Relación 1

header {* R1: Deducción natural proposicional *}

theory Rel_1
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 1:"p ⟶ q" and  
          2:"p"
  shows "q"


-- Solución M.Cumplido

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

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

lemma ejercicio_2:
  assumes 1:"p ⟶ q" and
          2:"q ⟶ r" and
          3:"p" 
  shows "r"

-- Solución M.Cumplido
 proof -
   have 4: "q" using 1 3 by (rule mp)
   show 5: "r" using 2 4 by (rule mp)
qed

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

lemma ejercicio_3:
  assumes 1:"p ⟶ (q ⟶ r)" and
          2:"p ⟶ q" and 
          3:"p"
  shows "r"

-- Solución M.Cumplido
proof -
  have 4:"q ⟶ r" using 1 3 by (rule mp)
  show "r" using 2 4 3 by (rule ejercicio_2)
qed

text {* --------------------------------------------------------------- 
  Ejercicio 4. Demostrar
     p ⟶ q, q ⟶ r ⊢ p ⟶ r
  ------------------------------------------------------------------ *}
lemma ejercicio_4:
  assumes 1:"p ⟶ q" and
          2:"q ⟶ r" 
  shows "p ⟶ r"


-- Solución M.Cumplido
proof -
   {assume 3:"p"
   have "r" using 1 2 3 by (rule ejercicio_2)}
   thus "p ⟶ r" by (rule impI)
qed

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

lemma ejercicio_5:
  assumes 1:"p ⟶ (q ⟶ r)" 
  shows   "q ⟶ (p ⟶ r)"

-- Solución M.Cumplido
proof -
  {assume 2: q
    {assume 3: p
     have 4:"q ⟶ r" using 1 3 by (rule mp) 
     have r using 4 2 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 1:"p ⟶ (q ⟶ r)" 
  shows   "(p ⟶ q) ⟶ (p ⟶ r)"

-- Solución M.Cumplido
proof -
{assume 2:"p ⟶ q"
    {assume 3: p 
    have 4: "q ⟶ r" using 1 3 by (rule mp)
    have 5: "p ⟶ r" using 2 4 by (rule ejercicio_4)
    have r using 5 3 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 1:"p"  
  shows   "q ⟶ p"

-- Solución M.Cumplido
proof -
{assume 2:q
  have 3: p
     proof (rule ccontr)
     assume 4:"¬p" 
     show False using 4 1 by (rule notE) 
     qed
}
thus "q⟶p" by (rule impI)
qed

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

lemma ejercicio_8:
  "p ⟶ (q ⟶ p)"

-- Solución M.Cumplido
proof -
{assume 1: p
 {assume 2:q
  have 3: p
     proof (rule ccontr)
     assume 4:"¬p" 
     show False using 4 1 by (rule notE) 
     qed}
 hence "q⟶p" by (rule impI)}
thus  "p ⟶ (q ⟶ p)"  by (rule impI)
qed

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

lemma ejercicio_9:
  assumes 1:"p ⟶ q" 
  shows   "(q ⟶ r) ⟶ (p ⟶ r)"

-- Solución M.Cumplido
proof -
{assume 2:"q⟶r"
have 3:"p⟶r" using 1 2 by (rule ejercicio_4)}
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 1:"p ⟶ (q ⟶ (r ⟶ s))" 
  shows   "r ⟶ (q ⟶ (p ⟶ s))"

-- Solución M.Cumplido
proof -
{assume 2: r
  {assume 3: q
    {assume 4: p
     have 5: "q ⟶ (r ⟶ s)" using 1 4 by (rule mp) 
     have 6:"r⟶s" using 5 3 by (rule mp)
     have s using 6 2 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))"


-- Solución M.Cumplido
proof -

{assume "p ⟶ (q ⟶ r)"
 hence  "(p ⟶ q) ⟶ (p ⟶ r)" by (rule ejercicio_6)
}
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 1:"(p ⟶ q) ⟶ r" 
  shows   "p ⟶ (q ⟶ r)"

-- Solución M.Cumplido
proof -
{assume 2:p
    {assume 3:q
     hence 4:"p⟶q"  by (rule ejercicio_7)
     have r using 1 4 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 1:"p" and
          2:"q" 
  shows "p ∧ q"

-- Solución M.Cumplido
proof -
show "p ∧ q" using 1 2 by (rule conjI)
qed 

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

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

-- Solución M.Cumplido
proof - 
show p using assms(1) by (rule conjunct1)
qed

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

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

-- Solución M.Cumplido
proof -
show q using assms(1) 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"

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

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

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

lemma ejercicio_18:
  assumes "p ∧ q" 
  shows   "p ⟶ q"

-- Solución M.Cumplido
proof -
{assume 1: p
 have 2: q using assms(1) 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"

-- Solución M.Cumplido
proof -
have 1: "p⟶q" using assms(1) by (rule conjunct1)
have 2: "p⟶r" using assms(1) by (rule conjunct2)
{assume 3: p
 have 4: q using 1 3 by (rule mp)
 have 5: r using 2 3 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)"

-- Solución M.Cumplido
proof -
{assume 1:p
have 2:"q & r" using assms(1) 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(1) 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"

-- Solución M.Cumplido
proof -

{assume 1:"p & q"
 have 2: p using 1 by (rule conjunct1)
 have 3: "q⟶r" using assms(1) 2 by (rule mp)
 have 4: q using 1 by (rule conjunct2)
 have 5: r using 3 4 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)"

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


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

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

-- Solución M.Cumplido
proof -
{assume 1: "p & q"
 have 2: "p⟶ q" using 1 by (rule ejercicio_18)
 have r using assms(1) 2 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"

-- Solución M.Cumplido
proof -

have 1: p using assms(1) by (rule conjunct1)
have 2: "q⟶r" using assms(1) by (rule conjunct2)
{assume 3: "p⟶q"
 have 4: q using 3 1 by (rule mp)
 have r using 2 4 by (rule mp)
}
thus "(p⟶q)⟶r" by (rule impI)
qed

section {* Disyunciones *}

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

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

-- Solución M.Cumplido
proof -
show "p | q" using assms(1) by (rule disjI1)
qed

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

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

-- Solución M.Cumplido
proof -
show "p | q" using assms(1) by (rule disjI2)
qed



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

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


-- Solución M.Cumplido
using assms(1)
proof (rule disjE)

{assume 1:p
 show 2: "q | p" using 1 by (rule disjI2)}
next
{assume 3:q
 show 4: "q | p" using 3 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"

-- Solución M.Cumplido
proof -
{assume 1: "p | q"
 have 2:"p | r" using 1
  proof (rule disjE)
  {assume p 
   thus "p | r" by (rule disjI1)}
  next
  {assume 3: q
   have r using assms(1) 3 by (rule mp)
   thus "p | r" by (rule disjI2)}
  qed}
thus  "p ∨ q ⟶ p ∨ r" by (rule impI)
qed

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

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

-- Solución M.Cumplido
using assms(1)
proof (rule disjE)
{assume p
 thus p by this}
qed

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

lemma ejercicio_30:
  assumes "p" 
  shows   "p ∨ p"

-- Solución M.Cumplido
proof -
show "p | p" using assms(1) by (rule disjI1)
qed

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

lemma ejercicio_31:
  assumes "p ∨ (q ∨ r)" 
  shows   "(p ∨ q) ∨ r"

-- Solución M.Cumplido
using assms(1)
proof (rule disjE)
{assume p
 hence "p | q" by (rule disjI1)
 thus  "(p ∨ q) ∨ r" by (rule disjI1)}

next
{assume 1: "q | r"
 show  "(p ∨ q) ∨ r" using 1
   
    proof (rule disjE)
    {assume q
     hence "p | q" by (rule disjI2)
     thus "(p ∨ q) ∨ r" by (rule disjI1)}
    next
    {assume r
     thus "(p ∨ q) ∨ r" 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)"

-- Solución M.Cumplido
using assms(1)
proof (rule disjE)
{assume r
 hence "q | r" by (rule disjI2)
 thus  "p ∨ (q ∨ r)" by (rule disjI2)}

next
{assume 1: "p | q"
 show  "p ∨ (q ∨ r)" using 1
   
    proof (rule disjE)
    {assume q
     hence "q | r" by (rule disjI1)
     thus "p ∨ (q ∨ r)" by (rule disjI2)}
    next
    {assume p
     thus "p ∨ (q ∨ r)" by (rule disjI1) }
    qed
}
qed


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

lemma ejercicio_33:
  assumes "p ∧ (q ∨ r)" 
  shows   "(p ∧ q) ∨ (p ∧ r)"

-- Solución M.Cumplido
proof -
have 1: p using assms(1) by (rule conjunct1)
have 2: "q | r" using assms(1) by (rule conjunct2)
show "(p ∧ q) ∨ (p ∧ r)" using 2
    proof (rule disjE)
    {assume 3: q
     have 4: "p & q" using 1 3 by (rule conjI) 
     thus  "(p ∧ q) ∨ (p ∧ r)" by (rule disjI1)}
    next
    {assume 5:r
     have 6: "p & r" using 1 5 by (rule conjI)
     thus  "(p ∧ q) ∨ (p ∧ r)" 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)"

-- Solución M.Cumplido
using assms(1)
proof (rule disjE)
{assume 1:"p & q"
 hence 2: p by (rule conjunct1)
 have q using 1 by (rule conjunct2)
 hence 3:"q | r" by (rule disjI1)
 show  "p ∧ (q ∨ r)" using 2 3 by (rule conjI)}
next
{assume 4:"p & r"
 hence 5: p by (rule conjunct1)
 have r using 4 by (rule conjunct2)
 hence 6:"q | r" by (rule disjI2)
 show  "p ∧ (q ∨ r)" using 5 6 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)"

-- Solución M.Cumplido
using assms(1)
proof (rule disjE)
{assume 1: p
 have 2: "p | q" using 1 by (rule disjI1)
 have 3: "p | r" using 1 by (rule disjI1)
 show  "(p ∨ q) ∧ (p ∨ r)" using 2 3 by (rule conjI) }
next
{assume 4: "q & r"
 hence 5: q by (rule conjunct1)
 have 6: r using 4 by (rule conjunct2)
 have 7: "p | q" using 5 by (rule disjI2)
 have 8: "p | r" using 6 by (rule disjI2)
 show  "(p ∨ q) ∧ (p ∨ r)" using 7 8 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)"

-- Solución M.Cumplido
proof -
have 1: "p | q" using assms(1) by (rule conjunct1)
have 2: "p | r" using assms(1) by (rule conjunct2)
have  "p ∨ (q ∧ r)" using 1
    proof (rule disjE)
    {assume p
     thus  "p ∨ (q ∧ r)" by (rule disjI1)}
    next
    {assume 3: q
     show  "p ∨ (q ∧ r)" using 2
          proof (rule disjE)
          {assume p
           thus  "p ∨ (q ∧ r)" by (rule disjI1)}
          next
          {assume 4: r 
           have "q & r" using 3 4 by (rule conjI) 
           thus "p ∨ (q ∧ r)" by (rule disjI2)}
          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"

-- Solución M.Cumplido
proof -
have 1: "p ⟶ r" using assms(1) by (rule conjunct1)
have 2: "q ⟶ r" using assms(1) by (rule conjunct2)
{assume 3: "p | q"
 have r using 3
    proof (rule disjE)
    {assume 4: p
     show r using 1 4 by (rule mp)}
    next
    {assume 5: q
     show r using 2 5 by (rule mp) }
    qed
}
thus  "p ∨ q ⟶ r" by (rule impI)
qed

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

lemma ejercicio_38:
  assumes "p ∨ q ⟶ r" 
  shows   "(p ⟶ r) ∧ (q ⟶ r)"

-- Solución M.Cumplido
proof -
{assume p 
 hence 1:"p | q" by (rule disjI1)
 have r using assms(1) 1 by (rule mp)}
hence 2: "p ⟶ r" by (rule impI)

{assume q
 hence 3:"p | q" by (rule disjI2)
 have r using assms(1) 3 by (rule mp)}
hence 4: "q ⟶ r" by (rule impI)

show  "(p ⟶ r) ∧ (q ⟶ r)" using 2 4 by (rule conjI)
qed


section {* Negaciones *}
 
text {* --------------------------------------------------------------- 
  Ejercicio 39. Demostrar
     p ⊢ ¬¬p
  ------------------------------------------------------------------ *}
 
lemma ejercicio_39:
  assumes "p"
  shows   "¬¬p"

-- Solución M.Cumplido
proof -
show  "¬¬p" using assms(1) by (rule notnotI)
qed
 
text {* --------------------------------------------------------------- 
  Ejercicio 40. Demostrar
     ¬p ⊢ p ⟶ q
  ------------------------------------------------------------------ *}
 
lemma ejercicio_40:
  assumes "¬p" 
  shows   "p ⟶ q"

-- Solución M.Cumplido
proof -
{assume 1: p
 have q using assms(1) 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"

-- Solución M.Cumplido
proof -
{assume 1: "~q"
 have "~p" using assms(1) 1 by (rule mt) }
thus  "¬q ⟶ ¬p" by (rule impI)
qed
 
text {* --------------------------------------------------------------- 
  Ejercicio 42. Demostrar
     p∨q, ¬q ⊢ p
  ------------------------------------------------------------------ *}
 
lemma ejercicio_42:
  assumes "p∨q"
          "¬q" 
  shows   "p"

-- Solución M.Cumplido
using assms(1)
proof (rule disjE)
{assume p
 thus p by this}
next
{assume 1:q
 show p using assms(2) 1 by (rule notE)}
qed 


text {* --------------------------------------------------------------- 
  Ejercicio 43. Demostrar
     p ∨ q, ¬p ⊢ q
  ------------------------------------------------------------------ *}
 
lemma ejercicio_43:
  assumes "p ∨ q"
          "¬p" 
  shows   "q"

-- Solución M.Cumplido
using assms(1)
proof (rule disjE)
{assume q
 thus q by this}
next
{assume 1:p
 show q using assms(2) 1 by (rule notE)}
qed 
 
text {* --------------------------------------------------------------- 
  Ejercicio 44. Demostrar
     p ∨ q ⊢ ¬(¬p ∧ ¬q)
  ------------------------------------------------------------------ *}
 
lemma ejercicio_44:
  assumes "p ∨ q" 
  shows   "¬(¬p ∧ ¬q)"

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

text {* --------------------------------------------------------------- 
  Ejercicio 45. Demostrar
     p ∧ q ⊢ ¬(¬p ∨ ¬q)
  ------------------------------------------------------------------ *}
 
lemma ejercicio_45:
  assumes "p ∧ q" 
  shows   "¬(¬p ∨ ¬q)"

-- Solución M.Cumplido
proof (rule notI)
assume 1: "¬p ∨ ¬q"
show False using 1
    proof (rule disjE)
     {assume 2: "~p" 
      have 3: p using assms(1) by (rule conjunct1)
      show False using 2 3 by (rule notE)}
     next
     {assume 4: "~q" 
      have 5: q using assms(1) 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"
oops

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

lemma ejercicio_47:
  assumes "¬p ∧ ¬q" 
  shows   "¬(p ∨ q)"
oops

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

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

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

lemma ejercicio_49:
  "¬(p ∧ ¬p)"
oops

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

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

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

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

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

lemma ejercicio_52:
  "p ∨ ¬p"
oops

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

lemma ejercicio_53:
  "((p ⟶ q) ⟶ p) ⟶ p"
oops

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

lemma ejercicio_54:
  assumes "¬q ⟶ ¬p"
  shows   "p ⟶ q"
oops

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

lemma ejercicio_55:
  assumes "¬(¬p ∧ ¬q)"
  shows   "p ∨ q"
oops

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

lemma ejercicio_56:
  assumes "¬(¬p ∨ ¬q)" 
  shows   "p ∧ q"
oops

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

lemma ejercicio_57:
  assumes "¬(p ∧ q)"
  shows   "¬p ∨ ¬q"
oops

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

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

end