬q
------------------------------------------------------------------ *}

lemma ejercicio_46:
assumes "¬(p ∨ q)"
shows "¬p ∧ ¬q"
proof-
{assume 1 : "p"
have 2 : "p ∨ q" using 1 by (rule disjI1)
have 3 : "False" using assms 2 by (rule notE)}
hence 4 : "¬p" by (rule notI)
{assume 5 : "q"
have 6 : "p ∨ q" using 5 by (rule disjI2)
have 7 : "False" using assms 6 by (rule notE)}
hence 8 : "¬q" by (rule notI)
show "¬p ∧ ¬q" using 4 8 by (rule conjI)
qed

(*Otra opción para usar tercio excluso*)

lemma ejercicio_46_2:
assumes "¬(p ∨ q)"
shows "¬p ∧ ¬q"
proof -
have 1:"¬p | p" by (rule excluded_middle)
show "¬p & ¬q"
using 1
proof (rule disjE)
{assume 2:"¬p"
have 3:"¬q | q" by (rule excluded_middle)
show "¬p & ¬q"
using 3
proof (rule disjE)
{assume 4:"¬q"
with 2 show "¬p & ¬q" by (rule conjI)}
next
{assume "q"
then have "p | q" by (rule disjI2)
with assms show "¬p & ¬q" by (rule notE)}
qed
}
next
{assume 5:"p"
then have "p | q" by (rule disjI1)
with assms show "¬p & ¬q" by (rule notE)
}
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 47. Demostrar
¬p ∧ ¬q ⊢ ¬(p ∨ q)
------------------------------------------------------------------ *}

lemma ejercicio_47:
assumes "¬p ∧ ¬q"
shows "¬(p ∨ q)"
proof-
{assume 1 : "p ∨ q"
have "p ∨ q" using 1 by this
moreover
{assume 2 : "p"
have 3 : "¬p" using assms by (rule conjunct1)
have 4 : "False" using 3 2 by (rule notE)}
moreover
{assume 5 : "q"
have 6 : "¬q" using assms by (rule conjunct2)
have 6 : "False" using 6 5 by (rule notE)}
ultimately have "False" by (rule disjE)}
thus "¬(p ∨ q)" by (rule notI)
qed

---**

proof(rule notI)
assume 1: "p∨q"
show False
using 1
proof(rule disjE)
{assume 2:"p"
have 3:"¬p" using assms by (rule conjunct1)
show "False" using 3 2 by (rule notE)}
next
{assume 5:"q"
have 6:"¬q" using assms by (rule conjunct2)
show "False" using 6 5 by (rule notE)}
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 48. Demostrar
¬p ∨ ¬q ⊢ ¬(p ∧ q)
------------------------------------------------------------------ *}

lemma ejercicio_48:
assumes "¬p ∨ ¬q"
shows "¬(p ∧ q)"
proof-
{assume 1 : "p ∧ q"
have "¬p ∨ ¬q" using assms by this
moreover
{assume 2 : "¬p"
have 3 : "p" using 1 by (rule conjunct1)
have 4 : "False" using 2 3 by (rule notE)}
moreover
{assume 5 : "¬q"
have 6 : "q" using 1 by (rule conjunct2)
have 7 : "False" using 5 6 by (rule notE)}
ultimately have "False" by (rule disjE)}
thus "¬(p ∧ q)" by (rule notI)
qed

--**
lemma ejercicio_48:
assumes "¬p ∨ ¬q"
shows "¬(p ∧ q)"

proof(rule notI)
assume 1:"p∧q"
show False
using assms
proof(rule disjE)
assume 2:"¬p"
have 3: "p" using 1 by (rule conjunct1)
show False using 2 3 by (rule notE)
next
assume 4: "¬q"
have 5: "q" using 1 by (rule conjunct2)
show False using 4 5 by (rule notE)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 49. Demostrar
⊢ ¬(p ∧ ¬p)
------------------------------------------------------------------ *}

lemma ejercicio_49:
"¬(p ∧ ¬p)"
proof-
{assume 1 : "p ∧ ¬p"
have 2 : "p" using 1 by (rule conjunct1)
have 3 : "¬p" using 1 by (rule conjunct2)
have 4 : "False" using 3 2 by (rule notE)}
thus "¬(p ∧ ¬p)" by (rule notI)
qed

---**

proof(rule notI)
assume 1: "p∧¬p"
have 2: "p" using 1 by (rule conjunct1)
have 3: "¬p" using 1 by (rule conjunct2)
show False using 3 2 by (rule notE)
qed

text {* ---------------------------------------------------------------
Ejercicio 50. Demostrar
p ∧ ¬p ⊢ q
------------------------------------------------------------------ *}

lemma ejercicio_50:
assumes "p ∧ ¬p"
shows "q"

proof-
have 1 : "p" using assms by (rule conjunct1)
have 2 : "¬p" using assms by (rule conjunct2)
show "q" using 2 1 by (rule notE)
qed

text {* ---------------------------------------------------------------
Ejercicio 51. Demostrar
¬¬p ⊢ p
------------------------------------------------------------------ *}

lemma ejercicio_51:
assumes "¬¬p"
shows "p"

proof-
show "p" using assms by (rule notnotD)
qed

--**

lemma ejercicio_51:
assumes "¬¬p"
shows "p"
proof(rule ccontr)
assume 1: "¬p"
show False using assms 1 by (rule notE)
qed

text {* ---------------------------------------------------------------
Ejercicio 52. Demostrar
⊢ p ∨ ¬p
------------------------------------------------------------------ *}

lemma ejercicio_52:
"p ∨ ¬p"

proof-
have 1: "¬p∨p" by (rule excluded_middle)
have 2: "¬p∨p" using 1 by this
moreover
{assume 3: "¬p"
have 4: "p∨¬p" using 3 by (rule disjI2)}
moreover
{assume 5: "p"
have 6: "p∨¬p" using 5 by (rule disjI1)}
ultimately show "p∨¬p" by (rule disjE)
qed

text {* ---------------------------------------------------------------
Ejercicio 53. Demostrar
⊢ ((p ⟶ q) ⟶ p) ⟶ p
------------------------------------------------------------------ *}

lemma ejercicio_53:
"((p ⟶ q) ⟶ p) ⟶ p"
proof-
{assume 1 : "(p ⟶ q) ⟶ p"
{assume 2 : "¬p"
have 3 : "¬(p⟶q)" using 1 2 by (rule mt)
{assume 4 : "p"
have 5 : "q" using 2 4 by (rule notE)}
hence 6 : "p⟶q" by (rule impI)
have 7 : "False" using 3 6 by (rule notE)}
hence 8 : "p" by (rule ccontr)}
thus "((p ⟶ q) ⟶ p) ⟶ p" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 54. Demostrar
¬q ⟶ ¬p ⊢ p ⟶ q
------------------------------------------------------------------ *}

lemma ejercicio_54:
assumes "¬q ⟶ ¬p"
shows "p ⟶ q"
proof-
{assume 1 : "p"
{assume 2 : "¬q"
have 3 : "¬p" using assms 2 by (rule mp)
have 4 : "False" using 3 1 by (rule notE)}
hence 5 : "q" by (rule ccontr)}
thus "p⟶q" by (rule impI)
qed

---**

proof(rule impI)
assume 1: "p"
have 2 : "¬¬p" using 1 by (rule notnotI)
have 3: "¬¬q" using assms 2 by (rule mt)
show "q" using 3 by (rule notnotD)
qed

text {* ---------------------------------------------------------------
Ejercicio 55. Demostrar
¬(¬p ∧ ¬q) ⊢ p ∨ q
------------------------------------------------------------------ *}

lemma ejercicio_55:
assumes "¬(¬p ∧ ¬q)"
shows "p ∨ q"
proof-
have 1 : "¬p ∨ p" by (rule excluded_middle)
have "¬p ∨ p" using 1 by this
moreover
{assume 2 : "¬p"
have 3 : "¬q ∨ q" by (rule excluded_middle)
have "¬q ∨ q" using 3 by this
moreover
{assume 4 : "¬q"
have 5 : "¬p ∧ ¬q" using 2 4 by (rule conjI)
have 6 : "False" using assms 5 by (rule notE)
have 7 : "p ∨ q" using 6 by (rule FalseE)}
moreover
{assume 8 : "q"
have 9 : "p ∨ q" using 8 by (rule disjI2)}
ultimately have "p ∨ q" by (rule disjE)}
moreover
{assume 10 : "p"
have 11 : "p ∨ q" using 10 by (rule disjI1)}
ultimately show "p ∨ q" by (rule disjE)
qed

(*Otra forma*)

lemma ejercicio_55:
assumes "¬(¬p ∧ ¬q)"
shows "p ∨ q"
proof -
have 1:"¬p | p" by (rule excluded_middle)
show "p | q" using 1
proof (rule disjE)
{assume 2:"¬p"
have 3:"¬q | q" by (rule excluded_middle)
show "p | q" using 3
proof (rule disjE)
{assume "q"
thus "p | q" by (rule disjI2)}
next
{assume "¬q"
with 2 have 4:"¬p & ¬q" by (rule conjI)
show "p | q" using assms 4 by (rule notE)}
qed}
next
{assume "p"
thus "p|q" by (rule disjI1)}
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 56. Demostrar
¬(¬p ∨ ¬q) ⊢ p ∧ q
------------------------------------------------------------------ *}

lemma ejercicio_56:
assumes "¬(¬p ∨ ¬q)"
shows "p ∧ q"
proof-
{assume 1 : "¬p"
have 2 : "¬p ∨ ¬q" using 1 by (rule disjI1)
have 3 : "False" using assms 2 by (rule notE)}
hence 4 : "p" by (rule ccontr)
{assume 5 : "¬q"
have 6 : "¬p ∨ ¬q" using 5 by (rule disjI2)
have 7 : "False" using assms 6 by (rule notE)}
hence 8 : "q" by (rule ccontr)
show "p ∧ q" using 4 8 by (rule conjI)
qed

--**
proof-
{assume 1: "¬p"
have 2: "¬p∨¬q"using 1 by (rule disjI1)
have 3: "False" using assms 2 by (rule notE)}
hence 8: "p" by (rule ccontr)
{assume 4: "¬q"
have 5: "¬p∨¬q" using 4 by (rule disjI2)
have 6: "False" using assms 5 by (rule notE)}
hence 9: "q" by (rule ccontr)
show "p∧q" using 8 9 by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 57. Demostrar
¬(p ∧ q) ⊢ ¬p ∨ ¬q
------------------------------------------------------------------ *}

lemma ejercicio_57:
assumes "¬(p ∧ q)"
shows "¬p ∨ ¬q"
proof-
have 1 : "¬p ∨ p" by (rule excluded_middle)
have "¬p ∨ p" using 1 by this
moreover
{assume 2 : "¬p"
have 3 : "¬p ∨ ¬q" using 2 by (rule disjI1)}
moreover
{assume 4 : "p"
have 5 : "¬q ∨ q" by (rule excluded_middle)
have "¬q ∨ q" using 5 by this
moreover
{assume 6 : "¬q"
have 7 : "¬p ∨ ¬q" using 6 by (rule disjI2)}
moreover
{assume 8 : "q"
have 9 : "p ∧ q" using 4 8 by (rule conjI)
have 10 : "False" using assms 9 by (rule notE)
have 11 : "¬p ∨ ¬q" using 10 by (rule FalseE)}
ultimately have "¬p ∨ ¬q" by (rule disjE)}
ultimately show "¬p ∨ ¬q" by (rule disjE)
qed

text {* ---------------------------------------------------------------
Ejercicio 58. Demostrar
⊢ (p ⟶ q) ∨ (q ⟶ p)
------------------------------------------------------------------ *}

lemma ejercicio_58:
"(p ⟶ q) ∨ (q ⟶ p)"

proof-
have 1 : "¬(p ⟶ q) ∨ (p ⟶ q)" by (rule excluded_middle)
have "¬(p ⟶ q) ∨ (p ⟶ q)" using 1 by this
moreover
{assume 2 : "¬(p ⟶ q)"
{assume 3 : "q"
{assume 4 : "p"
have 5 : "q" using 3 .}
hence 6 : "p⟶q" by (rule impI)
have 7 : "p" using 2 6 by (rule notE)}
hence 8 : "q⟶p" by (rule impI)
have 9 : "(p ⟶ q) ∨ (q ⟶ p)" using 8 by (rule disjI2)}
moreover
{assume 10 : "p ⟶ q"
have 11 : "(p ⟶ q) ∨ (q ⟶ p)" using 10 by (rule disjI1)}
ultimately show "(p ⟶ q) ∨ (q ⟶ p)" by (rule disjE)
qed

end
</source>

Relación 3

2016-03-25T16:48:11Z

Albromgon:

<source lang = "isar">header {* R3: Deducción natural proposicional *}

theory R3
imports Main
begin

text {*
---------------------------------------------------------------------
El objetivo de esta relación es demostrar cada uno de los ejercicios
usando sólo las reglas básicas de deducción natural de la lógica
proposicional (sin usar el método auto).

Las reglas básicas de la deducción natural son las siguientes:
· conjI: ⟦P; Q⟧ ⟹ P ∧ Q
· conjunct1: P ∧ Q ⟹ P
· conjunct2: P ∧ Q ⟹ Q
· notnotD: ¬¬ P ⟹ P
· notnotI: P ⟹ ¬¬ P
· mp: ⟦P ⟶ Q; P⟧ ⟹ Q
· mt: ⟦F ⟶ G; ¬G⟧ ⟹ ¬F
· impI: (P ⟹ Q) ⟹ P ⟶ Q
· disjI1: P ⟹ P ∨ Q
· disjI2: Q ⟹ P ∨ Q
· disjE: ⟦P ∨ Q; P ⟹ R; Q ⟹ R⟧ ⟹ R
· FalseE: False ⟹ P
· notE: ⟦¬P; P⟧ ⟹ R
· notI: (P ⟹ False) ⟹ ¬P
· iffI: ⟦P ⟹ Q; Q ⟹ P⟧ ⟹ P = Q
· iffD1: ⟦Q = P; Q⟧ ⟹ P
· iffD2: ⟦P = Q; Q⟧ ⟹ P
· ccontr: (¬P ⟹ False) ⟹ P
---------------------------------------------------------------------
*}

text {*
Se usarán las reglas notnotI y mt que demostramos a continuación. *}

lemma notnotI: "P ⟹ ¬¬ P"
by auto

lemma mt: "⟦F ⟶ G; ¬G⟧ ⟹ ¬F"
by auto

section {* Implicaciones *}

text {* ---------------------------------------------------------------
Ejercicio 1. Demostrar
p ⟶ q, p ⊢ q
------------------------------------------------------------------ *}

lemma ejercicio_1:
assumes "p ⟶ q"
"p"
shows "q"

proof -
show "q" using assms(1,2) by (rule mp)
qed

lemma ejercicio_1_b:
assumes "p ⟶ q"
"p"
shows "q"
proof-
show "q" using assms(1) assms(2) by (rule mp)
qed

text {* ---------------------------------------------------------------
Ejercicio 2. Demostrar
p ⟶ q, q ⟶ r, p ⊢ r
------------------------------------------------------------------ *}

lemma ejercicio_2:
assumes "p ⟶ q"
"q ⟶ r"
"p"
shows "r"

proof-
have 1: "q" using assms(1,3) by (rule mp)
show "r" using assms(2) 1 by (rule mp)
qed

text {* ---------------------------------------------------------------
Ejercicio 3. Demostrar
p ⟶ (q ⟶ r), p ⟶ q, p ⊢ r
------------------------------------------------------------------ *}

lemma ejercicio_3:
assumes "p ⟶ (q ⟶ r)"
"p ⟶ q"
"p"
shows "r"

proof-
have 1: "q" using assms(2,3) by (rule mp)
have 2: "q⟶r" using assms(1,3) by (rule mp)
show "r" using 2 1 by (rule mp)
qed

text {* ---------------------------------------------------------------
Ejercicio 4. Demostrar
p ⟶ q, q ⟶ r ⊢ p ⟶ r
------------------------------------------------------------------ *}

lemma ejercicio_4:
assumes "p ⟶ q"
"q ⟶ r"
shows "p ⟶ r"

proof-
{assume 1: "p"
have 2: "q" using assms(1) 1 ..
have 3: "r" using assms(2) 2 ..}
thus "p⟶r" by (rule impI)
qed

lemma ejercicio_4:
assumes "p ⟶ q"
"q ⟶ r"
shows "p ⟶ r"
proof-
{assume 1 : "p"
have 2 : "q" using assms(1) 1 by (rule mp)
have 3 : "r" using assms(2) 2 by (rule mp)}
thus "p⟶r" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 5. Demostrar
p ⟶ (q ⟶ r) ⊢ q ⟶ (p ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_5:
assumes "p ⟶ (q ⟶ r)"
shows "q ⟶ (p ⟶ r)"

{assume 1: "q"
{assume 2: "p"
have 3: "q⟶r" using assms 2 by (rule mp)
have "r" using 3 1 by (rule mp)}
hence "p⟶r" by (rule impI)}
thus "(q⟶(p⟶r))" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 6. Demostrar
p ⟶ (q ⟶ r) ⊢ (p ⟶ q) ⟶ (p ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_6:
assumes "p ⟶ (q ⟶ r)"
shows "(p ⟶ q) ⟶ (p ⟶ r)"

proof-
{assume 1:"p⟶q"
{assume 2: "p"
have 3: "q⟶r" using assms 2 by (rule mp)
have 4: "q" using 1 2 by (rule mp)
have "r" using 3 4 by (rule mp)}
hence "p⟶r" by (rule impI)}
thus "(p⟶q)⟶(p⟶r)" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 7. Demostrar
p ⊢ q ⟶ p
------------------------------------------------------------------ *}

lemma ejercicio_7:
assumes "p"
shows "q ⟶ p"

proof-
{assume 1: "q"
have "p" using assms .}
thus "q⟶p" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 8. Demostrar
⊢ p ⟶ (q ⟶ p)
------------------------------------------------------------------ *}

lemma ejercicio_8:
"p ⟶ (q ⟶ p)"

proof-
{assume 1:"p"
{assume 2:"q"
have 3: "p" using 1 .}
hence "q⟶p" ..}
thus "p⟶(q⟶p)" ..
qed

lemma ejercicio_8:
"p ⟶ (q ⟶ p)"
proof-
{assume 1 : "p"
{assume 2 : "q"
have 3 : "p" using 1 .}
hence "q⟶p" by (rule impI)}
thus "p⟶(q⟶p)" by (rule impI)
qed

lemma ejercicio_8:
"p ⟶ (q ⟶ p)"
proof (rule impI)
assume 1: "p"
show "q ⟶ p"
proof (rule impI)
assume "q"
show "p" using 1 by this
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 9. Demostrar
p ⟶ q ⊢ (q ⟶ r) ⟶ (p ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_9:
assumes "p ⟶ q"
shows "(q ⟶ r) ⟶ (p ⟶ r)"

proof-
{assume 1:"q⟶r"
{assume 2:"p"
have 3:"q" using assms 2 ..
have "r" using 1 3 ..}
hence "p⟶r" ..}
thus "(q⟶r)⟶(p⟶r) " ..
qed

lemma ejercicio_9:
assumes "p ⟶ q"
shows "(q ⟶ r) ⟶ (p ⟶ r)"
proof-
{assume 1 : "q⟶r"
{assume 2 : "p"
have 3 : "q" using assms(1) 2 by (rule mp)
have 4 : "r" using 1 3 by (rule mp)}
hence "p⟶r" by (rule impI)}
thus "(q⟶r)⟶(p⟶r)" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 10. Demostrar
p ⟶ (q ⟶ (r ⟶ s)) ⊢ r ⟶ (q ⟶ (p ⟶ s))
------------------------------------------------------------------ *}

lemma ejercicio_10:
assumes "p ⟶ (q ⟶ (r ⟶ s))"
shows "r ⟶ (q ⟶ (p ⟶ s))"

proof-
{assume 1:"r"
{assume 2:"q"
{assume 3:"p"
have 4:"q⟶(r⟶s)" using assms 3 ..
have 5: "r⟶s" using 4 2 ..
have 6: "s" using 5 1 ..}
hence "p⟶s" ..}
hence "q⟶(p⟶s)" ..}
thus "r⟶(q⟶(p⟶s))" ..
qed

lemma ejercicio_10:
assumes "p ⟶ (q ⟶ (r ⟶ s))"
shows "r ⟶ (q ⟶ (p ⟶ s))"
proof-
{assume 1 : "r"
{assume 2 : "q"
{assume 3 : "p"
have 4 : "q ⟶ (r ⟶ s)" using assms(1) 3 by (rule mp)
have 5 : "r⟶s" using 4 2 by (rule mp)
have 6 : "s" using 5 1 by (rule mp)}
hence "p⟶s" by (rule impI)}
hence "q⟶(p⟶s)" by (rule impI)}
thus "r⟶(q⟶(p⟶s))" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 11. Demostrar
⊢ (p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))
------------------------------------------------------------------ *}

lemma ejercicio_11:
"(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))"

proof (rule impI)
assume 3: "p⟶(q⟶r)"
show "(p⟶q)⟶(p⟶r)"
proof(rule impI)
assume 2:"p⟶q"
show "p⟶r"
proof (rule impI)
assume 1: "p"
have 4:"q" using 2 1 ..
have 5:"q⟶r" using 3 1 ..
show "r" using 5 4 ..
qed
qed
qed

lemma ejercicio_11:
"(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))"
proof-
{assume 1 : "p ⟶ (q ⟶ r)"
{assume 2 : "p⟶q"
{assume 3 : "p"
have 4 : "q⟶r" using 1 3 by (rule mp)
have 5 : "q" using 2 3 by (rule mp)
have 6 : "r" using 4 5 by (rule mp)}
hence "p⟶r" by (rule impI)}
hence "(p ⟶ q) ⟶ (p ⟶ r)" by (rule impI)}
thus "(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 12. Demostrar
(p ⟶ q) ⟶ r ⊢ p ⟶ (q ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_12:
assumes "(p ⟶ q) ⟶ r"
shows "p ⟶ (q ⟶ r)"

proof-
{assume 1:"p"
{assume 3:"q"
have 4:"p" using 1 .
have 5:"q" using 3 .
then have 6:"p⟶q" ..
have "r" using assms 6 ..}
hence "q⟶r" ..}
thus "p⟶(q⟶r)" ..
qed

lemma ejercicio_12:
assumes "(p ⟶ q) ⟶ r"
shows "p ⟶ (q ⟶ r)"
proof-
{assume 1 : "p"
{assume 2 : "q"
{assume 3 : "p"
have 4 : "q" using 2 .}
hence 5 : "p⟶q" by (rule impI)
have 6 : "r" using assms(1) 5 by (rule mp)}
hence "q⟶r" by (rule impI)}
thus "p⟶(q⟶r)" by (rule impI)
qed

section {* Conjunciones *}

text {* ---------------------------------------------------------------
Ejercicio 13. Demostrar
p, q ⊢ p ∧ q
------------------------------------------------------------------ *}

lemma ejercicio_13:
assumes "p"
"q"
shows "p ∧ q"

proof-
show "p∧q" using assms(1) assms(2) by (rule conjI)
qed

lemma ejercicio_13:
assumes "p"
"q"
shows "p ∧ q"
proof-
show "p ∧ q" using assms(1,2) by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 14. Demostrar
p ∧ q ⊢ p
------------------------------------------------------------------ *}

lemma ejercicio_14:
assumes "p ∧ q"
shows "p"

proof-
show "p" using assms by (rule conjunct1)
qed

text {* ---------------------------------------------------------------
Ejercicio 15. Demostrar
p ∧ q ⊢ q
------------------------------------------------------------------ *}

lemma ejercicio_15:
assumes "p ∧ q"
shows "q"

proof-
show "q" using assms by (rule conjunct2)
qed

text {* ---------------------------------------------------------------
Ejercicio 16. Demostrar
p ∧ (q ∧ r) ⊢ (p ∧ q) ∧ r
------------------------------------------------------------------ *}

lemma ejercicio_16:
assumes "p ∧ (q ∧ r)"
shows "(p ∧ q) ∧ r"

proof-
have 1:"p" using assms by (rule conjunct1)
have 2:"q∧r" using assms by (rule conjunct2)
then have 3:"q" by (rule conjunct1)
have 4:"r" using 2 by (rule conjunct2)
have "p∧q" using 1 3 by (rule conjI)
then show "(p∧q)∧r" using 4 by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 17. Demostrar
(p ∧ q) ∧ r ⊢ p ∧ (q ∧ r)
------------------------------------------------------------------ *}

lemma ejercicio_17:
assumes "(p ∧ q) ∧ r"
shows "p ∧ (q ∧ r)"

proof-
have 1:"p∧q" using assms by (rule conjunct1)
have 2:"r" using assms by (rule conjunct2)
have 3:"p" using 1 by (rule conjunct1)
have 4:"q" using 1 by (rule conjunct2)
have 5:"q∧r" using 4 2 by (rule conjI)
show "p∧(q∧r)" using 3 5 by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 18. Demostrar
p ∧ q ⊢ p ⟶ q
------------------------------------------------------------------ *}

lemma ejercicio_18:
assumes "p ∧ q"
shows "p ⟶ q"
proof-
have 1:"p" using assms ..
have 2:"q" using assms ..
then show "p⟶q" ..
qed

lemma ejercicio_18:
assumes "p ∧ q"
shows "p ⟶ q"
proof-
{assume "p"
have "q" using assms by (rule conjunct2)}
thus "p⟶q" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 19. Demostrar
(p ⟶ q) ∧ (p ⟶ r) ⊢ p ⟶ q ∧ r
------------------------------------------------------------------ *}

lemma ejercicio_19:
assumes "(p ⟶ q) ∧ (p ⟶ r)"
shows "p ⟶ q ∧ r"
proof-
have 1:"p⟶q" using assms ..
have 2:"p⟶r" using assms ..
{assume 3:"p"
have 4:"q" using 1 3 ..
have 5:"r" using 2 3 ..
have 6:"q∧r" using 4 5 ..}
thus "p ⟶ q ∧ r" ..
qed

lemma ejercicio_19:
assumes "(p ⟶ q) ∧ (p ⟶ r)"
shows "p ⟶ q ∧ r"
proof-
{assume 1 : "p"
have 2 : "p⟶q" using assms by (rule conjunct1)
have 3 : "p⟶r" using assms by (rule conjunct2)
have 4 : "q" using 2 1 by (rule mp)
have 5 : "r" using 3 1 by (rule mp)
have 6 : "q ∧ r" using 4 5 by (rule conjI)}
thus "p⟶(q ∧ r)" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 20. Demostrar
p ⟶ q ∧ r ⊢ (p ⟶ q) ∧ (p ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_20:
assumes "p ⟶ q ∧ r"
shows "(p ⟶ q) ∧ (p ⟶ r)"

proof-
{assume 1: p
have 2:"q∧r" using assms 1 ..
have "r" using 2 ..}
hence 1:"p⟶r" ..
{assume 1: p
have 2:"q∧r" using assms 1 ..
have "q" using 2 ..}
hence 2:"p⟶q" ..
then show "(p ⟶ q) ∧ (p ⟶ r)" using 1 ..
qed

lemma ejercicio_20:
assumes "p ⟶ q ∧ r"
shows "(p ⟶ q) ∧ (p ⟶ r)"
proof-
{assume 1 : "p"
have 2 : "q ∧ r" using assms 1 by (rule mp)
have 3 : "q" using 2 by (rule conjunct1)}
hence 4 : "p⟶q" by (rule impI)
{assume 5 : "p"
have 6 : "q ∧ r" using assms 5 by (rule mp)
have 7 : "r" using 6 by (rule conjunct2)}
hence 8 : "p⟶r" by (rule impI)
show "(p ⟶ q) ∧ (p ⟶ r)" using 4 8 by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 21. Demostrar
p ⟶ (q ⟶ r) ⊢ p ∧ q ⟶ r
------------------------------------------------------------------ *}

lemma ejercicio_21:
assumes "p ⟶ (q ⟶ r)"
shows "p ∧ q ⟶ r"

proof-
{assume 1:"p∧q"
have 2:"p" using 1 ..
have 3:"q⟶r" using assms 2 ..
have 4: "q" using 1 ..
have "r" using 3 4 ..}
thus "p∧q ⟶r" ..
qed

lemma ejercicio_21:
assumes "p ⟶ (q ⟶ r)"
shows "p ∧ q ⟶ r"
proof-
{assume 1 : "p ∧ q"
have 2 : "p" using 1 by (rule conjunct1)
have 3 : "q" using 1 by (rule conjunct2)
have 4 : "q⟶r" using assms 2 by (rule mp)
have 5 : "r" using 4 3 by (rule mp)}
thus "p ∧ q ⟶ r" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 22. Demostrar
p ∧ q ⟶ r ⊢ p ⟶ (q ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_22:
assumes "p ∧ q ⟶ r"
shows "p ⟶ (q ⟶ r)"

proof-
{assume 1:"p"
{assume 2:"q"
have 3:"p∧q" using 1 2 ..
have "r" using assms 3 ..}
hence "q⟶r" ..}
thus "p⟶(q⟶r)" ..
qed

lemma ejercicio_22:
assumes "p ∧ q ⟶ r"
shows "p ⟶ (q ⟶ r)"
proof-
{assume 1 : "p"
{assume 2 : "q"
have 3 : "p ∧ q" using 1 2 by (rule conjI)
have 4 : "r" using assms 3 by (rule mp)}
hence "q⟶r" by (rule impI)}
thus "p⟶(q⟶r)" by (rule impI)
qed

lemma ejercicio_22:
assumes 1:"p ∧ q ⟶ r"
shows "p ⟶ (q ⟶ r)"
proof (rule impI)
assume 2:"p"
show "q ⟶ r"
proof (rule impI)
assume 3:"q"
have 4:"p ∧ q" using 2 3 by (rule conjI)
show "r" using 1 4 by (rule mp)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 23. Demostrar
(p ⟶ q) ⟶ r ⊢ p ∧ q ⟶ r
------------------------------------------------------------------ *}

lemma ejercicio_23:
assumes "(p ⟶ q) ⟶ r"
shows "p ∧ q ⟶ r"

proof-
{assume 1:"p∧q"
have 2:"p" using 1 ..
{assume "p"
have 3: "q" using 1 ..}
hence 4:"p⟶q" ..
have "r" using assms 4 ..}
thus "p∧q ⟶r" ..
qed

lemma ejercicio_23:
assumes "(p ⟶ q) ⟶ r"
shows "p ∧ q ⟶ r"
proof-
{assume 1 : "p ∧ q"
{assume 2 : "p"
have 3 : "q" using 1 by (rule conjunct2)}
hence 4 : "p⟶q" by (rule impI)
have 5 : "r" using assms 4 by (rule mp)}
thus "p ∧ q ⟶ r" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 24. Demostrar
p ∧ (q ⟶ r) ⊢ (p ⟶ q) ⟶ r
------------------------------------------------------------------ *}

lemma ejercicio_24:
assumes "p ∧ (q ⟶ r)"
shows "(p ⟶ q) ⟶ r"

proof-
{assume 1: "p⟶q"
have 2:"p" using assms ..
have 3:"q⟶r" using assms..
have 4:"q" using 1 2 ..
have "r" using 3 4 ..}
thus "(p⟶q)⟶r" ..
qed

lemma ejercicio_24:
assumes "p ∧ (q ⟶ r)"
shows "(p ⟶ q) ⟶ r"
proof-
{assume 1 : "p⟶q"
have 2 : "p" using assms by (rule conjunct1)
have 3 : "q⟶r" using assms by (rule conjunct2)
have 4 : "q" using 1 2 by (rule mp)
have 5 : "r" using 3 4 by (rule mp)}
thus "(p⟶q)⟶r" by (rule impI)
qed

lemma ejercicio_24:
assumes 1:"p ∧ (q ⟶ r)"
shows "(p ⟶ q) ⟶ r"
proof (rule impI)
assume 2:"p ⟶ q"
have 3:"p" using 1 by (rule conjunct1)
have 4:"q" using 2 3 by (rule mp)
have 5:"q ⟶ r" using 1 by (rule conjunct2)
show "r" using 5 4 by (rule mp)
qed

section {* Disyunciones *}

text {* ---------------------------------------------------------------
Ejercicio 25. Demostrar
p ⊢ p ∨ q
------------------------------------------------------------------ *}

lemma ejercicio_25:
assumes "p"
shows "p ∨ q"

proof-
show "p∨q" using assms by (rule disjI1)
qed

text {* ---------------------------------------------------------------
Ejercicio 26. Demostrar
q ⊢ p ∨ q
------------------------------------------------------------------ *}

lemma ejercicio_26:
assumes "q"
shows "p ∨ q"

proof-
show "p∨q" using assms by (rule disjI2)
qed

text {* ---------------------------------------------------------------
Ejercicio 27. Demostrar
p ∨ q ⊢ q ∨ p
------------------------------------------------------------------ *}

lemma ejercicio_27:
assumes "p ∨ q"
shows "q ∨ p"

proof -
have "p ∨ q" using assms by this
moreover
{ assume 2: "p"
have "q ∨ p" using 2 by (rule disjI2) }
moreover
{ assume 3: "q"
have "q ∨ p" using 3 by (rule disjI1) }
ultimately show "q ∨ p" by (rule disjE)
qed

lemma ejercicio_27:
assumes 1:"p ∨ q"
shows "q ∨ p"
using 1
proof (rule disjE)
{assume 2:"p"
show "q ∨ p" using 2 by (rule disjI2)}
next
{assume 4:"q"
show "q ∨ p" using 4 by (rule disjI1)}
qed

text {* ---------------------------------------------------------------
Ejercicio 28. Demostrar
q ⟶ r ⊢ p ∨ q ⟶ p ∨ r
------------------------------------------------------------------ *}

lemma ejercicio_28:
assumes "q ⟶ r"
shows "p ∨ q ⟶ p ∨ r"
proof-
{assume "p ∨ q"
moreover
{ assume 1:"p"
have "p ∨ r" using 1 by (rule disjI1)}
moreover
{ assume 2:"q"
have 3:"r" using assms 2 by (rule mp)
have "p ∨ r" using 3 by (rule disjI2) }
ultimately have " p ∨ r" by (rule disjE)}
thus "p ∨ q ⟶ p ∨ r" by (rule impI)
qed

lemma ejercicio_28:
assumes 1:"q ⟶ r"
shows "p ∨ q ⟶ p ∨ r"
proof (rule impI)
assume 2:"p ∨ q"
thus "p ∨ r"
proof (rule disjE)
{assume 3:"p"
show "p ∨ r" using 3 by (rule disjI1)}
next
{assume 4:"q"
have 5:"r" using 1 4 by (rule mp)
show "p ∨ r" using 5 by (rule disjI2)}
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 29. Demostrar
p ∨ p ⊢ p
------------------------------------------------------------------ *}

lemma ejercicio_29:
assumes "p ∨ p"
shows "p"

proof-
have 1: "p ∨ p" using assms by this
moreover
{assume 1:"p"
have "p" using 1 by this}
moreover
{assume 2:"p"
have "p" using 2 by this}
ultimately show "p" by (rule disjE)

qed
text{* Otra forma:
using assms
proof
assume "p"
thus "p" .
qed}

lemma ejercicio_29:
assumes "p ∨ p"
shows "p"
proof-
have "p ∨ p" using assms by this
moreover
{assume 1 : "p"
have "p" using 1 .}
moreover
{assume 2 : "p"
have "p" using 2 .}
ultimately show "p" by (rule disjE)
qed*}

lemma ejercicio_29:
assumes 1:"p ∨ p"
shows "p"
using 1
proof (rule disjE)
{assume 2:"p"
show "p" using 2 by this}
next
{assume 3:"p"
show "p" using 3 by this}
qed

text {* ---------------------------------------------------------------
Ejercicio 30. Demostrar
p ⊢ p ∨ p
------------------------------------------------------------------ *}

lemma ejercicio_30:
assumes "p"
shows "p ∨ p"
proof-
show "p ∨ p" using assms by (rule disjI1)
qed

text{* Otra forma:
using assms
proof
qed}

text {* ---------------------------------------------------------------
Ejercicio 31. Demostrar
p ∨ (q ∨ r) ⊢ (p ∨ q) ∨ r
------------------------------------------------------------------ *}

lemma ejercicio_31:
assumes "p ∨ (q ∨ r)"
shows "(p ∨ q) ∨ r"
proof -
have "p ∨ (q ∨ r)" using assms(1) by this
moreover
{ assume 1: "p"
have 2: "p ∨ q" using 1 by (rule disjI1)
have 3: "(p ∨ q) ∨ r" using 2 by (rule disjI1)}
moreover
{ assume 4: "q ∨ r"
moreover
{ assume 5: "q"
have 6: "p ∨ q" using 5 by (rule disjI2)
have 7: "(p ∨ q) ∨ r" using 6 by (rule disjI1)}
moreover
{ assume 8: "r"
have 9: "(p ∨ q) ∨ r" using 8 by (rule disjI2)}
ultimately have "(p ∨ q) ∨ r" by (rule disjE)}
ultimately show "(p ∨ q) ∨ r" by (rule disjE)
qed

text{* Otra forma:
using assms
proof
assume "p"
then have "p∨q" ..
thus "(p∨q)∨r" ..
next
assume 1:"q∨r"
thus "(p ∨ q) ∨ r"
proof
assume "q"
then have "p∨q" ..
thus 1: "(p ∨ q) ∨ r" ..
next
assume "r"
then have 2:"(p∨q)∨r" ..
then have "q∨r⟶(p ∨ q) ∨ r" ..
thus "(p ∨ q) ∨ r" using 1 ..
qed}*}

lemma ejercicio_31:
assumes 1:"p ∨ (q ∨ r)"
shows "(p ∨ q) ∨ r"
using 1
proof (rule disjE)
{assume 2:"p"
have 3:"p ∨ q" using 2 by (rule disjI1)
show "(p ∨ q) ∨ r" using 3 by (rule disjI1)}
next
{assume 4:"q ∨ r"
thus "(p ∨ q) ∨ r"
proof (rule disjE)
{assume 5:"q"
have 6:"p ∨ q" using 5 by (rule disjI2)
show "(p ∨ q) ∨ r" using 6 by (rule disjI1)}
next
{assume 7:"r"
show "(p ∨ q) ∨ r" using 7 by (rule disjI2)}
qed}
qed

text {* ---------------------------------------------------------------
Ejercicio 32. Demostrar
(p ∨ q) ∨ r ⊢ p ∨ (q ∨ r)
------------------------------------------------------------------ *}

lemma ejercicio_32:
assumes "(p ∨ q) ∨ r"
shows "p ∨ (q ∨ r)"
proof -
have "(p ∨ q) ∨ r" using assms(1) by this
moreover
{ assume 1: "p ∨ q"
moreover
{ assume 2: "p"
have 3: "p ∨ (q ∨ r)" using 2 by (rule disjI1)}
moreover
{ assume 4: "q"
have 5: "q ∨ r" using 4 by (rule disjI1)
have 6: "p ∨ (q ∨ r)" using 5 by (rule disjI2)}
ultimately have "p ∨ (q ∨ r)" by (rule disjE)}
moreover
{ assume 7: "r"
have 8: "q ∨ r" using 7 by (rule disjI2)
have 9: "p ∨ (q ∨ r)" using 8 by (rule disjI2)}
ultimately show "p ∨ (q ∨ r)" by (rule disjE)
qed

lemma ejercicio_32:
assumes 1:"(p ∨ q) ∨ r"
shows "p ∨ (q ∨ r)"
using 1
proof (rule disjE)
{assume 2:"p ∨ q"
thus "p ∨ (q ∨ r)"
proof (rule disjE)
{assume 3:"p"
show "p ∨ (q ∨ r)" using 3 by (rule disjI1)}
next
{assume 4:"q"
have 5:"q ∨ r" using 4 by (rule disjI1)
show "p ∨ (q ∨ r)" using 5 by (rule disjI2)}
qed}
next
{assume 6:"r"
have 7:"q ∨ r" using 6 by (rule disjI2)
show "p ∨ (q ∨ r)" using 7 by (rule disjI2)}
qed

text {* ---------------------------------------------------------------
Ejercicio 33. Demostrar
p ∧ (q ∨ r) ⊢ (p ∧ q) ∨ (p ∧ r)
------------------------------------------------------------------ *}

lemma ejercicio_33:
assumes "p ∧ (q ∨ r)"
shows "(p ∧ q) ∨ (p ∧ r)"
proof -
have 1: "p" using assms(1) by (rule conjunct1)
have 2: "q ∨ r" using assms(1) by (rule conjunct2)
moreover
{ assume 3: "q"
have 4: "p ∧ q" using 1 3 by (rule conjI)
have 5: "(p ∧ q) ∨ (p ∧ r)" using 4 by (rule disjI1)}
moreover
{ assume 6: "r"
have 7: "p ∧ r" using 1 6 by (rule conjI)
have 8: "(p ∧ q) ∨ (p ∧ r)" using 7 by (rule disjI2)}
ultimately show "(p ∧ q) ∨ (p ∧ r)" by (rule disjE)
qed

lemma ejercicio_33:
assumes 1:"p ∧ (q ∨ r)"
shows "(p ∧ q) ∨ (p ∧ r)"
proof -
have 2:"p" using 1 by (rule conjunct1)
have 3:"q ∨ r" using 1 by (rule conjunct2)
thus "(p ∧ q) ∨ (p ∧ r)"
proof (rule disjE)
{assume 4:"q"
have 5:"p ∧ q" using 2 4 by (rule conjI)
show "(p ∧ q) ∨ (p ∧ r)" using 5 by (rule disjI1)}
next
{assume 6:"r"
have 7:"p ∧ r" using 2 6 by (rule conjI)
show "(p ∧ q) ∨ (p ∧ r)" using 7 by (rule disjI2)}
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 34. Demostrar
(p ∧ q) ∨ (p ∧ r) ⊢ p ∧ (q ∨ r)
------------------------------------------------------------------ *}

lemma ejercicio_34:
assumes "(p ∧ q) ∨ (p ∧ r)"
shows "p ∧ (q ∨ r)"
proof -
have "(p ∧ q) ∨ (p ∧ r)" using assms(1) by this
moreover
{ assume 1: "p ∧ q"
have 2: "q" using 1 by (rule conjunct2)
have 3: "q ∨ r" using 2 by (rule disjI1)
have 4: "p" using 1 by (rule conjunct1)
have 5: "p ∧ (q ∨ r)" using 4 3 by (rule conjI)}
moreover
{ assume 6: "p ∧ r"
have 7: "r" using 6 by (rule conjunct2)
have 8: "q ∨ r" using 7 by (rule disjI2)
have 9: "p" using 6 by (rule conjunct1)
have 10: "p ∧ (q ∨ r)" using 9 8 by (rule conjI)}
ultimately show "p ∧ (q ∨ r)" by (rule disjE)
qed

lemma ejercicio_34:
assumes 1:"(p ∧ q) ∨ (p ∧ r)"
shows "p ∧ (q ∨ r)"
using 1
proof (rule disjE)
{assume 2:"p ∧ q"
have 3:"p" using 2 by (rule conjunct1)
have 4:"q" using 2 by (rule conjunct2)
have 5:"q ∨ r" using 4 by (rule disjI1)
show "p ∧ (q ∨ r)" using 3 5 by (rule conjI)}
next
{assume 6:"p ∧ r"
have 7:"p" using 6 by (rule conjunct1)
have 8:"r" using 6 by (rule conjunct2)
have 9:"q ∨ r" using 8 by (rule disjI2)
show "p ∧ (q ∨ r)" using 7 9 by (rule conjI)}
qed

text {* ---------------------------------------------------------------
Ejercicio 35. Demostrar
p ∨ (q ∧ r) ⊢ (p ∨ q) ∧ (p ∨ r)
------------------------------------------------------------------ *}

lemma ejercicio_35:
assumes "p ∨ (q ∧ r)"
shows "(p ∨ q) ∧ (p ∨ r)"
proof -
have "p ∨ (q ∧ r)" using assms(1) by this
moreover
{ assume 1: "p"
have 2: "p ∨ q" using 1 by (rule disjI1)
have 3: "p ∨ r" using 1 by (rule disjI1)
have 4: "(p ∨ q) ∧ (p ∨ r)" using 2 3 by (rule conjI)}
moreover
{ assume 5: "q ∧ r"
have 6: "q" using 5 by (rule conjunct1)
have 7: "p ∨ q" using 6 by (rule disjI2)
have 8: "r" using 5 by (rule conjunct2)
have 9: "p ∨ r" using 8 by (rule disjI2)
have 10: "(p ∨ q) ∧ (p ∨ r)" using 7 9 by (rule conjI)}
ultimately show "(p ∨ q) ∧ (p ∨ r)" by (rule disjE)
qed

lemma ejercicio_35:
assumes 1:"p ∨ (q ∧ r)"
shows "(p ∨ q) ∧ (p ∨ r)"
using 1
proof (rule disjE)
{assume 2:"p"
have 3:"p ∨ q" using 2 by (rule disjI1)
have 4:"p ∨ r" using 2 by (rule disjI1)
show "(p ∨ q) ∧ (p ∨ r)" using 3 4 by (rule conjI)}
next
{assume 5:"q ∧ r"
have 6:"q" using 5 by (rule conjunct1)
have 7:"r" using 5 by (rule conjunct2)
have 8:"p ∨ q" using 6 by (rule disjI2)
have 9:"p ∨ r" using 7 by (rule disjI2)
show "(p ∨ q) ∧ (p ∨ r)" using 8 9 by (rule conjI)}
qed

text {* ---------------------------------------------------------------
Ejercicio 36. Demostrar
(p ∨ q) ∧ (p ∨ r) ⊢ p ∨ (q ∧ r)
------------------------------------------------------------------ *}

lemma ejercicio_36:
assumes "(p ∨ q) ∧ (p ∨ r)"
shows "p ∨ (q ∧ r)"
proof-
have "(p ∨ q)" using assms by (rule conjunct1)
moreover
{assume 1:"p"
have "p ∨ (q ∧ r)" using 1 by (rule disjI1) }
moreover
{assume 1:"q"
have "(p ∨ r)" using assms by (rule conjunct2)
moreover
{assume 2:"p"
have "p ∨ (q ∧ r)" using 2 by (rule disjI1)}
moreover
{assume 2:"r"
have 3:"(q ∧ r)" using 1 2 by (rule conjI)
have "p ∨ (q ∧ r)" using 3 by (rule disjI2)}
ultimately have "p ∨ (q ∧ r)" by (rule disjE)}
ultimately show "p ∨ (q ∧ r)" by (rule disjE)
qed

------------------- ----------

proof-
have 1: "p∨q" using assms by (rule conjunct1)
thus "p∨(q∧r)"
proof(rule disjE)
{assume 3: "p"
show "p∨(q∧r)" using 3 by ( rule disjI1)}
next
{assume 4: "q"
have 5:"p∨r" using assms by (rule conjunct2)
thus "p∨(q∧r)"
proof(rule disjE)
{assume 6: "p"
show "p∨(q∧r)" using 6 by (rule disjI1)}
next
{assume 7: "r"
have 8: "q∧r" using 4 7 by (rule conjI)
show "p∨(q∧r)" using 8 by (rule disjI2)}
qed}
qed
qed
text {* ---------------------------------------------------------------
Ejercicio 37. Demostrar
(p ⟶ r) ∧ (q ⟶ r) ⊢ p ∨ q ⟶ r
------------------------------------------------------------------ *}

lemma ejercicio_37:
assumes "(p ⟶ r) ∧ (q ⟶ r)"
shows "p ∨ q ⟶ r"
proof-
{assume "p ∨ q"
moreover
{assume 1:"p"
have 2:"(p ⟶ r)" using assms by (rule conjunct1)
have "r" using 2 1 by (rule mp)}
moreover
{assume 1:"q"
have 2:"(q ⟶ r)" using assms by (rule conjunct2)
have "r" using 2 1 by (rule mp)}
ultimately have "r" by (rule disjE)}
thus "p ∨ q ⟶ r" by (rule impI)
qed

lemma ejercicio_37:
assumes 1:"(p ⟶ r) ∧ (q ⟶ r)"
shows "p ∨ q ⟶ r"
proof (rule impI)
assume 2:"p ∨ q"
thus "r"
proof (rule disjE)
{assume 3:"p"
have 4:"p ⟶ r" using 1 by (rule conjunct1)
show "r" using 4 3 by (rule mp)}
next
{assume 5:"q"
have 6:"q ⟶ r" using 1 by (rule conjunct2)
show "r" using 6 5 by (rule mp)}
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 38. Demostrar
p ∨ q ⟶ r ⊢ (p ⟶ r) ∧ (q ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_38:
assumes "p ∨ q ⟶ r"
shows "(p ⟶ r) ∧ (q ⟶ r)"
proof-
{assume 1 : "p"
have 2 : "p ∨ q" using 1 by (rule disjI1)
have 3 : "r" using assms 2 by (rule mp)}
hence 4 : "p⟶r" by (rule impI)
{assume 5 : "q"
have 6 : "p ∨ q" using 5 by (rule disjI2)
have 7 : "r" using assms 6 by (rule mp)}
hence 8 : "q⟶r" by (rule impI)
show "(p ⟶ r) ∧ (q ⟶ r)" using 4 8 by (rule conjI)
qed

section {* Negaciones *}

text {* ---------------------------------------------------------------
Ejercicio 39. Demostrar
p ⊢ ¬¬p
------------------------------------------------------------------ *}

lemma ejercicio_39:
assumes "p"
shows "¬¬p"
proof-
show "¬¬p" using assms by (rule notnotI)
qed

text {* ---------------------------------------------------------------
Ejercicio 40. Demostrar
¬p ⊢ p ⟶ q
------------------------------------------------------------------ *}

lemma ejercicio_40:
assumes "¬p"
shows "p ⟶ q"

proof-
{assume 1: "p"
have "q" using assms 1 ..}
thus "p⟶q" ..
qed
-------------------------------
lemma ejercicio_40:
assumes "¬p"
shows "p ⟶ q"
proof-
{assume 1 : "p"
have 2 : "False" using assms 1 by (rule notE)
have 3 : "q" using 2 by (rule FalseE)}
thus "p⟶q" by (rule impI)
qed

------------------------------
lemma ejercicio_40:
assumes "¬p"
shows "p ⟶ q"

proof-
{assume 1: "p"
have 2: "q" using assms 1 by (rule notE)}
thus "p ⟶ q" by ( rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 41. Demostrar
p ⟶ q ⊢ ¬q ⟶ ¬p
------------------------------------------------------------------ *}

lemma ejercicio_41:
assumes "p ⟶ q"
shows "¬q ⟶ ¬p"
proof-
{assume 1:"¬q"
have "¬p" using assms 1 by (rule mt)}
thus "¬q ⟶ ¬p" ..
qed

text {* ---------------------------------------------------------------
Ejercicio 42. Demostrar
p∨q, ¬q ⊢ p
------------------------------------------------------------------ *}

lemma ejercicio_42:
assumes "p∨q"
"¬q"
shows "p"

using assms(1)
proof
assume "p"
thus "p" .
next
assume 1:"q"
show "p" using assms(2) 1 ..
qed

lemma ejercicio_42:
assumes "p∨q"
"¬q"
shows "p"
proof-
have "p ∨ q" using assms(1) by this
moreover
{assume 1 : "p"
have 2 : "p" using 1 .}
moreover
{assume 3 : "q"
have 4 : "¬q" using assms(2) .
have 5 : "False" using 4 3 by (rule notE)
have 6 : "p" using 5 by (rule FalseE)}
ultimately show "p" by (rule disjE)
qed

--------**---
using assms(1)
proof(rule disjE)
{assume 1: "p"
have "p" using 1 by this}
next
{assume 2: "q"
show "p" using assms(2) 2 by (rule notE)}
qed

text {* ---------------------------------------------------------------
Ejercicio 43. Demostrar
p ∨ q, ¬p ⊢ q
------------------------------------------------------------------ *}

lemma ejercicio_43:
assumes "p ∨ q"
"¬p"
shows "q"

using assms(1)
proof
assume "q"
thus "q" .
next
assume 1:"p"
show "q" using assms(2) 1 ..
qed

lemma ejercicio_43:
assumes "p ∨ q"
"¬p"
shows "q"
proof-
have "p ∨ q" using assms(1) by this
moreover
{assume 1 : "p"
have 2 : "¬p" using assms(2) .
have 3 : "False" using 2 1 by (rule notE)
have 4 : "q" using 3 by (rule FalseE)}
moreover
{assume 5 : "q"
have 6 : "q" using 5 .}
ultimately show "q" by (rule disjE)
qed

---**--
using assms(1)
proof(rule disjE)
assume 1: "p"
show "q" using assms(2) 1 by (rule notE)
next
assume 2:"p"
have "p" using 2 by this
qed

text {* ---------------------------------------------------------------
Ejercicio 44. Demostrar
p ∨ q ⊢ ¬(¬p ∧ ¬q)
------------------------------------------------------------------ *}

lemma ejercicio_44:
assumes "p ∨ q"
shows "¬(¬p ∧ ¬q)"
proof-
{assume 1 : "¬p ∧ ¬q"
have "p ∨ q" using assms by this
moreover
{assume 2 : "p"
have 3 : "¬p" using 1 by (rule conjunct1)
have 4 : "False" using 3 2 by (rule notE)}
moreover
{assume 5 : "q"
have 6 : "¬q" using 1 by (rule conjunct2)
have 7 : "False" using 6 5 by (rule notE)}
ultimately have 8 : "False" by (rule disjE)}
thus "¬(¬p ∧ ¬q)" by (rule notI)
qed

---**
proof(rule notI)
assume 2 :"¬p∧¬q"
show False
using assms
proof(rule disjE)
{assume 1: "p"
have 3: "¬p" using 2 by (rule conjunct1)
show False using 3 1 by (rule notE)}
{assume 4: "q"
have 5: "¬q" using 2 by (rule conjunct2)
show False using 5 4 by (rule notE)}
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 45. Demostrar
p ∧ q ⊢ ¬(¬p ∨ ¬q)
------------------------------------------------------------------ *}

lemma ejercicio_45:
assumes "p ∧ q"
shows "¬(¬p ∨ ¬q)"
proof-
{assume 1 : "¬p ∨ ¬q"
have "¬p ∨ ¬q" using 1 by this
moreover
{assume 2 : "¬p"
have 3 : "p" using assms by (rule conjunct1)
have 4 : "False" using 2 3 by (rule notE)}
moreover
{assume 5 : "¬q"
have 6 : "q" using assms by (rule conjunct2)
have 7 : "False" using 5 6 by (rule notE)}
ultimately have "False" by (rule disjE)}
thus "¬(¬p ∨ ¬q)" by (rule notI)
qed

---**

assumes "p ∧ q"
shows "¬(¬p ∨ ¬q)"

proof (rule notI)
assume 1: "¬p∨¬q"
show False
using 1
proof(rule disjE)
assume 2: "¬p"
have 3: "p" using assms by (rule conjunct1)
show False using 2 3 by (rule notE)
next
assume 4: "¬q"
have 5: "q" using assms by (rule conjunct2)
show False using 4 5 by (rule notE)
qed
qed
text {* ---------------------------------------------------------------
Ejercicio 46. Demostrar
¬(p ∨ q) ⊢ ¬p ∧ ¬q
------------------------------------------------------------------ *}

lemma ejercicio_46:
assumes "¬(p ∨ q)"
shows "¬p ∧ ¬q"
proof-
{assume 1 : "p"
have 2 : "p ∨ q" using 1 by (rule disjI1)
have 3 : "False" using assms 2 by (rule notE)}
hence 4 : "¬p" by (rule notI)
{assume 5 : "q"
have 6 : "p ∨ q" using 5 by (rule disjI2)
have 7 : "False" using assms 6 by (rule notE)}
hence 8 : "¬q" by (rule notI)
show "¬p ∧ ¬q" using 4 8 by (rule conjI)
qed

(*Otra opción para usar tercio excluso*)

lemma ejercicio_46_2:
assumes "¬(p ∨ q)"
shows "¬p ∧ ¬q"
proof -
have 1:"¬p | p" by (rule excluded_middle)
show "¬p & ¬q"
using 1
proof (rule disjE)
{assume 2:"¬p"
have 3:"¬q | q" by (rule excluded_middle)
show "¬p & ¬q"
using 3
proof (rule disjE)
{assume 4:"¬q"
with 2 show "¬p & ¬q" by (rule conjI)}
next
{assume "q"
then have "p | q" by (rule disjI2)
with assms show "¬p & ¬q" by (rule notE)}
qed
}
next
{assume 5:"p"
then have "p | q" by (rule disjI1)
with assms show "¬p & ¬q" by (rule notE)
}
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 47. Demostrar
¬p ∧ ¬q ⊢ ¬(p ∨ q)
------------------------------------------------------------------ *}

lemma ejercicio_47:
assumes "¬p ∧ ¬q"
shows "¬(p ∨ q)"
proof-
{assume 1 : "p ∨ q"
have "p ∨ q" using 1 by this
moreover
{assume 2 : "p"
have 3 : "¬p" using assms by (rule conjunct1)
have 4 : "False" using 3 2 by (rule notE)}
moreover
{assume 5 : "q"
have 6 : "¬q" using assms by (rule conjunct2)
have 6 : "False" using 6 5 by (rule notE)}
ultimately have "False" by (rule disjE)}
thus "¬(p ∨ q)" by (rule notI)
qed

---**

proof(rule notI)
assume 1: "p∨q"
show False
using 1
proof(rule disjE)
{assume 2:"p"
have 3:"¬p" using assms by (rule conjunct1)
show "False" using 3 2 by (rule notE)}
next
{assume 5:"q"
have 6:"¬q" using assms by (rule conjunct2)
show "False" using 6 5 by (rule notE)}
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 48. Demostrar
¬p ∨ ¬q ⊢ ¬(p ∧ q)
------------------------------------------------------------------ *}

lemma ejercicio_48:
assumes "¬p ∨ ¬q"
shows "¬(p ∧ q)"
proof-
{assume 1 : "p ∧ q"
have "¬p ∨ ¬q" using assms by this
moreover
{assume 2 : "¬p"
have 3 : "p" using 1 by (rule conjunct1)
have 4 : "False" using 2 3 by (rule notE)}
moreover
{assume 5 : "¬q"
have 6 : "q" using 1 by (rule conjunct2)
have 7 : "False" using 5 6 by (rule notE)}
ultimately have "False" by (rule disjE)}
thus "¬(p ∧ q)" by (rule notI)
qed

--**
lemma ejercicio_48:
assumes "¬p ∨ ¬q"
shows "¬(p ∧ q)"

proof(rule notI)
assume 1:"p∧q"
show False
using assms
proof(rule disjE)
assume 2:"¬p"
have 3: "p" using 1 by (rule conjunct1)
show False using 2 3 by (rule notE)
next
assume 4: "¬q"
have 5: "q" using 1 by (rule conjunct2)
show False using 4 5 by (rule notE)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 49. Demostrar
⊢ ¬(p ∧ ¬p)
------------------------------------------------------------------ *}

lemma ejercicio_49:
"¬(p ∧ ¬p)"
proof-
{assume 1 : "p ∧ ¬p"
have 2 : "p" using 1 by (rule conjunct1)
have 3 : "¬p" using 1 by (rule conjunct2)
have 4 : "False" using 3 2 by (rule notE)}
thus "¬(p ∧ ¬p)" by (rule notI)
qed

---**

proof(rule notI)
assume 1: "p∧¬p"
have 2: "p" using 1 by (rule conjunct1)
have 3: "¬p" using 1 by (rule conjunct2)
show False using 3 2 by (rule notE)
qed

text {* ---------------------------------------------------------------
Ejercicio 50. Demostrar
p ∧ ¬p ⊢ q
------------------------------------------------------------------ *}

lemma ejercicio_50:
assumes "p ∧ ¬p"
shows "q"

proof-
have 1 : "p" using assms by (rule conjunct1)
have 2 : "¬p" using assms by (rule conjunct2)
show "q" using 2 1 by (rule notE)
qed

text {* ---------------------------------------------------------------
Ejercicio 51. Demostrar
¬¬p ⊢ p
------------------------------------------------------------------ *}

lemma ejercicio_51:
assumes "¬¬p"
shows "p"

proof-
show "p" using assms by (rule notnotD)
qed

--**

lemma ejercicio_51:
assumes "¬¬p"
shows "p"
proof(rule ccontr)
assume 1: "¬p"
show False using assms 1 by (rule notE)
qed

text {* ---------------------------------------------------------------
Ejercicio 52. Demostrar
⊢ p ∨ ¬p
------------------------------------------------------------------ *}

lemma ejercicio_52:
"p ∨ ¬p"

proof-
have 1: "¬p∨p" by (rule excluded_middle)
have 2: "¬p∨p" using 1 by this
moreover
{assume 3: "¬p"
have 4: "p∨¬p" using 3 by (rule disjI2)}
moreover
{assume 5: "p"
have 6: "p∨¬p" using 5 by (rule disjI1)}
ultimately show "p∨¬p" by (rule disjE)
qed

text {* ---------------------------------------------------------------
Ejercicio 53. Demostrar
⊢ ((p ⟶ q) ⟶ p) ⟶ p
------------------------------------------------------------------ *}

lemma ejercicio_53:
"((p ⟶ q) ⟶ p) ⟶ p"
proof-
{assume 1 : "(p ⟶ q) ⟶ p"
{assume 2 : "¬p"
have 3 : "¬(p⟶q)" using 1 2 by (rule mt)
{assume 4 : "p"
have 5 : "q" using 2 4 by (rule notE)}
hence 6 : "p⟶q" by (rule impI)
have 7 : "False" using 3 6 by (rule notE)}
hence 8 : "p" by (rule ccontr)}
thus "((p ⟶ q) ⟶ p) ⟶ p" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 54. Demostrar
¬q ⟶ ¬p ⊢ p ⟶ q
------------------------------------------------------------------ *}

lemma ejercicio_54:
assumes "¬q ⟶ ¬p"
shows "p ⟶ q"
proof-
{assume 1 : "p"
{assume 2 : "¬q"
have 3 : "¬p" using assms 2 by (rule mp)
have 4 : "False" using 3 1 by (rule notE)}
hence 5 : "q" by (rule ccontr)}
thus "p⟶q" by (rule impI)
qed

---**

proof(rule impI)
assume 1: "p"
have 2 : "¬¬p" using 1 by (rule notnotI)
have 3: "¬¬q" using assms 2 by (rule mt)
show "q" using 3 by (rule notnotD)
qed

text {* ---------------------------------------------------------------
Ejercicio 55. Demostrar
¬(¬p ∧ ¬q) ⊢ p ∨ q
------------------------------------------------------------------ *}

lemma ejercicio_55:
assumes "¬(¬p ∧ ¬q)"
shows "p ∨ q"
proof-
have 1 : "¬p ∨ p" by (rule excluded_middle)
have "¬p ∨ p" using 1 by this
moreover
{assume 2 : "¬p"
have 3 : "¬q ∨ q" by (rule excluded_middle)
have "¬q ∨ q" using 3 by this
moreover
{assume 4 : "¬q"
have 5 : "¬p ∧ ¬q" using 2 4 by (rule conjI)
have 6 : "False" using assms 5 by (rule notE)
have 7 : "p ∨ q" using 6 by (rule FalseE)}
moreover
{assume 8 : "q"
have 9 : "p ∨ q" using 8 by (rule disjI2)}
ultimately have "p ∨ q" by (rule disjE)}
moreover
{assume 10 : "p"
have 11 : "p ∨ q" using 10 by (rule disjI1)}
ultimately show "p ∨ q" by (rule disjE)
qed

text {* ---------------------------------------------------------------
Ejercicio 56. Demostrar
¬(¬p ∨ ¬q) ⊢ p ∧ q
------------------------------------------------------------------ *}

lemma ejercicio_56:
assumes "¬(¬p ∨ ¬q)"
shows "p ∧ q"
proof-
{assume 1 : "¬p"
have 2 : "¬p ∨ ¬q" using 1 by (rule disjI1)
have 3 : "False" using assms 2 by (rule notE)}
hence 4 : "p" by (rule ccontr)
{assume 5 : "¬q"
have 6 : "¬p ∨ ¬q" using 5 by (rule disjI2)
have 7 : "False" using assms 6 by (rule notE)}
hence 8 : "q" by (rule ccontr)
show "p ∧ q" using 4 8 by (rule conjI)
qed

--**
proof-
{assume 1: "¬p"
have 2: "¬p∨¬q"using 1 by (rule disjI1)
have 3: "False" using assms 2 by (rule notE)}
hence 8: "p" by (rule ccontr)
{assume 4: "¬q"
have 5: "¬p∨¬q" using 4 by (rule disjI2)
have 6: "False" using assms 5 by (rule notE)}
hence 9: "q" by (rule ccontr)
show "p∧q" using 8 9 by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 57. Demostrar
¬(p ∧ q) ⊢ ¬p ∨ ¬q
------------------------------------------------------------------ *}

lemma ejercicio_57:
assumes "¬(p ∧ q)"
shows "¬p ∨ ¬q"
proof-
have 1 : "¬p ∨ p" by (rule excluded_middle)
have "¬p ∨ p" using 1 by this
moreover
{assume 2 : "¬p"
have 3 : "¬p ∨ ¬q" using 2 by (rule disjI1)}
moreover
{assume 4 : "p"
have 5 : "¬q ∨ q" by (rule excluded_middle)
have "¬q ∨ q" using 5 by this
moreover
{assume 6 : "¬q"
have 7 : "¬p ∨ ¬q" using 6 by (rule disjI2)}
moreover
{assume 8 : "q"
have 9 : "p ∧ q" using 4 8 by (rule conjI)
have 10 : "False" using assms 9 by (rule notE)
have 11 : "¬p ∨ ¬q" using 10 by (rule FalseE)}
ultimately have "¬p ∨ ¬q" by (rule disjE)}
ultimately show "¬p ∨ ¬q" by (rule disjE)
qed

text {* ---------------------------------------------------------------
Ejercicio 58. Demostrar
⊢ (p ⟶ q) ∨ (q ⟶ p)
------------------------------------------------------------------ *}

lemma ejercicio_58:
"(p ⟶ q) ∨ (q ⟶ p)"

proof-
have 1 : "¬(p ⟶ q) ∨ (p ⟶ q)" by (rule excluded_middle)
have "¬(p ⟶ q) ∨ (p ⟶ q)" using 1 by this
moreover
{assume 2 : "¬(p ⟶ q)"
{assume 3 : "q"
{assume 4 : "p"
have 5 : "q" using 3 .}
hence 6 : "p⟶q" by (rule impI)
have 7 : "p" using 2 6 by (rule notE)}
hence 8 : "q⟶p" by (rule impI)
have 9 : "(p ⟶ q) ∨ (q ⟶ p)" using 8 by (rule disjI2)}
moreover
{assume 10 : "p ⟶ q"
have 11 : "(p ⟶ q) ∨ (q ⟶ p)" using 10 by (rule disjI1)}
ultimately show "(p ⟶ q) ∨ (q ⟶ p)" by (rule disjE)
qed

end
</source>

Relación 3

2016-03-25T16:36:15Z

Albromgon:

<source lang = "isar">header {* R3: Deducción natural proposicional *}

theory R3
imports Main
begin

text {*
---------------------------------------------------------------------
El objetivo de esta relación es demostrar cada uno de los ejercicios
usando sólo las reglas básicas de deducción natural de la lógica
proposicional (sin usar el método auto).

Las reglas básicas de la deducción natural son las siguientes:
· conjI: ⟦P; Q⟧ ⟹ P ∧ Q
· conjunct1: P ∧ Q ⟹ P
· conjunct2: P ∧ Q ⟹ Q
· notnotD: ¬¬ P ⟹ P
· notnotI: P ⟹ ¬¬ P
· mp: ⟦P ⟶ Q; P⟧ ⟹ Q
· mt: ⟦F ⟶ G; ¬G⟧ ⟹ ¬F
· impI: (P ⟹ Q) ⟹ P ⟶ Q
· disjI1: P ⟹ P ∨ Q
· disjI2: Q ⟹ P ∨ Q
· disjE: ⟦P ∨ Q; P ⟹ R; Q ⟹ R⟧ ⟹ R
· FalseE: False ⟹ P
· notE: ⟦¬P; P⟧ ⟹ R
· notI: (P ⟹ False) ⟹ ¬P
· iffI: ⟦P ⟹ Q; Q ⟹ P⟧ ⟹ P = Q
· iffD1: ⟦Q = P; Q⟧ ⟹ P
· iffD2: ⟦P = Q; Q⟧ ⟹ P
· ccontr: (¬P ⟹ False) ⟹ P
---------------------------------------------------------------------
*}

text {*
Se usarán las reglas notnotI y mt que demostramos a continuación. *}

lemma notnotI: "P ⟹ ¬¬ P"
by auto

lemma mt: "⟦F ⟶ G; ¬G⟧ ⟹ ¬F"
by auto

section {* Implicaciones *}

text {* ---------------------------------------------------------------
Ejercicio 1. Demostrar
p ⟶ q, p ⊢ q
------------------------------------------------------------------ *}

lemma ejercicio_1:
assumes "p ⟶ q"
"p"
shows "q"

proof -
show "q" using assms(1,2) by (rule mp)
qed

lemma ejercicio_1_b:
assumes "p ⟶ q"
"p"
shows "q"
proof-
show "q" using assms(1) assms(2) by (rule mp)
qed

text {* ---------------------------------------------------------------
Ejercicio 2. Demostrar
p ⟶ q, q ⟶ r, p ⊢ r
------------------------------------------------------------------ *}

lemma ejercicio_2:
assumes "p ⟶ q"
"q ⟶ r"
"p"
shows "r"

proof-
have 1: "q" using assms(1,3) by (rule mp)
show "r" using assms(2) 1 by (rule mp)
qed

text {* ---------------------------------------------------------------
Ejercicio 3. Demostrar
p ⟶ (q ⟶ r), p ⟶ q, p ⊢ r
------------------------------------------------------------------ *}

lemma ejercicio_3:
assumes "p ⟶ (q ⟶ r)"
"p ⟶ q"
"p"
shows "r"

proof-
have 1: "q" using assms(2,3) by (rule mp)
have 2: "q⟶r" using assms(1,3) by (rule mp)
show "r" using 2 1 by (rule mp)
qed

text {* ---------------------------------------------------------------
Ejercicio 4. Demostrar
p ⟶ q, q ⟶ r ⊢ p ⟶ r
------------------------------------------------------------------ *}

lemma ejercicio_4:
assumes "p ⟶ q"
"q ⟶ r"
shows "p ⟶ r"

proof-
{assume 1: "p"
have 2: "q" using assms(1) 1 ..
have 3: "r" using assms(2) 2 ..}
thus "p⟶r" by (rule impI)
qed

lemma ejercicio_4:
assumes "p ⟶ q"
"q ⟶ r"
shows "p ⟶ r"
proof-
{assume 1 : "p"
have 2 : "q" using assms(1) 1 by (rule mp)
have 3 : "r" using assms(2) 2 by (rule mp)}
thus "p⟶r" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 5. Demostrar
p ⟶ (q ⟶ r) ⊢ q ⟶ (p ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_5:
assumes "p ⟶ (q ⟶ r)"
shows "q ⟶ (p ⟶ r)"

{assume 1: "q"
{assume 2: "p"
have 3: "q⟶r" using assms 2 by (rule mp)
have "r" using 3 1 by (rule mp)}
hence "p⟶r" by (rule impI)}
thus "(q⟶(p⟶r))" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 6. Demostrar
p ⟶ (q ⟶ r) ⊢ (p ⟶ q) ⟶ (p ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_6:
assumes "p ⟶ (q ⟶ r)"
shows "(p ⟶ q) ⟶ (p ⟶ r)"

proof-
{assume 1:"p⟶q"
{assume 2: "p"
have 3: "q⟶r" using assms 2 by (rule mp)
have 4: "q" using 1 2 by (rule mp)
have "r" using 3 4 by (rule mp)}
hence "p⟶r" by (rule impI)}
thus "(p⟶q)⟶(p⟶r)" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 7. Demostrar
p ⊢ q ⟶ p
------------------------------------------------------------------ *}

lemma ejercicio_7:
assumes "p"
shows "q ⟶ p"

proof-
{assume 1: "q"
have "p" using assms .}
thus "q⟶p" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 8. Demostrar
⊢ p ⟶ (q ⟶ p)
------------------------------------------------------------------ *}

lemma ejercicio_8:
"p ⟶ (q ⟶ p)"

proof-
{assume 1:"p"
{assume 2:"q"
have 3: "p" using 1 .}
hence "q⟶p" ..}
thus "p⟶(q⟶p)" ..
qed

lemma ejercicio_8:
"p ⟶ (q ⟶ p)"
proof-
{assume 1 : "p"
{assume 2 : "q"
have 3 : "p" using 1 .}
hence "q⟶p" by (rule impI)}
thus "p⟶(q⟶p)" by (rule impI)
qed

lemma ejercicio_8:
"p ⟶ (q ⟶ p)"
proof (rule impI)
assume 1: "p"
show "q ⟶ p"
proof (rule impI)
assume "q"
show "p" using 1 by this
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 9. Demostrar
p ⟶ q ⊢ (q ⟶ r) ⟶ (p ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_9:
assumes "p ⟶ q"
shows "(q ⟶ r) ⟶ (p ⟶ r)"

proof-
{assume 1:"q⟶r"
{assume 2:"p"
have 3:"q" using assms 2 ..
have "r" using 1 3 ..}
hence "p⟶r" ..}
thus "(q⟶r)⟶(p⟶r) " ..
qed

lemma ejercicio_9:
assumes "p ⟶ q"
shows "(q ⟶ r) ⟶ (p ⟶ r)"
proof-
{assume 1 : "q⟶r"
{assume 2 : "p"
have 3 : "q" using assms(1) 2 by (rule mp)
have 4 : "r" using 1 3 by (rule mp)}
hence "p⟶r" by (rule impI)}
thus "(q⟶r)⟶(p⟶r)" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 10. Demostrar
p ⟶ (q ⟶ (r ⟶ s)) ⊢ r ⟶ (q ⟶ (p ⟶ s))
------------------------------------------------------------------ *}

lemma ejercicio_10:
assumes "p ⟶ (q ⟶ (r ⟶ s))"
shows "r ⟶ (q ⟶ (p ⟶ s))"

proof-
{assume 1:"r"
{assume 2:"q"
{assume 3:"p"
have 4:"q⟶(r⟶s)" using assms 3 ..
have 5: "r⟶s" using 4 2 ..
have 6: "s" using 5 1 ..}
hence "p⟶s" ..}
hence "q⟶(p⟶s)" ..}
thus "r⟶(q⟶(p⟶s))" ..
qed

lemma ejercicio_10:
assumes "p ⟶ (q ⟶ (r ⟶ s))"
shows "r ⟶ (q ⟶ (p ⟶ s))"
proof-
{assume 1 : "r"
{assume 2 : "q"
{assume 3 : "p"
have 4 : "q ⟶ (r ⟶ s)" using assms(1) 3 by (rule mp)
have 5 : "r⟶s" using 4 2 by (rule mp)
have 6 : "s" using 5 1 by (rule mp)}
hence "p⟶s" by (rule impI)}
hence "q⟶(p⟶s)" by (rule impI)}
thus "r⟶(q⟶(p⟶s))" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 11. Demostrar
⊢ (p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))
------------------------------------------------------------------ *}

lemma ejercicio_11:
"(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))"

proof (rule impI)
assume 3: "p⟶(q⟶r)"
show "(p⟶q)⟶(p⟶r)"
proof(rule impI)
assume 2:"p⟶q"
show "p⟶r"
proof (rule impI)
assume 1: "p"
have 4:"q" using 2 1 ..
have 5:"q⟶r" using 3 1 ..
show "r" using 5 4 ..
qed
qed
qed

lemma ejercicio_11:
"(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))"
proof-
{assume 1 : "p ⟶ (q ⟶ r)"
{assume 2 : "p⟶q"
{assume 3 : "p"
have 4 : "q⟶r" using 1 3 by (rule mp)
have 5 : "q" using 2 3 by (rule mp)
have 6 : "r" using 4 5 by (rule mp)}
hence "p⟶r" by (rule impI)}
hence "(p ⟶ q) ⟶ (p ⟶ r)" by (rule impI)}
thus "(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 12. Demostrar
(p ⟶ q) ⟶ r ⊢ p ⟶ (q ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_12:
assumes "(p ⟶ q) ⟶ r"
shows "p ⟶ (q ⟶ r)"

proof-
{assume 1:"p"
{assume 3:"q"
have 4:"p" using 1 .
have 5:"q" using 3 .
then have 6:"p⟶q" ..
have "r" using assms 6 ..}
hence "q⟶r" ..}
thus "p⟶(q⟶r)" ..
qed

lemma ejercicio_12:
assumes "(p ⟶ q) ⟶ r"
shows "p ⟶ (q ⟶ r)"
proof-
{assume 1 : "p"
{assume 2 : "q"
{assume 3 : "p"
have 4 : "q" using 2 .}
hence 5 : "p⟶q" by (rule impI)
have 6 : "r" using assms(1) 5 by (rule mp)}
hence "q⟶r" by (rule impI)}
thus "p⟶(q⟶r)" by (rule impI)
qed

section {* Conjunciones *}

text {* ---------------------------------------------------------------
Ejercicio 13. Demostrar
p, q ⊢ p ∧ q
------------------------------------------------------------------ *}

lemma ejercicio_13:
assumes "p"
"q"
shows "p ∧ q"

proof-
show "p∧q" using assms(1) assms(2) by (rule conjI)
qed

lemma ejercicio_13:
assumes "p"
"q"
shows "p ∧ q"
proof-
show "p ∧ q" using assms(1,2) by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 14. Demostrar
p ∧ q ⊢ p
------------------------------------------------------------------ *}

lemma ejercicio_14:
assumes "p ∧ q"
shows "p"

proof-
show "p" using assms by (rule conjunct1)
qed

text {* ---------------------------------------------------------------
Ejercicio 15. Demostrar
p ∧ q ⊢ q
------------------------------------------------------------------ *}

lemma ejercicio_15:
assumes "p ∧ q"
shows "q"

proof-
show "q" using assms by (rule conjunct2)
qed

text {* ---------------------------------------------------------------
Ejercicio 16. Demostrar
p ∧ (q ∧ r) ⊢ (p ∧ q) ∧ r
------------------------------------------------------------------ *}

lemma ejercicio_16:
assumes "p ∧ (q ∧ r)"
shows "(p ∧ q) ∧ r"

proof-
have 1:"p" using assms by (rule conjunct1)
have 2:"q∧r" using assms by (rule conjunct2)
then have 3:"q" by (rule conjunct1)
have 4:"r" using 2 by (rule conjunct2)
have "p∧q" using 1 3 by (rule conjI)
then show "(p∧q)∧r" using 4 by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 17. Demostrar
(p ∧ q) ∧ r ⊢ p ∧ (q ∧ r)
------------------------------------------------------------------ *}

lemma ejercicio_17:
assumes "(p ∧ q) ∧ r"
shows "p ∧ (q ∧ r)"

proof-
have 1:"p∧q" using assms by (rule conjunct1)
have 2:"r" using assms by (rule conjunct2)
have 3:"p" using 1 by (rule conjunct1)
have 4:"q" using 1 by (rule conjunct2)
have 5:"q∧r" using 4 2 by (rule conjI)
show "p∧(q∧r)" using 3 5 by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 18. Demostrar
p ∧ q ⊢ p ⟶ q
------------------------------------------------------------------ *}

lemma ejercicio_18:
assumes "p ∧ q"
shows "p ⟶ q"
proof-
have 1:"p" using assms ..
have 2:"q" using assms ..
then show "p⟶q" ..
qed

lemma ejercicio_18:
assumes "p ∧ q"
shows "p ⟶ q"
proof-
{assume "p"
have "q" using assms by (rule conjunct2)}
thus "p⟶q" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 19. Demostrar
(p ⟶ q) ∧ (p ⟶ r) ⊢ p ⟶ q ∧ r
------------------------------------------------------------------ *}

lemma ejercicio_19:
assumes "(p ⟶ q) ∧ (p ⟶ r)"
shows "p ⟶ q ∧ r"
proof-
have 1:"p⟶q" using assms ..
have 2:"p⟶r" using assms ..
{assume 3:"p"
have 4:"q" using 1 3 ..
have 5:"r" using 2 3 ..
have 6:"q∧r" using 4 5 ..}
thus "p ⟶ q ∧ r" ..
qed

lemma ejercicio_19:
assumes "(p ⟶ q) ∧ (p ⟶ r)"
shows "p ⟶ q ∧ r"
proof-
{assume 1 : "p"
have 2 : "p⟶q" using assms by (rule conjunct1)
have 3 : "p⟶r" using assms by (rule conjunct2)
have 4 : "q" using 2 1 by (rule mp)
have 5 : "r" using 3 1 by (rule mp)
have 6 : "q ∧ r" using 4 5 by (rule conjI)}
thus "p⟶(q ∧ r)" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 20. Demostrar
p ⟶ q ∧ r ⊢ (p ⟶ q) ∧ (p ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_20:
assumes "p ⟶ q ∧ r"
shows "(p ⟶ q) ∧ (p ⟶ r)"

proof-
{assume 1: p
have 2:"q∧r" using assms 1 ..
have "r" using 2 ..}
hence 1:"p⟶r" ..
{assume 1: p
have 2:"q∧r" using assms 1 ..
have "q" using 2 ..}
hence 2:"p⟶q" ..
then show "(p ⟶ q) ∧ (p ⟶ r)" using 1 ..
qed

lemma ejercicio_20:
assumes "p ⟶ q ∧ r"
shows "(p ⟶ q) ∧ (p ⟶ r)"
proof-
{assume 1 : "p"
have 2 : "q ∧ r" using assms 1 by (rule mp)
have 3 : "q" using 2 by (rule conjunct1)}
hence 4 : "p⟶q" by (rule impI)
{assume 5 : "p"
have 6 : "q ∧ r" using assms 5 by (rule mp)
have 7 : "r" using 6 by (rule conjunct2)}
hence 8 : "p⟶r" by (rule impI)
show "(p ⟶ q) ∧ (p ⟶ r)" using 4 8 by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 21. Demostrar
p ⟶ (q ⟶ r) ⊢ p ∧ q ⟶ r
------------------------------------------------------------------ *}

lemma ejercicio_21:
assumes "p ⟶ (q ⟶ r)"
shows "p ∧ q ⟶ r"

proof-
{assume 1:"p∧q"
have 2:"p" using 1 ..
have 3:"q⟶r" using assms 2 ..
have 4: "q" using 1 ..
have "r" using 3 4 ..}
thus "p∧q ⟶r" ..
qed

lemma ejercicio_21:
assumes "p ⟶ (q ⟶ r)"
shows "p ∧ q ⟶ r"
proof-
{assume 1 : "p ∧ q"
have 2 : "p" using 1 by (rule conjunct1)
have 3 : "q" using 1 by (rule conjunct2)
have 4 : "q⟶r" using assms 2 by (rule mp)
have 5 : "r" using 4 3 by (rule mp)}
thus "p ∧ q ⟶ r" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 22. Demostrar
p ∧ q ⟶ r ⊢ p ⟶ (q ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_22:
assumes "p ∧ q ⟶ r"
shows "p ⟶ (q ⟶ r)"

proof-
{assume 1:"p"
{assume 2:"q"
have 3:"p∧q" using 1 2 ..
have "r" using assms 3 ..}
hence "q⟶r" ..}
thus "p⟶(q⟶r)" ..
qed

lemma ejercicio_22:
assumes "p ∧ q ⟶ r"
shows "p ⟶ (q ⟶ r)"
proof-
{assume 1 : "p"
{assume 2 : "q"
have 3 : "p ∧ q" using 1 2 by (rule conjI)
have 4 : "r" using assms 3 by (rule mp)}
hence "q⟶r" by (rule impI)}
thus "p⟶(q⟶r)" by (rule impI)
qed

lemma ejercicio_22:
assumes 1:"p ∧ q ⟶ r"
shows "p ⟶ (q ⟶ r)"
proof (rule impI)
assume 2:"p"
show "q ⟶ r"
proof (rule impI)
assume 3:"q"
have 4:"p ∧ q" using 2 3 by (rule conjI)
show "r" using 1 4 by (rule mp)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 23. Demostrar
(p ⟶ q) ⟶ r ⊢ p ∧ q ⟶ r
------------------------------------------------------------------ *}

lemma ejercicio_23:
assumes "(p ⟶ q) ⟶ r"
shows "p ∧ q ⟶ r"

proof-
{assume 1:"p∧q"
have 2:"p" using 1 ..
{assume "p"
have 3: "q" using 1 ..}
hence 4:"p⟶q" ..
have "r" using assms 4 ..}
thus "p∧q ⟶r" ..
qed

lemma ejercicio_23:
assumes "(p ⟶ q) ⟶ r"
shows "p ∧ q ⟶ r"
proof-
{assume 1 : "p ∧ q"
{assume 2 : "p"
have 3 : "q" using 1 by (rule conjunct2)}
hence 4 : "p⟶q" by (rule impI)
have 5 : "r" using assms 4 by (rule mp)}
thus "p ∧ q ⟶ r" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 24. Demostrar
p ∧ (q ⟶ r) ⊢ (p ⟶ q) ⟶ r
------------------------------------------------------------------ *}

lemma ejercicio_24:
assumes "p ∧ (q ⟶ r)"
shows "(p ⟶ q) ⟶ r"

proof-
{assume 1: "p⟶q"
have 2:"p" using assms ..
have 3:"q⟶r" using assms..
have 4:"q" using 1 2 ..
have "r" using 3 4 ..}
thus "(p⟶q)⟶r" ..
qed

lemma ejercicio_24:
assumes "p ∧ (q ⟶ r)"
shows "(p ⟶ q) ⟶ r"
proof-
{assume 1 : "p⟶q"
have 2 : "p" using assms by (rule conjunct1)
have 3 : "q⟶r" using assms by (rule conjunct2)
have 4 : "q" using 1 2 by (rule mp)
have 5 : "r" using 3 4 by (rule mp)}
thus "(p⟶q)⟶r" by (rule impI)
qed

lemma ejercicio_24:
assumes 1:"p ∧ (q ⟶ r)"
shows "(p ⟶ q) ⟶ r"
proof (rule impI)
assume 2:"p ⟶ q"
have 3:"p" using 1 by (rule conjunct1)
have 4:"q" using 2 3 by (rule mp)
have 5:"q ⟶ r" using 1 by (rule conjunct2)
show "r" using 5 4 by (rule mp)
qed

section {* Disyunciones *}

text {* ---------------------------------------------------------------
Ejercicio 25. Demostrar
p ⊢ p ∨ q
------------------------------------------------------------------ *}

lemma ejercicio_25:
assumes "p"
shows "p ∨ q"

proof-
show "p∨q" using assms by (rule disjI1)
qed

text {* ---------------------------------------------------------------
Ejercicio 26. Demostrar
q ⊢ p ∨ q
------------------------------------------------------------------ *}

lemma ejercicio_26:
assumes "q"
shows "p ∨ q"

proof-
show "p∨q" using assms by (rule disjI2)
qed

text {* ---------------------------------------------------------------
Ejercicio 27. Demostrar
p ∨ q ⊢ q ∨ p
------------------------------------------------------------------ *}

lemma ejercicio_27:
assumes "p ∨ q"
shows "q ∨ p"

proof -
have "p ∨ q" using assms by this
moreover
{ assume 2: "p"
have "q ∨ p" using 2 by (rule disjI2) }
moreover
{ assume 3: "q"
have "q ∨ p" using 3 by (rule disjI1) }
ultimately show "q ∨ p" by (rule disjE)
qed

lemma ejercicio_27:
assumes 1:"p ∨ q"
shows "q ∨ p"
using 1
proof (rule disjE)
{assume 2:"p"
show "q ∨ p" using 2 by (rule disjI2)}
next
{assume 4:"q"
show "q ∨ p" using 4 by (rule disjI1)}
qed

text {* ---------------------------------------------------------------
Ejercicio 28. Demostrar
q ⟶ r ⊢ p ∨ q ⟶ p ∨ r
------------------------------------------------------------------ *}

lemma ejercicio_28:
assumes "q ⟶ r"
shows "p ∨ q ⟶ p ∨ r"
proof-
{assume "p ∨ q"
moreover
{ assume 1:"p"
have "p ∨ r" using 1 by (rule disjI1)}
moreover
{ assume 2:"q"
have 3:"r" using assms 2 by (rule mp)
have "p ∨ r" using 3 by (rule disjI2) }
ultimately have " p ∨ r" by (rule disjE)}
thus "p ∨ q ⟶ p ∨ r" by (rule impI)
qed

lemma ejercicio_28:
assumes 1:"q ⟶ r"
shows "p ∨ q ⟶ p ∨ r"
proof (rule impI)
assume 2:"p ∨ q"
thus "p ∨ r"
proof (rule disjE)
{assume 3:"p"
show "p ∨ r" using 3 by (rule disjI1)}
next
{assume 4:"q"
have 5:"r" using 1 4 by (rule mp)
show "p ∨ r" using 5 by (rule disjI2)}
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 29. Demostrar
p ∨ p ⊢ p
------------------------------------------------------------------ *}

lemma ejercicio_29:
assumes "p ∨ p"
shows "p"

proof-
have 1: "p ∨ p" using assms by this
moreover
{assume 1:"p"
have "p" using 1 by this}
moreover
{assume 2:"p"
have "p" using 2 by this}
ultimately show "p" by (rule disjE)

qed
text{* Otra forma:
using assms
proof
assume "p"
thus "p" .
qed}

lemma ejercicio_29:
assumes "p ∨ p"
shows "p"
proof-
have "p ∨ p" using assms by this
moreover
{assume 1 : "p"
have "p" using 1 .}
moreover
{assume 2 : "p"
have "p" using 2 .}
ultimately show "p" by (rule disjE)
qed*}

lemma ejercicio_29:
assumes 1:"p ∨ p"
shows "p"
using 1
proof (rule disjE)
{assume 2:"p"
show "p" using 2 by this}
next
{assume 3:"p"
show "p" using 3 by this}
qed

text {* ---------------------------------------------------------------
Ejercicio 30. Demostrar
p ⊢ p ∨ p
------------------------------------------------------------------ *}

lemma ejercicio_30:
assumes "p"
shows "p ∨ p"
proof-
show "p ∨ p" using assms by (rule disjI1)
qed

text{* Otra forma:
using assms
proof
qed}

text {* ---------------------------------------------------------------
Ejercicio 31. Demostrar
p ∨ (q ∨ r) ⊢ (p ∨ q) ∨ r
------------------------------------------------------------------ *}

lemma ejercicio_31:
assumes "p ∨ (q ∨ r)"
shows "(p ∨ q) ∨ r"
proof -
have "p ∨ (q ∨ r)" using assms(1) by this
moreover
{ assume 1: "p"
have 2: "p ∨ q" using 1 by (rule disjI1)
have 3: "(p ∨ q) ∨ r" using 2 by (rule disjI1)}
moreover
{ assume 4: "q ∨ r"
moreover
{ assume 5: "q"
have 6: "p ∨ q" using 5 by (rule disjI2)
have 7: "(p ∨ q) ∨ r" using 6 by (rule disjI1)}
moreover
{ assume 8: "r"
have 9: "(p ∨ q) ∨ r" using 8 by (rule disjI2)}
ultimately have "(p ∨ q) ∨ r" by (rule disjE)}
ultimately show "(p ∨ q) ∨ r" by (rule disjE)
qed

text{* Otra forma:
using assms
proof
assume "p"
then have "p∨q" ..
thus "(p∨q)∨r" ..
next
assume 1:"q∨r"
thus "(p ∨ q) ∨ r"
proof
assume "q"
then have "p∨q" ..
thus 1: "(p ∨ q) ∨ r" ..
next
assume "r"
then have 2:"(p∨q)∨r" ..
then have "q∨r⟶(p ∨ q) ∨ r" ..
thus "(p ∨ q) ∨ r" using 1 ..
qed}*}

lemma ejercicio_31:
assumes 1:"p ∨ (q ∨ r)"
shows "(p ∨ q) ∨ r"
using 1
proof (rule disjE)
{assume 2:"p"
have 3:"p ∨ q" using 2 by (rule disjI1)
show "(p ∨ q) ∨ r" using 3 by (rule disjI1)}
next
{assume 4:"q ∨ r"
thus "(p ∨ q) ∨ r"
proof (rule disjE)
{assume 5:"q"
have 6:"p ∨ q" using 5 by (rule disjI2)
show "(p ∨ q) ∨ r" using 6 by (rule disjI1)}
next
{assume 7:"r"
show "(p ∨ q) ∨ r" using 7 by (rule disjI2)}
qed}
qed

text {* ---------------------------------------------------------------
Ejercicio 32. Demostrar
(p ∨ q) ∨ r ⊢ p ∨ (q ∨ r)
------------------------------------------------------------------ *}

lemma ejercicio_32:
assumes "(p ∨ q) ∨ r"
shows "p ∨ (q ∨ r)"
proof -
have "(p ∨ q) ∨ r" using assms(1) by this
moreover
{ assume 1: "p ∨ q"
moreover
{ assume 2: "p"
have 3: "p ∨ (q ∨ r)" using 2 by (rule disjI1)}
moreover
{ assume 4: "q"
have 5: "q ∨ r" using 4 by (rule disjI1)
have 6: "p ∨ (q ∨ r)" using 5 by (rule disjI2)}
ultimately have "p ∨ (q ∨ r)" by (rule disjE)}
moreover
{ assume 7: "r"
have 8: "q ∨ r" using 7 by (rule disjI2)
have 9: "p ∨ (q ∨ r)" using 8 by (rule disjI2)}
ultimately show "p ∨ (q ∨ r)" by (rule disjE)
qed

lemma ejercicio_32:
assumes 1:"(p ∨ q) ∨ r"
shows "p ∨ (q ∨ r)"
using 1
proof (rule disjE)
{assume 2:"p ∨ q"
thus "p ∨ (q ∨ r)"
proof (rule disjE)
{assume 3:"p"
show "p ∨ (q ∨ r)" using 3 by (rule disjI1)}
next
{assume 4:"q"
have 5:"q ∨ r" using 4 by (rule disjI1)
show "p ∨ (q ∨ r)" using 5 by (rule disjI2)}
qed}
next
{assume 6:"r"
have 7:"q ∨ r" using 6 by (rule disjI2)
show "p ∨ (q ∨ r)" using 7 by (rule disjI2)}
qed

text {* ---------------------------------------------------------------
Ejercicio 33. Demostrar
p ∧ (q ∨ r) ⊢ (p ∧ q) ∨ (p ∧ r)
------------------------------------------------------------------ *}

lemma ejercicio_33:
assumes "p ∧ (q ∨ r)"
shows "(p ∧ q) ∨ (p ∧ r)"
proof -
have 1: "p" using assms(1) by (rule conjunct1)
have 2: "q ∨ r" using assms(1) by (rule conjunct2)
moreover
{ assume 3: "q"
have 4: "p ∧ q" using 1 3 by (rule conjI)
have 5: "(p ∧ q) ∨ (p ∧ r)" using 4 by (rule disjI1)}
moreover
{ assume 6: "r"
have 7: "p ∧ r" using 1 6 by (rule conjI)
have 8: "(p ∧ q) ∨ (p ∧ r)" using 7 by (rule disjI2)}
ultimately show "(p ∧ q) ∨ (p ∧ r)" by (rule disjE)
qed

lemma ejercicio_33:
assumes 1:"p ∧ (q ∨ r)"
shows "(p ∧ q) ∨ (p ∧ r)"
proof -
have 2:"p" using 1 by (rule conjunct1)
have 3:"q ∨ r" using 1 by (rule conjunct2)
thus "(p ∧ q) ∨ (p ∧ r)"
proof (rule disjE)
{assume 4:"q"
have 5:"p ∧ q" using 2 4 by (rule conjI)
show "(p ∧ q) ∨ (p ∧ r)" using 5 by (rule disjI1)}
next
{assume 6:"r"
have 7:"p ∧ r" using 2 6 by (rule conjI)
show "(p ∧ q) ∨ (p ∧ r)" using 7 by (rule disjI2)}
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 34. Demostrar
(p ∧ q) ∨ (p ∧ r) ⊢ p ∧ (q ∨ r)
------------------------------------------------------------------ *}

lemma ejercicio_34:
assumes "(p ∧ q) ∨ (p ∧ r)"
shows "p ∧ (q ∨ r)"
proof -
have "(p ∧ q) ∨ (p ∧ r)" using assms(1) by this
moreover
{ assume 1: "p ∧ q"
have 2: "q" using 1 by (rule conjunct2)
have 3: "q ∨ r" using 2 by (rule disjI1)
have 4: "p" using 1 by (rule conjunct1)
have 5: "p ∧ (q ∨ r)" using 4 3 by (rule conjI)}
moreover
{ assume 6: "p ∧ r"
have 7: "r" using 6 by (rule conjunct2)
have 8: "q ∨ r" using 7 by (rule disjI2)
have 9: "p" using 6 by (rule conjunct1)
have 10: "p ∧ (q ∨ r)" using 9 8 by (rule conjI)}
ultimately show "p ∧ (q ∨ r)" by (rule disjE)
qed

lemma ejercicio_34:
assumes 1:"(p ∧ q) ∨ (p ∧ r)"
shows "p ∧ (q ∨ r)"
using 1
proof (rule disjE)
{assume 2:"p ∧ q"
have 3:"p" using 2 by (rule conjunct1)
have 4:"q" using 2 by (rule conjunct2)
have 5:"q ∨ r" using 4 by (rule disjI1)
show "p ∧ (q ∨ r)" using 3 5 by (rule conjI)}
next
{assume 6:"p ∧ r"
have 7:"p" using 6 by (rule conjunct1)
have 8:"r" using 6 by (rule conjunct2)
have 9:"q ∨ r" using 8 by (rule disjI2)
show "p ∧ (q ∨ r)" using 7 9 by (rule conjI)}
qed

text {* ---------------------------------------------------------------
Ejercicio 35. Demostrar
p ∨ (q ∧ r) ⊢ (p ∨ q) ∧ (p ∨ r)
------------------------------------------------------------------ *}

lemma ejercicio_35:
assumes "p ∨ (q ∧ r)"
shows "(p ∨ q) ∧ (p ∨ r)"
proof -
have "p ∨ (q ∧ r)" using assms(1) by this
moreover
{ assume 1: "p"
have 2: "p ∨ q" using 1 by (rule disjI1)
have 3: "p ∨ r" using 1 by (rule disjI1)
have 4: "(p ∨ q) ∧ (p ∨ r)" using 2 3 by (rule conjI)}
moreover
{ assume 5: "q ∧ r"
have 6: "q" using 5 by (rule conjunct1)
have 7: "p ∨ q" using 6 by (rule disjI2)
have 8: "r" using 5 by (rule conjunct2)
have 9: "p ∨ r" using 8 by (rule disjI2)
have 10: "(p ∨ q) ∧ (p ∨ r)" using 7 9 by (rule conjI)}
ultimately show "(p ∨ q) ∧ (p ∨ r)" by (rule disjE)
qed

lemma ejercicio_35:
assumes 1:"p ∨ (q ∧ r)"
shows "(p ∨ q) ∧ (p ∨ r)"
using 1
proof (rule disjE)
{assume 2:"p"
have 3:"p ∨ q" using 2 by (rule disjI1)
have 4:"p ∨ r" using 2 by (rule disjI1)
show "(p ∨ q) ∧ (p ∨ r)" using 3 4 by (rule conjI)}
next
{assume 5:"q ∧ r"
have 6:"q" using 5 by (rule conjunct1)
have 7:"r" using 5 by (rule conjunct2)
have 8:"p ∨ q" using 6 by (rule disjI2)
have 9:"p ∨ r" using 7 by (rule disjI2)
show "(p ∨ q) ∧ (p ∨ r)" using 8 9 by (rule conjI)}
qed

text {* ---------------------------------------------------------------
Ejercicio 36. Demostrar
(p ∨ q) ∧ (p ∨ r) ⊢ p ∨ (q ∧ r)
------------------------------------------------------------------ *}

lemma ejercicio_36:
assumes "(p ∨ q) ∧ (p ∨ r)"
shows "p ∨ (q ∧ r)"
proof-
have "(p ∨ q)" using assms by (rule conjunct1)
moreover
{assume 1:"p"
have "p ∨ (q ∧ r)" using 1 by (rule disjI1) }
moreover
{assume 1:"q"
have "(p ∨ r)" using assms by (rule conjunct2)
moreover
{assume 2:"p"
have "p ∨ (q ∧ r)" using 2 by (rule disjI1)}
moreover
{assume 2:"r"
have 3:"(q ∧ r)" using 1 2 by (rule conjI)
have "p ∨ (q ∧ r)" using 3 by (rule disjI2)}
ultimately have "p ∨ (q ∧ r)" by (rule disjE)}
ultimately show "p ∨ (q ∧ r)" by (rule disjE)
qed

------------------- ----------

proof-
have 1: "p∨q" using assms by (rule conjunct1)
thus "p∨(q∧r)"
proof(rule disjE)
{assume 3: "p"
show "p∨(q∧r)" using 3 by ( rule disjI1)}
next
{assume 4: "q"
have 5:"p∨r" using assms by (rule conjunct2)
thus "p∨(q∧r)"
proof(rule disjE)
{assume 6: "p"
show "p∨(q∧r)" using 6 by (rule disjI1)}
next
{assume 7: "r"
have 8: "q∧r" using 4 7 by (rule conjI)
show "p∨(q∧r)" using 8 by (rule disjI2)}
qed}
qed
qed
text {* ---------------------------------------------------------------
Ejercicio 37. Demostrar
(p ⟶ r) ∧ (q ⟶ r) ⊢ p ∨ q ⟶ r
------------------------------------------------------------------ *}

lemma ejercicio_37:
assumes "(p ⟶ r) ∧ (q ⟶ r)"
shows "p ∨ q ⟶ r"
proof-
{assume "p ∨ q"
moreover
{assume 1:"p"
have 2:"(p ⟶ r)" using assms by (rule conjunct1)
have "r" using 2 1 by (rule mp)}
moreover
{assume 1:"q"
have 2:"(q ⟶ r)" using assms by (rule conjunct2)
have "r" using 2 1 by (rule mp)}
ultimately have "r" by (rule disjE)}
thus "p ∨ q ⟶ r" by (rule impI)
qed

lemma ejercicio_37:
assumes 1:"(p ⟶ r) ∧ (q ⟶ r)"
shows "p ∨ q ⟶ r"
proof (rule impI)
assume 2:"p ∨ q"
thus "r"
proof (rule disjE)
{assume 3:"p"
have 4:"p ⟶ r" using 1 by (rule conjunct1)
show "r" using 4 3 by (rule mp)}
next
{assume 5:"q"
have 6:"q ⟶ r" using 1 by (rule conjunct2)
show "r" using 6 5 by (rule mp)}
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 38. Demostrar
p ∨ q ⟶ r ⊢ (p ⟶ r) ∧ (q ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_38:
assumes "p ∨ q ⟶ r"
shows "(p ⟶ r) ∧ (q ⟶ r)"
proof-
{assume 1 : "p"
have 2 : "p ∨ q" using 1 by (rule disjI1)
have 3 : "r" using assms 2 by (rule mp)}
hence 4 : "p⟶r" by (rule impI)
{assume 5 : "q"
have 6 : "p ∨ q" using 5 by (rule disjI2)
have 7 : "r" using assms 6 by (rule mp)}
hence 8 : "q⟶r" by (rule impI)
show "(p ⟶ r) ∧ (q ⟶ r)" using 4 8 by (rule conjI)
qed

section {* Negaciones *}

text {* ---------------------------------------------------------------
Ejercicio 39. Demostrar
p ⊢ ¬¬p
------------------------------------------------------------------ *}

lemma ejercicio_39:
assumes "p"
shows "¬¬p"
proof-
show "¬¬p" using assms by (rule notnotI)
qed

text {* ---------------------------------------------------------------
Ejercicio 40. Demostrar
¬p ⊢ p ⟶ q
------------------------------------------------------------------ *}

lemma ejercicio_40:
assumes "¬p"
shows "p ⟶ q"

proof-
{assume 1: "p"
have "q" using assms 1 ..}
thus "p⟶q" ..
qed
-------------------------------
lemma ejercicio_40:
assumes "¬p"
shows "p ⟶ q"
proof-
{assume 1 : "p"
have 2 : "False" using assms 1 by (rule notE)
have 3 : "q" using 2 by (rule FalseE)}
thus "p⟶q" by (rule impI)
qed

------------------------------
lemma ejercicio_40:
assumes "¬p"
shows "p ⟶ q"

proof-
{assume 1: "p"
have 2: "q" using assms 1 by (rule notE)}
thus "p ⟶ q" by ( rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 41. Demostrar
p ⟶ q ⊢ ¬q ⟶ ¬p
------------------------------------------------------------------ *}

lemma ejercicio_41:
assumes "p ⟶ q"
shows "¬q ⟶ ¬p"
proof-
{assume 1:"¬q"
have "¬p" using assms 1 by (rule mt)}
thus "¬q ⟶ ¬p" ..
qed

text {* ---------------------------------------------------------------
Ejercicio 42. Demostrar
p∨q, ¬q ⊢ p
------------------------------------------------------------------ *}

lemma ejercicio_42:
assumes "p∨q"
"¬q"
shows "p"

using assms(1)
proof
assume "p"
thus "p" .
next
assume 1:"q"
show "p" using assms(2) 1 ..
qed

lemma ejercicio_42:
assumes "p∨q"
"¬q"
shows "p"
proof-
have "p ∨ q" using assms(1) by this
moreover
{assume 1 : "p"
have 2 : "p" using 1 .}
moreover
{assume 3 : "q"
have 4 : "¬q" using assms(2) .
have 5 : "False" using 4 3 by (rule notE)
have 6 : "p" using 5 by (rule FalseE)}
ultimately show "p" by (rule disjE)
qed

--------**---
using assms(1)
proof(rule disjE)
{assume 1: "p"
have "p" using 1 by this}
next
{assume 2: "q"
show "p" using assms(2) 2 by (rule notE)}
qed

text {* ---------------------------------------------------------------
Ejercicio 43. Demostrar
p ∨ q, ¬p ⊢ q
------------------------------------------------------------------ *}

lemma ejercicio_43:
assumes "p ∨ q"
"¬p"
shows "q"

using assms(1)
proof
assume "q"
thus "q" .
next
assume 1:"p"
show "q" using assms(2) 1 ..
qed

lemma ejercicio_43:
assumes "p ∨ q"
"¬p"
shows "q"
proof-
have "p ∨ q" using assms(1) by this
moreover
{assume 1 : "p"
have 2 : "¬p" using assms(2) .
have 3 : "False" using 2 1 by (rule notE)
have 4 : "q" using 3 by (rule FalseE)}
moreover
{assume 5 : "q"
have 6 : "q" using 5 .}
ultimately show "q" by (rule disjE)
qed

---**--
using assms(1)
proof(rule disjE)
assume 1: "p"
show "q" using assms(2) 1 by (rule notE)
next
assume 2:"p"
have "p" using 2 by this
qed

text {* ---------------------------------------------------------------
Ejercicio 44. Demostrar
p ∨ q ⊢ ¬(¬p ∧ ¬q)
------------------------------------------------------------------ *}

lemma ejercicio_44:
assumes "p ∨ q"
shows "¬(¬p ∧ ¬q)"
proof-
{assume 1 : "¬p ∧ ¬q"
have "p ∨ q" using assms by this
moreover
{assume 2 : "p"
have 3 : "¬p" using 1 by (rule conjunct1)
have 4 : "False" using 3 2 by (rule notE)}
moreover
{assume 5 : "q"
have 6 : "¬q" using 1 by (rule conjunct2)
have 7 : "False" using 6 5 by (rule notE)}
ultimately have 8 : "False" by (rule disjE)}
thus "¬(¬p ∧ ¬q)" by (rule notI)
qed

---**
proof(rule notI)
assume 2 :"¬p∧¬q"
show False
using assms
proof(rule disjE)
{assume 1: "p"
have 3: "¬p" using 2 by (rule conjunct1)
show False using 3 1 by (rule notE)}
{assume 4: "q"
have 5: "¬q" using 2 by (rule conjunct2)
show False using 5 4 by (rule notE)}
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 45. Demostrar
p ∧ q ⊢ ¬(¬p ∨ ¬q)
------------------------------------------------------------------ *}

lemma ejercicio_45:
assumes "p ∧ q"
shows "¬(¬p ∨ ¬q)"
proof-
{assume 1 : "¬p ∨ ¬q"
have "¬p ∨ ¬q" using 1 by this
moreover
{assume 2 : "¬p"
have 3 : "p" using assms by (rule conjunct1)
have 4 : "False" using 2 3 by (rule notE)}
moreover
{assume 5 : "¬q"
have 6 : "q" using assms by (rule conjunct2)
have 7 : "False" using 5 6 by (rule notE)}
ultimately have "False" by (rule disjE)}
thus "¬(¬p ∨ ¬q)" by (rule notI)
qed

---**

assumes "p ∧ q"
shows "¬(¬p ∨ ¬q)"

proof (rule notI)
assume 1: "¬p∨¬q"
show False
using 1
proof(rule disjE)
assume 2: "¬p"
have 3: "p" using assms by (rule conjunct1)
show False using 2 3 by (rule notE)
next
assume 4: "¬q"
have 5: "q" using assms by (rule conjunct2)
show False using 4 5 by (rule notE)
qed
qed
text {* ---------------------------------------------------------------
Ejercicio 46. Demostrar
¬(p ∨ q) ⊢ ¬p ∧ ¬q
------------------------------------------------------------------ *}

lemma ejercicio_46:
assumes "¬(p ∨ q)"
shows "¬p ∧ ¬q"
proof-
{assume 1 : "p"
have 2 : "p ∨ q" using 1 by (rule disjI1)
have 3 : "False" using assms 2 by (rule notE)}
hence 4 : "¬p" by (rule notI)
{assume 5 : "q"
have 6 : "p ∨ q" using 5 by (rule disjI2)
have 7 : "False" using assms 6 by (rule notE)}
hence 8 : "¬q" by (rule notI)
show "¬p ∧ ¬q" using 4 8 by (rule conjI)
qed

(*Otra opción para usar tercio excluso*)

lemma ejercicio_46_2:
assumes "¬(p ∨ q)"
shows "¬p ∧ ¬q"
proof (rule disjE)(rule disjE)
have 1:"¬p | p" by (rule excluded_middle)
show "¬p & ¬q"
using 1
proof (rule disjE)
{assume 2:"¬p"
have 3:"¬q | q" by (rule excluded_middle)
show "¬p & ¬q"
using 3
proof (rule disjE)
{assume 4:"¬q"
with 2 show "¬p & ¬q" by (rule conjI)}
next
{assume "q"
then have "p | q" by (rule disjI2)
with assms show "¬p & ¬q" by (rule notE)}
qed
}
next
{assume 5:"p"
then have "p | q" by (rule disjI1)
with assms show "¬p & ¬q" by (rule notE)
}
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 47. Demostrar
¬p ∧ ¬q ⊢ ¬(p ∨ q)
------------------------------------------------------------------ *}

lemma ejercicio_47:
assumes "¬p ∧ ¬q"
shows "¬(p ∨ q)"
proof-
{assume 1 : "p ∨ q"
have "p ∨ q" using 1 by this
moreover
{assume 2 : "p"
have 3 : "¬p" using assms by (rule conjunct1)
have 4 : "False" using 3 2 by (rule notE)}
moreover
{assume 5 : "q"
have 6 : "¬q" using assms by (rule conjunct2)
have 6 : "False" using 6 5 by (rule notE)}
ultimately have "False" by (rule disjE)}
thus "¬(p ∨ q)" by (rule notI)
qed

---**

proof(rule notI)
assume 1: "p∨q"
show False
using 1
proof(rule disjE)
{assume 2:"p"
have 3:"¬p" using assms by (rule conjunct1)
show "False" using 3 2 by (rule notE)}
next
{assume 5:"q"
have 6:"¬q" using assms by (rule conjunct2)
show "False" using 6 5 by (rule notE)}
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 48. Demostrar
¬p ∨ ¬q ⊢ ¬(p ∧ q)
------------------------------------------------------------------ *}

lemma ejercicio_48:
assumes "¬p ∨ ¬q"
shows "¬(p ∧ q)"
proof-
{assume 1 : "p ∧ q"
have "¬p ∨ ¬q" using assms by this
moreover
{assume 2 : "¬p"
have 3 : "p" using 1 by (rule conjunct1)
have 4 : "False" using 2 3 by (rule notE)}
moreover
{assume 5 : "¬q"
have 6 : "q" using 1 by (rule conjunct2)
have 7 : "False" using 5 6 by (rule notE)}
ultimately have "False" by (rule disjE)}
thus "¬(p ∧ q)" by (rule notI)
qed

--**
lemma ejercicio_48:
assumes "¬p ∨ ¬q"
shows "¬(p ∧ q)"

proof(rule notI)
assume 1:"p∧q"
show False
using assms
proof(rule disjE)
assume 2:"¬p"
have 3: "p" using 1 by (rule conjunct1)
show False using 2 3 by (rule notE)
next
assume 4: "¬q"
have 5: "q" using 1 by (rule conjunct2)
show False using 4 5 by (rule notE)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 49. Demostrar
⊢ ¬(p ∧ ¬p)
------------------------------------------------------------------ *}

lemma ejercicio_49:
"¬(p ∧ ¬p)"
proof-
{assume 1 : "p ∧ ¬p"
have 2 : "p" using 1 by (rule conjunct1)
have 3 : "¬p" using 1 by (rule conjunct2)
have 4 : "False" using 3 2 by (rule notE)}
thus "¬(p ∧ ¬p)" by (rule notI)
qed

---**

proof(rule notI)
assume 1: "p∧¬p"
have 2: "p" using 1 by (rule conjunct1)
have 3: "¬p" using 1 by (rule conjunct2)
show False using 3 2 by (rule notE)
qed

text {* ---------------------------------------------------------------
Ejercicio 50. Demostrar
p ∧ ¬p ⊢ q
------------------------------------------------------------------ *}

lemma ejercicio_50:
assumes "p ∧ ¬p"
shows "q"

proof-
have 1 : "p" using assms by (rule conjunct1)
have 2 : "¬p" using assms by (rule conjunct2)
show "q" using 2 1 by (rule notE)
qed

text {* ---------------------------------------------------------------
Ejercicio 51. Demostrar
¬¬p ⊢ p
------------------------------------------------------------------ *}

lemma ejercicio_51:
assumes "¬¬p"
shows "p"

proof-
show "p" using assms by (rule notnotD)
qed

--**

lemma ejercicio_51:
assumes "¬¬p"
shows "p"
proof(rule ccontr)
assume 1: "¬p"
show False using assms 1 by (rule notE)
qed

text {* ---------------------------------------------------------------
Ejercicio 52. Demostrar
⊢ p ∨ ¬p
------------------------------------------------------------------ *}

lemma ejercicio_52:
"p ∨ ¬p"

proof-
have 1: "¬p∨p" by (rule excluded_middle)
have 2: "¬p∨p" using 1 by this
moreover
{assume 3: "¬p"
have 4: "p∨¬p" using 3 by (rule disjI2)}
moreover
{assume 5: "p"
have 6: "p∨¬p" using 5 by (rule disjI1)}
ultimately show "p∨¬p" by (rule disjE)
qed

text {* ---------------------------------------------------------------
Ejercicio 53. Demostrar
⊢ ((p ⟶ q) ⟶ p) ⟶ p
------------------------------------------------------------------ *}

lemma ejercicio_53:
"((p ⟶ q) ⟶ p) ⟶ p"
proof-
{assume 1 : "(p ⟶ q) ⟶ p"
{assume 2 : "¬p"
have 3 : "¬(p⟶q)" using 1 2 by (rule mt)
{assume 4 : "p"
have 5 : "q" using 2 4 by (rule notE)}
hence 6 : "p⟶q" by (rule impI)
have 7 : "False" using 3 6 by (rule notE)}
hence 8 : "p" by (rule ccontr)}
thus "((p ⟶ q) ⟶ p) ⟶ p" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 54. Demostrar
¬q ⟶ ¬p ⊢ p ⟶ q
------------------------------------------------------------------ *}

lemma ejercicio_54:
assumes "¬q ⟶ ¬p"
shows "p ⟶ q"
proof-
{assume 1 : "p"
{assume 2 : "¬q"
have 3 : "¬p" using assms 2 by (rule mp)
have 4 : "False" using 3 1 by (rule notE)}
hence 5 : "q" by (rule ccontr)}
thus "p⟶q" by (rule impI)
qed

---**

proof(rule impI)
assume 1: "p"
have 2 : "¬¬p" using 1 by (rule notnotI)
have 3: "¬¬q" using assms 2 by (rule mt)
show "q" using 3 by (rule notnotD)
qed

text {* ---------------------------------------------------------------
Ejercicio 55. Demostrar
¬(¬p ∧ ¬q) ⊢ p ∨ q
------------------------------------------------------------------ *}

lemma ejercicio_55:
assumes "¬(¬p ∧ ¬q)"
shows "p ∨ q"
proof-
have 1 : "¬p ∨ p" by (rule excluded_middle)
have "¬p ∨ p" using 1 by this
moreover
{assume 2 : "¬p"
have 3 : "¬q ∨ q" by (rule excluded_middle)
have "¬q ∨ q" using 3 by this
moreover
{assume 4 : "¬q"
have 5 : "¬p ∧ ¬q" using 2 4 by (rule conjI)
have 6 : "False" using assms 5 by (rule notE)
have 7 : "p ∨ q" using 6 by (rule FalseE)}
moreover
{assume 8 : "q"
have 9 : "p ∨ q" using 8 by (rule disjI2)}
ultimately have "p ∨ q" by (rule disjE)}
moreover
{assume 10 : "p"
have 11 : "p ∨ q" using 10 by (rule disjI1)}
ultimately show "p ∨ q" by (rule disjE)
qed

text {* ---------------------------------------------------------------
Ejercicio 56. Demostrar
¬(¬p ∨ ¬q) ⊢ p ∧ q
------------------------------------------------------------------ *}

lemma ejercicio_56:
assumes "¬(¬p ∨ ¬q)"
shows "p ∧ q"
proof-
{assume 1 : "¬p"
have 2 : "¬p ∨ ¬q" using 1 by (rule disjI1)
have 3 : "False" using assms 2 by (rule notE)}
hence 4 : "p" by (rule ccontr)
{assume 5 : "¬q"
have 6 : "¬p ∨ ¬q" using 5 by (rule disjI2)
have 7 : "False" using assms 6 by (rule notE)}
hence 8 : "q" by (rule ccontr)
show "p ∧ q" using 4 8 by (rule conjI)
qed

--**
proof-
{assume 1: "¬p"
have 2: "¬p∨¬q"using 1 by (rule disjI1)
have 3: "False" using assms 2 by (rule notE)}
hence 8: "p" by (rule ccontr)
{assume 4: "¬q"
have 5: "¬p∨¬q" using 4 by (rule disjI2)
have 6: "False" using assms 5 by (rule notE)}
hence 9: "q" by (rule ccontr)
show "p∧q" using 8 9 by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 57. Demostrar
¬(p ∧ q) ⊢ ¬p ∨ ¬q
------------------------------------------------------------------ *}

lemma ejercicio_57:
assumes "¬(p ∧ q)"
shows "¬p ∨ ¬q"
proof-
have 1 : "¬p ∨ p" by (rule excluded_middle)
have "¬p ∨ p" using 1 by this
moreover
{assume 2 : "¬p"
have 3 : "¬p ∨ ¬q" using 2 by (rule disjI1)}
moreover
{assume 4 : "p"
have 5 : "¬q ∨ q" by (rule excluded_middle)
have "¬q ∨ q" using 5 by this
moreover
{assume 6 : "¬q"
have 7 : "¬p ∨ ¬q" using 6 by (rule disjI2)}
moreover
{assume 8 : "q"
have 9 : "p ∧ q" using 4 8 by (rule conjI)
have 10 : "False" using assms 9 by (rule notE)
have 11 : "¬p ∨ ¬q" using 10 by (rule FalseE)}
ultimately have "¬p ∨ ¬q" by (rule disjE)}
ultimately show "¬p ∨ ¬q" by (rule disjE)
qed

text {* ---------------------------------------------------------------
Ejercicio 58. Demostrar
⊢ (p ⟶ q) ∨ (q ⟶ p)
------------------------------------------------------------------ *}

lemma ejercicio_58:
"(p ⟶ q) ∨ (q ⟶ p)"

proof-
have 1 : "¬(p ⟶ q) ∨ (p ⟶ q)" by (rule excluded_middle)
have "¬(p ⟶ q) ∨ (p ⟶ q)" using 1 by this
moreover
{assume 2 : "¬(p ⟶ q)"
{assume 3 : "q"
{assume 4 : "p"
have 5 : "q" using 3 .}
hence 6 : "p⟶q" by (rule impI)
have 7 : "p" using 2 6 by (rule notE)}
hence 8 : "q⟶p" by (rule impI)
have 9 : "(p ⟶ q) ∨ (q ⟶ p)" using 8 by (rule disjI2)}
moreover
{assume 10 : "p ⟶ q"
have 11 : "(p ⟶ q) ∨ (q ⟶ p)" using 10 by (rule disjI1)}
ultimately show "(p ⟶ q) ∨ (q ⟶ p)" by (rule disjE)
qed

end
</source>

Relación 3

2016-03-25T15:46:52Z

Albromgon: Añado otra solución ejercicio 46

<source lang = "isar">header {* R3: Deducción natural proposicional *}

theory R3
imports Main
begin

text {*
---------------------------------------------------------------------
El objetivo de esta relación es demostrar cada uno de los ejercicios
usando sólo las reglas básicas de deducción natural de la lógica
proposicional (sin usar el método auto).

Las reglas básicas de la deducción natural son las siguientes:
· conjI: ⟦P; Q⟧ ⟹ P ∧ Q
· conjunct1: P ∧ Q ⟹ P
· conjunct2: P ∧ Q ⟹ Q
· notnotD: ¬¬ P ⟹ P
· notnotI: P ⟹ ¬¬ P
· mp: ⟦P ⟶ Q; P⟧ ⟹ Q
· mt: ⟦F ⟶ G; ¬G⟧ ⟹ ¬F
· impI: (P ⟹ Q) ⟹ P ⟶ Q
· disjI1: P ⟹ P ∨ Q
· disjI2: Q ⟹ P ∨ Q
· disjE: ⟦P ∨ Q; P ⟹ R; Q ⟹ R⟧ ⟹ R
· FalseE: False ⟹ P
· notE: ⟦¬P; P⟧ ⟹ R
· notI: (P ⟹ False) ⟹ ¬P
· iffI: ⟦P ⟹ Q; Q ⟹ P⟧ ⟹ P = Q
· iffD1: ⟦Q = P; Q⟧ ⟹ P
· iffD2: ⟦P = Q; Q⟧ ⟹ P
· ccontr: (¬P ⟹ False) ⟹ P
---------------------------------------------------------------------
*}

text {*
Se usarán las reglas notnotI y mt que demostramos a continuación. *}

lemma notnotI: "P ⟹ ¬¬ P"
by auto

lemma mt: "⟦F ⟶ G; ¬G⟧ ⟹ ¬F"
by auto

section {* Implicaciones *}

text {* ---------------------------------------------------------------
Ejercicio 1. Demostrar
p ⟶ q, p ⊢ q
------------------------------------------------------------------ *}

lemma ejercicio_1:
assumes "p ⟶ q"
"p"
shows "q"

proof -
show "q" using assms(1,2) by (rule mp)
qed

lemma ejercicio_1_b:
assumes "p ⟶ q"
"p"
shows "q"
proof-
show "q" using assms(1) assms(2) by (rule mp)
qed

text {* ---------------------------------------------------------------
Ejercicio 2. Demostrar
p ⟶ q, q ⟶ r, p ⊢ r
------------------------------------------------------------------ *}

lemma ejercicio_2:
assumes "p ⟶ q"
"q ⟶ r"
"p"
shows "r"

proof-
have 1: "q" using assms(1,3) by (rule mp)
show "r" using assms(2) 1 by (rule mp)
qed

text {* ---------------------------------------------------------------
Ejercicio 3. Demostrar
p ⟶ (q ⟶ r), p ⟶ q, p ⊢ r
------------------------------------------------------------------ *}

lemma ejercicio_3:
assumes "p ⟶ (q ⟶ r)"
"p ⟶ q"
"p"
shows "r"

proof-
have 1: "q" using assms(2,3) by (rule mp)
have 2: "q⟶r" using assms(1,3) by (rule mp)
show "r" using 2 1 by (rule mp)
qed

text {* ---------------------------------------------------------------
Ejercicio 4. Demostrar
p ⟶ q, q ⟶ r ⊢ p ⟶ r
------------------------------------------------------------------ *}

lemma ejercicio_4:
assumes "p ⟶ q"
"q ⟶ r"
shows "p ⟶ r"

proof-
{assume 1: "p"
have 2: "q" using assms(1) 1 ..
have 3: "r" using assms(2) 2 ..}
thus "p⟶r" by (rule impI)
qed

lemma ejercicio_4:
assumes "p ⟶ q"
"q ⟶ r"
shows "p ⟶ r"
proof-
{assume 1 : "p"
have 2 : "q" using assms(1) 1 by (rule mp)
have 3 : "r" using assms(2) 2 by (rule mp)}
thus "p⟶r" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 5. Demostrar
p ⟶ (q ⟶ r) ⊢ q ⟶ (p ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_5:
assumes "p ⟶ (q ⟶ r)"
shows "q ⟶ (p ⟶ r)"

{assume 1: "q"
{assume 2: "p"
have 3: "q⟶r" using assms 2 by (rule mp)
have "r" using 3 1 by (rule mp)}
hence "p⟶r" by (rule impI)}
thus "(q⟶(p⟶r))" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 6. Demostrar
p ⟶ (q ⟶ r) ⊢ (p ⟶ q) ⟶ (p ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_6:
assumes "p ⟶ (q ⟶ r)"
shows "(p ⟶ q) ⟶ (p ⟶ r)"

proof-
{assume 1:"p⟶q"
{assume 2: "p"
have 3: "q⟶r" using assms 2 by (rule mp)
have 4: "q" using 1 2 by (rule mp)
have "r" using 3 4 by (rule mp)}
hence "p⟶r" by (rule impI)}
thus "(p⟶q)⟶(p⟶r)" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 7. Demostrar
p ⊢ q ⟶ p
------------------------------------------------------------------ *}

lemma ejercicio_7:
assumes "p"
shows "q ⟶ p"

proof-
{assume 1: "q"
have "p" using assms .}
thus "q⟶p" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 8. Demostrar
⊢ p ⟶ (q ⟶ p)
------------------------------------------------------------------ *}

lemma ejercicio_8:
"p ⟶ (q ⟶ p)"

proof-
{assume 1:"p"
{assume 2:"q"
have 3: "p" using 1 .}
hence "q⟶p" ..}
thus "p⟶(q⟶p)" ..
qed

lemma ejercicio_8:
"p ⟶ (q ⟶ p)"
proof-
{assume 1 : "p"
{assume 2 : "q"
have 3 : "p" using 1 .}
hence "q⟶p" by (rule impI)}
thus "p⟶(q⟶p)" by (rule impI)
qed

lemma ejercicio_8:
"p ⟶ (q ⟶ p)"
proof (rule impI)
assume 1: "p"
show "q ⟶ p"
proof (rule impI)
assume "q"
show "p" using 1 by this
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 9. Demostrar
p ⟶ q ⊢ (q ⟶ r) ⟶ (p ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_9:
assumes "p ⟶ q"
shows "(q ⟶ r) ⟶ (p ⟶ r)"

proof-
{assume 1:"q⟶r"
{assume 2:"p"
have 3:"q" using assms 2 ..
have "r" using 1 3 ..}
hence "p⟶r" ..}
thus "(q⟶r)⟶(p⟶r) " ..
qed

lemma ejercicio_9:
assumes "p ⟶ q"
shows "(q ⟶ r) ⟶ (p ⟶ r)"
proof-
{assume 1 : "q⟶r"
{assume 2 : "p"
have 3 : "q" using assms(1) 2 by (rule mp)
have 4 : "r" using 1 3 by (rule mp)}
hence "p⟶r" by (rule impI)}
thus "(q⟶r)⟶(p⟶r)" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 10. Demostrar
p ⟶ (q ⟶ (r ⟶ s)) ⊢ r ⟶ (q ⟶ (p ⟶ s))
------------------------------------------------------------------ *}

lemma ejercicio_10:
assumes "p ⟶ (q ⟶ (r ⟶ s))"
shows "r ⟶ (q ⟶ (p ⟶ s))"

proof-
{assume 1:"r"
{assume 2:"q"
{assume 3:"p"
have 4:"q⟶(r⟶s)" using assms 3 ..
have 5: "r⟶s" using 4 2 ..
have 6: "s" using 5 1 ..}
hence "p⟶s" ..}
hence "q⟶(p⟶s)" ..}
thus "r⟶(q⟶(p⟶s))" ..
qed

lemma ejercicio_10:
assumes "p ⟶ (q ⟶ (r ⟶ s))"
shows "r ⟶ (q ⟶ (p ⟶ s))"
proof-
{assume 1 : "r"
{assume 2 : "q"
{assume 3 : "p"
have 4 : "q ⟶ (r ⟶ s)" using assms(1) 3 by (rule mp)
have 5 : "r⟶s" using 4 2 by (rule mp)
have 6 : "s" using 5 1 by (rule mp)}
hence "p⟶s" by (rule impI)}
hence "q⟶(p⟶s)" by (rule impI)}
thus "r⟶(q⟶(p⟶s))" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 11. Demostrar
⊢ (p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))
------------------------------------------------------------------ *}

lemma ejercicio_11:
"(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))"

proof (rule impI)
assume 3: "p⟶(q⟶r)"
show "(p⟶q)⟶(p⟶r)"
proof(rule impI)
assume 2:"p⟶q"
show "p⟶r"
proof (rule impI)
assume 1: "p"
have 4:"q" using 2 1 ..
have 5:"q⟶r" using 3 1 ..
show "r" using 5 4 ..
qed
qed
qed

lemma ejercicio_11:
"(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))"
proof-
{assume 1 : "p ⟶ (q ⟶ r)"
{assume 2 : "p⟶q"
{assume 3 : "p"
have 4 : "q⟶r" using 1 3 by (rule mp)
have 5 : "q" using 2 3 by (rule mp)
have 6 : "r" using 4 5 by (rule mp)}
hence "p⟶r" by (rule impI)}
hence "(p ⟶ q) ⟶ (p ⟶ r)" by (rule impI)}
thus "(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 12. Demostrar
(p ⟶ q) ⟶ r ⊢ p ⟶ (q ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_12:
assumes "(p ⟶ q) ⟶ r"
shows "p ⟶ (q ⟶ r)"

proof-
{assume 1:"p"
{assume 3:"q"
have 4:"p" using 1 .
have 5:"q" using 3 .
then have 6:"p⟶q" ..
have "r" using assms 6 ..}
hence "q⟶r" ..}
thus "p⟶(q⟶r)" ..
qed

lemma ejercicio_12:
assumes "(p ⟶ q) ⟶ r"
shows "p ⟶ (q ⟶ r)"
proof-
{assume 1 : "p"
{assume 2 : "q"
{assume 3 : "p"
have 4 : "q" using 2 .}
hence 5 : "p⟶q" by (rule impI)
have 6 : "r" using assms(1) 5 by (rule mp)}
hence "q⟶r" by (rule impI)}
thus "p⟶(q⟶r)" by (rule impI)
qed

section {* Conjunciones *}

text {* ---------------------------------------------------------------
Ejercicio 13. Demostrar
p, q ⊢ p ∧ q
------------------------------------------------------------------ *}

lemma ejercicio_13:
assumes "p"
"q"
shows "p ∧ q"

proof-
show "p∧q" using assms(1) assms(2) by (rule conjI)
qed

lemma ejercicio_13:
assumes "p"
"q"
shows "p ∧ q"
proof-
show "p ∧ q" using assms(1,2) by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 14. Demostrar
p ∧ q ⊢ p
------------------------------------------------------------------ *}

lemma ejercicio_14:
assumes "p ∧ q"
shows "p"

proof-
show "p" using assms by (rule conjunct1)
qed

text {* ---------------------------------------------------------------
Ejercicio 15. Demostrar
p ∧ q ⊢ q
------------------------------------------------------------------ *}

lemma ejercicio_15:
assumes "p ∧ q"
shows "q"

proof-
show "q" using assms by (rule conjunct2)
qed

text {* ---------------------------------------------------------------
Ejercicio 16. Demostrar
p ∧ (q ∧ r) ⊢ (p ∧ q) ∧ r
------------------------------------------------------------------ *}

lemma ejercicio_16:
assumes "p ∧ (q ∧ r)"
shows "(p ∧ q) ∧ r"

proof-
have 1:"p" using assms by (rule conjunct1)
have 2:"q∧r" using assms by (rule conjunct2)
then have 3:"q" by (rule conjunct1)
have 4:"r" using 2 by (rule conjunct2)
have "p∧q" using 1 3 by (rule conjI)
then show "(p∧q)∧r" using 4 by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 17. Demostrar
(p ∧ q) ∧ r ⊢ p ∧ (q ∧ r)
------------------------------------------------------------------ *}

lemma ejercicio_17:
assumes "(p ∧ q) ∧ r"
shows "p ∧ (q ∧ r)"

proof-
have 1:"p∧q" using assms by (rule conjunct1)
have 2:"r" using assms by (rule conjunct2)
have 3:"p" using 1 by (rule conjunct1)
have 4:"q" using 1 by (rule conjunct2)
have 5:"q∧r" using 4 2 by (rule conjI)
show "p∧(q∧r)" using 3 5 by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 18. Demostrar
p ∧ q ⊢ p ⟶ q
------------------------------------------------------------------ *}

lemma ejercicio_18:
assumes "p ∧ q"
shows "p ⟶ q"
proof-
have 1:"p" using assms ..
have 2:"q" using assms ..
then show "p⟶q" ..
qed

lemma ejercicio_18:
assumes "p ∧ q"
shows "p ⟶ q"
proof-
{assume "p"
have "q" using assms by (rule conjunct2)}
thus "p⟶q" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 19. Demostrar
(p ⟶ q) ∧ (p ⟶ r) ⊢ p ⟶ q ∧ r
------------------------------------------------------------------ *}

lemma ejercicio_19:
assumes "(p ⟶ q) ∧ (p ⟶ r)"
shows "p ⟶ q ∧ r"
proof-
have 1:"p⟶q" using assms ..
have 2:"p⟶r" using assms ..
{assume 3:"p"
have 4:"q" using 1 3 ..
have 5:"r" using 2 3 ..
have 6:"q∧r" using 4 5 ..}
thus "p ⟶ q ∧ r" ..
qed

lemma ejercicio_19:
assumes "(p ⟶ q) ∧ (p ⟶ r)"
shows "p ⟶ q ∧ r"
proof-
{assume 1 : "p"
have 2 : "p⟶q" using assms by (rule conjunct1)
have 3 : "p⟶r" using assms by (rule conjunct2)
have 4 : "q" using 2 1 by (rule mp)
have 5 : "r" using 3 1 by (rule mp)
have 6 : "q ∧ r" using 4 5 by (rule conjI)}
thus "p⟶(q ∧ r)" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 20. Demostrar
p ⟶ q ∧ r ⊢ (p ⟶ q) ∧ (p ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_20:
assumes "p ⟶ q ∧ r"
shows "(p ⟶ q) ∧ (p ⟶ r)"

proof-
{assume 1: p
have 2:"q∧r" using assms 1 ..
have "r" using 2 ..}
hence 1:"p⟶r" ..
{assume 1: p
have 2:"q∧r" using assms 1 ..
have "q" using 2 ..}
hence 2:"p⟶q" ..
then show "(p ⟶ q) ∧ (p ⟶ r)" using 1 ..
qed

lemma ejercicio_20:
assumes "p ⟶ q ∧ r"
shows "(p ⟶ q) ∧ (p ⟶ r)"
proof-
{assume 1 : "p"
have 2 : "q ∧ r" using assms 1 by (rule mp)
have 3 : "q" using 2 by (rule conjunct1)}
hence 4 : "p⟶q" by (rule impI)
{assume 5 : "p"
have 6 : "q ∧ r" using assms 5 by (rule mp)
have 7 : "r" using 6 by (rule conjunct2)}
hence 8 : "p⟶r" by (rule impI)
show "(p ⟶ q) ∧ (p ⟶ r)" using 4 8 by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 21. Demostrar
p ⟶ (q ⟶ r) ⊢ p ∧ q ⟶ r
------------------------------------------------------------------ *}

lemma ejercicio_21:
assumes "p ⟶ (q ⟶ r)"
shows "p ∧ q ⟶ r"

proof-
{assume 1:"p∧q"
have 2:"p" using 1 ..
have 3:"q⟶r" using assms 2 ..
have 4: "q" using 1 ..
have "r" using 3 4 ..}
thus "p∧q ⟶r" ..
qed

lemma ejercicio_21:
assumes "p ⟶ (q ⟶ r)"
shows "p ∧ q ⟶ r"
proof-
{assume 1 : "p ∧ q"
have 2 : "p" using 1 by (rule conjunct1)
have 3 : "q" using 1 by (rule conjunct2)
have 4 : "q⟶r" using assms 2 by (rule mp)
have 5 : "r" using 4 3 by (rule mp)}
thus "p ∧ q ⟶ r" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 22. Demostrar
p ∧ q ⟶ r ⊢ p ⟶ (q ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_22:
assumes "p ∧ q ⟶ r"
shows "p ⟶ (q ⟶ r)"

proof-
{assume 1:"p"
{assume 2:"q"
have 3:"p∧q" using 1 2 ..
have "r" using assms 3 ..}
hence "q⟶r" ..}
thus "p⟶(q⟶r)" ..
qed

lemma ejercicio_22:
assumes "p ∧ q ⟶ r"
shows "p ⟶ (q ⟶ r)"
proof-
{assume 1 : "p"
{assume 2 : "q"
have 3 : "p ∧ q" using 1 2 by (rule conjI)
have 4 : "r" using assms 3 by (rule mp)}
hence "q⟶r" by (rule impI)}
thus "p⟶(q⟶r)" by (rule impI)
qed

lemma ejercicio_22:
assumes 1:"p ∧ q ⟶ r"
shows "p ⟶ (q ⟶ r)"
proof (rule impI)
assume 2:"p"
show "q ⟶ r"
proof (rule impI)
assume 3:"q"
have 4:"p ∧ q" using 2 3 by (rule conjI)
show "r" using 1 4 by (rule mp)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 23. Demostrar
(p ⟶ q) ⟶ r ⊢ p ∧ q ⟶ r
------------------------------------------------------------------ *}

lemma ejercicio_23:
assumes "(p ⟶ q) ⟶ r"
shows "p ∧ q ⟶ r"

proof-
{assume 1:"p∧q"
have 2:"p" using 1 ..
{assume "p"
have 3: "q" using 1 ..}
hence 4:"p⟶q" ..
have "r" using assms 4 ..}
thus "p∧q ⟶r" ..
qed

lemma ejercicio_23:
assumes "(p ⟶ q) ⟶ r"
shows "p ∧ q ⟶ r"
proof-
{assume 1 : "p ∧ q"
{assume 2 : "p"
have 3 : "q" using 1 by (rule conjunct2)}
hence 4 : "p⟶q" by (rule impI)
have 5 : "r" using assms 4 by (rule mp)}
thus "p ∧ q ⟶ r" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 24. Demostrar
p ∧ (q ⟶ r) ⊢ (p ⟶ q) ⟶ r
------------------------------------------------------------------ *}

lemma ejercicio_24:
assumes "p ∧ (q ⟶ r)"
shows "(p ⟶ q) ⟶ r"

proof-
{assume 1: "p⟶q"
have 2:"p" using assms ..
have 3:"q⟶r" using assms..
have 4:"q" using 1 2 ..
have "r" using 3 4 ..}
thus "(p⟶q)⟶r" ..
qed

lemma ejercicio_24:
assumes "p ∧ (q ⟶ r)"
shows "(p ⟶ q) ⟶ r"
proof-
{assume 1 : "p⟶q"
have 2 : "p" using assms by (rule conjunct1)
have 3 : "q⟶r" using assms by (rule conjunct2)
have 4 : "q" using 1 2 by (rule mp)
have 5 : "r" using 3 4 by (rule mp)}
thus "(p⟶q)⟶r" by (rule impI)
qed

lemma ejercicio_24:
assumes 1:"p ∧ (q ⟶ r)"
shows "(p ⟶ q) ⟶ r"
proof (rule impI)
assume 2:"p ⟶ q"
have 3:"p" using 1 by (rule conjunct1)
have 4:"q" using 2 3 by (rule mp)
have 5:"q ⟶ r" using 1 by (rule conjunct2)
show "r" using 5 4 by (rule mp)
qed

section {* Disyunciones *}

text {* ---------------------------------------------------------------
Ejercicio 25. Demostrar
p ⊢ p ∨ q
------------------------------------------------------------------ *}

lemma ejercicio_25:
assumes "p"
shows "p ∨ q"

proof-
show "p∨q" using assms by (rule disjI1)
qed

text {* ---------------------------------------------------------------
Ejercicio 26. Demostrar
q ⊢ p ∨ q
------------------------------------------------------------------ *}

lemma ejercicio_26:
assumes "q"
shows "p ∨ q"

proof-
show "p∨q" using assms by (rule disjI2)
qed

text {* ---------------------------------------------------------------
Ejercicio 27. Demostrar
p ∨ q ⊢ q ∨ p
------------------------------------------------------------------ *}

lemma ejercicio_27:
assumes "p ∨ q"
shows "q ∨ p"

proof -
have "p ∨ q" using assms by this
moreover
{ assume 2: "p"
have "q ∨ p" using 2 by (rule disjI2) }
moreover
{ assume 3: "q"
have "q ∨ p" using 3 by (rule disjI1) }
ultimately show "q ∨ p" by (rule disjE)
qed

lemma ejercicio_27:
assumes 1:"p ∨ q"
shows "q ∨ p"
using 1
proof (rule disjE)
{assume 2:"p"
show "q ∨ p" using 2 by (rule disjI2)}
next
{assume 4:"q"
show "q ∨ p" using 4 by (rule disjI1)}
qed

text {* ---------------------------------------------------------------
Ejercicio 28. Demostrar
q ⟶ r ⊢ p ∨ q ⟶ p ∨ r
------------------------------------------------------------------ *}

lemma ejercicio_28:
assumes "q ⟶ r"
shows "p ∨ q ⟶ p ∨ r"
proof-
{assume "p ∨ q"
moreover
{ assume 1:"p"
have "p ∨ r" using 1 by (rule disjI1)}
moreover
{ assume 2:"q"
have 3:"r" using assms 2 by (rule mp)
have "p ∨ r" using 3 by (rule disjI2) }
ultimately have " p ∨ r" by (rule disjE)}
thus "p ∨ q ⟶ p ∨ r" by (rule impI)
qed

lemma ejercicio_28:
assumes 1:"q ⟶ r"
shows "p ∨ q ⟶ p ∨ r"
proof (rule impI)
assume 2:"p ∨ q"
thus "p ∨ r"
proof (rule disjE)
{assume 3:"p"
show "p ∨ r" using 3 by (rule disjI1)}
next
{assume 4:"q"
have 5:"r" using 1 4 by (rule mp)
show "p ∨ r" using 5 by (rule disjI2)}
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 29. Demostrar
p ∨ p ⊢ p
------------------------------------------------------------------ *}

lemma ejercicio_29:
assumes "p ∨ p"
shows "p"

proof-
have 1: "p ∨ p" using assms by this
moreover
{assume 1:"p"
have "p" using 1 by this}
moreover
{assume 2:"p"
have "p" using 2 by this}
ultimately show "p" by (rule disjE)

qed
text{* Otra forma:
using assms
proof
assume "p"
thus "p" .
qed}

lemma ejercicio_29:
assumes "p ∨ p"
shows "p"
proof-
have "p ∨ p" using assms by this
moreover
{assume 1 : "p"
have "p" using 1 .}
moreover
{assume 2 : "p"
have "p" using 2 .}
ultimately show "p" by (rule disjE)
qed*}

lemma ejercicio_29:
assumes 1:"p ∨ p"
shows "p"
using 1
proof (rule disjE)
{assume 2:"p"
show "p" using 2 by this}
next
{assume 3:"p"
show "p" using 3 by this}
qed

text {* ---------------------------------------------------------------
Ejercicio 30. Demostrar
p ⊢ p ∨ p
------------------------------------------------------------------ *}

lemma ejercicio_30:
assumes "p"
shows "p ∨ p"
proof-
show "p ∨ p" using assms by (rule disjI1)
qed

text{* Otra forma:
using assms
proof
qed}

text {* ---------------------------------------------------------------
Ejercicio 31. Demostrar
p ∨ (q ∨ r) ⊢ (p ∨ q) ∨ r
------------------------------------------------------------------ *}

lemma ejercicio_31:
assumes "p ∨ (q ∨ r)"
shows "(p ∨ q) ∨ r"
proof -
have "p ∨ (q ∨ r)" using assms(1) by this
moreover
{ assume 1: "p"
have 2: "p ∨ q" using 1 by (rule disjI1)
have 3: "(p ∨ q) ∨ r" using 2 by (rule disjI1)}
moreover
{ assume 4: "q ∨ r"
moreover
{ assume 5: "q"
have 6: "p ∨ q" using 5 by (rule disjI2)
have 7: "(p ∨ q) ∨ r" using 6 by (rule disjI1)}
moreover
{ assume 8: "r"
have 9: "(p ∨ q) ∨ r" using 8 by (rule disjI2)}
ultimately have "(p ∨ q) ∨ r" by (rule disjE)}
ultimately show "(p ∨ q) ∨ r" by (rule disjE)
qed

text{* Otra forma:
using assms
proof
assume "p"
then have "p∨q" ..
thus "(p∨q)∨r" ..
next
assume 1:"q∨r"
thus "(p ∨ q) ∨ r"
proof
assume "q"
then have "p∨q" ..
thus 1: "(p ∨ q) ∨ r" ..
next
assume "r"
then have 2:"(p∨q)∨r" ..
then have "q∨r⟶(p ∨ q) ∨ r" ..
thus "(p ∨ q) ∨ r" using 1 ..
qed}*}

lemma ejercicio_31:
assumes 1:"p ∨ (q ∨ r)"
shows "(p ∨ q) ∨ r"
using 1
proof (rule disjE)
{assume 2:"p"
have 3:"p ∨ q" using 2 by (rule disjI1)
show "(p ∨ q) ∨ r" using 3 by (rule disjI1)}
next
{assume 4:"q ∨ r"
thus "(p ∨ q) ∨ r"
proof (rule disjE)
{assume 5:"q"
have 6:"p ∨ q" using 5 by (rule disjI2)
show "(p ∨ q) ∨ r" using 6 by (rule disjI1)}
next
{assume 7:"r"
show "(p ∨ q) ∨ r" using 7 by (rule disjI2)}
qed}
qed

text {* ---------------------------------------------------------------
Ejercicio 32. Demostrar
(p ∨ q) ∨ r ⊢ p ∨ (q ∨ r)
------------------------------------------------------------------ *}

lemma ejercicio_32:
assumes "(p ∨ q) ∨ r"
shows "p ∨ (q ∨ r)"
proof -
have "(p ∨ q) ∨ r" using assms(1) by this
moreover
{ assume 1: "p ∨ q"
moreover
{ assume 2: "p"
have 3: "p ∨ (q ∨ r)" using 2 by (rule disjI1)}
moreover
{ assume 4: "q"
have 5: "q ∨ r" using 4 by (rule disjI1)
have 6: "p ∨ (q ∨ r)" using 5 by (rule disjI2)}
ultimately have "p ∨ (q ∨ r)" by (rule disjE)}
moreover
{ assume 7: "r"
have 8: "q ∨ r" using 7 by (rule disjI2)
have 9: "p ∨ (q ∨ r)" using 8 by (rule disjI2)}
ultimately show "p ∨ (q ∨ r)" by (rule disjE)
qed

lemma ejercicio_32:
assumes 1:"(p ∨ q) ∨ r"
shows "p ∨ (q ∨ r)"
using 1
proof (rule disjE)
{assume 2:"p ∨ q"
thus "p ∨ (q ∨ r)"
proof (rule disjE)
{assume 3:"p"
show "p ∨ (q ∨ r)" using 3 by (rule disjI1)}
next
{assume 4:"q"
have 5:"q ∨ r" using 4 by (rule disjI1)
show "p ∨ (q ∨ r)" using 5 by (rule disjI2)}
qed}
next
{assume 6:"r"
have 7:"q ∨ r" using 6 by (rule disjI2)
show "p ∨ (q ∨ r)" using 7 by (rule disjI2)}
qed

text {* ---------------------------------------------------------------
Ejercicio 33. Demostrar
p ∧ (q ∨ r) ⊢ (p ∧ q) ∨ (p ∧ r)
------------------------------------------------------------------ *}

lemma ejercicio_33:
assumes "p ∧ (q ∨ r)"
shows "(p ∧ q) ∨ (p ∧ r)"
proof -
have 1: "p" using assms(1) by (rule conjunct1)
have 2: "q ∨ r" using assms(1) by (rule conjunct2)
moreover
{ assume 3: "q"
have 4: "p ∧ q" using 1 3 by (rule conjI)
have 5: "(p ∧ q) ∨ (p ∧ r)" using 4 by (rule disjI1)}
moreover
{ assume 6: "r"
have 7: "p ∧ r" using 1 6 by (rule conjI)
have 8: "(p ∧ q) ∨ (p ∧ r)" using 7 by (rule disjI2)}
ultimately show "(p ∧ q) ∨ (p ∧ r)" by (rule disjE)
qed

lemma ejercicio_33:
assumes 1:"p ∧ (q ∨ r)"
shows "(p ∧ q) ∨ (p ∧ r)"
proof -
have 2:"p" using 1 by (rule conjunct1)
have 3:"q ∨ r" using 1 by (rule conjunct2)
thus "(p ∧ q) ∨ (p ∧ r)"
proof (rule disjE)
{assume 4:"q"
have 5:"p ∧ q" using 2 4 by (rule conjI)
show "(p ∧ q) ∨ (p ∧ r)" using 5 by (rule disjI1)}
next
{assume 6:"r"
have 7:"p ∧ r" using 2 6 by (rule conjI)
show "(p ∧ q) ∨ (p ∧ r)" using 7 by (rule disjI2)}
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 34. Demostrar
(p ∧ q) ∨ (p ∧ r) ⊢ p ∧ (q ∨ r)
------------------------------------------------------------------ *}

lemma ejercicio_34:
assumes "(p ∧ q) ∨ (p ∧ r)"
shows "p ∧ (q ∨ r)"
proof -
have "(p ∧ q) ∨ (p ∧ r)" using assms(1) by this
moreover
{ assume 1: "p ∧ q"
have 2: "q" using 1 by (rule conjunct2)
have 3: "q ∨ r" using 2 by (rule disjI1)
have 4: "p" using 1 by (rule conjunct1)
have 5: "p ∧ (q ∨ r)" using 4 3 by (rule conjI)}
moreover
{ assume 6: "p ∧ r"
have 7: "r" using 6 by (rule conjunct2)
have 8: "q ∨ r" using 7 by (rule disjI2)
have 9: "p" using 6 by (rule conjunct1)
have 10: "p ∧ (q ∨ r)" using 9 8 by (rule conjI)}
ultimately show "p ∧ (q ∨ r)" by (rule disjE)
qed

lemma ejercicio_34:
assumes 1:"(p ∧ q) ∨ (p ∧ r)"
shows "p ∧ (q ∨ r)"
using 1
proof (rule disjE)
{assume 2:"p ∧ q"
have 3:"p" using 2 by (rule conjunct1)
have 4:"q" using 2 by (rule conjunct2)
have 5:"q ∨ r" using 4 by (rule disjI1)
show "p ∧ (q ∨ r)" using 3 5 by (rule conjI)}
next
{assume 6:"p ∧ r"
have 7:"p" using 6 by (rule conjunct1)
have 8:"r" using 6 by (rule conjunct2)
have 9:"q ∨ r" using 8 by (rule disjI2)
show "p ∧ (q ∨ r)" using 7 9 by (rule conjI)}
qed

text {* ---------------------------------------------------------------
Ejercicio 35. Demostrar
p ∨ (q ∧ r) ⊢ (p ∨ q) ∧ (p ∨ r)
------------------------------------------------------------------ *}

lemma ejercicio_35:
assumes "p ∨ (q ∧ r)"
shows "(p ∨ q) ∧ (p ∨ r)"
proof -
have "p ∨ (q ∧ r)" using assms(1) by this
moreover
{ assume 1: "p"
have 2: "p ∨ q" using 1 by (rule disjI1)
have 3: "p ∨ r" using 1 by (rule disjI1)
have 4: "(p ∨ q) ∧ (p ∨ r)" using 2 3 by (rule conjI)}
moreover
{ assume 5: "q ∧ r"
have 6: "q" using 5 by (rule conjunct1)
have 7: "p ∨ q" using 6 by (rule disjI2)
have 8: "r" using 5 by (rule conjunct2)
have 9: "p ∨ r" using 8 by (rule disjI2)
have 10: "(p ∨ q) ∧ (p ∨ r)" using 7 9 by (rule conjI)}
ultimately show "(p ∨ q) ∧ (p ∨ r)" by (rule disjE)
qed

lemma ejercicio_35:
assumes 1:"p ∨ (q ∧ r)"
shows "(p ∨ q) ∧ (p ∨ r)"
using 1
proof (rule disjE)
{assume 2:"p"
have 3:"p ∨ q" using 2 by (rule disjI1)
have 4:"p ∨ r" using 2 by (rule disjI1)
show "(p ∨ q) ∧ (p ∨ r)" using 3 4 by (rule conjI)}
next
{assume 5:"q ∧ r"
have 6:"q" using 5 by (rule conjunct1)
have 7:"r" using 5 by (rule conjunct2)
have 8:"p ∨ q" using 6 by (rule disjI2)
have 9:"p ∨ r" using 7 by (rule disjI2)
show "(p ∨ q) ∧ (p ∨ r)" using 8 9 by (rule conjI)}
qed

text {* ---------------------------------------------------------------
Ejercicio 36. Demostrar
(p ∨ q) ∧ (p ∨ r) ⊢ p ∨ (q ∧ r)
------------------------------------------------------------------ *}

lemma ejercicio_36:
assumes "(p ∨ q) ∧ (p ∨ r)"
shows "p ∨ (q ∧ r)"
proof-
have "(p ∨ q)" using assms by (rule conjunct1)
moreover
{assume 1:"p"
have "p ∨ (q ∧ r)" using 1 by (rule disjI1) }
moreover
{assume 1:"q"
have "(p ∨ r)" using assms by (rule conjunct2)
moreover
{assume 2:"p"
have "p ∨ (q ∧ r)" using 2 by (rule disjI1)}
moreover
{assume 2:"r"
have 3:"(q ∧ r)" using 1 2 by (rule conjI)
have "p ∨ (q ∧ r)" using 3 by (rule disjI2)}
ultimately have "p ∨ (q ∧ r)" by (rule disjE)}
ultimately show "p ∨ (q ∧ r)" by (rule disjE)
qed

------------------- ----------

proof-
have 1: "p∨q" using assms by (rule conjunct1)
thus "p∨(q∧r)"
proof(rule disjE)
{assume 3: "p"
show "p∨(q∧r)" using 3 by ( rule disjI1)}
next
{assume 4: "q"
have 5:"p∨r" using assms by (rule conjunct2)
thus "p∨(q∧r)"
proof(rule disjE)
{assume 6: "p"
show "p∨(q∧r)" using 6 by (rule disjI1)}
next
{assume 7: "r"
have 8: "q∧r" using 4 7 by (rule conjI)
show "p∨(q∧r)" using 8 by (rule disjI2)}
qed}
qed
qed
text {* ---------------------------------------------------------------
Ejercicio 37. Demostrar
(p ⟶ r) ∧ (q ⟶ r) ⊢ p ∨ q ⟶ r
------------------------------------------------------------------ *}

lemma ejercicio_37:
assumes "(p ⟶ r) ∧ (q ⟶ r)"
shows "p ∨ q ⟶ r"
proof-
{assume "p ∨ q"
moreover
{assume 1:"p"
have 2:"(p ⟶ r)" using assms by (rule conjunct1)
have "r" using 2 1 by (rule mp)}
moreover
{assume 1:"q"
have 2:"(q ⟶ r)" using assms by (rule conjunct2)
have "r" using 2 1 by (rule mp)}
ultimately have "r" by (rule disjE)}
thus "p ∨ q ⟶ r" by (rule impI)
qed

lemma ejercicio_37:
assumes 1:"(p ⟶ r) ∧ (q ⟶ r)"
shows "p ∨ q ⟶ r"
proof (rule impI)
assume 2:"p ∨ q"
thus "r"
proof (rule disjE)
{assume 3:"p"
have 4:"p ⟶ r" using 1 by (rule conjunct1)
show "r" using 4 3 by (rule mp)}
next
{assume 5:"q"
have 6:"q ⟶ r" using 1 by (rule conjunct2)
show "r" using 6 5 by (rule mp)}
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 38. Demostrar
p ∨ q ⟶ r ⊢ (p ⟶ r) ∧ (q ⟶ r)
------------------------------------------------------------------ *}

lemma ejercicio_38:
assumes "p ∨ q ⟶ r"
shows "(p ⟶ r) ∧ (q ⟶ r)"
proof-
{assume 1 : "p"
have 2 : "p ∨ q" using 1 by (rule disjI1)
have 3 : "r" using assms 2 by (rule mp)}
hence 4 : "p⟶r" by (rule impI)
{assume 5 : "q"
have 6 : "p ∨ q" using 5 by (rule disjI2)
have 7 : "r" using assms 6 by (rule mp)}
hence 8 : "q⟶r" by (rule impI)
show "(p ⟶ r) ∧ (q ⟶ r)" using 4 8 by (rule conjI)
qed

section {* Negaciones *}

text {* ---------------------------------------------------------------
Ejercicio 39. Demostrar
p ⊢ ¬¬p
------------------------------------------------------------------ *}

lemma ejercicio_39:
assumes "p"
shows "¬¬p"
proof-
show "¬¬p" using assms by (rule notnotI)
qed

text {* ---------------------------------------------------------------
Ejercicio 40. Demostrar
¬p ⊢ p ⟶ q
------------------------------------------------------------------ *}

lemma ejercicio_40:
assumes "¬p"
shows "p ⟶ q"

proof-
{assume 1: "p"
have "q" using assms 1 ..}
thus "p⟶q" ..
qed
-------------------------------
lemma ejercicio_40:
assumes "¬p"
shows "p ⟶ q"
proof-
{assume 1 : "p"
have 2 : "False" using assms 1 by (rule notE)
have 3 : "q" using 2 by (rule FalseE)}
thus "p⟶q" by (rule impI)
qed

------------------------------
lemma ejercicio_40:
assumes "¬p"
shows "p ⟶ q"

proof-
{assume 1: "p"
have 2: "q" using assms 1 by (rule notE)}
thus "p ⟶ q" by ( rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 41. Demostrar
p ⟶ q ⊢ ¬q ⟶ ¬p
------------------------------------------------------------------ *}

lemma ejercicio_41:
assumes "p ⟶ q"
shows "¬q ⟶ ¬p"
proof-
{assume 1:"¬q"
have "¬p" using assms 1 by (rule mt)}
thus "¬q ⟶ ¬p" ..
qed

text {* ---------------------------------------------------------------
Ejercicio 42. Demostrar
p∨q, ¬q ⊢ p
------------------------------------------------------------------ *}

lemma ejercicio_42:
assumes "p∨q"
"¬q"
shows "p"

using assms(1)
proof
assume "p"
thus "p" .
next
assume 1:"q"
show "p" using assms(2) 1 ..
qed

lemma ejercicio_42:
assumes "p∨q"
"¬q"
shows "p"
proof-
have "p ∨ q" using assms(1) by this
moreover
{assume 1 : "p"
have 2 : "p" using 1 .}
moreover
{assume 3 : "q"
have 4 : "¬q" using assms(2) .
have 5 : "False" using 4 3 by (rule notE)
have 6 : "p" using 5 by (rule FalseE)}
ultimately show "p" by (rule disjE)
qed

--------**---
using assms(1)
proof(rule disjE)
{assume 1: "p"
have "p" using 1 by this}
next
{assume 2: "q"
show "p" using assms(2) 2 by (rule notE)}
qed

text {* ---------------------------------------------------------------
Ejercicio 43. Demostrar
p ∨ q, ¬p ⊢ q
------------------------------------------------------------------ *}

lemma ejercicio_43:
assumes "p ∨ q"
"¬p"
shows "q"

using assms(1)
proof
assume "q"
thus "q" .
next
assume 1:"p"
show "q" using assms(2) 1 ..
qed

lemma ejercicio_43:
assumes "p ∨ q"
"¬p"
shows "q"
proof-
have "p ∨ q" using assms(1) by this
moreover
{assume 1 : "p"
have 2 : "¬p" using assms(2) .
have 3 : "False" using 2 1 by (rule notE)
have 4 : "q" using 3 by (rule FalseE)}
moreover
{assume 5 : "q"
have 6 : "q" using 5 .}
ultimately show "q" by (rule disjE)
qed

---**--
using assms(1)
proof(rule disjE)
assume 1: "p"
show "q" using assms(2) 1 by (rule notE)
next
assume 2:"p"
have "p" using 2 by this
qed

text {* ---------------------------------------------------------------
Ejercicio 44. Demostrar
p ∨ q ⊢ ¬(¬p ∧ ¬q)
------------------------------------------------------------------ *}

lemma ejercicio_44:
assumes "p ∨ q"
shows "¬(¬p ∧ ¬q)"
proof-
{assume 1 : "¬p ∧ ¬q"
have "p ∨ q" using assms by this
moreover
{assume 2 : "p"
have 3 : "¬p" using 1 by (rule conjunct1)
have 4 : "False" using 3 2 by (rule notE)}
moreover
{assume 5 : "q"
have 6 : "¬q" using 1 by (rule conjunct2)
have 7 : "False" using 6 5 by (rule notE)}
ultimately have 8 : "False" by (rule disjE)}
thus "¬(¬p ∧ ¬q)" by (rule notI)
qed

---**
proof(rule notI)
assume 2 :"¬p∧¬q"
show False
using assms
proof(rule disjE)
{assume 1: "p"
have 3: "¬p" using 2 by (rule conjunct1)
show False using 3 1 by (rule notE)}
{assume 4: "q"
have 5: "¬q" using 2 by (rule conjunct2)
show False using 5 4 by (rule notE)}
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 45. Demostrar
p ∧ q ⊢ ¬(¬p ∨ ¬q)
------------------------------------------------------------------ *}

lemma ejercicio_45:
assumes "p ∧ q"
shows "¬(¬p ∨ ¬q)"
proof-
{assume 1 : "¬p ∨ ¬q"
have "¬p ∨ ¬q" using 1 by this
moreover
{assume 2 : "¬p"
have 3 : "p" using assms by (rule conjunct1)
have 4 : "False" using 2 3 by (rule notE)}
moreover
{assume 5 : "¬q"
have 6 : "q" using assms by (rule conjunct2)
have 7 : "False" using 5 6 by (rule notE)}
ultimately have "False" by (rule disjE)}
thus "¬(¬p ∨ ¬q)" by (rule notI)
qed

---**

assumes "p ∧ q"
shows "¬(¬p ∨ ¬q)"

proof (rule notI)
assume 1: "¬p∨¬q"
show False
using 1
proof(rule disjE)
assume 2: "¬p"
have 3: "p" using assms by (rule conjunct1)
show False using 2 3 by (rule notE)
next
assume 4: "¬q"
have 5: "q" using assms by (rule conjunct2)
show False using 4 5 by (rule notE)
qed
qed
text {* ---------------------------------------------------------------
Ejercicio 46. Demostrar
¬(p ∨ q) ⊢ ¬p ∧ ¬q
------------------------------------------------------------------ *}

lemma ejercicio_46:
assumes "¬(p ∨ q)"
shows "¬p ∧ ¬q"
proof-
{assume 1 : "p"
have 2 : "p ∨ q" using 1 by (rule disjI1)
have 3 : "False" using assms 2 by (rule notE)}
hence 4 : "¬p" by (rule notI)
{assume 5 : "q"
have 6 : "p ∨ q" using 5 by (rule disjI2)
have 7 : "False" using assms 6 by (rule notE)}
hence 8 : "¬q" by (rule notI)
show "¬p ∧ ¬q" using 4 8 by (rule conjI)
qed

(*Otra opción para usar tercio excluso*)

lemma ejercicio_46_2:
assumes "¬(p ∨ q)"
shows "¬p ∧ ¬q"
proof -
have 1:"¬p | p" by (rule excluded_middle)
show "¬p & ¬q"
using 1
proof (rule disjE)
{assume 2:"¬p"
have 3:"¬q | q" by (rule excluded_middle)
show "¬p & ¬q"
using 3
proof (rule disjE)
{assume 4:"¬q"
with 2 show "¬p & ¬q" by (rule conjI)}
next
{assume "q"
then have "p | q" by (rule disjI2)
with assms show "¬p & ¬q" by (rule notE)}
qed
}
next
{assume 5:"p"
then have "p | q" by (rule disjI1)
with assms show "¬p & ¬q" by (rule notE)
}
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 47. Demostrar
¬p ∧ ¬q ⊢ ¬(p ∨ q)
------------------------------------------------------------------ *}

lemma ejercicio_47:
assumes "¬p ∧ ¬q"
shows "¬(p ∨ q)"
proof-
{assume 1 : "p ∨ q"
have "p ∨ q" using 1 by this
moreover
{assume 2 : "p"
have 3 : "¬p" using assms by (rule conjunct1)
have 4 : "False" using 3 2 by (rule notE)}
moreover
{assume 5 : "q"
have 6 : "¬q" using assms by (rule conjunct2)
have 6 : "False" using 6 5 by (rule notE)}
ultimately have "False" by (rule disjE)}
thus "¬(p ∨ q)" by (rule notI)
qed

---**

proof(rule notI)
assume 1: "p∨q"
show False
using 1
proof(rule disjE)
{assume 2:"p"
have 3:"¬p" using assms by (rule conjunct1)
show "False" using 3 2 by (rule notE)}
next
{assume 5:"q"
have 6:"¬q" using assms by (rule conjunct2)
show "False" using 6 5 by (rule notE)}
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 48. Demostrar
¬p ∨ ¬q ⊢ ¬(p ∧ q)
------------------------------------------------------------------ *}

lemma ejercicio_48:
assumes "¬p ∨ ¬q"
shows "¬(p ∧ q)"
proof-
{assume 1 : "p ∧ q"
have "¬p ∨ ¬q" using assms by this
moreover
{assume 2 : "¬p"
have 3 : "p" using 1 by (rule conjunct1)
have 4 : "False" using 2 3 by (rule notE)}
moreover
{assume 5 : "¬q"
have 6 : "q" using 1 by (rule conjunct2)
have 7 : "False" using 5 6 by (rule notE)}
ultimately have "False" by (rule disjE)}
thus "¬(p ∧ q)" by (rule notI)
qed

--**
lemma ejercicio_48:
assumes "¬p ∨ ¬q"
shows "¬(p ∧ q)"

proof(rule notI)
assume 1:"p∧q"
show False
using assms
proof(rule disjE)
assume 2:"¬p"
have 3: "p" using 1 by (rule conjunct1)
show False using 2 3 by (rule notE)
next
assume 4: "¬q"
have 5: "q" using 1 by (rule conjunct2)
show False using 4 5 by (rule notE)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 49. Demostrar
⊢ ¬(p ∧ ¬p)
------------------------------------------------------------------ *}

lemma ejercicio_49:
"¬(p ∧ ¬p)"
proof-
{assume 1 : "p ∧ ¬p"
have 2 : "p" using 1 by (rule conjunct1)
have 3 : "¬p" using 1 by (rule conjunct2)
have 4 : "False" using 3 2 by (rule notE)}
thus "¬(p ∧ ¬p)" by (rule notI)
qed

---**

proof(rule notI)
assume 1: "p∧¬p"
have 2: "p" using 1 by (rule conjunct1)
have 3: "¬p" using 1 by (rule conjunct2)
show False using 3 2 by (rule notE)
qed

text {* ---------------------------------------------------------------
Ejercicio 50. Demostrar
p ∧ ¬p ⊢ q
------------------------------------------------------------------ *}

lemma ejercicio_50:
assumes "p ∧ ¬p"
shows "q"

proof-
have 1 : "p" using assms by (rule conjunct1)
have 2 : "¬p" using assms by (rule conjunct2)
show "q" using 2 1 by (rule notE)
qed

text {* ---------------------------------------------------------------
Ejercicio 51. Demostrar
¬¬p ⊢ p
------------------------------------------------------------------ *}

lemma ejercicio_51:
assumes "¬¬p"
shows "p"

proof-
show "p" using assms by (rule notnotD)
qed

--**

lemma ejercicio_51:
assumes "¬¬p"
shows "p"
proof(rule ccontr)
assume 1: "¬p"
show False using assms 1 by (rule notE)
qed

text {* ---------------------------------------------------------------
Ejercicio 52. Demostrar
⊢ p ∨ ¬p
------------------------------------------------------------------ *}

lemma ejercicio_52:
"p ∨ ¬p"

proof-
have 1: "¬p∨p" by (rule excluded_middle)
have 2: "¬p∨p" using 1 by this
moreover
{assume 3: "¬p"
have 4: "p∨¬p" using 3 by (rule disjI2)}
moreover
{assume 5: "p"
have 6: "p∨¬p" using 5 by (rule disjI1)}
ultimately show "p∨¬p" by (rule disjE)
qed

text {* ---------------------------------------------------------------
Ejercicio 53. Demostrar
⊢ ((p ⟶ q) ⟶ p) ⟶ p
------------------------------------------------------------------ *}

lemma ejercicio_53:
"((p ⟶ q) ⟶ p) ⟶ p"
proof-
{assume 1 : "(p ⟶ q) ⟶ p"
{assume 2 : "¬p"
have 3 : "¬(p⟶q)" using 1 2 by (rule mt)
{assume 4 : "p"
have 5 : "q" using 2 4 by (rule notE)}
hence 6 : "p⟶q" by (rule impI)
have 7 : "False" using 3 6 by (rule notE)}
hence 8 : "p" by (rule ccontr)}
thus "((p ⟶ q) ⟶ p) ⟶ p" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 54. Demostrar
¬q ⟶ ¬p ⊢ p ⟶ q
------------------------------------------------------------------ *}

lemma ejercicio_54:
assumes "¬q ⟶ ¬p"
shows "p ⟶ q"
proof-
{assume 1 : "p"
{assume 2 : "¬q"
have 3 : "¬p" using assms 2 by (rule mp)
have 4 : "False" using 3 1 by (rule notE)}
hence 5 : "q" by (rule ccontr)}
thus "p⟶q" by (rule impI)
qed

---**

proof(rule impI)
assume 1: "p"
have 2 : "¬¬p" using 1 by (rule notnotI)
have 3: "¬¬q" using assms 2 by (rule mt)
show "q" using 3 by (rule notnotD)
qed

text {* ---------------------------------------------------------------
Ejercicio 55. Demostrar
¬(¬p ∧ ¬q) ⊢ p ∨ q
------------------------------------------------------------------ *}

lemma ejercicio_55:
assumes "¬(¬p ∧ ¬q)"
shows "p ∨ q"
proof-
have 1 : "¬p ∨ p" by (rule excluded_middle)
have "¬p ∨ p" using 1 by this
moreover
{assume 2 : "¬p"
have 3 : "¬q ∨ q" by (rule excluded_middle)
have "¬q ∨ q" using 3 by this
moreover
{assume 4 : "¬q"
have 5 : "¬p ∧ ¬q" using 2 4 by (rule conjI)
have 6 : "False" using assms 5 by (rule notE)
have 7 : "p ∨ q" using 6 by (rule FalseE)}
moreover
{assume 8 : "q"
have 9 : "p ∨ q" using 8 by (rule disjI2)}
ultimately have "p ∨ q" by (rule disjE)}
moreover
{assume 10 : "p"
have 11 : "p ∨ q" using 10 by (rule disjI1)}
ultimately show "p ∨ q" by (rule disjE)
qed

text {* ---------------------------------------------------------------
Ejercicio 56. Demostrar
¬(¬p ∨ ¬q) ⊢ p ∧ q
------------------------------------------------------------------ *}

lemma ejercicio_56:
assumes "¬(¬p ∨ ¬q)"
shows "p ∧ q"
proof-
{assume 1 : "¬p"
have 2 : "¬p ∨ ¬q" using 1 by (rule disjI1)
have 3 : "False" using assms 2 by (rule notE)}
hence 4 : "p" by (rule ccontr)
{assume 5 : "¬q"
have 6 : "¬p ∨ ¬q" using 5 by (rule disjI2)
have 7 : "False" using assms 6 by (rule notE)}
hence 8 : "q" by (rule ccontr)
show "p ∧ q" using 4 8 by (rule conjI)
qed

--**
proof-
{assume 1: "¬p"
have 2: "¬p∨¬q"using 1 by (rule disjI1)
have 3: "False" using assms 2 by (rule notE)}
hence 8: "p" by (rule ccontr)
{assume 4: "¬q"
have 5: "¬p∨¬q" using 4 by (rule disjI2)
have 6: "False" using assms 5 by (rule notE)}
hence 9: "q" by (rule ccontr)
show "p∧q" using 8 9 by (rule conjI)
qed

text {* ---------------------------------------------------------------
Ejercicio 57. Demostrar
¬(p ∧ q) ⊢ ¬p ∨ ¬q
------------------------------------------------------------------ *}

lemma ejercicio_57:
assumes "¬(p ∧ q)"
shows "¬p ∨ ¬q"
proof-
have 1 : "¬p ∨ p" by (rule excluded_middle)
have "¬p ∨ p" using 1 by this
moreover
{assume 2 : "¬p"
have 3 : "¬p ∨ ¬q" using 2 by (rule disjI1)}
moreover
{assume 4 : "p"
have 5 : "¬q ∨ q" by (rule excluded_middle)
have "¬q ∨ q" using 5 by this
moreover
{assume 6 : "¬q"
have 7 : "¬p ∨ ¬q" using 6 by (rule disjI2)}
moreover
{assume 8 : "q"
have 9 : "p ∧ q" using 4 8 by (rule conjI)
have 10 : "False" using assms 9 by (rule notE)
have 11 : "¬p ∨ ¬q" using 10 by (rule FalseE)}
ultimately have "¬p ∨ ¬q" by (rule disjE)}
ultimately show "¬p ∨ ¬q" by (rule disjE)
qed

text {* ---------------------------------------------------------------
Ejercicio 58. Demostrar
⊢ (p ⟶ q) ∨ (q ⟶ p)
------------------------------------------------------------------ *}

lemma ejercicio_58:
"(p ⟶ q) ∨ (q ⟶ p)"

proof-
have 1 : "¬(p ⟶ q) ∨ (p ⟶ q)" by (rule excluded_middle)
have "¬(p ⟶ q) ∨ (p ⟶ q)" using 1 by this
moreover
{assume 2 : "¬(p ⟶ q)"
{assume 3 : "q"
{assume 4 : "p"
have 5 : "q" using 3 .}
hence 6 : "p⟶q" by (rule impI)
have 7 : "p" using 2 6 by (rule notE)}
hence 8 : "q⟶p" by (rule impI)
have 9 : "(p ⟶ q) ∨ (q ⟶ p)" using 8 by (rule disjI2)}
moreover
{assume 10 : "p ⟶ q"
have 11 : "(p ⟶ q) ∨ (q ⟶ p)" using 10 by (rule disjI1)}
ultimately show "(p ⟶ q) ∨ (q ⟶ p)" by (rule disjE)
qed

end
</source>

Relación 2

2016-02-23T13:16:34Z

Albromgon: Añado contraejemplo al ejercicio 7.

=== Relación 1(b): Representación del conocimiento proposicional ===

----
'''Ejercicio 1.''' Formalizar el siguiente argumento
<blockquote>
''Siempre que un número x es divisible por 10, acaba en 0. El número x no acaba en 0. Por lo tanto, x no es divisible por 10.''
</blockquote>
Usando los símbolos D: el número es divisible por 10 y C: el número acaba en cero.
----

'''Solución:'''
<math>D \to C \qquad \lnot C \to \lnot D</math>
----
'''Ejercicio 2.''' Formalizar el siguiente argumento
<blockquote>
''Si la válvula está abierta o la monitorización está preparada, entonces se envía una señal de reconocimiento y un mensaje de funcionamiento al controlador del ordenador. Si se envía un mensaje de funcionamiento al controlador del ordenador o el sistema está en estado normal, entonces se aceptan las órdenes del operador. Por lo tanto, si la válvula está abierta, entonces se aceptan las órdenes del operador.''
</blockquote>
Usando los símbolos V: La válvula está abierta, P: La monitorización está preparada, R: Envía una señal de reconocimiento, F: Envía un mensaje de funcionamiento, A: Se aceptan órdenes del operador y N: El sistema está en estado normal.
----

'''Solución:'''
<math>
(V \vee P \to R \wedge F) \quad
(F \vee N \to A) \quad \vDash V\to A</math>

----
'''Ejercicio 3.''' Formalizar el siguiente argumento
<blockquote>
''Cuando tanto la temperatura como la presión atmosférica permanecen contantes, no llueve. La temperatura permanece constante. Por lo tanto, en caso de que llueva, la presión atmosférica no permanece constante.''
</blockquote>
Usando los símbolos T: La temperatura permanece constante, P: La presión atmosférica permanece constante y L: Llueve
----

'''Solución:'''
<math> (T \wedge P) \to \lnot L </math> 
<math> T \vDash L \to \lnot P </math>

Falta un detalle en la solución

<math> ((T \wedge P) \to \lnot L) \quad T \quad \vDash L \to \lnot P </math>
----
'''Ejercicio 4.''' Formalizar el siguiente argumento
<blockquote>
''En cierto experimento, cuando hemos empleado un fármaco A, el paciente ha mejorado considerablemente en el caso, y sólo en el caso, en que no se haya empleado también un fármaco B. Además, o se ha empleado el fármaco A o se ha empleado el fármaco B. En consecuencia, podemos afirmar que si no hemos empleado el fármaco B, el paciente ha mejorado considerablemente.''
</blockquote>
Usando los símbolos A: Hemos empleado el fármaco A, B: Hemos empleado el fármaco B y M: El paciente ha mejorado notablemente.
----

'''Solución:'''
<math> (A \to (M \leftrightarrow \lnot B)) \wedge (A \vee B) \quad \vDash \lnot B \to M </math>
----
'''Ejercicio 5.''' Definir por recursión sobre fórmulas las siguientes funciones

* nv(F) que calcula el número variables proposicionales que ocurren en la fórmula F. Por ejemplo,
: nv(p → p ∨ q) = 3.
* prof(F) que calcula la profundidad del árbol de análisis de la fórmula F. Por ejemplo,
: prof(p → p ∨ q) = 2.

Demostrar por inducción, que para toda fórmula F,
: nv(F) ≤ 2^prof(F)
----
<math>
nv(F)=
\begin{cases}
1 \quad si \quad F Atomica \\
nv(G) \quad si \quad F= \lnot G \\
nv(G)+nv(H) \quad si \quad F=(G*H)
\end{cases}
</math>

<math>
prof(F)=
\begin{cases}
1 \quad si \quad F Atomica \\
prof(G) \quad si \quad F= \lnot G \\
1+ max(prof(G),prof(H)) \quad si \quad F=(G*H)
\end{cases}
</math>

Demostración por inducción: 
Se comprueba para lo que consideramos nuestro caso base, cuando F es atómica 
<math>
F \quad atomica \to nv(F)=1 \leqslant 2^{prof(F)}=2^1
</math> 
Posteriormente suponemos que la desigualdad se da para <math>nv(F) < n+1</math>, y lo comprobamos para <math>nv(F)=n+1</math>: 
<math>nv(F)=n+1 \to F=(G*H) \to n(F)=nv(G)+nv(H) \quad </math> .Pero <math>(nv(G)<n+1) \wedge (nv(H)<n+1) \to nv(F)=nv(G)+nv(H) \leqslant 2^{prof(G)}+2^{prof(H)} \leqslant 2^{prof(F)} </math>

'''Solución:'''

----
'''Ejercicio 6.''' ¿Existe un conjunto S de tres fórmulas tal que de todos los subconjuntos de S sólo uno es consistente?
----

'''Solución:'''
No entiendo bien la pregunta, pero doy una "posible" respuesta: 
Si <math>S=\{ F_1 , F_2,F_3 \}, \quad </math> tendríamos los subconjuntos siguientes <math>\{F_1\},\{F_2\},\{F_3\},\{F_1,F_2\},\{F_1,F_3\},\{F_2,F_3\},S</math> 
Luego, si tomamos <math>F_1 \quad</math> una tautología y <math>F_2,F_3 \quad</math> contradicciones, sólo el subconjunto <math>\{F_1\} \quad</math> sería consistente. 
Por ejemplo: <math>F_1=(p\wedge q)\to (p\vee r), \quad F_2=(p\vee q) \leftrightarrow \lnot (p \vee q), \quad F_3=(p\to q)\wedge \lnot (p \to q)</math>

----
'''Ejercicio 7.''' ¿Es cierto que si F → G y F son satisfacibles, entonces G es satisfacible? Si es cierto, dar una explicación. Si no es cierto, dar un contraejemplo.
----

'''Solución:'''
Dado que <math>F \quad</math> es satisfacible y <math>F\to F</math> también lo es, para cada una de ellas existe una interpretación para la que sean ciertas. Luego, la tabla de verdad de todas las combinaciones posibles (para las que <math>F\to G</math> es cierta) serán: 
<math>
\begin{array}{|c|c||c|} F & G & F\to G \\
\hline
1&1&1\\
0&0&1\\
0&1&1\\
\end{array}
</math>
Por lo tanto, no necesariamente <math>G \quad</math> es satisfacible, pues si es una contradicción, aún existen interpretaciones para las que <math>F\to G \quad </math> es cierta. 
Si <math>F\quad</math> y <math>F\to G \quad</math> deben ser satisfacibles de forma simultánea, entonces sí tendríamos que <math>G \quad</math> es satisfacible.

Como contraejemplo podemos tomar las fórmulas <math>F = \neg p</math> y <math>G = (p \wedge \neg p)</math>. Tenemos que <math>F</math> es satisfacible (<math>I(p)=0</math> es modelo de <math>F</math>) y que <math>F \to G</math> también es satisfacible (<math>I(p)=1</math> es modelo de <math>\neg p \to (p \wedge \neg p)</math>) pero <math>G</math> es insatisfacible.
----
'''Ejercicio 8.''' Demostrar o refutar las siguientes proposiciones:

# Si F es una fórmula satisfacible, entonces todas las subfórmulas de F son satisfacibles.
# Existen fórmulas válidas tales que todas sus subfórmulas son válidas.
----

'''Solución:'''
# Damos un contraejemplo para probar que no se cumple. La fórmula <math>F =\neg (p \wedge \neg p )</math> es una tautología, es decir, toda interpretación es modelo de <math>F</math>. En cambio, su subfórmula <math>G = p \wedge \neg p</math>, por la definición de interpretación (<math>I(F)=H_{\neg}(I(G))</math>) es una contradicción, o sea, que ninguna interpretación es modelo de <math>G</math> y por lo tanto es una subfórmula de <math>F</math> insatisfacible.
# Una fórmula válida es una tautología. Luego queda probado si encontramos algun ejemplo de tautología en la que toda subfórmula sea tautología o refutado si demostramos que no es posible. La considero falsa pues en primer lugar, las fórmulas atómicas son subfórmulas y pueden tomar interpretaciones ciertas o falsas. Y en segundo lugar, las subfórmulas más simples que pueden componer la tautología (que no sean atómicas), están formadas por la combinación de fórmulas atómicas y conectivas lógicas, las cuales no son tautologías (si formamos una fórmula empleando una única conectiva).

Relación 2

2016-02-23T09:08:26Z

Albromgon:

=== Relación 1(b): Representación del conocimiento proposicional ===

----
'''Ejercicio 1.''' Formalizar el siguiente argumento
<blockquote>
''Siempre que un número x es divisible por 10, acaba en 0. El número x no acaba en 0. Por lo tanto, x no es divisible por 10.''
</blockquote>
Usando los símbolos D: el número es divisible por 10 y C: el número acaba en cero.
----

'''Solución:'''
<math>D \to C \qquad \lnot C \to \lnot D</math>
----
'''Ejercicio 2.''' Formalizar el siguiente argumento
<blockquote>
''Si la válvula está abierta o la monitorización está preparada, entonces se envía una señal de reconocimiento y un mensaje de funcionamiento al controlador del ordenador. Si se envía un mensaje de funcionamiento al controlador del ordenador o el sistema está en estado normal, entonces se aceptan las órdenes del operador. Por lo tanto, si la válvula está abierta, entonces se aceptan las órdenes del operador.''
</blockquote>
Usando los símbolos V: La válvula está abierta, P: La monitorización está preparada, R: Envía una señal de reconocimiento, F: Envía un mensaje de funcionamiento, A: Se aceptan órdenes del operador y N: El sistema está en estado normal.
----

'''Solución:'''
<math>
(V \vee P \to R \wedge F) \quad
(F \vee N \to A) \quad \vDash V\to A</math>

----
'''Ejercicio 3.''' Formalizar el siguiente argumento
<blockquote>
''Cuando tanto la temperatura como la presión atmosférica permanecen contantes, no llueve. La temperatura permanece constante. Por lo tanto, en caso de que llueva, la presión atmosférica no permanece constante.''
</blockquote>
Usando los símbolos T: La temperatura permanece constante, P: La presión atmosférica permanece constante y L: Llueve
----

'''Solución:'''
<math> (T \wedge P) \to \lnot L </math> 
<math> T \vDash L \to \lnot P </math>

Falta un detalle en la solución

<math> ((T \wedge P) \to \lnot L) \quad T \quad \vDash L \to \lnot P </math>
----
'''Ejercicio 4.''' Formalizar el siguiente argumento
<blockquote>
''En cierto experimento, cuando hemos empleado un fármaco A, el paciente ha mejorado considerablemente en el caso, y sólo en el caso, en que no se haya empleado también un fármaco B. Además, o se ha empleado el fármaco A o se ha empleado el fármaco B. En consecuencia, podemos afirmar que si no hemos empleado el fármaco B, el paciente ha mejorado considerablemente.''
</blockquote>
Usando los símbolos A: Hemos empleado el fármaco A, B: Hemos empleado el fármaco B y M: El paciente ha mejorado notablemente.
----

'''Solución:'''
<math> (A \to (M \leftrightarrow \lnot B)) \wedge (A \vee B) \quad \vDash \lnot B \to M </math>
----
'''Ejercicio 5.''' Definir por recursión sobre fórmulas las siguientes funciones

* nv(F) que calcula el número variables proposicionales que ocurren en la fórmula F. Por ejemplo,
: nv(p → p ∨ q) = 3.
* prof(F) que calcula la profundidad del árbol de análisis de la fórmula F. Por ejemplo,
: prof(p → p ∨ q) = 2.

Demostrar por inducción, que para toda fórmula F,
: nv(F) ≤ 2^prof(F)
----
<math>
nv(F)=
\begin{cases}
1 \quad si \quad F Atomica \\
nv(G) \quad si \quad F= \lnot G \\
nv(G)+nv(H) \quad si \quad F=(G*H)
\end{cases}
</math>

<math>
prof(F)=
\begin{cases}
1 \quad si \quad F Atomica \\
prof(G) \quad si \quad F= \lnot G \\
1+ max(prof(G),prof(H)) \quad si \quad F=(G*H)
\end{cases}
</math>

Demostración por inducción: 
Se comprueba para lo que consideramos nuestro caso base, cuando F es atómica 
<math>
F \quad atomica \to nv(F)=1 \leqslant 2^{prof(F)}=2^1
</math> 
Posteriormente suponemos que la desigualdad se da para <math>nv(F) < n+1</math>, y lo comprobamos para <math>nv(F)=n+1</math>: 
<math>nv(F)=n+1 \to F=(G*H) \to n(F)=nv(G)+nv(H) \quad </math> .Pero <math>(nv(G)<n+1) \wedge (nv(H)<n+1) \to nv(F)=nv(G)+nv(H) \leqslant 2^{prof(G)}+2^{prof(H)} \leqslant 2^{prof(F)} </math>

'''Solución:'''

----
'''Ejercicio 6.''' ¿Existe un conjunto S de tres fórmulas tal que de todos los subconjuntos de S sólo uno es consistente?
----

'''Solución:'''
No entiendo bien la pregunta, pero doy una "posible" respuesta: 
Si <math>S=\{ F_1 , F_2,F_3 \}, \quad </math> tendríamos los subconjuntos siguientes <math>\{F_1\},\{F_2\},\{F_3\},\{F_1,F_2\},\{F_1,F_3\},\{F_2,F_3\},S</math> 
Luego, si tomamos <math>F_1 \quad</math> una tautología y <math>F_2,F_3 \quad</math> contradicciones, sólo el subconjunto <math>\{F_1\} \quad</math> sería consistente. 
Por ejemplo: <math>F_1=(p\wedge q)\to (p\vee r), \quad F_2=(p\vee q) \leftrightarrow \lnot (p \vee q), \quad F_3=(p\to q)\wedge \lnot (p \to q)</math>

----
'''Ejercicio 7.''' ¿Es cierto que si F → G y F son satisfacibles, entonces G es satisfacible? Si es cierto, dar una explicación. Si no es cierto, dar un contraejemplo.
----

'''Solución:'''
Dado que <math>F \quad</math> es satisfacible y <math>F\to F</math> también lo es, para cada una de ellas existe una interpretación para la que sean ciertas. Luego, la tabla de verdad de todas las combinaciones posibles (para las que <math>F\to G</math> es cierta) serán: 
<math>
\begin{array}{|c|c||c|} F & G & F\to G \\
\hline
1&1&1\\
0&0&1\\
0&1&1\\
\end{array}
</math>
Por lo tanto, no necesariamente <math>G \quad</math> es satisfacible, pues si es una contradicción, aún existen interpretaciones para las que <math>F\to G \quad </math> es cierta. 
Si <math>F\quad</math> y <math>F\to G \quad</math> deben ser satisfacibles de forma simultánea, entonces sí tendríamos que <math>G \quad</math> es satisfacible.
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'''Ejercicio 8.''' Demostrar o refutar las siguientes proposiciones:

# Si F es una fórmula satisfacible, entonces todas las subfórmulas de F son satisfacibles.
# Existen fórmulas válidas tales que todas sus subfórmulas son válidas.
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'''Solución:'''
# Damos un contraejemplo para probar que no se cumple. La fórmula <math>F =\neg (p \wedge \neg p )</math> es una tautología, es decir, toda interpretación es modelo de <math>F</math>. En cambio, su subfórmula <math>G = p \wedge \neg p</math>, por la definición de interpretación (<math>I(F)=H_{\neg}(I(G))</math>) es una contradicción, o sea, que ninguna interpretación es modelo de <math>G</math> y por lo tanto es una subfórmula de <math>F</math> insatisfacible.
# Una fórmula válida es una tautología. Luego queda probado si encontramos algun ejemplo de tautología en la que toda subfórmula sea tautología o refutado si demostramos que no es posible. La considero falsa pues en primer lugar, las fórmulas atómicas son subfórmulas y pueden tomar interpretaciones ciertas o falsas. Y en segundo lugar, las subfórmulas más simples que pueden componer la tautología (que no sean atómicas), están formadas por la combinación de fórmulas atómicas y conectivas lógicas, las cuales no son tautologías (si formamos una fórmula empleando una única conectiva).