Relación 6
De Razonamiento automático (2018-19)
Revisión del 01:16 4 feb 2019 de Antramhur (discusión | contribuciones)
chapter {* R6: Deducción natural proposicional *}
theory R6_Deduccion_natural_proposicional
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
------------------------------------------------------------------ *}
(* pabalagon manperjim josgomrom4 cammonagu raffergon2 chrgencar gleherlop giafus1
marfruman1 enrparalv pabbergue antramhur*)
lemma ejercicio_1:
assumes 1: "p ⟶ q" and
2: "p"
shows "q"
proof -
show "q" using 1 2 by (rule mp)
qed
(* benber alfmarcua *)
lemma ejercicio_1_1:
assumes "p ⟶ q"
"p"
shows "q"
using assms by (rule mp)
text {* ---------------------------------------------------------------
Ejercicio 2. Demostrar
p ⟶ q, q ⟶ r, p ⊢ r
------------------------------------------------------------------ *}
(* pabalagon manperjim josgomrom4 cammonagu gleherlop raffergon2 chrgencar giafus1
marfruman1 alfmarcua enrparalv pabbergue antramhur *)
lemma ejercicio_2:
assumes 1: "p ⟶ q" and
2: "q ⟶ r" and
3: "p"
shows "r"
proof -
have 4: "q" using 1 3 by (rule mp)
show "r" using 2 4 by (rule mp)
qed
(* benber *)
lemma ejercicio_2_1:
assumes "p ⟶ q"
"q ⟶ r"
"p"
shows "r"
proof -
have "q" using `p ⟶ q` `p` by (rule mp)
with `q ⟶ r` show "r" by (rule mp)
qed
text {* ---------------------------------------------------------------
Ejercicio 3. Demostrar
p ⟶ (q ⟶ r), p ⟶ q, p ⊢ r
------------------------------------------------------------------ *}
(* pabalagon manperjim josgomrom4 cammonagu gleherlop raffergon2 chrgencar giafus1
marfruman1 alfmarcua enrparalv pabbergue antramhur *)
lemma ejercicio_3:
assumes 1: "p ⟶ (q ⟶ r)" and
2: "p ⟶ q" and
3: "p"
shows "r"
proof -
have 4: "q ⟶ r" using 1 3 by (rule mp)
have 5: "q" using 2 3 by (rule mp)
show "r" using 4 5 by (rule mp)
qed
(* benber *)
lemma ejercicio_3_1:
assumes "p ⟶ (q ⟶ r)"
"p ⟶ q"
"p"
shows "r"
proof -
have "q ⟶ r" using `p ⟶ (q ⟶ r)` `p` by (rule mp)
moreover have "q" using `p ⟶ q` `p` by (rule mp)
ultimately show "r" by (rule mp)
qed
text {* ---------------------------------------------------------------
Ejercicio 4. Demostrar
p ⟶ q, q ⟶ r ⊢ p ⟶ r
------------------------------------------------------------------ *}
(* pabalagon manperjim josgomrom4 cammonagu chrgencar raffergon2 gleherlop giafus1
marfruman1 alfmarcua enrparalv pabbergue antramhur*)
lemma ejercicio_4:
assumes 1: "p ⟶ q" and
2: "q ⟶ r"
shows "p ⟶ r"
proof -
{ assume 3: "p"
have 4: "q" using 1 3 by (rule mp)
have 5: "r" using 2 4 by (rule mp)}
thus "p ⟶ r" by (rule impI)
qed
(* benber *)
lemma ejercicio_4_1:
assumes "p ⟶ q"
"q ⟶ r"
shows "p ⟶ r"
proof
assume p
with `p ⟶ q` have "q" by (rule mp)
with `q ⟶ r` show "r" by (rule mp)
qed
text {* ---------------------------------------------------------------
Ejercicio 5. Demostrar
p ⟶ (q ⟶ r) ⊢ q ⟶ (p ⟶ r)
------------------------------------------------------------------ *}
(* pabalagon manperjim raffergon2 giafus1
marfruman1 alfmarcua enrparalv chrgencar pabbergue antramhur*)
lemma ejercicio_5:
assumes 1: "p ⟶ (q ⟶ r)"
shows "q ⟶ (p ⟶ r)"
proof (rule impI)
assume 2: "q"
show "p ⟶ r"
proof (rule impI)
assume 3: "p"
have 4: "q ⟶ r" using 1 3 by (rule mp)
show "r" using 4 2 by (rule mp)
qed
qed
(* benber josgomrom4 cammonagu *)
lemma ejercicio_5_1:
assumes "p ⟶ (q ⟶ r)"
shows "q ⟶ (p ⟶ r)"
proof
assume "q"
show "p ⟶ r"
proof
assume "p"
with `p ⟶ (q ⟶ r)` have "q ⟶ r" by (rule mp)
thus "r" using `q` by (rule mp)
qed
qed
(* pabalagon *)
lemma ejercicio_5_2:
assumes 1: "p ⟶ (q ⟶ r)"
shows "q ⟶ (p ⟶ r)"
proof -
{ assume 2: "q"
{ assume 3: "p"
have 4: "q ⟶ r" using 1 3 by (rule mp)
have 5: "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)
------------------------------------------------------------------ *}
(* pabalagon manperjim cammonagu chrgencar raffergon2 gleherlop giafus1
marfruman1 alfmarcua enrparalv pabbergue antramhur*)
lemma ejercicio_6:
assumes 1: "p ⟶ (q ⟶ r)"
shows "(p ⟶ q) ⟶ (p ⟶ r)"
proof (rule impI)
assume 2: "p ⟶ q"
show "p ⟶ r"
proof (rule impI)
assume 3: "p"
have 4: "q ⟶ r" using 1 3 by (rule mp)
have 5: "q" using 2 3 by (rule mp)
show "r" using 4 5 by (rule mp)
qed
qed
(* benber josgomrom4 *)
lemma ejercicio_6_1:
assumes "p ⟶ (q ⟶ r)"
shows "(p ⟶ q) ⟶ (p ⟶ r)"
proof
assume "p ⟶ q"
show "p ⟶ r"
proof
assume "p"
with `p ⟶ (q ⟶ r)` have "q ⟶ r" by (rule mp)
moreover from `p ⟶ q` `p` have "q" by (rule mp)
ultimately show "r" by (rule mp)
qed
qed
text {* ---------------------------------------------------------------
Ejercicio 7. Demostrar
p ⊢ q ⟶ p
------------------------------------------------------------------ *}
(* pabalagon manperjim josgomrom4 raffergon2 giafus1 marfruman1 enrparalv pabbergue antramhur*)
lemma ejercicio_7:
assumes 1: "p"
shows "q ⟶ p"
proof (rule impI)
assume 2: "q"
show "p" using 1 by this
qed
(* benber cammonagu chrgencar gleherlop *)
lemma ejercicio_7_1:
assumes "p"
shows "q ⟶ p"
proof
show "p" using `p` .
qed
(* alfmarcua *)
lemma ejercicio_7_2:
assumes "p"
shows "q ⟶ p"
using assms by (rule impI)
text {* ---------------------------------------------------------------
Ejercicio 8. Demostrar
⊢ p ⟶ (q ⟶ p)
------------------------------------------------------------------ *}
(* pabalagon manperjim josgomrom4 raffergon2 giafus1 marfruman1 enrparalv pabbergue antramhur *)
lemma ejercicio_8:
"p ⟶ (q ⟶ p)"
proof (rule impI)
assume 1: "p"
show "q ⟶ p"
proof (rule impI)
assume 2: "q"
show "p" using 1 by this
qed
qed
(* benber cammonagu gleherlop chrgencar alfmarcua *)
lemma ejercicio_8_1:
"p ⟶ (q ⟶ p)"
using ejercicio_7_1 by (rule impI)
text {* ---------------------------------------------------------------
Ejercicio 9. Demostrar
p ⟶ q ⊢ (q ⟶ r) ⟶ (p ⟶ r)
------------------------------------------------------------------ *}
(* pabalagon manperjim raffergon2 giafus1 marfruman1 alfmarcua enrparalv antramhur*)
lemma ejercicio_9:
assumes 1: "p ⟶ q"
shows "(q ⟶ r) ⟶ (p ⟶ r)"
proof (rule impI)
assume 2: "q ⟶ r"
show "p ⟶ r"
proof (rule impI)
assume 3: "p"
have 4: "q" using 1 3 by (rule mp)
show "r" using 2 4 by (rule mp)
qed
qed
(* benber josgomrom4 cammonagu chrgencar gleherlop pabbergue *)
lemma ejercicio_9_1:
assumes "p ⟶ q"
shows "(q ⟶ r) ⟶ (p ⟶ r)"
proof
assume "q ⟶ r"
show "p ⟶ r"
proof
assume "p"
with `p ⟶ q` have "q" by (rule mp)
with `q ⟶ r` show "r" by (rule mp)
qed
qed
text {* ---------------------------------------------------------------
Ejercicio 10. Demostrar
p ⟶ (q ⟶ (r ⟶ s)) ⊢ r ⟶ (q ⟶ (p ⟶ s))
------------------------------------------------------------------ *}
(* pabalagon manperjim raffergon2 marfruman1 alfmarcua enrparalv pabbergue antramhur *)
lemma ejercicio_10:
assumes 1: "p ⟶ (q ⟶ (r ⟶ s))"
shows "r ⟶ (q ⟶ (p ⟶ s))"
proof (rule impI)
assume 2: "r"
show "q ⟶ (p ⟶ s)"
proof (rule impI)
assume 3: "q"
show "p ⟶ s"
proof (rule impI)
assume 4: "p"
have 5: "q ⟶ (r ⟶ s)" using 1 4 by (rule mp)
have 6: "r ⟶ s" using 5 3 by (rule mp)
show "s" using 6 2 by (rule mp)
qed
qed
qed
(* benber josgomrom4 cammonagu gleherlop chrgencar giafus1 *)
lemma ejercicio_10_1:
assumes "p ⟶ (q ⟶ (r ⟶ s))"
shows "r ⟶ (q ⟶ (p ⟶ s))"
proof
assume "r"
show "q ⟶ (p ⟶ s)"
proof
assume "q"
show "p ⟶ s"
proof
assume "p"
with `p ⟶ (q ⟶ (r ⟶ s))`
have "q ⟶ (r ⟶ s)" by (rule mp)
hence "r ⟶ s" using `q` by (rule mp)
thus "s" using `r` by (rule mp)
qed
qed
qed
text {* ---------------------------------------------------------------
Ejercicio 11. Demostrar
⊢ (p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))
------------------------------------------------------------------ *}
(* pabalagon manperjim josgomrom4 cammonagu giafus1 alfmarcua chrgencar pabbergue *)
lemma ejercicio_11:
"(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))"
proof (rule impI)
assume 1: "p ⟶ (q ⟶ r)"
show "(p ⟶ q) ⟶ (p ⟶ r)" using 1 ejercicio_6 by simp
qed
(* pabalagon raffergon2 marfruman1 enrparalv antramhur *)
lemma ejercicio_11_2:
"(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))"
proof (rule impI)
assume 1: "p ⟶ (q ⟶ r)"
show "(p ⟶ q) ⟶ (p ⟶ r)"
proof (rule impI)
assume 2: "p ⟶ q"
show "p ⟶ r"
proof (rule impI)
assume 3: "p"
have 4: "q ⟶ r" using 1 3 by (rule mp)
have 5: "q" using 2 3 by (rule mp)
show "r" using 4 5 by (rule mp)
qed
qed
qed
(* benber *)
lemma ejercicio_11_1:
"(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))"
proof
assume "p ⟶ (q ⟶ r)"
show "(p ⟶ q) ⟶ (p ⟶ r)"
proof
assume "p ⟶ q"
show "p ⟶ r"
proof
assume p
with `p ⟶ (q ⟶ r)` have "q ⟶ r" by (rule mp)
moreover have "q" using `p ⟶ q` `p` by (rule mp)
ultimately show r by (rule mp)
qed
qed
qed
text {* ---------------------------------------------------------------
Ejercicio 12. Demostrar
(p ⟶ q) ⟶ r ⊢ p ⟶ (q ⟶ r)
------------------------------------------------------------------ *}
(* pabalagon manperjim raffergon2 marfruman1 enrparalv pabbergue antramhur *)
lemma ejercicio_12:
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"
proof (rule impI)
assume 5: "p"
show "q" using 3 by this
qed
show "r" using 1 4 by (rule mp)
qed
qed
(* benber josgomrom4 cammonagu giafus1 alfmarcua chrgencar *)
lemma ejercicio_12_1:
assumes "(p ⟶ q) ⟶ r"
shows "p ⟶ (q ⟶ r)"
proof
assume "p"
show "q ⟶ r"
proof
assume "q"
hence "p ⟶ q" by (rule impI)
with `(p ⟶ q) ⟶ r` show "r" by (rule mp)
qed
qed
section {* Conjunciones *}
text {* ---------------------------------------------------------------
Ejercicio 13. Demostrar
p, q ⊢ p ∧ q
------------------------------------------------------------------ *}
(* pabalagon *)
lemma ejercicio_13:
assumes "p"
"q"
shows "p ∧ q"
using assms(1, 2) by (rule conjI)
(* benber manperjim josgomrom4 cammonagu raffergon2 marfruman1 alfmarcua enrparalv chrgencar pabbergue antramhur*)
lemma ejercicio_13_1:
assumes "p"
"q"
shows "p ∧ q"
using assms by (rule conjI)
text {* ---------------------------------------------------------------
Ejercicio 14. Demostrar
p ∧ q ⊢ p
------------------------------------------------------------------ *}
(* pabalagon josgomrom4 *)
lemma ejercicio_14:
assumes "p ∧ q"
shows "p"
using assms(1) by (rule conjunct1)
(* benber cammonagu raffergon2 marfruman1 alfmarcua enrparalv chrgencar pabbergue antramhur *)
lemma ejercicio_14_1:
assumes "p ∧ q"
shows "p"
using assms by (rule conjunct1)
text {* ---------------------------------------------------------------
Ejercicio 15. Demostrar
p ∧ q ⊢ q
------------------------------------------------------------------ *}
(* pabalagon manperjim josgomrom4 *)
lemma ejercicio_15:
assumes "p ∧ q"
shows "q"
using assms(1) by (rule conjunct2)
(* benber cammonagu raffergon2 marfruman1 alfmarcua enrparalv chrgencar pabbergue antramhur *)
lemma ejercicio_15_1:
assumes "p ∧ q"
shows "q"
using assms by (rule conjunct2)
text {* ---------------------------------------------------------------
Ejercicio 16. Demostrar
p ∧ (q ∧ r) ⊢ (p ∧ q) ∧ r
------------------------------------------------------------------ *}
(* pabalagon manperjim josgomrom4 raffergon2 marfruman1 alfmarcua cammmonagu enrparalv chrgencar pabbergue antramhur *)
lemma ejercicio_16:
assumes "p ∧ (q ∧ r)"
shows "(p ∧ q) ∧ r"
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
(* benber *)
lemma ejercicio_16_1:
assumes "p ∧ (q ∧ r)"
shows "(p ∧ q) ∧ r"
proof - (* TODO? *)
have "q ∧ r" using assms by (rule conjunct2)
have "p" using assms by (rule conjunct1)
moreover have "q" using `q ∧ r` by (rule conjunct1)
ultimately have "p ∧ q" by (rule conjI)
moreover have "r" using `q ∧ r` by (rule conjunct2)
ultimately show "(p ∧ q) ∧ r" by (rule conjI)
qed
text {* ---------------------------------------------------------------
Ejercicio 17. Demostrar
(p ∧ q) ∧ r ⊢ p ∧ (q ∧ r)
------------------------------------------------------------------ *}
(* pabalagon manperjim josgomrom4 raffergon2 marfruman1 alfmarcua cammonagu chrgencar pabbergue antramhur *)
lemma ejercicio_17:
assumes 1: "(p ∧ q) ∧ r"
shows "p ∧ (q ∧ r)"
proof -
have 2: "r" using 1 by (rule conjunct2)
have 3: "p ∧ q" using 1 by (rule conjunct1)
have 4: "p" using 3 by (rule conjunct1)
have 5: "q" using 3 by (rule conjunct2)
have 6: "q ∧ r" using 5 2 by (rule conjI)
show ?thesis using 4 6 by (rule conjI)
qed
(* benber *)
lemma ejercicio_17_1:
assumes "(p ∧ q) ∧ r"
shows "p ∧ (q ∧ r)"
proof -
have "p ∧ q" using assms by (rule conjunct1)
have "p" using `p ∧ q` by (rule conjunct1)
moreover have "q ∧ r"
proof (rule conjI)
show "q" using `p ∧ q` by (rule conjunct2)
next
show "r" using assms by (rule conjunct2)
qed
ultimately show ?thesis by (rule conjI)
qed
text {* ---------------------------------------------------------------
Ejercicio 18. Demostrar
p ∧ q ⊢ p ⟶ q
------------------------------------------------------------------ *}
(* pabalagon raffergon2 marfruman1*)
lemma ejercicio_18:
assumes "p ∧ q"
shows "p ⟶ q"
proof (rule impI)
assume "p"
show "q" using assms(1) by (rule conjunct2)
qed
(* benber manperjim josgomrom4 cammonagu alfmarcua chrgencar pabbergue antramhur *)
lemma ejercicio_18_1:
assumes "p ∧ q"
shows "p ⟶ q"
proof
show "q" using assms by (rule conjunct2)
qed
text {* ---------------------------------------------------------------
Ejercicio 19. Demostrar
(p ⟶ q) ∧ (p ⟶ r) ⊢ p ⟶ q ∧ r
------------------------------------------------------------------ *}
(* pabalagon raffergon2 marfruman1 alfmarcua cammonagu chrgencar pabbergue antramhur *)
lemma ejercicio_19:
assumes 1: "(p ⟶ q) ∧ (p ⟶ r)"
shows "p ⟶ q ∧ r"
proof (rule impI)
assume 2: "p"
have 3: "p ⟶ q" using 1 by (rule conjunct1)
have 4: "p ⟶ r" using 1 by (rule conjunct2)
have 5: "q" using 3 2 by (rule mp)
have 6: "r" using 4 2 by (rule mp)
show "q ∧ r" using 5 6 by (rule conjI)
qed
(* benber cammonagu *)
lemma ejercicio_19_1:
assumes "(p ⟶ q) ∧ (p ⟶ r)"
shows "p ⟶ q ∧ r"
proof
assume p
show "q ∧ r"
proof
have "p ⟶ q" using assms by (rule conjunct1)
thus "q" using `p` by (rule mp)
next
have "p ⟶ r" using assms by (rule conjunct2)
thus "r" using `p` by (rule mp)
qed
qed
text {* ---------------------------------------------------------------
Ejercicio 20. Demostrar
p ⟶ q ∧ r ⊢ (p ⟶ q) ∧ (p ⟶ r)
------------------------------------------------------------------ *}
(* pabalagon raffergon2 marfruman1 alfmarcua cammonagu chrgencar pabbergue antramhur*)
lemma ejercicio_20:
assumes 1: "p ⟶ q ∧ r"
shows "(p ⟶ q) ∧ (p ⟶ r)"
proof (rule conjI)
show "p ⟶ q"
proof (rule impI)
assume 2: "p"
have 3: "q ∧ r" using 1 2 by (rule mp)
show 4: "q" using 3 by (rule conjunct1)
qed
show "p ⟶ r"
proof (rule impI)
assume 2: "p"
have 3: "q ∧ r" using 1 2 by (rule mp)
show 4: "r" using 3 by (rule conjunct2)
qed
qed
(* benber *)
lemma ejercicio_20_1:
assumes "p ⟶ q ∧ r"
shows "(p ⟶ q) ∧ (p ⟶ r)"
proof
show "p ⟶ q"
proof
assume "p"
with assms have "q ∧ r" by (rule mp)
thus "q" by (rule conjunct1)
qed
next
show "p ⟶ r"
proof
assume "p"
with assms have "q ∧ r" by (rule mp)
thus "r" by (rule conjunct2)
qed
qed
text {* ---------------------------------------------------------------
Ejercicio 21. Demostrar
p ⟶ (q ⟶ r) ⊢ p ∧ q ⟶ r
------------------------------------------------------------------ *}
(* pabalagon raffergon2 marfruman1 alfmarcua cammonagu chrgencar pabbergue antramhur *)
lemma ejercicio_21:
assumes 1: "p ⟶ (q ⟶ r)"
shows "p ∧ q ⟶ r"
proof (rule impI)
assume 2: "p ∧ q"
have 3: "p" using 2 by (rule conjunct1)
have 4: "q ⟶ r" using 1 3 by (rule mp)
have 5: "q" using 2 by (rule conjunct2)
show "r" using 4 5 by (rule mp)
qed
(* benber *)
lemma ejercicio_21_1:
assumes "p ⟶ (q ⟶ r)"
shows "p ∧ q ⟶ r"
proof
assume "p ∧ q"
hence "p" by (rule conjunct1)
with `p ⟶ (q ⟶ r)` have "q ⟶ r" by (rule mp)
moreover from `p ∧ q` have "q" by (rule conjunct2)
ultimately show "r" by (rule mp)
qed
text {* ---------------------------------------------------------------
Ejercicio 22. Demostrar
p ∧ q ⟶ r ⊢ p ⟶ (q ⟶ r)
------------------------------------------------------------------ *}
(* pabalagon raffergon2 marfruman1 alfmarcua chrgencar pabbergue antramhur *)
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
(* benber *)
lemma ejercicio_22_1:
assumes "p ∧ q ⟶ r"
shows "p ⟶ (q ⟶ r)"
proof
assume "p"
show "q ⟶ r"
proof
assume "q"
with `p` have "p ∧ q" by (rule conjI)
with `p ∧ q ⟶ r` show "r" by (rule mp)
qed
qed
text {* ---------------------------------------------------------------
Ejercicio 23. Demostrar
(p ⟶ q) ⟶ r ⊢ p ∧ q ⟶ r
------------------------------------------------------------------ *}
(* pabalagon raffergon2 marfruman1 alfmarcua chrgencar pabbergue antramhur*)
lemma ejercicio_23:
assumes 1: "(p ⟶ q) ⟶ r"
shows "p ∧ q ⟶ r"
proof (rule impI)
assume 2: "p ∧ q"
have 3: "p ⟶ q"
proof (rule impI)
assume "p"
show "q" using 2 by (rule conjunct2)
qed
show "r" using 1 3 by (rule mp)
qed
(* benber *)
lemma ejercicio_23_1:
assumes "(p ⟶ q) ⟶ r"
shows "p ∧ q ⟶ r"
proof
assume "p ∧ q"
hence "q" by (rule conjunct2)
hence "p ⟶ q" by (rule impI)
with `(p ⟶ q) ⟶ r` show "r" by (rule mp)
qed
text {* ---------------------------------------------------------------
Ejercicio 24. Demostrar
p ∧ (q ⟶ r) ⊢ (p ⟶ q) ⟶ r
------------------------------------------------------------------ *}
(* pabalagon raffergon2 marfruman1 alfmarcua chrgencar pabbergue antramhur *)
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 ⟶ r" using 1 by (rule conjunct2)
have 5: "q" using 2 3 by (rule mp)
show 6: "r" using 4 5 by (rule mp)
qed
(* benber *)
lemma ejercicio_24_1:
assumes "p ∧ (q ⟶ r)"
shows "(p ⟶ q) ⟶ r"
proof
have "q ⟶ r" using assms by (rule conjunct2)
assume "p ⟶ q"
moreover have "p" using assms by (rule conjunct1)
ultimately have "q" by (rule mp)
with `q ⟶ r` show r by (rule mp)
qed
section {* Disyunciones *}
text {* ---------------------------------------------------------------
Ejercicio 25. Demostrar
p ⊢ p ∨ q
------------------------------------------------------------------ *}
(* pabalagon manperjim benber cammonagu raffergon2 marfruman1 alfmarcua chrgencar pabbergue *)
lemma ejercicio_25:
assumes "p"
shows "p ∨ q"
using assms(1) by (rule disjI1)
text {* ---------------------------------------------------------------
Ejercicio 26. Demostrar
q ⊢ p ∨ q
------------------------------------------------------------------ *}
(* pabalagon manperjim benber cammonagu marfruman1 alfmarcua chrgencar pabbergue *)
lemma ejercicio_26:
assumes "q"
shows "p ∨ q"
using assms(1) by (rule disjI2)
text {* ---------------------------------------------------------------
Ejercicio 27. Demostrar
p ∨ q ⊢ q ∨ p
------------------------------------------------------------------ *}
(* pabalagon raffergon2 marfruman1 chrgencar pabbergue *)
lemma ejercicio_27:
assumes 1: "p ∨ q"
shows "q ∨ p"
using 1 proof (rule disjE)
assume 2: "p" thus "q ∨ p" by (rule disjI2)
next
assume 3: "q" thus "q ∨ p" by (rule disjI1)
qed
(* benber alfmarcua *)
lemma ejercicio_27_1:
assumes "p ∨ q"
shows "q ∨ p"
proof -
have "p ∨ q" using assms .
moreover have "p ⟹ q ∨ p" by (rule disjI2)
moreover have "q ⟹ q ∨ p" by (rule disjI1)
ultimately show "q ∨ p" by (rule disjE)
qed
text {* ---------------------------------------------------------------
Ejercicio 28. Demostrar
q ⟶ r ⊢ p ∨ q ⟶ p ∨ r
------------------------------------------------------------------ *}
(* pabalagon raffergon2 marfruman1 chrgencar alfmarcua pabbergue *)
lemma ejercicio_28:
assumes 1: "q ⟶ r"
shows "p ∨ q ⟶ p ∨ r"
proof (rule impI)
assume 2: "p ∨ q" show "p ∨ r" using 2
proof (rule disjE)
assume 3: p thus "p ∨ r" by (rule disjI1)
next
assume 4: q have r using 1 4 by (rule mp)
thus "p ∨ r" by (rule disjI2)
qed
qed
(* benber *)
lemma ejercicio_28_1:
assumes "q ⟶ r"
shows "p ∨ q ⟶ p ∨ r"
proof
assume "p ∨ q"
moreover have "p ⟹ p ∨ r" by (rule disjI1)
moreover have "q ⟹ p ∨ r"
proof (rule disjI2)
assume "q"
with `q ⟶ r` show "r" by (rule mp)
qed
ultimately show "p ∨ r" by (rule disjE)
qed
text {* ---------------------------------------------------------------
Ejercicio 29. Demostrar
p ∨ p ⊢ p
------------------------------------------------------------------ *}
(* pabalagon marfruman1 *)
lemma ejercicio_29:
assumes 1: "p ∨ p"
shows "p"
using 1 proof (rule disjE)
assume "p" thus "p" .
next
assume "p" thus "p" .
qed
(* benber cammonagu raffergon2 chrgencar alfmarcua pabbergue *)
lemma ejercicio_29_1:
assumes "p ∨ p"
shows "p"
using assms by (rule disjE)
text {* ---------------------------------------------------------------
Ejercicio 30. Demostrar
p ⊢ p ∨ p
------------------------------------------------------------------ *}
(* pabalagon manperjim benber raffergon2 marfruman1 chrgencar alfmarcua pabbergue *)
lemma ejercicio_30:
assumes "p"
shows "p ∨ p"
using assms(1) by (rule disjI1)
text {* ---------------------------------------------------------------
Ejercicio 31. Demostrar
p ∨ (q ∨ r) ⊢ (p ∨ q) ∨ r
------------------------------------------------------------------ *}
(* pabalagon marfruman1 alfmarcua pabbergue *)
lemma ejercicio_31:
assumes 1: "p ∨ (q ∨ r)"
shows "(p ∨ q) ∨ r" (is "?R")
using 1 proof (rule disjE)
assume "p" hence "p ∨ q" by (rule disjI1)
thus ?R by (rule disjI1)
next
assume "q ∨ r" thus ?R
proof (rule disjE)
assume "q" hence "p ∨ q" by (rule disjI2)
thus "(p ∨ q) ∨ r" by (rule disjI1)
next
assume "r" thus ?thesis by (rule disjI2)
qed
qed
(* benber *)
lemma ejercicio_31_1:
assumes "p ∨ (q ∨ r)"
shows "(p ∨ q) ∨ r"
proof -
have "p ∨ (q ∨ r)" using assms .
moreover {
assume "p"
hence "p ∨ q" by (rule disjI1)
hence "(p ∨ q) ∨ r" by (rule disjI1)
}
moreover {
assume "q ∨ r"
moreover {
assume "q"
hence "p ∨ q" by (rule disjI2)
hence "(p ∨ q) ∨ r" by (rule disjI1)
}
moreover {
assume "r"
hence "(p ∨ q) ∨ r" by (rule disjI2)
}
ultimately have "(p ∨ q) ∨ r" by (rule disjE)
}
ultimately show "(p ∨ q) ∨ r" by (rule disjE)
qed
text {* ---------------------------------------------------------------
Ejercicio 32. Demostrar
(p ∨ q) ∨ r ⊢ p ∨ (q ∨ r)
------------------------------------------------------------------ *}
(* pabalagon marfruman1 alfmarcua pabbergue *)
lemma ejercicio_32:
assumes 1: "(p ∨ q) ∨ r"
shows "p ∨ (q ∨ r)"
using 1 proof (rule disjE)
assume "p ∨ q" thus ?thesis
proof (rule disjE)
assume p thus ?thesis by (rule disjI1)
next
assume q hence "q ∨ r" by (rule disjI1)
thus ?thesis by (rule disjI2)
qed
next
assume r hence "q ∨ r" by (rule disjI2)
thus ?thesis by (rule disjI2)
qed
(* benber *)
lemma ejercicio_32_1:
assumes "(p ∨ q) ∨ r"
shows "p ∨ (q ∨ r)"
proof -
have "(p ∨ q) ∨ r" using assms .
moreover {
assume "p ∨ q"
moreover {
assume "p"
hence ?thesis by (rule disjI1)
}
moreover {
assume "q"
hence "q ∨ r" by (rule disjI1)
hence ?thesis by (rule disjI2)
}
ultimately have ?thesis by (rule disjE)
}
moreover {
assume "r"
hence "q ∨ r" by (rule disjI2)
hence ?thesis by (rule disjI2)
}
ultimately show ?thesis by (rule disjE)
qed
text {* ---------------------------------------------------------------
Ejercicio 33. Demostrar
p ∧ (q ∨ r) ⊢ (p ∧ q) ∨ (p ∧ r)
------------------------------------------------------------------ *}
(* pabalagon marfruman1 chrgencar alfmarcua pabbergue *)
lemma ejercicio_33:
assumes 1: "p ∧ (q ∨ r)"
shows "(p ∧ q) ∨ (p ∧ r)"
proof -
have 2: p using 1 by (rule conjunct1)
show ?thesis
proof (rule disjE)
assume 3: q have "p ∧ q" using 2 3 by (rule conjI)
thus ?thesis by (rule disjI1)
next
assume 4: r have "p ∧ r" using 2 4 by (rule conjI)
thus ?thesis by (rule disjI2)
next
show "q ∨ r" using 1 by (rule conjunct2)
qed
qed
(* benber *)
lemma ejercicio_33_1:
assumes "p ∧ (q ∨ r)"
shows "(p ∧ q) ∨ (p ∧ r)"
proof -
have "p" using assms by (rule conjunct1)
have "q ∨ r" using assms by (rule conjunct2)
moreover {
assume "q"
with `p` have "p ∧ q" by (rule conjI)
hence ?thesis by (rule disjI1)
}
moreover {
assume "r"
with `p` have "p ∧ r" by (rule conjI)
hence ?thesis by (rule disjI2)
}
ultimately show ?thesis by (rule disjE)
qed
text {* ---------------------------------------------------------------
Ejercicio 34. Demostrar
(p ∧ q) ∨ (p ∧ r) ⊢ p ∧ (q ∨ r)
------------------------------------------------------------------ *}
(* pabalagon marfruman1 chrgencar alfmarcua pabbergue *)
lemma ejercicio_34:
assumes "(p ∧ q) ∨ (p ∧ r)"
shows "p ∧ (q ∨ r)"
using assms(1) proof (rule disjE)
assume 2: "p ∧ q" hence q by (rule conjunct2)
hence 3: "q ∨ r" by (rule disjI1)
have p using 2 by (rule conjunct1)
thus ?thesis using 3 by (rule conjI)
next
assume 4: "p ∧ r" hence r by (rule conjunct2)
hence 5: "q ∨ r" by (rule disjI2)
have p using 4 by (rule conjunct1)
thus ?thesis using 5 by (rule conjI)
qed
(* benber *)
lemma ejercicio_34_1:
assumes "(p ∧ q) ∨ (p ∧ r)"
shows "p ∧ (q ∨ r)"
proof -
have "(p ∧ q) ∨ (p ∧ r)" using assms .
moreover {
assume "p ∧ q"
hence "p" by (rule conjunct1)
moreover {
have "q" using `p ∧ q` by (rule conjunct2)
hence "q ∨ r" by (rule disjI1)
}
ultimately have ?thesis by (rule conjI)
}
moreover {
assume "p ∧ r"
hence "p" by (rule conjunct1)
moreover {
have "r" using `p ∧ r` by (rule conjunct2)
hence "q ∨ r" by (rule disjI2)
}
ultimately have ?thesis by (rule conjI)
}
ultimately show ?thesis by (rule disjE)
qed
text {* ---------------------------------------------------------------
Ejercicio 35. Demostrar
p ∨ (q ∧ r) ⊢ (p ∨ q) ∧ (p ∨ r)
------------------------------------------------------------------ *}
(* pabalagon marfruman1 chrgencar alfmarcua pabbergue *)
lemma ejercicio_35:
assumes "p ∨ (q ∧ r)"
shows "(p ∨ q) ∧ (p ∨ r)"
using assms(1) proof (rule disjE)
assume 1: p hence 2: "p ∨ r" by (rule disjI1)
have "p ∨ q" using 1 by (rule disjI1)
thus ?thesis using 2 by (rule conjI)
next
assume 3: "q ∧ r" hence r by (rule conjunct2)
hence 4: "p ∨ r" by (rule disjI2)
have q using 3 by (rule conjunct1)
hence "p ∨ q" by (rule disjI2)
thus ?thesis using 4 by (rule conjI)
qed
(* benber *)
lemma ejercicio_35_1:
assumes "p ∨ (q ∧ r)"
shows "(p ∨ q) ∧ (p ∨ r)"
proof -
have "p ∨ (q ∧ r)" using assms .
moreover {
assume "p"
hence "p ∨ q" by (rule disjI1)
moreover have "p ∨ r" using `p` by (rule disjI1)
ultimately have ?thesis by (rule conjI)
}
moreover {
assume "q ∧ r"
{
have "q" using `q ∧ r` by (rule conjunct1)
hence "p ∨ q" by (rule disjI2)
}
moreover {
have "r" using `q ∧ r` by (rule conjunct2)
hence "p ∨ r" by (rule disjI2)
}
ultimately have ?thesis by (rule conjI)
}
ultimately show ?thesis by (rule disjE)
qed
text {* ---------------------------------------------------------------
Ejercicio 36. Demostrar
(p ∨ q) ∧ (p ∨ r) ⊢ p ∨ (q ∧ r)
------------------------------------------------------------------ *}
(* pabalagon marfruman1 chrgencar alfmarcua pabbergue *)
lemma ejercicio_36:
assumes 1: "(p ∨ q) ∧ (p ∨ r)"
shows "p ∨ (q ∧ r)"
proof -
have 2: "p ∨ q" using 1 by (rule conjunct1)
have 3: "p ∨ r" using 1 by (rule conjunct2)
show ?thesis using 2
proof (rule disjE)
assume p thus ?thesis by (rule disjI1)
next
assume 4: q show ?thesis using 3
proof (rule disjE)
assume p thus ?thesis by (rule disjI1)
next
assume 5: r have "q ∧ r" using 4 5 by (rule conjI)
thus ?thesis by (rule disjI2)
qed
qed
qed
(* benber *)
lemma ejercicio_36_1:
assumes "(p ∨ q) ∧ (p ∨ r)"
shows "p ∨ (q ∧ r)"
proof -
have "p ∨ q" using assms by (rule conjunct1)
moreover {
assume "p"
hence ?thesis by (rule disjI1)
}
moreover {
assume "q"
have "p ∨ r" using assms by (rule conjunct2)
moreover {
assume "p"
hence ?thesis by (rule disjI1)
}
moreover {
assume "r"
with `q` have "q ∧ r" by (rule conjI)
hence ?thesis by (rule disjI2)
}
ultimately have ?thesis by (rule disjE)
}
ultimately show ?thesis by (rule disjE)
qed
text {* ---------------------------------------------------------------
Ejercicio 37. Demostrar
(p ⟶ r) ∧ (q ⟶ r) ⊢ p ∨ q ⟶ r
------------------------------------------------------------------ *}
(* pabalagon marfruman1 alfmarcua pabbergue*)
lemma ejercicio_37:
assumes 1: "(p ⟶ r) ∧ (q ⟶ r)"
shows "p ∨ q ⟶ r"
proof (rule impI)
have 2: "p ⟶ r" using 1 by (rule conjunct1)
have 3: "q ⟶ r" using 1 by (rule conjunct2)
assume 4: "p ∨ q" show "r" using 4
proof (rule disjE)
assume 5: "p" show "r" using 2 5 by (rule mp)
next
assume 6: "q" show "r" using 3 6 by (rule mp)
qed
qed
(* benber *)
lemma ejercicio_37_1:
assumes "(p ⟶ r) ∧ (q ⟶ r)"
shows "p ∨ q ⟶ r"
proof
assume "p ∨ q"
moreover {
have "p ⟶ r" using assms by (rule conjunct1)
moreover assume "p"
ultimately have "r" by (rule mp)
}
moreover {
have "q ⟶ r" using assms by (rule conjunct2)
moreover assume "q"
ultimately have "r" by (rule mp)
}
ultimately show "r" by (rule disjE)
qed
text {* ---------------------------------------------------------------
Ejercicio 38. Demostrar
p ∨ q ⟶ r ⊢ (p ⟶ r) ∧ (q ⟶ r)
------------------------------------------------------------------ *}
(* pabalagon marfruman1 alfmarcua pabbergue*)
lemma ejercicio_38:
assumes "p ∨ q ⟶ r"
shows "(p ⟶ r) ∧ (q ⟶ r)"
proof (rule conjI)
show "p ⟶ r"
proof (rule impI)
assume "p" hence 1: "p ∨ q" by (rule disjI1)
show "r" using assms(1) 1 by (rule mp)
qed
next
show "q ⟶ r"
proof (rule impI)
assume q hence 2: "p ∨ q" by (rule disjI2)
show r using assms(1) 2 by (rule mp)
qed
qed
(* benber *)
lemma ejercicio_38_1:
assumes "p ∨ q ⟶ r"
shows "(p ⟶ r) ∧ (q ⟶ r)"
proof
show "p ⟶ r"
proof
assume "p"
hence "p ∨ q" by (rule disjI1)
with assms show "r" by (rule mp)
qed
next
show "q ⟶ r"
proof
assume "q"
hence "p ∨ q" by (rule disjI2)
with assms show "r" by (rule mp)
qed
qed
section {* Negaciones *}
text {* ---------------------------------------------------------------
Ejercicio 39. Demostrar
p ⊢ ¬¬p
------------------------------------------------------------------ *}
(* pabalagon manperjim benber raffergon2 marfruman1 chrgencar alfmarcua pabbergue *)
lemma ejercicio_39:
assumes "p"
shows "¬¬p"
using assms(1) by (rule notnotI)
text {* ---------------------------------------------------------------
Ejercicio 40. Demostrar
¬p ⊢ p ⟶ q
------------------------------------------------------------------ *}
(* pabalagon marfruman1 alfmarcua pabbergue *)
lemma ejercicio_40:
assumes 1: "¬p"
shows "p ⟶ q"
proof (rule impI)
assume 2: p show q using 1 2 by (rule notE)
qed
(* benber *)
lemma ejercicio_40_1:
assumes "¬p"
shows "p ⟶ q"
proof
assume "p"
with `¬p` show "q" by (rule notE)
qed
text {* ---------------------------------------------------------------
Ejercicio 41. Demostrar
p ⟶ q ⊢ ¬q ⟶ ¬p
------------------------------------------------------------------ *}
(* pabalagon raffergon2 marfruman1 alfmarcua pabbergue *)
lemma ejercicio_41:
assumes 1: "p ⟶ q"
shows "¬q ⟶ ¬p"
proof (rule impI)
assume 2: "¬q" show "¬p" using 1 2 by (rule mt)
qed
(* benber *)
lemma ejercicio_41_1:
assumes "p ⟶ q"
shows "¬q ⟶ ¬p"
proof
assume "¬q"
with `p ⟶ q` show "¬p" by (rule mt)
qed
text {* ---------------------------------------------------------------
Ejercicio 42. Demostrar
p∨q, ¬q ⊢ p
------------------------------------------------------------------ *}
(* pabalagon raffergon2 marfruman1 alfmarcua pabbergue *)
lemma ejercicio_42:
assumes "p∨q"
"¬q"
shows "p"
using assms(1) proof (rule disjE)
assume "p" thus "p" .
next
assume 2: "q" show "p" using assms(2) 2 by (rule notE)
qed
(* benber *)
lemma ejercicio_42_1:
assumes "p∨q"
"¬q"
shows "p"
proof -
note `p ∨ q`
moreover have "p ⟹ p" .
moreover {
assume "q"
with `¬q` have "p" by (rule notE)
}
ultimately show "p" by (rule disjE)
qed
text {* ---------------------------------------------------------------
Ejercicio 42. Demostrar
p ∨ q, ¬p ⊢ q
------------------------------------------------------------------ *}
(* pabalagon raffergon2 marfruman1 alfmarcua pabbergue *)
lemma ejercicio_43:
assumes "p ∨ q"
"¬p"
shows "q"
using assms(1) proof (rule disjE)
assume 1: "p" show "q" using assms(2) 1 by (rule notE)
next
assume "q" thus "q" .
qed
(* benber *)
lemma ejercicio_43_1:
assumes "p ∨ q"
"¬p"
shows "q"
proof -
note `p ∨ q`
moreover {
assume "p"
with `¬p` have "q" by (rule notE)
}
moreover have "q ⟹ q" .
ultimately show "q" by (rule disjE)
qed
text {* ---------------------------------------------------------------
Ejercicio 44. Demostrar
p ∨ q ⊢ ¬(¬p ∧ ¬q)
------------------------------------------------------------------ *}
(* pabalagon marfruman1*)
lemma ejercicio_44:
assumes "p ∨ q"
shows "¬(¬p ∧ ¬q)"
proof (rule notI)
assume 1: "¬p ∧ ¬q" hence 2: "¬p" by (rule conjunct1)
have 3: "¬q" using 1 by (rule conjunct2)
show "False"
using assms(1) proof (rule disjE)
assume 4: "p" show ?thesis using 2 4 by (rule notE)
next
assume 5: "q" show ?thesis using 3 5 by (rule notE)
qed
qed
(* benber alfmarcua *)
lemma ejercicio_44_1:
assumes "p ∨ q"
shows "¬(¬p ∧ ¬q)"
proof
assume "¬p ∧ ¬q"
note `p ∨ q`
moreover {
from `¬p ∧ ¬q` have "¬p" by (rule conjunct1)
moreover assume "p"
ultimately have "False" by (rule notE)
}
moreover {
from `¬p ∧ ¬q` have "¬q" by (rule conjunct2)
moreover assume "q"
ultimately have "False" by (rule notE)
}
ultimately show "False" by (rule disjE)
qed
text {* ---------------------------------------------------------------
Ejercicio 45. Demostrar
p ∧ q ⊢ ¬(¬p ∨ ¬q)
------------------------------------------------------------------ *}
(* pabalagon marfruman1*)
lemma ejercicio_45:
assumes 1: "p ∧ q"
shows "¬(¬p ∨ ¬q)"
proof (rule notI)
assume 2: "¬p ∨ ¬q" have 3: "p" using 1 by (rule conjunct1)
have 4: "q" using 1 by (rule conjunct2)
show "False" using 2
proof (rule disjE)
assume "¬p" thus ?thesis using 3 by (rule notE)
next
assume "¬q" thus ?thesis using 4 by (rule notE)
qed
qed
(* benber alfmarcua *)
lemma ejercicio_45_1:
assumes "p ∧ q"
shows "¬(¬p ∨ ¬q)"
proof
assume "¬p ∨ ¬q"
moreover {
assume "¬p"
moreover have "p" using `p ∧ q` by (rule conjunct1)
ultimately have "False" by (rule notE)
}
moreover {
assume "¬q"
moreover have "q" using `p ∧ q` by (rule conjunct2)
ultimately have "False" by (rule notE)
}
ultimately show False by (rule disjE)
qed
text {* ---------------------------------------------------------------
Ejercicio 46. Demostrar
¬(p ∨ q) ⊢ ¬p ∧ ¬q
------------------------------------------------------------------ *}
(* pabalagon marfruman1*)
lemma ejercicio_46:
assumes 1: "¬(p ∨ q)"
shows "¬p ∧ ¬q"
proof (rule conjI)
show "¬p"
proof (rule notI)
assume p hence 2: "p ∨ q" by (rule disjI1)
show False using 1 2 by (rule notE)
qed
show "¬q"
proof (rule notI)
assume q hence 3: "p ∨ q" by (rule disjI2)
show False using 1 3 by (rule notE)
qed
qed
(* benber *)
lemma ejercicio_46_1:
assumes "¬(p ∨ q)"
shows "¬p ∧ ¬q"
proof
show "¬p"
proof (rule ccontr)
assume "¬¬p"
hence "p" by (rule notnotD)
hence "p ∨ q" by (rule disjI1)
with assms show "False" by (rule notE)
qed
next
show "¬q"
proof (rule ccontr)
assume "¬¬q"
hence "q" by (rule notnotD)
hence "p ∨ q" by (rule disjI2)
with assms show "False" by (rule notE)
qed
qed
(* alfmarcua *)
lemma ejercicio_46_2:
assumes "¬(p ∨ q)"
shows "¬p ∧ ¬q"
proof (rule conjI)
have "p ⟹ p ∨ q" by (rule disjI1)
then have "p ⟶ p ∨ q" by (rule impI)
then show "¬ p" using assms by (rule mt)
next
have "q ⟹ p ∨ q" by (rule disjI2)
then have "q ⟶ p ∨ q" by (rule impI)
then show "¬ q" using assms by (rule mt)
qed
text {* ---------------------------------------------------------------
Ejercicio 47. Demostrar
¬p ∧ ¬q ⊢ ¬(p ∨ q)
------------------------------------------------------------------ *}
(* pabalagon marfruman1*)
lemma ejercicio_47:
assumes 1: "¬p ∧ ¬q"
shows "¬(p ∨ q)"
proof (rule notI)
have 2: "¬p" using 1 by (rule conjunct1)
have 3: "¬q" using 1 by (rule conjunct2)
assume 4: "p ∨ q"
show False
using 4 proof (rule disjE)
assume 5: p show ?thesis using 2 5 by (rule notE)
next
assume 6: q show ?thesis using 3 6 by (rule notE)
qed
qed
(* benber *)
lemma ejercicio_47_1:
assumes "¬p ∧ ¬q"
shows "¬(p ∨ q)"
proof
assume "p ∨ q"
hence "¬(¬p ∧ ¬q)" by (rule ejercicio_44_1)
thus "False" using assms by (rule notE)
qed
(* alfmarcua *)
lemma ejercicio_47_2:
assumes "¬p ∧ ¬q"
shows "¬(p ∨ q)"
proof (rule notI)
have "¬ p" using assms by (rule conjunct1)
have "¬ q" using assms by (rule conjunct2)
assume "p ∨ q"
then have "q" using `¬ p` by (rule ejercicio_43)
show False using `¬ q` `q` by (rule notE)
qed
text {* ---------------------------------------------------------------
Ejercicio 48. Demostrar
¬p ∨ ¬q ⊢ ¬(p ∧ q)
------------------------------------------------------------------ *}
(* pabalagon marfruman1*)
lemma ejercicio_48:
assumes 1: "¬p ∨ ¬q"
shows "¬(p ∧ q)"
proof (rule notI)
assume 2: "p ∧ q" hence 3: p by (rule conjunct1)
have 4: q using 2 by (rule conjunct2)
show False
using 1 proof (rule disjE)
assume "¬p" thus ?thesis using 3 by (rule notE)
next
assume "¬q" thus ?thesis using 4 by (rule notE)
qed
qed
(* benber alfmarcua *)
lemma ejercicio_48_1:
assumes "¬p ∨ ¬q"
shows "¬(p ∧ q)"
proof
assume "p ∧ q"
hence "¬(¬p ∨ ¬q)" by (rule ejercicio_45_1)
thus "False" using assms by (rule notE)
qed
text {* ---------------------------------------------------------------
Ejercicio 49. Demostrar
⊢ ¬(p ∧ ¬p)
------------------------------------------------------------------ *}
(* pabalagon marfruman1 alfmarcua *)
lemma ejercicio_49:
"¬(p ∧ ¬p)"
proof (rule notI)
assume 1: "p ∧ ¬p" hence 2: p by (rule conjunct1)
have "¬p" using 1 by (rule conjunct2)
thus "False" using 2 by (rule notE)
qed
(* benber *)
lemma ejercicio_49_1:
"¬(p ∧ ¬p)"
proof
assume "p ∧ ¬p"
hence "¬p" by (rule conjunct2)
moreover have "p" using `p ∧ ¬p` by (rule conjunct1)
ultimately show "False" by (rule notE)
qed
text {* ---------------------------------------------------------------
Ejercicio 50. Demostrar
p ∧ ¬p ⊢ q
------------------------------------------------------------------ *}
(* pabalagon raffergon2 marfruman1 alfmarcua *)
lemma ejercicio_50:
assumes 1: "p ∧ ¬p"
shows "q"
proof (rule notE)
show p using 1 by (rule conjunct1)
show "¬p" using 1 by (rule conjunct2)
qed
(* benber *)
lemma ejercicio_50_1:
assumes "p ∧ ¬p"
shows "q"
proof -
have "¬p" using `p ∧ ¬p` by (rule conjunct2)
moreover have "p" using `p ∧ ¬p` by (rule conjunct1)
ultimately show "q" by (rule notE)
qed
text {* ---------------------------------------------------------------
Ejercicio 51. Demostrar
¬¬p ⊢ p
------------------------------------------------------------------ *}
(* pabalagon manperjim benber raffergon2 marfruman1 alfmarcua *)
lemma ejercicio_51:
assumes "¬¬p"
shows "p"
using assms(1) by (rule notnotD)
text {* ---------------------------------------------------------------
Ejercicio 52. Demostrar
⊢ p ∨ ¬p
------------------------------------------------------------------ *}
(* pabalagon marfruman1*)
lemma ejercicio_52:
"p ∨ ¬p"
proof (rule ccontr)
assume 1: "¬(p ∨ ¬p)"
have 2: "¬p" proof (rule notI)
assume p hence 3: "p ∨ ¬p" by (rule disjI1)
show "False" using 1 3 by (rule notE)
qed
have 4: "p ∨ ¬p" using 2 by (rule disjI2)
show "False" using 1 4 by (rule notE)
qed
(* benber *)
lemma ejercicio_52_1:
"p ∨ ¬p"
proof (rule ccontr)
assume "¬ (p ∨ ¬ p)"
hence "¬p ∧ ¬¬p" by (rule ejercicio_46_1)
hence "¬p" by (rule conjunct1)
moreover {
have "¬¬p" using `¬p ∧ ¬¬p` by (rule conjunct2)
hence "p" by (rule notnotD)
}
ultimately show "False" by (rule notE)
qed
(* alfmarcua *)
lemma ejercicio_52_2:
"p ∨ ¬p"
proof (rule ccontr)
assume "¬ (p ∨ ¬ p)"
then have "¬ p ∧ ¬¬ p" by (rule ejercicio_46)
then show False by (rule ejercicio_50)
qed
text {* ---------------------------------------------------------------
Ejercicio 53. Demostrar
⊢ ((p ⟶ q) ⟶ p) ⟶ p
------------------------------------------------------------------ *}
(* pabalagon marfruman1 alfmarcua *)
lemma ejercicio_53:
"((p ⟶ q) ⟶ p) ⟶ p"
proof (rule impI)
assume 1: "(p ⟶ q) ⟶ p"
show p proof (rule ccontr)
assume 2: "¬p"
have 3: "¬(p ⟶ q)" using 1 2 by (rule mt)
have 4: "p ⟶ q" proof (rule impI)
assume 5: p show q using 2 5 by (rule notE)
qed
show False using 3 4 by (rule notE)
qed
qed
(* benber *)
lemma ejercicio_53_1:
"((p ⟶ q) ⟶ p) ⟶ p"
proof
assume "(p ⟶ q) ⟶ p"
have "p ∨ ¬p" by (rule ejercicio_52_1)
moreover have "p ⟹ p" .
moreover {
assume "¬ p"
hence "p ⟶ q" by (rule ejercicio_40_1)
with `(p ⟶ q) ⟶ p` have "p" by (rule mp)
}
ultimately show "p" by (rule disjE)
qed
text {* ---------------------------------------------------------------
Ejercicio 54. Demostrar
¬q ⟶ ¬p ⊢ p ⟶ q
------------------------------------------------------------------ *}
(* pabalagon marfruman1*)
lemma ejercicio_54:
assumes 1: "¬q ⟶ ¬p"
shows "p ⟶ q"
proof (rule impI)
assume 2: "p" hence 3: "¬¬p" by (rule notnotI)
have "¬¬q" using 1 3 by (rule mt)
thus "q" by (rule notnotD)
qed
(* benber alfmarcua *)
lemma ejercicio_54_1:
assumes "¬q ⟶ ¬p"
shows "p ⟶ q"
proof
assume "p"
hence "¬¬p" by (rule notnotI)
with assms have "¬¬q" by (rule mt)
thus "q" by (rule notnotD)
qed
text {* ---------------------------------------------------------------
Ejercicio 55. Demostrar
¬(¬p ∧ ¬q) ⊢ p ∨ q
------------------------------------------------------------------ *}
(* pabalagon marfruman1*)
lemma ejercicio_55:
assumes 1: "¬(¬p ∧ ¬q)"
shows "p ∨ q"
proof (rule ccontr)
assume 2: "¬(p ∨ q)"
have 3: "p"
proof (rule ccontr)
assume 4: "¬p"
have 5: "q"
proof (rule ccontr)
assume 6: "¬q" have 7: "¬p ∧ ¬q" using 4 6 by (rule conjI)
show False using 1 7 by (rule notE)
qed
have 8: "p ∨ q" using 5 by (rule disjI2)
show False using 2 8 by (rule notE)
qed
have 9: "p ∨ q" using 3 by (rule disjI1)
show False using 2 9 by (rule notE)
qed
(* benber alfmarcua*)
lemma ejercicio_55_1:
assumes "¬(¬p ∧ ¬q)"
shows "p ∨ q"
proof (rule ccontr)
assume "¬(p ∨ q)"
hence "¬p ∧ ¬q" by (rule ejercicio_46_1)
with assms show False by (rule notE)
qed
text {* ---------------------------------------------------------------
Ejercicio 56. Demostrar
¬(¬p ∨ ¬q) ⊢ p ∧ q
------------------------------------------------------------------ *}
(* pabalagon marfruman1*)
lemma ejercicio_56:
assumes 1: "¬(¬p ∨ ¬q)"
shows "p ∧ q"
proof (rule conjI)
show 3: p
proof (rule ccontr)
assume "¬p" hence 4: "¬p ∨ ¬q" by (rule disjI1)
show False using 1 4 by (rule notE)
qed
show 5: q
proof (rule ccontr)
assume "¬q" hence 6: "¬p ∨ ¬q" by (rule disjI2)
show False using 1 6 by (rule notE)
qed
qed
(* benber alfmarcua *)
lemma ejercicio_56_1:
assumes "¬(¬p ∨ ¬q)"
shows "p ∧ q"
proof -
have "¬¬p ∧ ¬¬q" using assms by (rule ejercicio_46_1)
hence "¬¬p" by (rule conjunct1)
hence "p" by (rule notnotD)
moreover {
have "¬¬q" using `¬¬p ∧ ¬¬q` by (rule conjunct2)
hence "q" by (rule notnotD)
}
ultimately show "p ∧ q" by (rule conjI)
qed
text {* ---------------------------------------------------------------
Ejercicio 57. Demostrar
¬(p ∧ q) ⊢ ¬p ∨ ¬q
------------------------------------------------------------------ *}
(* pabalagon marfruman1*)
lemma ejercicio_57:
assumes 1: "¬(p ∧ q)"
shows "¬p ∨ ¬q"
proof (rule ccontr)
assume 2: "¬(¬p ∨ ¬q)"
show False using 1
proof (rule notE)
show 3: "p ∧ q"
proof (rule conjI)
show p
proof (rule ccontr)
assume "¬p" hence 4: "¬p ∨ ¬q" by (rule disjI1)
show False using 2 4 by (rule notE)
qed
next
show q
proof (rule ccontr)
assume "¬q" hence 5: "¬p ∨ ¬q" by (rule disjI2)
show False using 2 5 by (rule notE)
qed
qed
qed
qed
(* benber alfmarcua *)
lemma ejercicio_57_1:
assumes "¬(p ∧ q)"
shows "¬p ∨ ¬q"
proof (rule ccontr)
assume "¬(¬p ∨ ¬q)"
hence "p ∧ q" by (rule ejercicio_56_1)
with assms show "False" by (rule notE)
qed
text {* ---------------------------------------------------------------
Ejercicio 58. Demostrar
⊢ (p ⟶ q) ∨ (q ⟶ p)
------------------------------------------------------------------ *}
(* pabalagon marfruman1 *)
lemma ejercicio_58:
"(p ⟶ q) ∨ (q ⟶ p)"
proof -
have "(p ⟶ q) ∨ ¬(p ⟶ q)" proof (rule ccontr)
assume 1: "¬((p ⟶ q) ∨ ¬(p ⟶ q))"
have 2: "¬(p ⟶ q)" proof (rule notI)
assume "p ⟶ q"
hence 3: "(p ⟶ q) ∨ ¬(p ⟶ q)" by (rule disjI1)
show False using 1 3 by (rule notE)
qed
hence 4: "(p ⟶ q) ∨ ¬(p ⟶ q)" by (rule disjI2)
show "False" using 1 4 by (rule notE)
qed
thus ?thesis proof (rule disjE)
assume "p ⟶ q" thus ?thesis by (rule disjI1)
next
assume 1: "¬(p ⟶ q)"
have "q ⟶ p" proof (rule impI)
assume 2: q
have 3: "p ⟶ q" proof (rule impI)
assume p show q using 2 .
qed
show p using 1 3 by (rule notE)
qed
thus ?thesis by (rule disjI2)
qed
qed
(* pabalagon *)
lemma ejercicio_58_2:
"(p ⟶ q) ∨ (q ⟶ p)"
proof (rule ccontr)
assume 1: "¬((p ⟶ q) ∨ (q ⟶ p))"
hence 1: "(p ∧ ¬q) ∧ (q ∧ ¬p)" by simp
hence "p ∧ ¬q" ..
hence 2: p ..
have "q ∧ ¬p" using 1 ..
hence 3: "¬p" ..
show False using 3 2 by (rule notE)
qed
(* benber *)
lemma ejercicio_58_1:
"(p ⟶ q) ∨ (q ⟶ p)"
proof (rule ccontr)
assume "¬ ((p ⟶ q) ∨ (q ⟶ p))"
hence 1: "¬(p ⟶ q) ∧ ¬(q ⟶ p)" by (rule ejercicio_46_1)
have "p ∨ ¬p" by (rule ejercicio_52_1)
moreover {
assume "p"
hence "q ⟶ p" by (rule ejercicio_7_1)
have "¬(q ⟶ p)" using 1 by (rule conjunct2)
hence "False" using `q ⟶ p` by (rule notE)
}
moreover {
assume "¬p"
hence "p ⟶ q" by (rule ejercicio_40_1)
have "¬(p ⟶ q)" using 1 by (rule conjunct1)
hence "False" using `p ⟶ q` by (rule notE)
}
ultimately show "False" by (rule disjE)
moreover {
assume "p"
hence "q ⟶ p" by (rule ejercicio_7_1)
have "¬(q ⟶ p)" using 1 by (rule conjunct2)
hence "False" using `q ⟶ p` by (rule notE)
}
qed
(* alfmarcua *)
lemma ejercicio_58_3:
"(p ⟶ q) ∨ (q ⟶ p)"
proof (rule disjE)
show "p ∨ ¬ p" by (rule ejercicio_52)
next
assume "¬p"
then have "p ⟶ q" by (rule ejercicio_40)
then show ?thesis by (rule disjI1)
next
assume "p"
then have "q ⟶ p" by (rule ejercicio_7)
then show ?thesis by (rule disjI2)
qed
end