Relación 4
De Lógica matemática y fundamentos (2017-18)
Revisión del 17:09 18 mar 2018 de Marmedmar3 (discusión | contribuciones)
chapter {* R4: Deducción natural proposicional *}
theory R4
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: josrodjim2 carmarria inmbenber marmedmar3
assumes 1: "p ⟶ q" and
2: "p"
shows "q"
proof -
show 1: "q" using 1 2 by (rule mp)
qed
text {* ---------------------------------------------------------------
Ejercicio 2. Demostrar
p ⟶ q, q ⟶ r, p ⊢ r
------------------------------------------------------------------ *}
lemma ejercicio_2: josrodjim2 carmarria inmbenber marmedmar3
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
text {* ---------------------------------------------------------------
Ejercicio 3. Demostrar
p ⟶ (q ⟶ r), p ⟶ q, p ⊢ r
------------------------------------------------------------------ *}
lemma ejercicio_3: josrodjim2 carmarria inmbenber marmedmar3
assumes 1: "p ⟶ (q ⟶ r)" and
2: "p ⟶ q" and
3: "p"
shows "r"
proof-
have 4: "q" using 2 3 by (rule mp)
have 5: "(q ⟶ r)" using 1 3 by (rule mp)
show "r" using 5 4 by (rule mp)
qed
text {* ---------------------------------------------------------------
Ejercicio 4. Demostrar
p ⟶ q, q ⟶ r ⊢ p ⟶ r
------------------------------------------------------------------ *}
lemma ejercicio_4: josrodjim2
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)}
hence 6: "p ⟶ r" using 3 5 by (rule impI)
show "p⟶r" using 6 by this
oops
lemma ejercicio_4: carmarria
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
lemma ejercicio_4: inmbenber
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
lemma ejercicio_4:marmedmar3
assumes "p ⟶ q"
"q ⟶ r"
shows "p ⟶ r"
proof
assume "p"
with assms(1) have "q" by (rule mp)
with assms(2) show "r" by (rule mp)
qed
text {* ---------------------------------------------------------------
Ejercicio 5. Demostrar
p ⟶ (q ⟶ r) ⊢ q ⟶ (p ⟶ r)
------------------------------------------------------------------ *}
lemma ejercicio_5: josrodjim2 (NO DA ERROR, PERO SI ALERTA)
assumes "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 6: "p⟶r" using 3 5 by (rule impI)}
hence 7: "q ⟶ (p ⟶ r)" using 2 6 by (rule impI)
show "q ⟶ (p ⟶ r)" using 7 by this
oops
lemma ejercicio_5: carmarria
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
lemma ejercicio_5: inmbenber
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
lemma ejercicio_5: marmedmar3
assumes "p ⟶ (q ⟶ r)"
shows "q ⟶ (p ⟶ r)"
proof
assume "q"
show "(p ⟶ r)"
proof
assume "p"
with assms(1) have "(q ⟶ r)" by (rule mp)
thus "r" using `q` by (rule mp)
qed
qed
text {* ---------------------------------------------------------------
Ejercicio 6. Demostrar
p ⟶ (q ⟶ r) ⊢ (p ⟶ q) ⟶ (p ⟶ r)
------------------------------------------------------------------ *}
lemma ejercicio_6:josrodjim2
assumes 1: "p ⟶ (q ⟶ r)"
shows "(p ⟶ q) ⟶ (p ⟶ r)"
proof-
{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 7: "p⟶r" using 3 6 by (rule impI)}
hence 8: "(p ⟶ q) ⟶ (p ⟶ r)" using 2 7 by (rule impI)
show "(p ⟶ q) ⟶ (p ⟶ r)" using 8 by this
oops
lemma ejercicio_6: carmarria, inmbenber
assumes 1: "p ⟶ (q ⟶ r)"
shows "(p ⟶ q) ⟶ (p ⟶ r)"
proof -
{assume 2: "p ⟶ q"
{assume 3: "p"
have 4: "q" using 2 3 by (rule mp)
have 5: "q ⟶ r" using 1 3 by (rule mp)
have 6: "r" using 5 4 by (rule mp)
}
hence 7: "p ⟶ r" by (rule impI)
}
thus "(p ⟶ q) ⟶ (p ⟶ r)" by (rule impI)
qed
lemma ejercicio_6:marmedmar3
assumes 1: "p ⟶ (q ⟶ r)"
shows "(p ⟶ q) ⟶ (p ⟶ r)"
proof
assume 2: "(p ⟶ q)"
show "(p ⟶ r)"
proof
assume 3: "p"
have 4: "q" using 2 3 by (rule mp)
have 5: "(q ⟶ r)" using 1 3 by (rule mp)
show "r" using 5 4 by (rule mp)
qed
qed
text {* ---------------------------------------------------------------
Ejercicio 7. Demostrar
p ⊢ q ⟶ p
------------------------------------------------------------------ *}
lemma ejercicio_7: (NO SE QUE ESTA MAL)
assumes "p"
shows "q ⟶ p"
proof -
{ assume 2: "q"
have 3: "p" using 1
}
hence 4: "q⟶p" using 2 3 by (rule impI)
show "q⟶p" using 4 by this
oops
lemma ejercicio_7: carmarria, inmbenber
assumes 1: "p"
shows "q ⟶ p"
proof -
{assume 2: "q"
have 3: "p" using 1 by this}
thus "q ⟶ p" by (rule impI)
qed
lemma ejercicio_7:marmedmar3
assumes "p"
shows "q ⟶ p"
proof
assume "q"
show "p" using assms(1) by this
qed
text {* ---------------------------------------------------------------
Ejercicio 8. Demostrar
⊢ p ⟶ (q ⟶ p)
------------------------------------------------------------------ *}
lemma ejercicio_8: carmarria
"p ⟶ (q ⟶ p)"
proof -
{assume 1: p
{assume 2: q
have 3: p using 1 by this
}
hence 4: "q ⟶ p" by (rule impI)
}
thus "p ⟶ (q ⟶ p)" by (rule impI)
qed
lemma ejercicio_8: marmedmar3
"p ⟶ (q ⟶ p)"
proof
assume 1: "p"
show "(q ⟶ p)"
proof
assume 2: "q"
show "p" using 1 by this
qed
qed
text {* ---------------------------------------------------------------
Ejercicio 9. Demostrar
p ⟶ q ⊢ (q ⟶ r) ⟶ (p ⟶ r)
------------------------------------------------------------------ *}
lemma ejercicio_9: carmarria
assumes 1: "p ⟶ q"
shows "(q ⟶ r) ⟶ (p ⟶ r)"
proof -
{assume 2: "q ⟶ r"
{assume 3: "p"
have 4: "q" using 1 3 by (rule mp)
have 5: "r" using 2 4 by (rule mp)
}
hence 6: "p ⟶ r" by (rule impI)
}
thus 7: "(q ⟶ r) ⟶ (p ⟶ r)" by (rule impI)
qed
lemma ejercicio_9: marmedmar3
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))
------------------------------------------------------------------ *}
lemma ejercicio_10: carmarria
assumes 1: "p ⟶ (q ⟶ (r ⟶ s))"
shows "r ⟶ (q ⟶ (p ⟶ s))"
proof-
{assume 2: r
{assume 3: q
{assume 4: p
have 5: "q ⟶ (r ⟶ s)" using 1 4 by (rule mp)
have 6: "r ⟶ s" using 5 3 by (rule mp)
have 7: s using 6 2 by (rule mp)
}
hence 8: "p ⟶ s" by (rule impI)
}
hence 9: "q ⟶ (p ⟶ s)" by (rule impI)
}
thus "r ⟶ (q ⟶ (p ⟶ s))" by (rule impI)
qed
lemma ejercicio_10: marmedmar3
assumes 1: "p ⟶ (q ⟶ (r ⟶ s))"
shows "r ⟶ (q ⟶ (p ⟶ s))"
proof
assume 2: "r"
show "(q ⟶ (p ⟶ s))"
proof
assume 3: "q"
show "(p ⟶ s)"
proof
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
text {* ---------------------------------------------------------------
Ejercicio 11. Demostrar
⊢ (p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))
------------------------------------------------------------------ *}
lemma ejercicio_11: carmarria
"(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))"
proof -
{assume 1: "p ⟶ (q ⟶ r)"
{assume 2: "p ⟶ q"
{assume 3: p
have 4: q using 2 3 by (rule mp)
have 5: "q ⟶ r" using 1 3 by (rule mp)
have 6: r using 5 4 by (rule mp)
}
hence 7: "p ⟶ r" by (rule impI)
}
hence 8: "(p ⟶ q) ⟶ (p ⟶ r)" by (rule impI)
}
thus "(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))" by (rule impI)
qed
lemma ejercicio_11: marmedmar3
"(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))"
proof
assume 1: "(p ⟶ (q ⟶ r))"
show "((p ⟶ q) ⟶ (p ⟶ r))"
proof
assume 2: "(p ⟶ q)"
show "(p ⟶ r)"
proof
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
text {* ---------------------------------------------------------------
Ejercicio 12. Demostrar
(p ⟶ q) ⟶ r ⊢ p ⟶ (q ⟶ r)
------------------------------------------------------------------ *}
lemma ejercicio_12: carmarria
assumes 1: "(p ⟶ q) ⟶ r"
shows "p ⟶ (q ⟶ r)"
proof -
{assume 2: p
{assume 3: q
{assume 4: p
have 5: q using 3 by this
}
hence 6: "p ⟶ q" by (rule impI)
have 7: r using 1 6 by (rule mp)
}
hence 8: "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: carmarria marmedmar3
assumes "p"
"q"
shows "p ∧ q"
proof-
show "p \<and> q" using assms(1) assms(2) by (rule conjI)
qed
text {* ---------------------------------------------------------------
Ejercicio 14. Demostrar
p ∧ q ⊢ p
------------------------------------------------------------------ *}
lemma ejercicio_14: carmarria marmedmar3
assumes "p ∧ q"
shows "p"
proof -
show "p" using assms by (rule conjunct1)
qed
text {* ---------------------------------------------------------------
Ejercicio 15. Demostrar
p ∧ q ⊢ q
------------------------------------------------------------------ *}
lemma ejercicio_15: carmarria marmedmar3
assumes "p ∧ q"
shows "q"
proof -
show q using assms(1) by (rule conjunct2)
qed
text {* ---------------------------------------------------------------
Ejercicio 16. Demostrar
p ∧ (q ∧ r) ⊢ (p ∧ q) ∧ r
------------------------------------------------------------------ *}
lemma ejercicio_16: carmarria marmedmar3
assumes 1: "p ∧ (q ∧ r)"
shows "(p ∧ q) ∧ r"
proof -
have 2: "q \<and> r" using 1 by (rule conjunct2)
have 3: p using 1 by (rule conjunct1)
have 4: q using 2 by (rule conjunct1)
have 5: r using 2 by (rule conjunct2)
have 6: "p \<and> q" using 3 4 by (rule conjI)
show "(p \<and> q) \<and> r" using 6 5 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 (rule conjI)
have "p ∧ q" using assms ..
thus "p" ..
next
show "q ∧ r"
proof (rule conjI)
have "p ∧ q" using assms ..
thus "q" ..
next
show "r" using assms ..
qed
qed
lemma ejercicio_17: carmarria marmedmar3
assumes 1:"(p ∧ q) ∧ r"
shows "p ∧ (q ∧ r)"
proof-
have 2:"p \<and> q" using 1 by (rule conjunct1)
have 3: r using 1 by (rule conjunct2)
have 4: p using 2 by (rule conjunct1)
have 5: q using 2 by (rule conjunct2)
have 6: "q \<and> r" using 5 3 by (rule conjI)
show "p \<and> (q \<and> r)" using 4 6 by (rule conjI)
qed
text {* ---------------------------------------------------------------
Ejercicio 18. Demostrar
p ∧ q ⊢ p ⟶ q
------------------------------------------------------------------ *}
lemma ejercicio_18: carmarria
assumes "p ∧ q"
shows "p ⟶ q"
proof -
{assume p
have q using assms(1) by (rule conjunct2)
}
thus "p ⟶ q" by (rule impI)
qed
lemma ejercicio_18: marmedmar3
assumes "p ∧ q"
shows "p ⟶ q"
proof
assume "p"
show "q" using assms by (rule conjunct2)
qed
text {* ---------------------------------------------------------------
Ejercicio 19. Demostrar
(p ⟶ q) ∧ (p ⟶ r) ⊢ p ⟶ q ∧ r
------------------------------------------------------------------ *}
lemma ejercicio_19: carmarria
assumes 1: "(p ⟶ q) ∧ (p ⟶ r)"
shows "p ⟶ q ∧ r"
proof -
{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)
have 7: "q \<and> r" using 5 6 by (rule conjI)
}
thus "p ⟶ q \<and> r" by (rule impI)
qed
lemma ejercicio_19: marmedmar3
assumes 1: "(p ⟶ q) ∧ (p ⟶ r)"
shows "p ⟶ q ∧ r"
proof
have 2: "(p ⟶ q)" using assms by (rule conjunct1)
have 3: "(p ⟶ r)" using assms by (rule conjunct2)
assume 4: "p"
with `(p ⟶ q)` have 5: "q" by (rule mp)
have 6: "r" using 3 4 by (rule mp)
show "q ∧ r" using 5 6 by (rule conjI)
qed
text {* ---------------------------------------------------------------
Ejercicio 20. Demostrar
p ⟶ q ∧ r ⊢ (p ⟶ q) ∧ (p ⟶ r)
------------------------------------------------------------------ *}
lemma ejercicio_20: carmarria
assumes 1: "p ⟶ q ∧ r"
shows "(p ⟶ q) ∧ (p ⟶ r)"
proof -
{assume 2: p
have 3: "q \<and> r" using 1 2 by (rule mp)
have 4: q using 3 by (rule conjunct1)
}
hence 5: "p ⟶ q" by (rule impI)
{assume 6: p
have 7: "q \<and> r" using 1 6 by (rule mp)
have 8: r using 7 by (rule conjunct2)
}
hence 9: "p ⟶ r" by (rule impI)
show "(p ⟶ q) \<and> (p ⟶ r)" using 5 9 by (rule conjI)
qed
text {* ---------------------------------------------------------------
Ejercicio 21. Demostrar
p ⟶ (q ⟶ r) ⊢ p ∧ q ⟶ r
------------------------------------------------------------------ *}
lemma ejercicio_21: carmarria
assumes 1: "p ⟶ (q ⟶ r)"
shows "p ∧ q ⟶ r"
proof -
{assume 2: "p \<and> q"
have 3: p using 2 by (rule conjunct1)
have 4: q using 2 by (rule conjunct2)
have 5: "q ⟶ r" using 1 3 by (rule mp)
have 6: r using 5 4 by (rule mp)
}
thus "p ∧ q ⟶ r" by (rule impI)
qed
lemma ejercicio_21: marmedmar3
assumes "p ⟶ (q ⟶ r)"
shows "p ∧ q ⟶ r"
proof
assume "p ∧ q"
hence "p" by (rule conjunct1)
with assms have "(q ⟶ r)" by (rule mp)
have "q" using `p ∧ q` by (rule conjunct2)
with `(q ⟶ r)` show "r" by (rule mp)
qed
text {* ---------------------------------------------------------------
Ejercicio 22. Demostrar
p ∧ q ⟶ r ⊢ p ⟶ (q ⟶ r)
------------------------------------------------------------------ *}
lemma ejercicio_22: carmarria
assumes "p ∧ q ⟶ r"
shows "p ⟶ (q ⟶ r)"
proof -
{assume p
{assume q
have "p \<and> q" using `p` `q` by (rule conjI)
have r using assms(1) `p \<and> q` by (rule mp)
}
hence "q ⟶ r" by (rule impI)
}
thus "p ⟶ (q ⟶ r)" by (rule impI)
qed
lemma ejercicio_22: marmedmar3
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 assms show "r" by (rule mp)
qed
qed
text {* ---------------------------------------------------------------
Ejercicio 23. Demostrar
(p ⟶ q) ⟶ r ⊢ p ∧ q ⟶ r
------------------------------------------------------------------ *}
lemma ejercicio_23: carmarria
assumes "(p ⟶ q) ⟶ r"
shows "p ∧ q ⟶ r"
proof -
{assume "p \<and> q"
{assume p
have q using `p \<and> q` by (rule conjunct2)
}
hence "p ⟶ q" by (rule impI)
have r using assms(1) `p ⟶ q` by (rule mp)
}
thus "p \<and> q ⟶ r" by (rule impI)
qed
lemma ejercicio_23: marmedmar3
assumes "(p ⟶ q) ⟶ r"
shows "p ∧ q ⟶ r"
proof
assume "p ∧ q"
have "p" using `p ∧ q` by (rule conjunct1)
have "q" using `p ∧ q` by (rule conjunct2)
hence "(p ⟶ q)" by (rule impI)
with assms show "r" by (rule mp)
qed
text {* ---------------------------------------------------------------
Ejercicio 24. Demostrar
p ∧ (q ⟶ r) ⊢ (p ⟶ q) ⟶ r
------------------------------------------------------------------ *}
lemma ejercicio_24: carmarria
assumes 1: "p ∧ (q ⟶ r)"
shows "(p ⟶ q) ⟶ r"
proof -
{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)
have 6: r using 4 5 by (rule mp)
}
thus "(p ⟶ q) ⟶ r" by (rule impI)
qed
lemma ejercicio_24: marmedmar3
assumes "p ∧ (q ⟶ r)"
shows "(p ⟶ q) ⟶ r"
proof
assume "(p ⟶ q)"
have "(q ⟶ r)" using assms by (rule conjunct2)
have "p" using assms by (rule conjunct1)
with `(p ⟶ q)` have "q" by (rule mp)
with `(q ⟶ r)` show "r" by (rule mp)
qed
section {* Disyunciones *}
text {* ---------------------------------------------------------------
Ejercicio 25. Demostrar
p ⊢ p ∨ q
------------------------------------------------------------------ *}
lemma ejercicio_25: carmarria
assumes "p"
shows "p ∨ q"
proof -
show "p | q" using assms(1) by (rule disjI1)
qed
text {* ---------------------------------------------------------------
Ejercicio 26. Demostrar
q ⊢ p ∨ q
------------------------------------------------------------------ *}
lemma ejercicio_26: carmarria
assumes "q"
shows "p ∨ q"
proof -
show "p | q" using assms(1) by (rule disjI2)
qed
text {* ---------------------------------------------------------------
Ejercicio 27. Demostrar
p ∨ q ⊢ q ∨ p
------------------------------------------------------------------ *}
lemma ejercicio_27: carmarria
assumes "p ∨ q"
shows "q ∨ p"
proof -
have "p | q" using assms(1) by this
moreover
{assume p
have "q | p" using `p` by (rule disjI2)
}
moreover
{assume q
have "q | p" using `q` by (rule disjI1)
}
ultimately show "q | p" by (rule disjE)
qed
text {* ---------------------------------------------------------------
Ejercicio 28. Demostrar
q ⟶ r ⊢ p ∨ q ⟶ p ∨ r
------------------------------------------------------------------ *}
lemma ejercicio_28: carmarria
assumes "q ⟶ r"
shows "p ∨ q ⟶ p ∨ r"
proof -
{assume "p | q"
moreover
{assume p
have "p | r" using `p` by (rule disjI1)
}
moreover
{assume q
have r using assms(1) `q` by (rule mp)
have "p | r" using `r` 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: carmarria
assumes "p ∨ p"
shows "p"
proof -
have "p | p" using assms(1) by this
moreover
{assume p
}
moreover
{assume p
}
ultimately show p by (rule disjE)
qed
text {* ---------------------------------------------------------------
Ejercicio 30. Demostrar
p ⊢ p ∨ p
------------------------------------------------------------------ *}
lemma ejercicio_30: carmarria
assumes "p"
shows "p ∨ p"
proof -
show "p | p" using assms(1) by (rule disjI1)
qed
text {* ---------------------------------------------------------------
Ejercicio 31. Demostrar
p ∨ (q ∨ r) ⊢ (p ∨ q) ∨ r
------------------------------------------------------------------ *}
lemma ejercicio_31:
assumes "p ∨ (q ∨ r)"
shows "(p ∨ q) ∨ r"
oops
text {* ---------------------------------------------------------------
Ejercicio 32. Demostrar
(p ∨ q) ∨ r ⊢ p ∨ (q ∨ r)
------------------------------------------------------------------ *}
lemma ejercicio_32:
assumes "(p ∨ q) ∨ r"
shows "p ∨ (q ∨ r)"
oops
text {* ---------------------------------------------------------------
Ejercicio 33. Demostrar
p ∧ (q ∨ r) ⊢ (p ∧ q) ∨ (p ∧ r)
------------------------------------------------------------------ *}
lemma ejercicio_33:
assumes "p ∧ (q ∨ r)"
shows "(p ∧ q) ∨ (p ∧ r)"
oops
text {* ---------------------------------------------------------------
Ejercicio 34. Demostrar
(p ∧ q) ∨ (p ∧ r) ⊢ p ∧ (q ∨ r)
------------------------------------------------------------------ *}
lemma ejercicio_34:
assumes "(p ∧ q) ∨ (p ∧ r)"
shows "p ∧ (q ∨ r)"
oops
text {* ---------------------------------------------------------------
Ejercicio 35. Demostrar
p ∨ (q ∧ r) ⊢ (p ∨ q) ∧ (p ∨ r)
------------------------------------------------------------------ *}
lemma ejercicio_35:
assumes "p ∨ (q ∧ r)"
shows "(p ∨ q) ∧ (p ∨ r)"
oops
text {* ---------------------------------------------------------------
Ejercicio 36. Demostrar
(p ∨ q) ∧ (p ∨ r) ⊢ p ∨ (q ∧ r)
------------------------------------------------------------------ *}
lemma ejercicio_36:
assumes "(p ∨ q) ∧ (p ∨ r)"
shows "p ∨ (q ∧ r)"
oops
text {* ---------------------------------------------------------------
Ejercicio 37. Demostrar
(p ⟶ r) ∧ (q ⟶ r) ⊢ p ∨ q ⟶ r
------------------------------------------------------------------ *}
lemma ejercicio_37:
assumes "(p ⟶ r) ∧ (q ⟶ r)"
shows "p ∨ q ⟶ r"
oops
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"
oops
text {* ---------------------------------------------------------------
Ejercicio 40. Demostrar
¬p ⊢ p ⟶ q
------------------------------------------------------------------ *}
lemma ejercicio_40:
assumes "¬p"
shows "p ⟶ q"
oops
text {* ---------------------------------------------------------------
Ejercicio 41. Demostrar
p ⟶ q ⊢ ¬q ⟶ ¬p
------------------------------------------------------------------ *}
lemma ejercicio_41:
assumes "p ⟶ q"
shows "¬q ⟶ ¬p"
oops
text {* ---------------------------------------------------------------
Ejercicio 42. Demostrar
p∨q, ¬q ⊢ p
------------------------------------------------------------------ *}
lemma ejercicio_42:
assumes "p∨q"
"¬q"
shows "p"
oops
text {* ---------------------------------------------------------------
Ejercicio 42. Demostrar
p ∨ q, ¬p ⊢ q
------------------------------------------------------------------ *}
lemma ejercicio_43:
assumes "p ∨ q"
"¬p"
shows "q"
oops
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"
oops
text {* ---------------------------------------------------------------
Ejercicio 51. Demostrar
¬¬p ⊢ p
------------------------------------------------------------------ *}
lemma ejercicio_51:
assumes "¬¬p"
shows "p"
oops
text {* ---------------------------------------------------------------
Ejercicio 52. Demostrar
⊢ p ∨ ¬p
------------------------------------------------------------------ *}
lemma ejercicio_52:
"p ∨ ¬p"
oops
text {* ---------------------------------------------------------------
Ejercicio 53. Demostrar
⊢ ((p ⟶ q) ⟶ p) ⟶ p
------------------------------------------------------------------ *}
lemma ejercicio_53:
"((p ⟶ q) ⟶ p) ⟶ p"
oops
text {* ---------------------------------------------------------------
Ejercicio 54. Demostrar
¬q ⟶ ¬p ⊢ p ⟶ q
------------------------------------------------------------------ *}
lemma ejercicio_54:
assumes "¬q ⟶ ¬p"
shows "p ⟶ q"
oops
text {* ---------------------------------------------------------------
Ejercicio 55. Demostrar
¬(¬p ∧ ¬q) ⊢ p ∨ q
------------------------------------------------------------------ *}
lemma ejercicio_55:
assumes "¬(¬p ∧ ¬q)"
shows "p ∨ q"
oops
text {* ---------------------------------------------------------------
Ejercicio 56. Demostrar
¬(¬p ∨ ¬q) ⊢ p ∧ q
------------------------------------------------------------------ *}
lemma ejercicio_56:
assumes "¬(¬p ∨ ¬q)"
shows "p ∧ q"
oops
text {* ---------------------------------------------------------------
Ejercicio 57. Demostrar
¬(p ∧ q) ⊢ ¬p ∨ ¬q
------------------------------------------------------------------ *}
lemma ejercicio_57:
assumes "¬(p ∧ q)"
shows "¬p ∨ ¬q"
oops
text {* ---------------------------------------------------------------
Ejercicio 58. Demostrar
⊢ (p ⟶ q) ∨ (q ⟶ p)
------------------------------------------------------------------ *}
lemma ejercicio_58:
"(p ⟶ q) ∨ (q ⟶ p)"
oops
end