Relación 3

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

---**

proof(rule impI)
assume 1: "p"
have 2: "q" using assms(1) 1 by (rule mp)
show "r" using assms(2) 2 by (rule mp)
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 

---**
proof(rule impI)
assume 1: "p⟶q"
{assume 2: "p"
have 3: "q⟶r" using assms(1) 2 by (rule mp)
have 4: "q" using 1 2 by (rule mp)
have 5: "r" using 3 4 by (rule mp)}
then show"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
 
--**
proof(rule impI)
assume 1: "q⟶r"
{assume 2: "p"
have 3: "q" using assms 2 by (rule mp)
have 4: "r" using 1 3 by (rule mp)}
then show "p⟶r" by (rule impI)
qed

--**
(* aunque sé que hay algunos pasos innecesarios pero es otra forma de verlo.*)
proof-
{assume 1 : "q⟶r" 
{assume 2 : "p"
{assume 3: "¬r"
have 4: "¬q" using 1 3 by (rule mt)
have 5: "q" using assms(1) 2 by (rule mp)
have 7: "False" using 4 5 by (rule notE)}
hence 6: "r"  by (rule ccontr)}
hence 8 : "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

--**
 
lemma ejercicio_18:
  assumes "p ∧ q" 
  shows   "p ⟶ q"
proof(rule impI)
assume 1: "p"
show "q" using assms by (rule conjunct2)
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