Relación 3
De Lógica Matemática y fundamentos (2015-16)
Revisión del 11:09 20 mar 2016 de Marherfer6 (discusión | contribuciones)
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
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
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
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
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
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
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
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}
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}
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
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
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
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
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
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
text {* ---------------------------------------------------------------
Ejercicio 38. Demostrar
p ∨ q ⟶ r ⊢ (p ⟶ r) ∧ (q ⟶ r)
------------------------------------------------------------------ *}
lemma ejercicio_38:
assumes "p ∨ q ⟶ r"
shows "(p ⟶ r) ∧ (q ⟶ r)"
oops
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
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
text {* ---------------------------------------------------------------
Ejercicio 42. 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
text {* ---------------------------------------------------------------
Ejercicio 40. Demostrar
p ∨ q ⊢ ¬(¬p ∧ ¬q)
------------------------------------------------------------------ *}
lemma ejercicio_44:
assumes "p ∨ q"
shows "¬(¬p ∧ ¬q)"
oops
text {* ---------------------------------------------------------------
Ejercicio 45. Demostrar
p ∧ q ⊢ ¬(¬p ∨ ¬q)
------------------------------------------------------------------ *}
lemma ejercicio_45:
assumes "p ∧ q"
shows "¬(¬p ∨ ¬q)"
oops
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"
proof-
have 1: "p" using assms by (rule conjunct1)
have 2: "¬p" using assms by (rule conjunct2)
have 3: "q" using 2 1 by (rule notE)
show "q" using 3 by this
qed
text {* ---------------------------------------------------------------
Ejercicio 51. Demostrar
¬¬p ⊢ p
------------------------------------------------------------------ *}
lemma ejercicio_51:
assumes "¬¬p"
shows "p"
proof-
show "p" using assms by (rule notnotD)
qed
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