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Lógica matemática y fundamentos (2017-18) - Contribuciones del usuario [es]
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Contribuciones del usuario
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https://www.glc.us.es/~jalonso/LMF2018/index.php?title=Relaci%C3%B3n_7&diff=181
Relación 7
2018-04-08T22:15:40Z
<p>Joslopjim4: </p>
<hr />
<div><source lang = "isar"><br />
chapter {* R7: Deducción natural en lógica de primer orden *}<br />
<br />
theory R7<br />
imports Main <br />
begin<br />
<br />
text {*<br />
Demostrar o refutar los siguientes lemas usando sólo las reglas<br />
básicas de deducción natural de la lógica proposicional, de los<br />
cuantificadores y de la igualdad: <br />
· conjI: ⟦P; Q⟧ ⟹ P ∧ Q<br />
· conjunct1: P ∧ Q ⟹ P<br />
· conjunct2: P ∧ Q ⟹ Q <br />
· notnotD: ¬¬ P ⟹ P<br />
· mp: ⟦P ⟶ Q; P⟧ ⟹ Q <br />
· impI: (P ⟹ Q) ⟹ P ⟶ Q<br />
· disjI1: P ⟹ P ∨ Q<br />
· disjI2: Q ⟹ P ∨ Q<br />
· disjE: ⟦P ∨ Q; P ⟹ R; Q ⟹ R⟧ ⟹ R <br />
· FalseE: False ⟹ P<br />
· notE: ⟦¬P; P⟧ ⟹ R<br />
· notI: (P ⟹ False) ⟹ ¬P<br />
· iffI: ⟦P ⟹ Q; Q ⟹ P⟧ ⟹ P = Q<br />
· iffD1: ⟦Q = P; Q⟧ ⟹ P <br />
· iffD2: ⟦P = Q; Q⟧ ⟹ P<br />
· ccontr: (¬P ⟹ False) ⟹ P<br />
· excluded_middel:(¬P ∨ P) <br />
<br />
· allI: ⟦∀x. P x; P x ⟹ R⟧ ⟹ R<br />
· allE: (⋀x. P x) ⟹ ∀x. P x<br />
· exI: P x ⟹ ∃x. P x<br />
· exE: ⟦∃x. P x; ⋀x. P x ⟹ Q⟧ ⟹ Q<br />
<br />
· refl: t = t<br />
· subst: ⟦s = t; P s⟧ ⟹ P t<br />
· trans: ⟦r = s; s = t⟧ ⟹ r = t<br />
· sym: s = t ⟹ t = s<br />
· not_sym: t ≠ s ⟹ s ≠ t<br />
· ssubst: ⟦t = s; P s⟧ ⟹ P t<br />
· box_equals: ⟦a = b; a = c; b = d⟧ ⟹ a: = d<br />
· arg_cong: x = y ⟹ f x = f y<br />
· fun_cong: f = g ⟹ f x = g x<br />
· cong: ⟦f = g; x = y⟧ ⟹ f x = g y<br />
*}<br />
<br />
text {*<br />
Se usarán las reglas notnotI y mt que demostramos a continuación.<br />
*}<br />
<br />
lemma notnotI: "P ⟹ ¬¬ P"<br />
by auto<br />
<br />
lemma mt: "⟦F ⟶ G; ¬G⟧ ⟹ ¬F"<br />
by auto<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 1. Demostrar<br />
∀x. P x ⟶ Q x ⊢ (∀x. P x) ⟶ (∀x. Q x)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_1a: carmarria<br />
assumes 1: "∀x. P x ⟶ Q x"<br />
shows "(∀x. P x) ⟶ (∀x. Q x)"<br />
proof -<br />
{assume 2: "! x. P x"<br />
{fix a<br />
have 3: "P a ⟶ Q a" using 1 by (rule allE)<br />
have 4: "P a" using 2 by (rule allE)<br />
have 5: "Q a" using 3 4 by (rule mp)<br />
}<br />
hence 6: "! x. Q x" by (rule allI)<br />
}<br />
thus "(! x. P x) ⟶ (! x. Q x)" by (rule impI)<br />
qed<br />
<br />
joslopjim4<br />
<br />
assumes "∀x. P x ⟶ Q x"<br />
shows "(∀x. P x) ⟶ (∀x. Q x)"<br />
proof (rule impI)<br />
assume "∀x. P x"<br />
show "∀x. Q x"<br />
proof (rule allI)<br />
fix a<br />
have "P a ⟶ Q a" using assms by (rule allE)<br />
have "P a" using `∀x. P x` by (rule allE)<br />
show "Q a" using `P a ⟶ Q a` `P a` by (rule mp)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 2. Demostrar<br />
∃x. ¬(P x) ⊢ ¬(∀x. P x)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_2a: carmarria<br />
assumes 1: "∃x. ¬(P x)"<br />
shows "¬(∀x. P x)"<br />
proof -<br />
obtain a where 2: "¬(P a)" using 1 by (rule exE)<br />
{assume 3: "! x. P x"<br />
have 4: "P a" using 3 by (rule allE)<br />
have 5: False using 2 4 by (rule notE)<br />
}<br />
thus "¬(! x. P x)" by (rule notI)<br />
qed<br />
<br />
joslopjim4<br />
<br />
assumes "∃x. ¬(P x)"<br />
shows "¬(∀x. P x)"<br />
proof (rule ccontr)<br />
assume "¬¬(∀x. P x)"<br />
then have "∀x. P x" by (rule notnotD)<br />
obtain a where "¬(P a)" using assms by (rule exE)<br />
have "P a" using `∀x. P x` by (rule allE)<br />
show False using `¬(P a)` `P a` by (rule notE)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 3. Demostrar<br />
∀x. P x ⊢ ∀y. P y<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_3a: carmarria<br />
assumes 1: "∀x. P x"<br />
shows "∀y. P y"<br />
proof -<br />
{fix a<br />
have 2: "P a" using 1 by (rule allE)<br />
}<br />
thus "! y. P y" by (rule allI)<br />
qed<br />
<br />
joslopjim4<br />
<br />
assumes "∀x. P x"<br />
shows "∀y. P y"<br />
proof (rule allI)<br />
fix a<br />
show "P a" using assms by (rule allE)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 4. Demostrar<br />
∀x. P x ⟶ Q x ⊢ (∀x. ¬(Q x)) ⟶ (∀x. ¬ (P x))<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_4: carmarria<br />
assumes 1: "∀x. P x ⟶ Q x"<br />
shows "(∀x. ¬(Q x)) ⟶ (∀x. ¬ (P x))"<br />
proof -<br />
{assume 2: "!x. ¬(Q x)"<br />
{fix a<br />
have 3: "¬(Q a)" using 2 by (rule allE)<br />
have 4: "P a ⟶ Q a" using 1 by (rule allE)<br />
have 5: "¬(P a)" using 4 3 by (rule mt)<br />
}<br />
hence 6: "!x. ¬ (P x)" by (rule allI)<br />
}<br />
thus "(!x. ¬(Q x)) ⟶ (!x. ¬ (P x))" by (rule impI)<br />
qed<br />
<br />
joslopjim4<br />
<br />
assumes "∀x. P x ⟶ Q x"<br />
shows "(∀x. ¬(Q x)) ⟶ (∀x. ¬ (P x))"<br />
proof (rule impI)<br />
assume "∀x. ¬(Q x)"<br />
show "∀x. ¬ (P x)"<br />
proof (rule allI)<br />
fix a<br />
have "¬(Q a)" using `∀x. ¬(Q x)` by (rule allE)<br />
have "P a ⟶ Q a" using assms by (rule allE)<br />
thus "¬(P a)" using `¬(Q a)` by (rule mt)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 5. Demostrar<br />
∀x. P x ⟶ ¬(Q x) ⊢ ¬(∃x. P x ∧ Q x)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_5: carmarria<br />
assumes 1: "∀x. P x ⟶ ¬(Q x)"<br />
shows "¬(∃x. P x ∧ Q x)"<br />
proof -<br />
{assume 2: "? x. P x & Q x"<br />
obtain a where 3: "P a & Q a" using 2 by (rule exE)<br />
have 4: "P a ⟶ ¬(Q a)" using 1 by (rule allE)<br />
have 5: "P a" using 3 by (rule conjunct1)<br />
have 6: "Q a" using 3 by (rule conjunct2)<br />
have 7: "¬(Q a)" using 4 5 by (rule mp)<br />
have 8: False using 7 6 by (rule notE)<br />
}<br />
thus "¬(? x. P x & Q x)" by (rule notI)<br />
qed<br />
<br />
joslopjim4<br />
<br />
assumes "∀x. P x ⟶ ¬(Q x)"<br />
shows "¬(∃x. P x ∧ Q x)"<br />
proof (rule ccontr)<br />
assume "¬¬(∃x. P x ∧ Q x)"<br />
then have "∃x. P x ∧ Q x" by (rule notnotD)<br />
obtain b where "P b ∧ Q b" using `∃x. P x ∧ Q x` by (rule exE)<br />
have "P b" using `P b ∧ Q b` by (rule conjunct1)<br />
have "Q b" using `P b ∧ Q b` by (rule conjunct2)<br />
have "P b ⟶ ¬(Q b)"<br />
proof -<br />
fix b<br />
show "P b ⟶ ¬(Q b)" using assms by (rule allE)<br />
qed<br />
have "¬(Q b)" using `P b ⟶ ¬(Q b)` `P b` by (rule mp)<br />
show False using `¬(Q b)` `Q b` by (rule notE)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 6. Demostrar<br />
∀x y. P x y ⊢ ∀u v. P u v<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_6: carmarria<br />
assumes 1: "∀x y. P x y"<br />
shows "∀u v. P u v"<br />
proof -<br />
{fix a<br />
have 2: "! y. P a y" using 1 by (rule allE)<br />
{fix b<br />
have 3: "P a b" using 2 by (rule allE)<br />
}<br />
hence 4: "!v. P a v" by (rule allI)<br />
}<br />
thus "!u v. P u v" by (rule allI)<br />
qed<br />
<br />
joslopjim4<br />
<br />
assumes "∀x y. P x y"<br />
shows "∀u v. P u v"<br />
proof (rule allI)<br />
fix a<br />
have "∀y. P a y" using assms by (rule allE)<br />
show "∀v. P a v"<br />
proof (rule allI)<br />
fix b<br />
show "P a b" using `∀y. P a y` by (rule allE)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 7. Demostrar<br />
∃x y. P x y ⟹ ∃u v. P u v<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_7: carmarria<br />
assumes 1: "∃x y. P x y"<br />
shows "∃u v. P u v"<br />
proof -<br />
obtain a where 2: "? y. P a y" using 1 by (rule exE)<br />
obtain b where 3: "P a b" using 2 by (rule exE)<br />
have 4: "? v. P a v" using 3 by (rule exI)<br />
show 5: "? u v. P u v" using 4 by (rule exI)<br />
qed<br />
<br />
joslopjim4<br />
<br />
assumes "∃x y. P x y"<br />
shows "∃u v. P u v"<br />
proof-<br />
obtain a where "∃y. P a y" using assms by (rule exE)<br />
obtain b where "P a b" using `∃y. P a y` by (rule exE)<br />
then have "∃v. P a v" by (rule exI)<br />
thus "∃u v. P u v" by (rule exI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 8. Demostrar<br />
∃x. ∀y. P x y ⊢ ∀y. ∃x. P x y<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_8: carmarria<br />
assumes 1: "∃x. ∀y. P x y"<br />
shows "∀y. ∃x. P x y"<br />
proof -<br />
obtain a where 2: "! y. P a y" using 1 by (rule exE)<br />
{fix b<br />
have 3: "P a b" using 2 by (rule allE)<br />
have 4: "? x. P x b" using 3 by (rule exI)<br />
}<br />
thus "! y. ? x. P x y" by (rule allI)<br />
qed<br />
<br />
joslopjim4 <br />
<br />
assumes "∃x. ∀y. P x y"<br />
shows "∀y. ∃x. P x y"<br />
proof (rule allI)<br />
fix b<br />
obtain a where "∀y. P a y" using assms by (rule exE)<br />
then have "P a b" by (rule allE)<br />
then show "∃x. P x b" by (rule exI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 9. Demostrar<br />
∃x. P a ⟶ Q x ⊢ P a ⟶ (∃x. Q x)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_9: carmarria<br />
assumes 1: "∃x. P a ⟶ Q x"<br />
shows "P a ⟶ (∃x. Q x)"<br />
proof -<br />
{assume 2: "P a"<br />
obtain b where 3: "P a ⟶ Q b" using 1 by (rule exE)<br />
have 4: "Q b" using 3 2 by (rule mp)<br />
have 5: "? x. Q x" using 4 by (rule exI)<br />
}<br />
thus "P a ⟶ (? x. Q x)" by (rule impI)<br />
qed<br />
<br />
joslopjim4<br />
<br />
assumes "∃x. P a ⟶ Q x"<br />
shows "P a ⟶ (∃x. Q x)"<br />
proof (rule impI)<br />
assume "P a"<br />
obtain b where "P a ⟶ Q b" using assms by (rule exE)<br />
then have "Q b" using `P a` by (rule mp)<br />
thus "∃x. Q x" by (rule exI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 10. Demostrar<br />
P a ⟶ (∃x. Q x) ⊢ ∃x. P a ⟶ Q x <br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_10a: carmarria<br />
fixes P Q :: "'b ⇒ bool" <br />
assumes 1: "P a ⟶ (∃x. Q x)"<br />
shows "∃x. P a ⟶ Q x"<br />
proof -<br />
have 2: "¬(P a) ∨ (P a)" by (rule excluded_middle)<br />
moreover {assume 3: "¬(P a)" <br />
{assume 4: "P a"<br />
have 5: False using 3 4 by (rule notE)<br />
have 6: "Q b" using 5 by (rule FalseE)<br />
}<br />
hence 7: "P a ⟶Q b" by (rule impI)<br />
have 8: "? x. P a ⟶ Q x" using 7 by (rule exI)<br />
}<br />
moreover {assume 9: "P a"<br />
have 10: "? x. Q x" using 1 9 by (rule mp)<br />
obtain b where 11: "Q b" using 10 by (rule exE)<br />
{assume 12: "P a"<br />
have 13: "Q b" using 11 by this<br />
}<br />
hence 13: "P a ⟶ Q b" by (rule impI)<br />
have 14: "? x. P a ⟶ Q x" using 13 by (rule exI)<br />
}<br />
ultimately show "? x. P a ⟶ Q x" by (rule disjE)<br />
qed <br />
<br />
joslopjim4<br />
<br />
fixes P Q :: "'b ⇒ bool" <br />
assumes "P a ⟶ (∃x. Q x)"<br />
shows "∃x. P a ⟶ Q x"<br />
proof-<br />
have "¬(P a) ∨ P a" by (rule excluded_middle)<br />
thus "∃x. P a ⟶ Q x"<br />
proof <br />
assume "P a"<br />
have "∃x. Q x" using assms `P a` by (rule mp)<br />
obtain b where "Q b" using `∃x. Q x` by (rule exE)<br />
have "P a ⟶ Q b" <br />
proof<br />
assume "P a"<br />
show "Q b" using `Q b` by this<br />
qed<br />
show "∃x. P a ⟶ Q x" using `P a ⟶ Q b` by (rule exI)<br />
next<br />
assume "¬(P a)"<br />
have "P a ⟶ Q a"<br />
proof<br />
assume "P a"<br />
have False using `¬(P a)` `P a` by (rule notE)<br />
thus "Q a" by (rule FalseE)<br />
qed<br />
show "∃x. P a ⟶ Q x" using `P a ⟶ Q a` by (rule exI)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 11. Demostrar<br />
(∃x. P x) ⟶ Q a ⊢ ∀x. P x ⟶ Q a<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_11a: carmarria<br />
assumes 1:"(∃x. P x) ⟶ Q a"<br />
shows "∀x. P x ⟶ Q a"<br />
proof-<br />
{fix b<br />
{assume 2: "P b"<br />
have 3: "? x. P x" using 2 by (rule exI)<br />
have 4: "Q a" using 1 3 by (rule mp)<br />
}<br />
hence 5: "P b ⟶ Q a" by (rule impI)<br />
}<br />
thus "! x. P x ⟶ Q a" by (rule allI)<br />
qed<br />
<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 12. Demostrar<br />
∀x. P x ⟶ Q a ⊢ ∃ x. P x ⟶ Q a<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_12a: carmarria<br />
assumes 1: "∀x. P x ⟶ Q a"<br />
shows "∃x. P x ⟶ Q a"<br />
proof -<br />
have 2: "P b ⟶ Q a" using 1 by (rule allE)<br />
show 3: "? x. P x ⟶ Q a" using 2 by (rule exI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 13. Demostrar<br />
(∀x. P x) ∨ (∀x. Q x) ⊢ ∀x. P x ∨ Q x<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_13a: carmarria<br />
assumes 1: "(∀x. P x) ∨ (∀x. Q x)"<br />
shows "∀x. P x ∨ Q x"<br />
proof -<br />
{fix a<br />
have 2: "(! x. P x) ∨ (! x. Q x)" using 1 by this<br />
moreover {assume 3:"! x. P x"<br />
have 4: "P a" using 3 by (rule allE)<br />
have 5: "P a | Q a" using 4 by (rule disjI1)<br />
}<br />
moreover {assume 6: "! x. Q x" <br />
have 7: "Q a" using 6 by (rule allE)<br />
have 8: "P a | Q a" using 7 by (rule disjI2)<br />
}<br />
ultimately have 9: "P a | Q a" by (rule disjE)<br />
}<br />
thus "! x. P x | Q x" by (rule allI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 14. Demostrar<br />
∃x. P x ∧ Q x ⊢ (∃x. P x) ∧ (∃x. Q x)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_14a: carmarria<br />
assumes 1: "∃x. P x ∧ Q x"<br />
shows "(∃x. P x) ∧ (∃x. Q x)"<br />
proof -<br />
obtain a where 2: "P a & Q a" using 1 by (rule exE)<br />
have 3: "P a" using 2 by (rule conjunct1)<br />
have 4: "Q a" using 2 by (rule conjunct2)<br />
have 5: "? x. P x" using 3 by (rule exI)<br />
have 6: "? x. Q x" using 4 by (rule exI)<br />
show "(? x. P x) ∧ (? x. Q x)" using 5 6 by (rule conjI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 15. Demostrar<br />
∀x y. P y ⟶ Q x ⊢ (∃y. P y) ⟶ (∀x. Q x)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_15a: carmarria<br />
assumes 1: "∀x y. P y ⟶ Q x"<br />
shows "(∃y. P y) ⟶ (∀x. Q x)"<br />
proof -<br />
{assume 2: "? y. P y"<br />
{fix a<br />
have 3: "! y. P y ⟶ Q a" using 1 by (rule allE)<br />
obtain b where 4: "P b" using 2 by (rule exE)<br />
have 5: "P b ⟶ Q a" using 3 by (rule allE)<br />
have 6: "Q a" using 5 4 by (rule mp)<br />
}<br />
hence 7: "! x. Q x" by (rule allI)<br />
}<br />
thus "(? y. P y) ⟶ (! x. Q x)" by (rule impI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 16. Demostrar<br />
¬(∀x. ¬(P x)) ⊢ ∃x. P x<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_16a: carmarria<br />
assumes 1: "¬(∀x. ¬(P x))"<br />
shows "∃x. P x"<br />
proof -<br />
{assume 2: "¬(? x. P x)"<br />
{fix b<br />
{assume 3: "P b"<br />
have 4: "? x. P x" using 3 by (rule exI)<br />
have 5: False using 2 4 by (rule notE)<br />
}<br />
hence 6: "¬(P b)" by (rule notI)<br />
}<br />
hence 7: "! x. ¬(P x)" by (rule allI)<br />
have 8: False using 1 7 by (rule notE)<br />
}<br />
hence 9: "¬¬(? x. P x)" by (rule notI)<br />
show "? x. P x" using 9 by (rule notnotD)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 17. Demostrar<br />
∀x. ¬(P x) ⊢ ¬(∃x. P x)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_17a: carmarria<br />
assumes 1: "∀x. ¬(P x)"<br />
shows "¬(∃x. P x)"<br />
proof -<br />
{assume 2: "? x. P x"<br />
obtain a where 3: "P a" using 2 by (rule exE)<br />
have 4: "¬(P a)" using 1 by (rule allE)<br />
have 5: False using 4 3 by (rule notE)<br />
}<br />
thus "¬(? x. P x)" by (rule notI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 18. Demostrar<br />
∃x. P x ⊢ ¬(∀x. ¬(P x))<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_18a: carmarria<br />
assumes 1: "∃x. P x"<br />
shows "¬(∀x. ¬(P x))"<br />
proof-<br />
{assume 2: "! x. ¬(P x)"<br />
obtain a where 3: "P a" using 1 by (rule exE)<br />
have 4: "¬(P a)" using 2 by (rule allE)<br />
have 5: False using 4 3 by (rule notE)<br />
}<br />
thus "¬(! x. ¬(P x))" by (rule notI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 19. Demostrar<br />
P a ⟶ (∀x. Q x) ⊢ ∀x. P a ⟶ Q x<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_19a: carmarria<br />
assumes 1: "P a ⟶ (∀x. Q x)"<br />
shows "∀x. P a ⟶ Q x"<br />
proof -<br />
{fix b<br />
{assume 2: "P a"<br />
have 3: "! x. Q x" using 1 2 by (rule mp)<br />
have 4: "Q b" using 3 by (rule allE)<br />
}<br />
hence 5: "P a ⟶ Q b" by (rule impI)<br />
}<br />
thus "! x. P a ⟶ Q x" by (rule allI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 20. Demostrar<br />
{∀x y z. R x y ∧ R y z ⟶ R x z, <br />
∀x. ¬(R x x)}<br />
⊢ ∀x y. R x y ⟶ ¬(R y x)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_20a: carmarria<br />
assumes 1: "∀x y z. R x y ∧ R y z ⟶ R x z" and<br />
2: "∀x. ¬(R x x)" <br />
shows "∀x y. R x y ⟶ ¬(R y x)"<br />
proof -<br />
{fix a<br />
{fix b<br />
{assume 3: "R a b"<br />
have 4: "! y z. R a y & R y z ⟶ R a z" using 1 by (rule allE)<br />
have 5: "! z. R a b & R b z ⟶ R a z" using 4 by (rule allE)<br />
have 6: "R a b & R b a ⟶ R a a" using 5 by (rule allE)<br />
have 7: "¬(R a a)" using 2 by (rule allE)<br />
have 8: "¬(R a b & R b a)" using 6 7 by (rule mt)<br />
{assume 9: "R b a"<br />
have 10: "R a b & R b a" using 3 9 by (rule conjI)<br />
have 11: False using 8 10 by (rule notE)<br />
}<br />
hence 12: "¬(R b a)" by (rule notI)<br />
}<br />
hence 13: "R a b ⟶ ¬(R b a)" by (rule impI)<br />
}<br />
hence 14: "! y. R a y ⟶ ¬(R y a)" by (rule allI)<br />
}<br />
thus "! x y. R x y ⟶ ¬(R y x)" by (rule allI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 21. Demostrar<br />
{∀x. P x ∨ Q x, ∃x. ¬(Q x), ∀x. R x ⟶ ¬(P x)} ⊢ ∃x. ¬(R x)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_21a: carmarria<br />
assumes 1: "∀x. P x ∨ Q x" and<br />
2: "∃x. ¬(Q x)" and<br />
3: "∀x. R x ⟶ ¬(P x)"<br />
shows "∃x. ¬(R x)" <br />
proof -<br />
obtain a where 4: "¬(Q a)" using 2 by (rule exE)<br />
have 5: "P a | Q a" using 1 by (rule allE)<br />
moreover {assume 6: "P a"<br />
have 7: "¬¬(P a)" using 6 by (rule notnotI)<br />
have 8: "R a ⟶ ¬(P a)" using 3 by (rule allE)<br />
have 9: "¬(R a)" using 8 7 by (rule mt)<br />
have 10: "? x. ¬(R x)" using 9 by (rule exI)<br />
}<br />
moreover {assume 10: "Q a"<br />
have 11: False using 4 10 by (rule notE)<br />
have 12: "? x. ¬(R x)" using 11 by (rule FalseE)<br />
}<br />
ultimately show "? x. ¬(R x)" by (rule disjE)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 22. Demostrar<br />
{∀x. P x ⟶ Q x ∨ R x, ¬(∃x. P x ∧ R x)} ⊢ ∀x. P x ⟶ Q x<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_22a: carmarria<br />
assumes 1: "∀x. P x ⟶ Q x ∨ R x" and<br />
2: "¬(∃x. P x ∧ R x)"<br />
shows "∀x. P x ⟶ Q x"<br />
proof -<br />
{fix a<br />
{assume 3: "P a"<br />
have 4: "P a ⟶ Q a | R a" using 1 by (rule allE)<br />
have 5: "Q a | R a" using 4 3 by (rule mp)<br />
moreover {assume 6: "Q a"<br />
}<br />
moreover {assume 7: "R a" <br />
have 8: "P a & R a" using 3 7 by (rule conjI)<br />
have 9: "? x. P x & R x" using 8 by (rule exI)<br />
have 10: False using 2 9 by (rule notE)<br />
have 11: "Q a" using 10 by (rule FalseE)<br />
}<br />
ultimately have 12: "Q a" by (rule disjE)<br />
}<br />
hence 13: "P a ⟶ Q a" by (rule impI)<br />
}<br />
thus "! x. P x ⟶ Q x" by (rule allI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 23. Demostrar<br />
∃x y. R x y ∨ R y x ⊢ ∃x y. R x y<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_23a: carmarria<br />
assumes 1: "∃x y. R x y ∨ R y x"<br />
shows "∃x y. R x y"<br />
proof -<br />
obtain a where 2: "? y. R a y | R y a" using 1 by (rule exE)<br />
obtain b where 3: "R a b | R b a" using 2 by (rule exE)<br />
moreover {assume 4: "R a b"<br />
have 5: "? y. R a y" using 4 by (rule exI)<br />
have 6: "? x y. R x y" using 5 by (rule exI)<br />
}<br />
moreover {assume 7: "R b a"<br />
have 8: "? y. R b y" using 7 by (rule exI)<br />
have 9: "? x y. R x y" using 8 by (rule exI)<br />
}<br />
ultimately show "? x y. R x y" by (rule disjE)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 24. Demostrar<br />
(∃x. ∀y. P x y) ⟶ (∀y. ∃x. P x y)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_24a: carmarria<br />
"(∃x. ∀y. P x y) ⟶ (∀y. ∃x. P x y)"<br />
proof -<br />
{assume 1: "? x. ! y. P x y"<br />
{fix b<br />
obtain a where 2: "! y. P a y" using 1 by (rule exE)<br />
have 3: "P a b" using 2 by (rule allE)<br />
have 4: "? x. P x b" using 3 by (rule exI)<br />
}<br />
hence 5: "! y. ? x. P x y" by (rule allI)<br />
}<br />
thus "(? x. ! y. P x y) ⟶ (! y. ? x. P x y)" by (rule impI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 25. Demostrar<br />
(∀x. P x ⟶ Q) ⟷ ((∃x. P x) ⟶ Q)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_25a: carmarria<br />
"(∀x. P x ⟶ Q) ⟷ ((∃x. P x) ⟶ Q)"<br />
proof (rule iffI)<br />
{assume 1: "! x. P x ⟶ Q"<br />
{assume 2: "? x. P x" <br />
obtain a where 3: "P a" using 2 by (rule exE)<br />
have 4: "P a ⟶ Q" using 1 by (rule allE)<br />
have 5: Q using 4 3 by (rule mp)<br />
}<br />
thus 6: "(? x. P x) ⟶ Q" by (rule impI)<br />
}<br />
next<br />
{assume 7: "(? x. P x) ⟶ Q" <br />
{fix a<br />
{assume 8: "P a"<br />
have 9: "? x. P x" using 8 by (rule exI)<br />
have 10: Q using 7 9 by (rule mp)<br />
}<br />
hence 11: "P a ⟶ Q" by (rule impI)<br />
}<br />
thus 12: "! x. P x ⟶ Q" by (rule allI)<br />
}<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 26. Demostrar<br />
((∀x. P x) ∧ (∀x. Q x)) ⟷ (∀x. P x ∧ Q x)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_26a: carmarria<br />
"((∀x. P x) ∧ (∀x. Q x)) ⟷ (∀x. P x ∧ Q x)"<br />
proof (rule iffI)<br />
{assume 1: "((! x. P x) ∧ (! x. Q x))"<br />
have 2: "! x. P x" using 1 by (rule conjunct1)<br />
have 3: "! x. Q x" using 1 by (rule conjunct2)<br />
{fix a <br />
have 4: "P a" using 2 by (rule allE)<br />
have 5: "Q a" using 3 by (rule allE)<br />
have 6: "P a & Q a" using 4 5 by (rule conjI)<br />
}<br />
thus 7: "! x. P x & Q x" by (rule allI)<br />
}<br />
next<br />
{assume 8: "! x. P x & Q x"<br />
{fix a<br />
have 9: "P a & Q a" using 8 by (rule allE)<br />
have 10: "P a" using 9 by (rule conjunct1)<br />
}<br />
hence 11: "! x. P x" by (rule allI)<br />
{fix a<br />
have 12: "P a & Q a" using 8 by (rule allE)<br />
have 13: "Q a" using 12 by (rule conjunct2)<br />
}<br />
hence 14: "! x. Q x" by (rule allI)<br />
show "(! x. P x) & (! x. Q x)" using 11 14 by (rule conjI)<br />
}<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 27. Demostrar o refutar<br />
((∀x. P x) ∨ (∀x. Q x)) ⟷ (∀x. P x ∨ Q x)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_27a: carmarria<br />
"((∀x. P x) ∨ (∀x. Q x)) ⟷ (∀x. P x ∨ Q x)"<br />
proof (rule iffI)<br />
{assume 1: "(! x. P x) | (! x. Q x)" <br />
moreover {assume 2: "! x. P x"<br />
{fix a<br />
have 3: "P a" using 2 by (rule allE)<br />
have 4: "P a | Q a" using 3 by (rule disjI1)<br />
}<br />
hence 5: "! x. P x | Q x" by (rule allI)<br />
}<br />
moreover {assume 6: "! x. Q x"<br />
{fix a<br />
have 7: "Q a" using 6 by (rule allE)<br />
have 8:"P a | Q a" using 7 by (rule disjI2)<br />
}<br />
hence 9: "! x. P x | Q x" by (rule allI)<br />
}<br />
ultimately show "! x. P x | Q x" by (rule disjE)<br />
}<br />
next<br />
{assume 10: "! x. P x | Q x"<br />
oops<br />
<br />
(* Si tomamos el universo {1,2}, P(x): x=1, Q(x): x=2. Tenemos que es verdad que <br />
(∀x. x=1 ∨ x=2), pero no se cumple ni (∀x. x=1) ni (∀x. x=2), por lo tanto el bicondicional es falso *)<br />
<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 28. Demostrar o refutar<br />
((∃x. P x) ∨ (∃x. Q x)) ⟷ (∃x. P x ∨ Q x)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_28a: carmarria<br />
"((∃x. P x) ∨ (∃x. Q x)) ⟷ (∃x. P x ∨ Q x)"<br />
proof (rule iffI)<br />
{assume 1:"(? x. P x) | (? x. Q x)"<br />
moreover {assume 2: "? x. P x"<br />
obtain a where 3: "P a" using 2 by (rule exE)<br />
have 4: "P a | Q a" using 3 by (rule disjI1)<br />
have 5: "? x. P x | Q x" using 4 by (rule exI)<br />
}<br />
moreover {assume 6: "? x. Q x"<br />
obtain a where 7: "Q a" using 6 by (rule exE)<br />
have 8: "P a | Q a" using 7 by (rule disjI2)<br />
have 9: "? x. P x | Q x" using 8 by (rule exI)<br />
}<br />
ultimately show "? x. P x | Q x" by (rule disjE)<br />
}<br />
next<br />
{assume 10: "? x. P x | Q x"<br />
obtain a where 11: "P a | Q a" using 10 by (rule exE)<br />
moreover {assume 12: "P a"<br />
have 13: "? x. P x" using 12 by (rule exI)<br />
have 14: "(? x. P x) | (? x. Q x)" using 13 by (rule disjI1)<br />
}<br />
moreover {assume 15: "Q a"<br />
have 16: "? x. Q x" using 15 by (rule exI)<br />
have 17: "(? x. P x) | (? x. Q x)" using 16 by (rule disjI2)<br />
}<br />
ultimately show "(? x. P x) | (? x. Q x)" by (rule disjE)<br />
}<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 29. Demostrar o refutar<br />
(∀x. ∃y. P x y) ⟶ (∃y. ∀x. P x y)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_29: carmarria<br />
"(∀x. ∃y. P x y) ⟶ (∃y. ∀x. P x y)"<br />
oops <br />
(* Si tomamos el universo {1,2} y P(x,y): x=y, entonces tenemos que (∀x. ∃y. x=y) pero no se cumple<br />
que (∃y. ∀x. x=y) *)<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 30. Demostrar o refutar<br />
(¬(∀x. P x)) ⟷ (∃x. ¬P x)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_30a: carmarria<br />
"(¬(∀x. P x)) ⟷ (∃x. ¬P x)"<br />
proof (rule iffI)<br />
{assume 1: "¬(! x. P x)" <br />
{assume 2: "¬(? x. ¬(P x))"<br />
{fix a<br />
{assume 3: "¬P a"<br />
have 4: "? x. ¬P x" using 3 by (rule exI)<br />
have 5: False using 2 4 by (rule notE)<br />
}<br />
hence 6: "¬(¬P a)" by (rule notI)<br />
have 7: "P a" using 6 by (rule notnotD)<br />
}<br />
hence 8: "! x. P x" by (rule allI)<br />
have 9: False using 1 8 by (rule notE)<br />
}<br />
hence 10: "¬¬(? x. ¬(P x))" by (rule notI)<br />
show "? x. ¬(P x)" using 10 by (rule notnotD)<br />
}<br />
next<br />
{assume 11: "? x. ¬P x"<br />
obtain a where 12: "¬P a" using 11 by (rule exE)<br />
{assume 13: "! x. P x" <br />
have 14: "P a" using 13 by (rule allE)<br />
have 15: False using 12 14 by (rule notE)<br />
}<br />
thus "¬(! x. P x)" by (rule notI)<br />
}<br />
qed<br />
<br />
section {* Ejercicios sobre igualdad *}<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 31. Demostrar o refutar<br />
P a ⟹ ∀x. x = a ⟶ P x<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_31a: carmarria<br />
assumes 1: "P a"<br />
shows "∀x. x = a ⟶ P x"<br />
proof -<br />
{fix b<br />
{assume 2: "b = a" <br />
have 3: "a = b" using 2 by (rule sym)<br />
have 4: "P b" using 3 1 by (rule subst)<br />
}<br />
hence 5: "b = a ⟶ P b" by (rule impI)<br />
}<br />
thus "! x. x = a ⟶ P x" by (rule allI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 32. Demostrar o refutar<br />
∃x y. R x y ∨ R y x; ¬(∃x. R x x)⟧ ⟹ ∃x y. x ≠ y<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_32a: carmarria<br />
fixes R :: "'c ⇒ 'c ⇒ bool"<br />
assumes 1: "∃x y. R x y ∨ R y x" and<br />
2: "¬(∃x. R x x)"<br />
shows "∃(x::'c) y. x ≠ y"<br />
proof -<br />
obtain a where 3: "? y. R a y | R y a" using 1 by (rule exE)<br />
obtain b where 4: "R a b | R b a" using 3 by (rule exE)<br />
{assume 5: "a = b"<br />
have 6: "R a b | R b a" using 4 by this<br />
moreover {assume 7: "R a b" <br />
have 8: "R b b" using 5 7 by (rule subst)<br />
}<br />
moreover {assume 9: "R b a" <br />
have 10: "R b b" using 5 9 by (rule subst)<br />
}<br />
ultimately have 11: "R b b" by (rule disjE)<br />
have 12: "? x. R x x" using 11 by (rule exI)<br />
have 13: False using 2 12 by (rule notE)<br />
}<br />
hence 14: " a ≠ b " by (rule notI)<br />
have 15: "? y. a ≠ y" using 14 by (rule exI)<br />
show "? x y. (x ≠ y)" (* using 15 by (rule exI) *)<br />
oops<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 33. Demostrar o refutar<br />
{∀x. P a x x, <br />
∀x y z. P x y z ⟶ P (f x) y (f z)} <br />
⊢ P (f a) a (f a)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_33a: carmarria<br />
assumes 1: "∀x. P a x x" and<br />
2: "∀x y z. P x y z ⟶ P (f x) y (f z)"<br />
shows "P (f a) a (f a)"<br />
proof -<br />
have 3: "P a a a" using 1 by (rule allE)<br />
have 4: "! y z. P a y z ⟶ P (f a) y (f z)" using 2 by (rule allE)<br />
have 5: "! z. P a a z ⟶ P (f a) a (f z)" using 4 by (rule allE)<br />
have 6: "P a a a ⟶ P (f a) a (f a)" using 5 by (rule allE)<br />
show "P (f a) a (f a)" using 6 3 by (rule mp)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 34. Demostrar o refutar<br />
{∀x. P a x x, <br />
∀x y z. P x y z ⟶ P (f x) y (f z)⟧<br />
⊢ ∃z. P (f a) z (f (f a))<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_34a: carmarria<br />
assumes 1: "∀x. P a x x" and<br />
2: "∀x y z. P x y z ⟶ P (f x) y (f z)"<br />
shows "∃z. P (f a) z (f (f a))"<br />
proof -<br />
have 3: "P a (f a) (f a)" using 1 by (rule allE)<br />
have 4: "! y z. P a y z ⟶ P (f a) y (f z)" using 2 by (rule allE)<br />
have 5: "! z. P a (f a) z ⟶ P (f a) (f a) (f z)" using 4 by (rule allE)<br />
have 6: "P a (f a) (f a) ⟶ P (f a) (f a) (f (f a))" using 5 by (rule allE)<br />
have 7: "P (f a) (f a) (f (f a))" using 6 3 by (rule mp)<br />
show 8: "? z. P (f a) z (f (f a))" using 7 by (rule exI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 35. Demostrar o refutar<br />
{∀y. Q a y, <br />
∀x y. Q x y ⟶ Q (s x) (s y)} <br />
⊢ ∃z. Qa z ∧ Q z (s (s a))<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_35a: carmarria<br />
assumes 1: "∀y. Q a y" and<br />
2: "∀x y. Q x y ⟶ Q (s x) (s y)" <br />
shows "∃z. Q a z ∧ Q z (s (s a))"<br />
proof-<br />
have 3: "! y. Q a y ⟶ Q (s a) (s y)" using 2 by (rule allE)<br />
have 4: "Q a (s a) ⟶ Q (s a) (s (s a))" using 3 by (rule allE)<br />
have 5: "Q a (s a)" using 1 by (rule allE)<br />
have 6: "Q (s a) (s (s a))" using 4 5 by (rule mp)<br />
have 7: "Q a (s a) & Q (s a) (s (s a))" using 5 6 by (rule conjI)<br />
show "? z. Q a z & Q z (s (s a))" using 7 by (rule exI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 36. Demostrar o refutar<br />
{x = f x, odd (f x)} ⊢ odd x<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_36a: carmarria<br />
assumes 1: "x = f x" and<br />
2: "odd (f x)"<br />
shows "odd x"<br />
proof -<br />
have 3: "f x = x" using 1 by (rule sym)<br />
show "odd x" using 3 2 by (rule subst)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 37. Demostrar o refutar<br />
{x = f x, triple (f x) (f x) x} ⊢ triple x x x<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_37a: carmarria<br />
assumes 1: "x = f x" and<br />
2: "triple (f x) (f x) x"<br />
shows "triple x x x"<br />
proof -<br />
have 3: "f x = x" using 1 by (rule sym)<br />
show 4: "triple x x x" using 3 2 by (rule subst)<br />
qed<br />
<br />
<br />
end<br />
<br />
<br />
</source></div>
Joslopjim4
https://www.glc.us.es/~jalonso/LMF2018/index.php?title=Relaci%C3%B3n_6&diff=180
Relación 6
2018-04-08T15:47:31Z
<p>Joslopjim4: </p>
<hr />
<div>=== Relación 6: Sintaxis y semántica de la Lógica de primer orden (b) ===<br />
<br />
<br />
----<br />
'''Ejercicio 1.''' Dar una fórmula F satisfacible, tal que todos sus modelos sean necesariamante infinitos.<br />
<br />
----<br />
<br />
'''Solución:'''<br />
<br />
----<br />
'''Ejercicio 2.''' ¿Cuántos elementos han de tener los modelos de la fórmula F = ∀x f(f(x))= x ∧ ∀x f(x) ≠ x?<br />
----<br />
<br />
'''Solución:'''<br />
<br />
<br />
----<br />
'''Ejercicio 3.''' <br />
<br />
* Dar un ejemplo de una fórmula F tal que todos sus modelos tengan al menos dos elementos.<br />
<br />
* Dar un ejemplo de una fórmula F tal que todos sus modelos tengan dos elementos como máximo.<br />
<br />
* Dar un ejemplo de una fórmula F tal que todos sus modelos tengan exactamente dos elementos.<br />
<br />
* Dar un ejemplo de una fórmula F tal que todos sus modelos tengan exactamente tres elementos.<br />
<br />
----<br />
<br />
'''Solución:'''<br />
<br />
joslopjim4<br />
<br />
* exists x exists y (x≠y)<br />
*<br />
* exists z exists y (x≠y & all z (z=x | x=y))<br />
* exists x exists y exists z (x≠y & x≠z & y≠z & all a (a=x | a=y | a=z))<br />
<br />
----<br />
'''Ejercicio 4.''' <br />
<br />
Sea L un lenguaje de primer orden con un símbolo de predicado, Q (de aridad 2) y un símbolo de función, f (de aridad 1). Se considera la estructura I dada por: Universo: {a,b}, Qᴵ={(a,b), (b,a)}, fᴵ(a)= a y fᴵ(b)=a. Decidir cuáles de las siguientes fórmulas se satisfacen en la estructura:<br />
* ∀x (Q(f(x),x) → Q(x,x))<br />
* ∃x (Q(f(x),x) → Q(x,x))<br />
<br />
----<br />
<br />
'''Solución:'''<br />
<br />
joslopjim4<br />
<br />
* No, no se satisface para x=b<br />
* Sí, se satisface para x=b<br />
<br />
----<br />
'''Ejercicio 5.''' <br />
<br />
Sea L un lenguaje de primer orden con un símbolo de predicado P de aridad 2. Probar que las fórmulas ∀x ∃y P(x,y) y ∃x ∀y P(x,y) no son equivalentes,<br />
dando una estructura que sea modelo de la primera pero no de la segunda.<br />
<br />
<br />
----<br />
<br />
'''Solución:'''<br />
<br />
joslopjim4<br />
<br />
U=Z<br />
I(P)=<<br />
<br />
En este caso, la primera formula siempre se satisface, sin embargo la segunda nunca.</div>
Joslopjim4
https://www.glc.us.es/~jalonso/LMF2018/index.php?title=Relaci%C3%B3n_6&diff=179
Relación 6
2018-04-08T15:35:42Z
<p>Joslopjim4: </p>
<hr />
<div>=== Relación 6: Sintaxis y semántica de la Lógica de primer orden (b) ===<br />
<br />
<br />
----<br />
'''Ejercicio 1.''' Dar una fórmula F satisfacible, tal que todos sus modelos sean necesariamante infinitos.<br />
<br />
----<br />
<br />
'''Solución:'''<br />
<br />
----<br />
'''Ejercicio 2.''' ¿Cuántos elementos han de tener los modelos de la fórmula F = ∀x f(f(x))= x ∧ ∀x f(x) ≠ x?<br />
----<br />
<br />
'''Solución:'''<br />
<br />
<br />
----<br />
'''Ejercicio 3.''' <br />
<br />
* Dar un ejemplo de una fórmula F tal que todos sus modelos tengan al menos dos elementos.<br />
<br />
* Dar un ejemplo de una fórmula F tal que todos sus modelos tengan dos elementos como máximo.<br />
<br />
* Dar un ejemplo de una fórmula F tal que todos sus modelos tengan exactamente dos elementos.<br />
<br />
* Dar un ejemplo de una fórmula F tal que todos sus modelos tengan exactamente tres elementos.<br />
<br />
----<br />
<br />
'''Solución:'''<br />
<br />
<br />
<br />
----<br />
'''Ejercicio 4.''' <br />
<br />
Sea L un lenguaje de primer orden con un símbolo de predicado, Q (de aridad 2) y un símbolo de función, f (de aridad 1). Se considera la estructura I dada por: Universo: {a,b}, Qᴵ={(a,b), (b,a)}, fᴵ(a)= a y fᴵ(b)=a. Decidir cuáles de las siguientes fórmulas se satisfacen en la estructura:<br />
* ∀x (Q(f(x),x) → Q(x,x))<br />
* ∃x (Q(f(x),x) → Q(x,x))<br />
<br />
----<br />
<br />
'''Solución:'''<br />
<br />
joslopjim4<br />
<br />
* No, no se satisface para x=b<br />
* Sí, se satisface para x=b<br />
<br />
----<br />
'''Ejercicio 5.''' <br />
<br />
Sea L un lenguaje de primer orden con un símbolo de predicado P de aridad 2. Probar que las fórmulas ∀x ∃y P(x,y) y ∃x ∀y P(x,y) no son equivalentes,<br />
dando una estructura que sea modelo de la primera pero no de la segunda.<br />
<br />
<br />
----<br />
<br />
'''Solución:'''<br />
<br />
joslopjim4<br />
<br />
U=Z<br />
I(P)=<<br />
<br />
En este caso, la primera formula siempre se satisface, sin embargo la segunda nunca.</div>
Joslopjim4
https://www.glc.us.es/~jalonso/LMF2018/index.php?title=Relaci%C3%B3n_5&diff=165
Relación 5
2018-03-27T12:28:23Z
<p>Joslopjim4: </p>
<hr />
<div>=== Relación 5: Sintaxis y semántica de la Lógica de primer orden ===<br />
<br />
<br />
----<br />
'''Ejercicio 1.''' Formalizar el siguiente argumento (ejercicio 5 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Todo aquel que entre en el país y no sea un VIP será cacheado por un aduanero.<br />
* Hay un contrabandista que entra en el país y que solo podrá ser cacheado por contrabandistas.<br />
* Ningún contrabandista es un VIP.<br />
* Por lo tanto, algún aduanero es contrabandista.<br />
----<br />
<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*all x ( ( entra(x) & ( -vip(x) ) ) -> ( exists y (aduanero(y) & cachea(y,x) ) ) )<br />
* exists x (contrabandista(x) & entra(x) & ( all y (cachea(y,x) & contrabandista(y))))<br />
* all x (contrabandista(x) -> (- vip(x)))<br />
*exists x (aduanero(x) & contrabandista(x))<br />
<br />
----<br />
'''Ejercicio 2.''' Formalizar el siguiente argumento (ejercicio 15 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Todos los robots obedecen a los amigos del programador jefe.<br />
* Alvaro es amigo del programador jefe.<br />
* Benito no obedece a Alvaro.<br />
* Benito no es un robot.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*all x all y ((robot(x) & amigo(y)) -> obedece(x,y))<br />
*amigo(Alvaro)<br />
*-obedece(Benito,Alvaro)<br />
*-robot(Benito)<br />
<br />
----<br />
'''Ejercicio 3.''' Formalizar el siguiente argumento (ejercicio 16 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Hay algún pez x que para cualquier pez y, si el pez x no se come al pez y entonces existe un pez z tal que z es un tiburón o bien z protege al pez y.<br />
* No hay ningún pez que se coma a todos los demás.<br />
* Ningún pez protege a ningún otro.<br />
* Por lo tanto, existe algún tiburón en la pecera.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*exists x ( all y ( -C(x,y) -> ( exists z ( T(z) | P(z,y) ) ) ) )<br />
*all x (all y (-C( x , y ) ) )<br />
*all x (all y (-P( x , y ) ) )<br />
*exists x ( T(x))<br />
<br />
----<br />
'''Ejercicio 4.''' Formalizar el siguiente argumento (ejercicio 21 de LPO de APPLI2), verificando la corrección de la solución:<br />
<br />
* Carlos afeita a todos los habitantes de Las Chinas que no se afeitan a sí mismo y sólo a ellos.<br />
* Carlos es un habitante de las Chinas.<br />
* Por lo tanto, Carlos no afeita a nadie.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*all x ( chino(x) & ( -afeita(x,x) ) <-> afeita(Carlos,x) )<br />
*chino(Carlos)<br />
*all y ( -afeita(Carlos,y))<br />
<br />
----<br />
'''Ejercicio 5.''' <br />
Sea F la fórmula P(x) → P (a) , donde a es un símbolo de constante. Dar un<br />
ejemplo de una interpretación en la que F sea verdadera. Y un ejemplo de una<br />
interpretación en la que F sea falsa.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*Verdadera<br />
<br />
U={1}<br />
I(P)={1}<br />
I(a)=1<br />
<br />
*Falsa<br />
<br />
U={1,2}<br />
I(P)={1}<br />
I(a)=2<br />
<br />
----<br />
'''Ejercicio 6.''' <br />
Sea L un lenguaje de primer orden con dos símbolos de predicado, P (de aridad<br />
1), Q (de aridad 2) y un símbolo de función, f , de aridad 1. Sea I = ( U, I )<br />
la estructura dada por:<br />
U = { a, b, c, d } ;<br />
I ( P ) = { a, b } ,<br />
I ( Q ) = {( a, b ) , ( b, b ) , ( c, b )} ,<br />
I ( f ) = {( a, b ) , ( b, b ) , ( c, a ) , ( d, c )} .<br />
¿Cuál es el valor de cada una de las siguientes fórmulas en dicha estructura? <br />
* P ( x ) → ∃ yQ ( y, x ) .<br />
*. ∀ xQ ( f ( x ) , x ) .<br />
* Q ( f ( x ) , x ) → Q ( x, x ) .<br />
* Q ( x, y ) → P ( x ) .<br />
----<br />
<br />
'''Solución:'''<br />
<br />
joslopjim4<br />
<br />
*<br />
A1(x)=a -> I(F)=0<br />
A2(x)=b -> I(F)=1<br />
A3(x)=c -> I(F)=1<br />
A4(x)=d -> I(F)=1<br />
<br />
*<br />
I(F)=0 si x=a<br />
<br />
*<br />
A1(x)=a -> I(F)=1<br />
A2(x)=b -> I(F)=1<br />
A3(x)=c -> I(F)=1<br />
A4(x)=d -> I(F)=1<br />
<br />
*<br />
A1(x)=a, A1(y)=a -> I(F)=1<br />
A1(x)=a, A2(y)=b -> I(F)=1<br />
A1(x)=a, A3(y)=c -> I(F)=1<br />
A1(x)=a, A4(y)=d -> I(F)=1<br />
<br />
A2(x)=b, A1(y)=a -> I(F)=1<br />
A2(x)=b, A2(y)=b -> I(F)=1<br />
A2(x)=b, A3(y)=c -> I(F)=1<br />
A2(x)=b, A4(y)=d -> I(F)=1<br />
<br />
A3(x)=c, A1(y)=a -> I(F)=1<br />
A3(x)=c, A2(y)=b -> I(F)=0<br />
A3(x)=c, A3(y)=c -> I(F)=1<br />
A3(x)=c, A4(y)=d -> I(F)=1<br />
<br />
A4(x)=d, A1(y)=a -> I(F)=1<br />
A4(x)=d, A2(y)=b -> I(F)=1<br />
A4(x)=d, A3(y)=c -> I(F)=1<br />
A4(x)=d, A4(y)=d -> I(F)=1<br />
<br />
----<br />
'''Ejercicio 7.'''<br />
<br />
En el lenguaje con igualdad L = { a, f } , siendo f un símbolo de función<br />
de aridad 1 y a una constante, se consideran las siguientes fórmulas:<br />
* F₁ : = ∀ x [ f ( x ) ≠ a ] ,<br />
* F₂ : = ∀ x ∀ y [ f ( x ) = f ( y ) → x = y ] ,<br />
* F₃ : = ∀ x [ x ≠ a → ∃ y [ f ( y ) = x ]] .<br />
<br />
Probar que ninguna de estas fórmulas es consecuencia lógica de las dos restantes.<br />
----<br />
<br />
'''Solución:'''<br />
<br />
joslopjim4<br />
<br />
* Caso 1: F3 no consecuencia lógica de F1 y F2<br />
<br />
U=naturales<br />
I(a)=0<br />
I(f)=x+2<br />
<br />
----<br />
'''Ejercicio 8.''' Dar una fórmula F, tal que todo modelo de F tenga al menos 3 elementos. Generalizarlo a n cualquiera.<br />
<br />
----<br />
<br />
'''Solución:'''<br />
<br />
joslopjim4<br />
<br />
exists x exists y exists z (x≠y & x≠z & y≠z)<br />
<br />
Para un n cualquiera, se ponen n exists con n variable diferentes y luego se hacen sus distinciones mediante una combinación sin repetición de n elementos cogidos de dos en dos</div>
Joslopjim4
https://www.glc.us.es/~jalonso/LMF2018/index.php?title=Relaci%C3%B3n_5&diff=164
Relación 5
2018-03-26T18:10:23Z
<p>Joslopjim4: </p>
<hr />
<div>=== Relación 5: Sintaxis y semántica de la Lógica de primer orden ===<br />
<br />
<br />
----<br />
'''Ejercicio 1.''' Formalizar el siguiente argumento (ejercicio 5 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Todo aquel que entre en el país y no sea un VIP será cacheado por un aduanero.<br />
* Hay un contrabandista que entra en el país y que solo podrá ser cacheado por contrabandistas.<br />
* Ningún contrabandista es un VIP.<br />
* Por lo tanto, algún aduanero es contrabandista.<br />
----<br />
<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*all x ( ( entra(x) & ( -vip(x) ) ) -> ( exists y (aduanero(y) & cachea(y,x) ) ) )<br />
* exists x (contrabandista(x) & entra(x) & ( all y (cachea(y,x) & contrabandista(y))))<br />
* all x (contrabandista(x) -> (- vip(x)))<br />
*exists x (aduanero(x) & contrabandista(x))<br />
<br />
----<br />
'''Ejercicio 2.''' Formalizar el siguiente argumento (ejercicio 15 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Todos los robots obedecen a los amigos del programador jefe.<br />
* Alvaro es amigo del programador jefe.<br />
* Benito no obedece a Alvaro.<br />
* Benito no es un robot.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*all x all y ((robot(x) & amigo(y)) -> obedece(x,y))<br />
*amigo(Alvaro)<br />
*-obedece(Benito,Alvaro)<br />
*-robot(Benito)<br />
<br />
----<br />
'''Ejercicio 3.''' Formalizar el siguiente argumento (ejercicio 16 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Hay algún pez x que para cualquier pez y, si el pez x no se come al pez y entonces existe un pez z tal que z es un tiburón o bien z protege al pez y.<br />
* No hay ningún pez que se coma a todos los demás.<br />
* Ningún pez protege a ningún otro.<br />
* Por lo tanto, existe algún tiburón en la pecera.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*exists x ( all y ( -C(x,y) -> ( exists z ( T(z) | P(z,y) ) ) ) )<br />
*all x (all y (-C( x , y ) ) )<br />
*all x (all y (-P( x , y ) ) )<br />
*exists x ( T(x))<br />
<br />
----<br />
'''Ejercicio 4.''' Formalizar el siguiente argumento (ejercicio 21 de LPO de APPLI2), verificando la corrección de la solución:<br />
<br />
* Carlos afeita a todos los habitantes de Las Chinas que no se afeitan a sí mismo y sólo a ellos.<br />
* Carlos es un habitante de las Chinas.<br />
* Por lo tanto, Carlos no afeita a nadie.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*all x ( chino(x) & ( -afeita(x,x) ) <-> afeita(Carlos,x) )<br />
*chino(Carlos)<br />
*all y ( -afeita(Carlos,y))<br />
<br />
----<br />
'''Ejercicio 5.''' <br />
Sea F la fórmula P(x) → P (a) , donde a es un símbolo de constante. Dar un<br />
ejemplo de una interpretación en la que F sea verdadera. Y un ejemplo de una<br />
interpretación en la que F sea falsa.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*Verdadera<br />
<br />
U={1}<br />
I(P)={1}<br />
I(a)=1<br />
<br />
*Falsa<br />
<br />
U={1,2}<br />
I(P)={1}<br />
I(a)=2<br />
<br />
----<br />
'''Ejercicio 6.''' <br />
Sea L un lenguaje de primer orden con dos símbolos de predicado, P (de aridad<br />
1), Q (de aridad 2) y un símbolo de función, f , de aridad 1. Sea I = ( U, I )<br />
la estructura dada por:<br />
U = { a, b, c, d } ;<br />
I ( P ) = { a, b } ,<br />
I ( Q ) = {( a, b ) , ( b, b ) , ( c, b )} ,<br />
I ( f ) = {( a, b ) , ( b, b ) , ( c, a ) , ( d, c )} .<br />
¿Cuál es el valor de cada una de las siguientes fórmulas en dicha estructura? <br />
* P ( x ) → ∃ yQ ( y, x ) .<br />
*. ∀ xQ ( f ( x ) , x ) .<br />
* Q ( f ( x ) , x ) → Q ( x, x ) .<br />
* Q ( x, y ) → P ( x ) .<br />
----<br />
<br />
'''Solución:'''<br />
<br />
joslopjim4<br />
<br />
*<br />
A1(x)=a -> I(F)=0<br />
A2(x)=b -> I(F)=1<br />
A3(x)=c -> I(F)=1<br />
A4(x)=d -> I(F)=1<br />
<br />
*<br />
I(F)=0 si x=a<br />
<br />
*<br />
A1(x)=a -> I(F)=1<br />
A2(x)=b -> I(F)=1<br />
A3(x)=c -> I(F)=1<br />
A4(x)=d -> I(F)=1<br />
<br />
*<br />
A1(x)=a, A1(y)=a -> I(F)=1<br />
A1(x)=a, A2(y)=b -> I(F)=1<br />
A1(x)=a, A3(y)=c -> I(F)=1<br />
A1(x)=a, A4(y)=d -> I(F)=1<br />
<br />
A2(x)=b, A1(y)=a -> I(F)=1<br />
A2(x)=b, A2(y)=b -> I(F)=1<br />
A2(x)=b, A3(y)=c -> I(F)=1<br />
A2(x)=b, A4(y)=d -> I(F)=1<br />
<br />
A3(x)=c, A1(y)=a -> I(F)=1<br />
A3(x)=c, A2(y)=b -> I(F)=0<br />
A3(x)=c, A3(y)=c -> I(F)=1<br />
A3(x)=c, A4(y)=d -> I(F)=1<br />
<br />
A4(x)=d, A1(y)=a -> I(F)=1<br />
A4(x)=d, A2(y)=b -> I(F)=1<br />
A4(x)=d, A3(y)=c -> I(F)=1<br />
A4(x)=d, A4(y)=d -> I(F)=1<br />
<br />
----<br />
'''Ejercicio 7.'''<br />
<br />
En el lenguaje con igualdad L = { a, f } , siendo f un símbolo de función<br />
de aridad 1 y a una constante, se consideran las siguientes fórmulas:<br />
* F₁ : = ∀ x [ f ( x ) ≠ a ] ,<br />
* F₂ : = ∀ x ∀ y [ f ( x ) = f ( y ) → x = y ] ,<br />
* F₃ : = ∀ x [ x ≠ a → ∃ y [ f ( y ) = x ]] .<br />
<br />
Probar que ninguna de estas fórmulas es consecuencia lógica de las dos restantes.<br />
----<br />
<br />
'''Solución:'''<br />
<br />
joslopjim4<br />
<br />
* Caso 1: F3 no consecuencia lógica de F1 y F2<br />
<br />
U=naturales<br />
I(a)=0<br />
I(f)=x+2<br />
<br />
----<br />
'''Ejercicio 8.''' Dar una fórmula F, tal que todo modelo de F tenga al menos 3 elementos. Generalizarlo a n cualquiera.<br />
<br />
----<br />
<br />
'''Solución:'''</div>
Joslopjim4
https://www.glc.us.es/~jalonso/LMF2018/index.php?title=Relaci%C3%B3n_5&diff=163
Relación 5
2018-03-26T18:08:00Z
<p>Joslopjim4: </p>
<hr />
<div>=== Relación 5: Sintaxis y semántica de la Lógica de primer orden ===<br />
<br />
<br />
----<br />
'''Ejercicio 1.''' Formalizar el siguiente argumento (ejercicio 5 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Todo aquel que entre en el país y no sea un VIP será cacheado por un aduanero.<br />
* Hay un contrabandista que entra en el país y que solo podrá ser cacheado por contrabandistas.<br />
* Ningún contrabandista es un VIP.<br />
* Por lo tanto, algún aduanero es contrabandista.<br />
----<br />
<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*all x ( ( entra(x) & ( -vip(x) ) ) -> ( exists y (aduanero(y) & cachea(y,x) ) ) )<br />
* exists x (contrabandista(x) & entra(x) & ( all y (cachea(y,x) & contrabandista(y))))<br />
* all x (contrabandista(x) -> (- vip(x)))<br />
*exists x (aduanero(x) & contrabandista(x))<br />
<br />
----<br />
'''Ejercicio 2.''' Formalizar el siguiente argumento (ejercicio 15 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Todos los robots obedecen a los amigos del programador jefe.<br />
* Alvaro es amigo del programador jefe.<br />
* Benito no obedece a Alvaro.<br />
* Benito no es un robot.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*all x all y ((robot(x) & amigo(y)) -> obedece(x,y))<br />
*amigo(Alvaro)<br />
*-obedece(Benito,Alvaro)<br />
*-robot(Benito)<br />
<br />
----<br />
'''Ejercicio 3.''' Formalizar el siguiente argumento (ejercicio 16 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Hay algún pez x que para cualquier pez y, si el pez x no se come al pez y entonces existe un pez z tal que z es un tiburón o bien z protege al pez y.<br />
* No hay ningún pez que se coma a todos los demás.<br />
* Ningún pez protege a ningún otro.<br />
* Por lo tanto, existe algún tiburón en la pecera.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*<br />
*all x (all y (-C( x , y ) ) )<br />
*all x (all y (-P( x , y ) ) )<br />
*exists x ( T(x))<br />
<br />
----<br />
'''Ejercicio 4.''' Formalizar el siguiente argumento (ejercicio 21 de LPO de APPLI2), verificando la corrección de la solución:<br />
<br />
* Carlos afeita a todos los habitantes de Las Chinas que no se afeitan a sí mismo y sólo a ellos.<br />
* Carlos es un habitante de las Chinas.<br />
* Por lo tanto, Carlos no afeita a nadie.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*all x ( chino(x) & ( -afeita(x,x) ) <-> afeita(Carlos,x) )<br />
*chino(Carlos)<br />
*all y ( -afeita(Carlos,y))<br />
<br />
----<br />
'''Ejercicio 5.''' <br />
Sea F la fórmula P(x) → P (a) , donde a es un símbolo de constante. Dar un<br />
ejemplo de una interpretación en la que F sea verdadera. Y un ejemplo de una<br />
interpretación en la que F sea falsa.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*Verdadera<br />
<br />
U={1}<br />
I(P)={1}<br />
I(a)=1<br />
<br />
*Falsa<br />
<br />
U={1,2}<br />
I(P)={1}<br />
I(a)=2<br />
<br />
----<br />
'''Ejercicio 6.''' <br />
Sea L un lenguaje de primer orden con dos símbolos de predicado, P (de aridad<br />
1), Q (de aridad 2) y un símbolo de función, f , de aridad 1. Sea I = ( U, I )<br />
la estructura dada por:<br />
U = { a, b, c, d } ;<br />
I ( P ) = { a, b } ,<br />
I ( Q ) = {( a, b ) , ( b, b ) , ( c, b )} ,<br />
I ( f ) = {( a, b ) , ( b, b ) , ( c, a ) , ( d, c )} .<br />
¿Cuál es el valor de cada una de las siguientes fórmulas en dicha estructura? <br />
* P ( x ) → ∃ yQ ( y, x ) .<br />
*. ∀ xQ ( f ( x ) , x ) .<br />
* Q ( f ( x ) , x ) → Q ( x, x ) .<br />
* Q ( x, y ) → P ( x ) .<br />
----<br />
<br />
'''Solución:'''<br />
<br />
joslopjim4<br />
<br />
*<br />
A1(x)=a -> I(F)=0<br />
A2(x)=b -> I(F)=1<br />
A3(x)=c -> I(F)=1<br />
A4(x)=d -> I(F)=1<br />
<br />
*<br />
I(F)=0 si x=a<br />
<br />
*<br />
A1(x)=a -> I(F)=1<br />
A2(x)=b -> I(F)=1<br />
A3(x)=c -> I(F)=1<br />
A4(x)=d -> I(F)=1<br />
<br />
*<br />
A1(x)=a, A1(y)=a -> I(F)=1<br />
A1(x)=a, A2(y)=b -> I(F)=1<br />
A1(x)=a, A3(y)=c -> I(F)=1<br />
A1(x)=a, A4(y)=d -> I(F)=1<br />
<br />
A2(x)=b, A1(y)=a -> I(F)=1<br />
A2(x)=b, A2(y)=b -> I(F)=1<br />
A2(x)=b, A3(y)=c -> I(F)=1<br />
A2(x)=b, A4(y)=d -> I(F)=1<br />
<br />
A3(x)=c, A1(y)=a -> I(F)=1<br />
A3(x)=c, A2(y)=b -> I(F)=0<br />
A3(x)=c, A3(y)=c -> I(F)=1<br />
A3(x)=c, A4(y)=d -> I(F)=1<br />
<br />
A4(x)=d, A1(y)=a -> I(F)=1<br />
A4(x)=d, A2(y)=b -> I(F)=1<br />
A4(x)=d, A3(y)=c -> I(F)=1<br />
A4(x)=d, A4(y)=d -> I(F)=1<br />
<br />
----<br />
'''Ejercicio 7.'''<br />
<br />
En el lenguaje con igualdad L = { a, f } , siendo f un símbolo de función<br />
de aridad 1 y a una constante, se consideran las siguientes fórmulas:<br />
* F₁ : = ∀ x [ f ( x ) ≠ a ] ,<br />
* F₂ : = ∀ x ∀ y [ f ( x ) = f ( y ) → x = y ] ,<br />
* F₃ : = ∀ x [ x ≠ a → ∃ y [ f ( y ) = x ]] .<br />
<br />
Probar que ninguna de estas fórmulas es consecuencia lógica de las dos restantes.<br />
----<br />
<br />
'''Solución:'''<br />
<br />
joslopjim4<br />
<br />
* Caso 1: F3 no consecuencia lógica de F1 y F2<br />
<br />
U=naturales<br />
I(a)=0<br />
I(f)=x+2<br />
<br />
----<br />
'''Ejercicio 8.''' Dar una fórmula F, tal que todo modelo de F tenga al menos 3 elementos. Generalizarlo a n cualquiera.<br />
<br />
----<br />
<br />
'''Solución:'''</div>
Joslopjim4
https://www.glc.us.es/~jalonso/LMF2018/index.php?title=Relaci%C3%B3n_5&diff=162
Relación 5
2018-03-26T14:47:54Z
<p>Joslopjim4: </p>
<hr />
<div>=== Relación 5: Sintaxis y semántica de la Lógica de primer orden ===<br />
<br />
<br />
----<br />
'''Ejercicio 1.''' Formalizar el siguiente argumento (ejercicio 5 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Todo aquel que entre en el país y no sea un VIP será cacheado por un aduanero.<br />
* Hay un contrabandista que entra en el país y que solo podrá ser cacheado por contrabandistas.<br />
* Ningún contrabandista es un VIP.<br />
* Por lo tanto, algún aduanero es contrabandista.<br />
----<br />
<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*all x ( ( entra(x) & ( -vip(x) ) ) -> ( exists y (aduanero(y) & cachea(y,x) ) ) )<br />
* exists x (contrabandista(x) & entra(x) & ( all y (cachea(y,x) & contrabandista(y))))<br />
* all x (contrabandista(x) -> (- vip(x)))<br />
*exists x (aduanero(x) & contrabandista(x))<br />
<br />
----<br />
'''Ejercicio 2.''' Formalizar el siguiente argumento (ejercicio 15 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Todos los robots obedecen a los amigos del programador jefe.<br />
* Alvaro es amigo del programador jefe.<br />
* Benito no obedece a Alvaro.<br />
* Benito no es un robot.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*all x all y ((robot(x) & amigo(y)) -> obedece(x,y))<br />
*amigo(Alvaro)<br />
*-obedece(Benito,Alvaro)<br />
*-robot(Benito)<br />
<br />
----<br />
'''Ejercicio 3.''' Formalizar el siguiente argumento (ejercicio 16 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Hay algún pez x que para cualquier pez y, si el pez x no se come al pez y entonces existe un pez z tal que z es un tiburón o bien z protege al pez y.<br />
* No hay ningún pez que se coma a todos los demás.<br />
* Ningún pez protege a ningún otro.<br />
* Por lo tanto, existe algún tiburón en la pecera.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*<br />
*all x (all y (-C( x , y ) ) )<br />
*all x (all y (-P( x , y ) ) )<br />
*exists x ( T(x))<br />
<br />
----<br />
'''Ejercicio 4.''' Formalizar el siguiente argumento (ejercicio 21 de LPO de APPLI2), verificando la corrección de la solución:<br />
<br />
* Carlos afeita a todos los habitantes de Las Chinas que no se afeitan a sí mismo y sólo a ellos.<br />
* Carlos es un habitante de las Chinas.<br />
* Por lo tanto, Carlos no afeita a nadie.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*<br />
*chino(Carlos)<br />
*all y ( -afeita(Carlos,y))<br />
<br />
----<br />
'''Ejercicio 5.''' <br />
Sea F la fórmula P(x) → P (a) , donde a es un símbolo de constante. Dar un<br />
ejemplo de una interpretación en la que F sea verdadera. Y un ejemplo de una<br />
interpretación en la que F sea falsa.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*Verdadera<br />
<br />
U={1}<br />
I(P)={1}<br />
I(a)=1<br />
<br />
*Falsa<br />
<br />
U={1,2}<br />
I(P)={1}<br />
I(a)=2<br />
<br />
----<br />
'''Ejercicio 6.''' <br />
Sea L un lenguaje de primer orden con dos símbolos de predicado, P (de aridad<br />
1), Q (de aridad 2) y un símbolo de función, f , de aridad 1. Sea I = ( U, I )<br />
la estructura dada por:<br />
U = { a, b, c, d } ;<br />
I ( P ) = { a, b } ,<br />
I ( Q ) = {( a, b ) , ( b, b ) , ( c, b )} ,<br />
I ( f ) = {( a, b ) , ( b, b ) , ( c, a ) , ( d, c )} .<br />
¿Cuál es el valor de cada una de las siguientes fórmulas en dicha estructura? <br />
* P ( x ) → ∃ yQ ( y, x ) .<br />
*. ∀ xQ ( f ( x ) , x ) .<br />
* Q ( f ( x ) , x ) → Q ( x, x ) .<br />
* Q ( x, y ) → P ( x ) .<br />
----<br />
<br />
'''Solución:'''<br />
<br />
joslopjim4<br />
<br />
*<br />
A1(x)=a -> I(F)=0<br />
A2(x)=b -> I(F)=1<br />
A3(x)=c -> I(F)=1<br />
A4(x)=d -> I(F)=1<br />
<br />
*<br />
I(F)=0 si x=a<br />
<br />
*<br />
A1(x)=a -> I(F)=1<br />
A2(x)=b -> I(F)=1<br />
A3(x)=c -> I(F)=1<br />
A4(x)=d -> I(F)=1<br />
<br />
*<br />
A1(x)=a, A1(y)=a -> I(F)=1<br />
A1(x)=a, A2(y)=b -> I(F)=1<br />
A1(x)=a, A3(y)=c -> I(F)=1<br />
A1(x)=a, A4(y)=d -> I(F)=1<br />
<br />
A2(x)=b, A1(y)=a -> I(F)=1<br />
A2(x)=b, A2(y)=b -> I(F)=1<br />
A2(x)=b, A3(y)=c -> I(F)=1<br />
A2(x)=b, A4(y)=d -> I(F)=1<br />
<br />
A3(x)=c, A1(y)=a -> I(F)=1<br />
A3(x)=c, A2(y)=b -> I(F)=0<br />
A3(x)=c, A3(y)=c -> I(F)=1<br />
A3(x)=c, A4(y)=d -> I(F)=1<br />
<br />
A4(x)=d, A1(y)=a -> I(F)=1<br />
A4(x)=d, A2(y)=b -> I(F)=1<br />
A4(x)=d, A3(y)=c -> I(F)=1<br />
A4(x)=d, A4(y)=d -> I(F)=1<br />
<br />
----<br />
'''Ejercicio 7.'''<br />
<br />
En el lenguaje con igualdad L = { a, f } , siendo f un símbolo de función<br />
de aridad 1 y a una constante, se consideran las siguientes fórmulas:<br />
* F₁ : = ∀ x [ f ( x ) ≠ a ] ,<br />
* F₂ : = ∀ x ∀ y [ f ( x ) = f ( y ) → x = y ] ,<br />
* F₃ : = ∀ x [ x ≠ a → ∃ y [ f ( y ) = x ]] .<br />
<br />
Probar que ninguna de estas fórmulas es consecuencia lógica de las dos restantes.<br />
----<br />
<br />
'''Solución:'''<br />
<br />
joslopjim4<br />
<br />
* Caso 1: F3 no consecuencia lógica de F1 y F2<br />
<br />
U=naturales<br />
I(a)=0<br />
I(f)=x+2<br />
<br />
----<br />
'''Ejercicio 8.''' Dar una fórmula F, tal que todo modelo de F tenga al menos 3 elementos. Generalizarlo a n cualquiera.<br />
<br />
----<br />
<br />
'''Solución:'''</div>
Joslopjim4
https://www.glc.us.es/~jalonso/LMF2018/index.php?title=Relaci%C3%B3n_5&diff=161
Relación 5
2018-03-26T13:39:55Z
<p>Joslopjim4: </p>
<hr />
<div>=== Relación 5: Sintaxis y semántica de la Lógica de primer orden ===<br />
<br />
<br />
----<br />
'''Ejercicio 1.''' Formalizar el siguiente argumento (ejercicio 5 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Todo aquel que entre en el país y no sea un VIP será cacheado por un aduanero.<br />
* Hay un contrabandista que entra en el país y que solo podrá ser cacheado por contrabandistas.<br />
* Ningún contrabandista es un VIP.<br />
* Por lo tanto, algún aduanero es contrabandista.<br />
----<br />
<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*all x ( entra ( x ) & ( -vip ( x ) ) ) -> ( exists y ( aduanero ( y ) & cachea ( y , x ) ) ) (Da error sintáctico)<br />
* exists x (contrabandista(x) & entra(x) & ( all y (cachea(y,x) & contrabandista(y))))<br />
* all x (contrabandista(x) -> (- vip(x)))<br />
*exists x (aduanero(x) & contrabandista(x))<br />
<br />
----<br />
'''Ejercicio 2.''' Formalizar el siguiente argumento (ejercicio 15 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Todos los robots obedecen a los amigos del programador jefe.<br />
* Alvaro es amigo del programador jefe.<br />
* Benito no obedece a Alvaro.<br />
* Benito no es un robot.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*all x all y ((robot(x) & amigo(y)) -> obedece(x,y))<br />
*amigo(Alvaro)<br />
*-obedece(Benito,Alvaro)<br />
*-robot(Benito)<br />
<br />
----<br />
'''Ejercicio 3.''' Formalizar el siguiente argumento (ejercicio 16 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Hay algún pez x que para cualquier pez y, si el pez x no se come al pez y entonces existe un pez z tal que z es un tiburón o bien z protege al pez y.<br />
* No hay ningún pez que se coma a todos los demás.<br />
* Ningún pez protege a ningún otro.<br />
* Por lo tanto, existe algún tiburón en la pecera.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*<br />
*all x (all y (-C( x , y ) ) )<br />
*all x (all y (-P( x , y ) ) )<br />
*exists x ( T(x))<br />
<br />
----<br />
'''Ejercicio 4.''' Formalizar el siguiente argumento (ejercicio 21 de LPO de APPLI2), verificando la corrección de la solución:<br />
<br />
* Carlos afeita a todos los habitantes de Las Chinas que no se afeitan a sí mismo y sólo a ellos.<br />
* Carlos es un habitante de las Chinas.<br />
* Por lo tanto, Carlos no afeita a nadie.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*<br />
*chino(Carlos)<br />
*all y ( -afeita(Carlos,y))<br />
<br />
----<br />
'''Ejercicio 5.''' <br />
Sea F la fórmula P(x) → P (a) , donde a es un símbolo de constante. Dar un<br />
ejemplo de una interpretación en la que F sea verdadera. Y un ejemplo de una<br />
interpretación en la que F sea falsa.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*Verdadera<br />
<br />
U={1}<br />
I(P)={1}<br />
I(a)=1<br />
<br />
*Falsa<br />
<br />
U={1,2}<br />
I(P)={1}<br />
I(a)=2<br />
<br />
----<br />
'''Ejercicio 6.''' <br />
Sea L un lenguaje de primer orden con dos símbolos de predicado, P (de aridad<br />
1), Q (de aridad 2) y un símbolo de función, f , de aridad 1. Sea I = ( U, I )<br />
la estructura dada por:<br />
U = { a, b, c, d } ;<br />
I ( P ) = { a, b } ,<br />
I ( Q ) = {( a, b ) , ( b, b ) , ( c, b )} ,<br />
I ( f ) = {( a, b ) , ( b, b ) , ( c, a ) , ( d, c )} .<br />
¿Cuál es el valor de cada una de las siguientes fórmulas en dicha estructura? <br />
* P ( x ) → ∃ yQ ( y, x ) .<br />
*. ∀ xQ ( f ( x ) , x ) .<br />
* Q ( f ( x ) , x ) → Q ( x, x ) .<br />
* Q ( x, y ) → P ( x ) .<br />
----<br />
<br />
'''Solución:'''<br />
<br />
joslopjim4<br />
<br />
*<br />
A1(x)=a -> I(F)=0<br />
A2(x)=b -> I(F)=1<br />
A3(x)=c -> I(F)=1<br />
A4(x)=d -> I(F)=1<br />
<br />
*<br />
I(F)=0 si x=a<br />
<br />
*<br />
A1(x)=a -> I(F)=1<br />
A2(x)=b -> I(F)=1<br />
A3(x)=c -> I(F)=1<br />
A4(x)=d -> I(F)=1<br />
<br />
*<br />
A1(x)=a, A1(y)=a -> I(F)=1<br />
A1(x)=a, A2(y)=b -> I(F)=1<br />
A1(x)=a, A3(y)=c -> I(F)=1<br />
A1(x)=a, A4(y)=d -> I(F)=1<br />
<br />
A2(x)=b, A1(y)=a -> I(F)=1<br />
A2(x)=b, A2(y)=b -> I(F)=1<br />
A2(x)=b, A3(y)=c -> I(F)=1<br />
A2(x)=b, A4(y)=d -> I(F)=1<br />
<br />
A3(x)=c, A1(y)=a -> I(F)=1<br />
A3(x)=c, A2(y)=b -> I(F)=0<br />
A3(x)=c, A3(y)=c -> I(F)=1<br />
A3(x)=c, A4(y)=d -> I(F)=1<br />
<br />
A4(x)=d, A1(y)=a -> I(F)=1<br />
A4(x)=d, A2(y)=b -> I(F)=1<br />
A4(x)=d, A3(y)=c -> I(F)=1<br />
A4(x)=d, A4(y)=d -> I(F)=1<br />
<br />
----<br />
'''Ejercicio 7.'''<br />
<br />
En el lenguaje con igualdad L = { a, f } , siendo f un símbolo de función<br />
de aridad 1 y a una constante, se consideran las siguientes fórmulas:<br />
* F₁ : = ∀ x [ f ( x ) ≠ a ] ,<br />
* F₂ : = ∀ x ∀ y [ f ( x ) = f ( y ) → x = y ] ,<br />
* F₃ : = ∀ x [ x ≠ a → ∃ y [ f ( y ) = x ]] .<br />
<br />
Probar que ninguna de estas fórmulas es consecuencia lógica de las dos restantes.<br />
----<br />
<br />
'''Solución:'''<br />
<br />
joslopjim4<br />
<br />
* Caso 1: F3 no consecuencia lógica de F1 y F2<br />
<br />
U=naturales<br />
I(a)=0<br />
I(f)=x+2<br />
<br />
----<br />
'''Ejercicio 8.''' Dar una fórmula F, tal que todo modelo de F tenga al menos 3 elementos. Generalizarlo a n cualquiera.<br />
<br />
----<br />
<br />
'''Solución:'''</div>
Joslopjim4
https://www.glc.us.es/~jalonso/LMF2018/index.php?title=Relaci%C3%B3n_5&diff=156
Relación 5
2018-03-24T23:28:21Z
<p>Joslopjim4: </p>
<hr />
<div>=== Relación 5: Sintaxis y semántica de la Lógica de primer orden ===<br />
<br />
<br />
----<br />
'''Ejercicio 1.''' Formalizar el siguiente argumento (ejercicio 5 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Todo aquel que entre en el país y no sea un VIP será cacheado por un aduanero.<br />
* Hay un contrabandista que entra en el país y que solo podrá ser cacheado por contrabandistas.<br />
* Ningún contrabandista es un VIP.<br />
* Por lo tanto, algún aduanero es contrabandista.<br />
----<br />
<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*all x ( entra ( x ) & ( -vip ( x ) ) ) -> ( exists y ( aduanero ( y ) & cachea ( y , x ) ) ) (Da error sintáctico)<br />
* exists x (contrabandista(x) & entra(x) & ( all y (cachea(y,x) & contrabandista(y))))<br />
* all x (contrabandista(x) -> (- vip(x)))<br />
*exists x (aduanero(x) & contrabandista(x))<br />
<br />
----<br />
'''Ejercicio 2.''' Formalizar el siguiente argumento (ejercicio 15 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Todos los robots obedecen a los amigos del programador jefe.<br />
* Alvaro es amigo del programador jefe.<br />
* Benito no obedece a Alvaro.<br />
* Benito no es un robot.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*<br />
*amigo(Alvaro)<br />
*-obedece(Benito,Alvaro)<br />
*-robot(Benito)<br />
<br />
----<br />
'''Ejercicio 3.''' Formalizar el siguiente argumento (ejercicio 16 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Hay algún pez x que para cualquier pez y, si el pez x no se come al pez y entonces existe un pez z tal que z es un tiburón o bien z protege al pez y.<br />
* No hay ningún pez que se coma a todos los demás.<br />
* Ningún pez protege a ningún otro.<br />
* Por lo tanto, existe algún tiburón en la pecera.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*<br />
*all x (all y (-C( x , y ) ) )<br />
*all x (all y (-P( x , y ) ) )<br />
*exists x ( T(x))<br />
<br />
----<br />
'''Ejercicio 4.''' Formalizar el siguiente argumento (ejercicio 21 de LPO de APPLI2), verificando la corrección de la solución:<br />
<br />
* Carlos afeita a todos los habitantes de Las Chinas que no se afeitan a sí mismo y sólo a ellos.<br />
* Carlos es un habitante de las Chinas.<br />
* Por lo tanto, Carlos no afeita a nadie.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*<br />
*chino(Carlos)<br />
*all y ( -afeita(Carlos,y))<br />
<br />
----<br />
'''Ejercicio 5.''' <br />
Sea F la fórmula P(x) → P (a) , donde a es un símbolo de constante. Dar un<br />
ejemplo de una interpretación en la que F sea verdadera. Y un ejemplo de una<br />
interpretación en la que F sea falsa.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*Verdadera<br />
<br />
U={1}<br />
I(P)={1}<br />
I(a)=1<br />
<br />
*Falsa<br />
<br />
U={1,2}<br />
I(P)={1}<br />
I(a)=2<br />
<br />
----<br />
'''Ejercicio 6.''' <br />
Sea L un lenguaje de primer orden con dos símbolos de predicado, P (de aridad<br />
1), Q (de aridad 2) y un símbolo de función, f , de aridad 1. Sea I = ( U, I )<br />
la estructura dada por:<br />
U = { a, b, c, d } ;<br />
I ( P ) = { a, b } ,<br />
I ( Q ) = {( a, b ) , ( b, b ) , ( c, b )} ,<br />
I ( f ) = {( a, b ) , ( b, b ) , ( c, a ) , ( d, c )} .<br />
¿Cuál es el valor de cada una de las siguientes fórmulas en dicha estructura? <br />
* P ( x ) → ∃ yQ ( y, x ) .<br />
*. ∀ xQ ( f ( x ) , x ) .<br />
* Q ( f ( x ) , x ) → Q ( x, x ) .<br />
* Q ( x, y ) → P ( x ) .<br />
----<br />
<br />
'''Solución:'''<br />
<br />
joslopjim4<br />
<br />
*<br />
A1(x)=a -> I(F)=0<br />
A2(x)=b -> I(F)=1<br />
A3(x)=c -> I(F)=1<br />
A4(x)=d -> I(F)=1<br />
<br />
*<br />
I(F)=0 si x=a<br />
<br />
*<br />
A1(x)=a -> I(F)=1<br />
A2(x)=b -> I(F)=1<br />
A3(x)=c -> I(F)=1<br />
A4(x)=d -> I(F)=1<br />
<br />
*<br />
A1(x)=a, A1(y)=a -> I(F)=1<br />
A1(x)=a, A2(y)=b -> I(F)=1<br />
A1(x)=a, A3(y)=c -> I(F)=1<br />
A1(x)=a, A4(y)=d -> I(F)=1<br />
<br />
A2(x)=b, A1(y)=a -> I(F)=1<br />
A2(x)=b, A2(y)=b -> I(F)=1<br />
A2(x)=b, A3(y)=c -> I(F)=1<br />
A2(x)=b, A4(y)=d -> I(F)=1<br />
<br />
A3(x)=c, A1(y)=a -> I(F)=1<br />
A3(x)=c, A2(y)=b -> I(F)=0<br />
A3(x)=c, A3(y)=c -> I(F)=1<br />
A3(x)=c, A4(y)=d -> I(F)=1<br />
<br />
A4(x)=d, A1(y)=a -> I(F)=1<br />
A4(x)=d, A2(y)=b -> I(F)=1<br />
A4(x)=d, A3(y)=c -> I(F)=1<br />
A4(x)=d, A4(y)=d -> I(F)=1<br />
<br />
----<br />
'''Ejercicio 7.'''<br />
<br />
En el lenguaje con igualdad L = { a, f } , siendo f un símbolo de función<br />
de aridad 1 y a una constante, se consideran las siguientes fórmulas:<br />
* F₁ : = ∀ x [ f ( x ) ≠ a ] ,<br />
* F₂ : = ∀ x ∀ y [ f ( x ) = f ( y ) → x = y ] ,<br />
* F₃ : = ∀ x [ x ≠ a → ∃ y [ f ( y ) = x ]] .<br />
<br />
Probar que ninguna de estas fórmulas es consecuencia lógica de las dos restantes.<br />
----<br />
<br />
'''Solución:'''<br />
<br />
joslopjim4<br />
<br />
* Caso 1: F3 no consecuencia lógica de F1 y F2<br />
<br />
U=naturales<br />
I(a)=0<br />
I(f)=x+2<br />
<br />
----<br />
'''Ejercicio 8.''' Dar una fórmula F, tal que todo modelo de F tenga al menos 3 elementos. Generalizarlo a n cualquiera.<br />
<br />
----<br />
<br />
'''Solución:'''</div>
Joslopjim4
https://www.glc.us.es/~jalonso/LMF2018/index.php?title=Relaci%C3%B3n_5&diff=155
Relación 5
2018-03-24T23:03:45Z
<p>Joslopjim4: </p>
<hr />
<div>=== Relación 5: Sintaxis y semántica de la Lógica de primer orden ===<br />
<br />
<br />
----<br />
'''Ejercicio 1.''' Formalizar el siguiente argumento (ejercicio 5 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Todo aquel que entre en el país y no sea un VIP será cacheado por un aduanero.<br />
* Hay un contrabandista que entra en el país y que solo podrá ser cacheado por contrabandistas.<br />
* Ningún contrabandista es un VIP.<br />
* Por lo tanto, algún aduanero es contrabandista.<br />
----<br />
<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*all x ( entra ( x ) & ( -vip ( x ) ) ) -> ( exists y ( aduanero ( y ) & cachea ( y , x ) ) ) (Da error sintáctico)<br />
* exists x (contrabandista(x) & entra(x) & ( all y (cachea(y,x) & contrabandista(y))))<br />
* all x (contrabandista(x) -> (- vip(x)))<br />
*exists x (aduanero(x) & contrabandista(x))<br />
<br />
----<br />
'''Ejercicio 2.''' Formalizar el siguiente argumento (ejercicio 15 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Todos los robots obedecen a los amigos del programador jefe.<br />
* Alvaro es amigo del programador jefe.<br />
* Benito no obedece a Alvaro.<br />
* Benito no es un robot.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*<br />
*amigo(Alvaro)<br />
*-obedece(Benito,Alvaro)<br />
*-robot(Benito)<br />
<br />
----<br />
'''Ejercicio 3.''' Formalizar el siguiente argumento (ejercicio 16 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Hay algún pez x que para cualquier pez y, si el pez x no se come al pez y entonces existe un pez z tal que z es un tiburón o bien z protege al pez y.<br />
* No hay ningún pez que se coma a todos los demás.<br />
* Ningún pez protege a ningún otro.<br />
* Por lo tanto, existe algún tiburón en la pecera.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*<br />
*all x (all y (-C( x , y ) ) )<br />
*all x (all y (-P( x , y ) ) )<br />
*exists x ( T(x))<br />
<br />
----<br />
'''Ejercicio 4.''' Formalizar el siguiente argumento (ejercicio 21 de LPO de APPLI2), verificando la corrección de la solución:<br />
<br />
* Carlos afeita a todos los habitantes de Las Chinas que no se afeitan a sí mismo y sólo a ellos.<br />
* Carlos es un habitante de las Chinas.<br />
* Por lo tanto, Carlos no afeita a nadie.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
<br />
----<br />
'''Ejercicio 5.''' <br />
Sea F la fórmula P(x) → P (a) , donde a es un símbolo de constante. Dar un<br />
ejemplo de una interpretación en la que F sea verdadera. Y un ejemplo de una<br />
interpretación en la que F sea falsa.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*Verdadera<br />
<br />
U={1}<br />
I(P)={1}<br />
I(a)=1<br />
<br />
*Falsa<br />
<br />
U={1,2}<br />
I(P)={1}<br />
I(a)=2<br />
<br />
----<br />
'''Ejercicio 6.''' <br />
Sea L un lenguaje de primer orden con dos símbolos de predicado, P (de aridad<br />
1), Q (de aridad 2) y un símbolo de función, f , de aridad 1. Sea I = ( U, I )<br />
la estructura dada por:<br />
U = { a, b, c, d } ;<br />
I ( P ) = { a, b } ,<br />
I ( Q ) = {( a, b ) , ( b, b ) , ( c, b )} ,<br />
I ( f ) = {( a, b ) , ( b, b ) , ( c, a ) , ( d, c )} .<br />
¿Cuál es el valor de cada una de las siguientes fórmulas en dicha estructura? <br />
* P ( x ) → ∃ yQ ( y, x ) .<br />
*. ∀ xQ ( f ( x ) , x ) .<br />
* Q ( f ( x ) , x ) → Q ( x, x ) .<br />
* Q ( x, y ) → P ( x ) .<br />
----<br />
<br />
'''Solución:'''<br />
<br />
joslopjim4<br />
<br />
*<br />
A1(x)=a -> I(F)=0<br />
A2(x)=b -> I(F)=1<br />
A3(x)=c -> I(F)=1<br />
A4(x)=d -> I(F)=1<br />
<br />
*<br />
I(F)=0 si x=a<br />
<br />
*<br />
A1(x)=a -> I(F)=1<br />
A2(x)=b -> I(F)=1<br />
A3(x)=c -> I(F)=1<br />
A4(x)=d -> I(F)=1<br />
<br />
*<br />
A1(x)=a, A1(y)=a -> I(F)=1<br />
A1(x)=a, A2(y)=b -> I(F)=1<br />
A1(x)=a, A3(y)=c -> I(F)=1<br />
A1(x)=a, A4(y)=d -> I(F)=1<br />
<br />
A2(x)=b, A1(y)=a -> I(F)=1<br />
A2(x)=b, A2(y)=b -> I(F)=1<br />
A2(x)=b, A3(y)=c -> I(F)=1<br />
A2(x)=b, A4(y)=d -> I(F)=1<br />
<br />
A3(x)=c, A1(y)=a -> I(F)=1<br />
A3(x)=c, A2(y)=b -> I(F)=0<br />
A3(x)=c, A3(y)=c -> I(F)=1<br />
A3(x)=c, A4(y)=d -> I(F)=1<br />
<br />
A4(x)=d, A1(y)=a -> I(F)=1<br />
A4(x)=d, A2(y)=b -> I(F)=1<br />
A4(x)=d, A3(y)=c -> I(F)=1<br />
A4(x)=d, A4(y)=d -> I(F)=1<br />
<br />
----<br />
'''Ejercicio 7.'''<br />
<br />
En el lenguaje con igualdad L = { a, f } , siendo f un símbolo de función<br />
de aridad 1 y a una constante, se consideran las siguientes fórmulas:<br />
* F₁ : = ∀ x [ f ( x ) ≠ a ] ,<br />
* F₂ : = ∀ x ∀ y [ f ( x ) = f ( y ) → x = y ] ,<br />
* F₃ : = ∀ x [ x ≠ a → ∃ y [ f ( y ) = x ]] .<br />
<br />
Probar que ninguna de estas fórmulas es consecuencia lógica de las dos restantes.<br />
----<br />
<br />
'''Solución:'''<br />
<br />
joslopjim4<br />
<br />
* Caso 1: F3 no consecuencia lógica de F1 y F2<br />
<br />
U=naturales<br />
I(a)=0<br />
I(f)=x+2<br />
<br />
----<br />
'''Ejercicio 8.''' Dar una fórmula F, tal que todo modelo de F tenga al menos 3 elementos. Generalizarlo a n cualquiera.<br />
<br />
----<br />
<br />
'''Solución:'''</div>
Joslopjim4
https://www.glc.us.es/~jalonso/LMF2018/index.php?title=Relaci%C3%B3n_5&diff=154
Relación 5
2018-03-24T22:58:56Z
<p>Joslopjim4: </p>
<hr />
<div>=== Relación 5: Sintaxis y semántica de la Lógica de primer orden ===<br />
<br />
<br />
----<br />
'''Ejercicio 1.''' Formalizar el siguiente argumento (ejercicio 5 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Todo aquel que entre en el país y no sea un VIP será cacheado por un aduanero.<br />
* Hay un contrabandista que entra en el país y que solo podrá ser cacheado por contrabandistas.<br />
* Ningún contrabandista es un VIP.<br />
* Por lo tanto, algún aduanero es contrabandista.<br />
----<br />
<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*all x ( entra ( x ) & ( -vip ( x ) ) ) -> ( exists y ( aduanero ( y ) & cachea ( y , x ) ) ) (Da error sintáctico)<br />
* exists x (contrabandista(x) & entra(x) & ( all y (cachea(y,x) & contrabandista(y))))<br />
* all x (contrabandista(x) -> (- vip(x)))<br />
*exists x (aduanero(x) & contrabandista(x))<br />
<br />
----<br />
'''Ejercicio 2.''' Formalizar el siguiente argumento (ejercicio 15 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Todos los robots obedecen a los amigos del programador jefe.<br />
* Alvaro es amigo del programador jefe.<br />
* Benito no obedece a Alvaro.<br />
* Benito no es un robot.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*<br />
*amigo(Alvaro)<br />
*-obedece(Benito,Alvaro)<br />
*-robot(Benito)<br />
<br />
----<br />
'''Ejercicio 3.''' Formalizar el siguiente argumento (ejercicio 16 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Hay algún pez x que para cualquier pez y, si el pez x no se come al pez y entonces existe un pez z tal que z es un tiburón o bien z protege al pez y.<br />
* No hay ningún pez que se coma a todos los demás.<br />
* Ningún pez protege a ningún otro.<br />
* Por lo tanto, existe algún tiburón en la pecera.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
----<br />
'''Ejercicio 4.''' Formalizar el siguiente argumento (ejercicio 21 de LPO de APPLI2), verificando la corrección de la solución:<br />
<br />
* Carlos afeita a todos los habitantes de Las Chinas que no se afeitan a sí mismo y sólo a ellos.<br />
* Carlos es un habitante de las Chinas.<br />
* Por lo tanto, Carlos no afeita a nadie.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
<br />
----<br />
'''Ejercicio 5.''' <br />
Sea F la fórmula P(x) → P (a) , donde a es un símbolo de constante. Dar un<br />
ejemplo de una interpretación en la que F sea verdadera. Y un ejemplo de una<br />
interpretación en la que F sea falsa.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*Verdadera<br />
<br />
U={1}<br />
I(P)={1}<br />
I(a)=1<br />
<br />
*Falsa<br />
<br />
U={1,2}<br />
I(P)={1}<br />
I(a)=2<br />
<br />
----<br />
'''Ejercicio 6.''' <br />
Sea L un lenguaje de primer orden con dos símbolos de predicado, P (de aridad<br />
1), Q (de aridad 2) y un símbolo de función, f , de aridad 1. Sea I = ( U, I )<br />
la estructura dada por:<br />
U = { a, b, c, d } ;<br />
I ( P ) = { a, b } ,<br />
I ( Q ) = {( a, b ) , ( b, b ) , ( c, b )} ,<br />
I ( f ) = {( a, b ) , ( b, b ) , ( c, a ) , ( d, c )} .<br />
¿Cuál es el valor de cada una de las siguientes fórmulas en dicha estructura? <br />
* P ( x ) → ∃ yQ ( y, x ) .<br />
*. ∀ xQ ( f ( x ) , x ) .<br />
* Q ( f ( x ) , x ) → Q ( x, x ) .<br />
* Q ( x, y ) → P ( x ) .<br />
----<br />
<br />
'''Solución:'''<br />
<br />
joslopjim4<br />
<br />
*<br />
A1(x)=a -> I(F)=0<br />
A2(x)=b -> I(F)=1<br />
A3(x)=c -> I(F)=1<br />
A4(x)=d -> I(F)=1<br />
<br />
*<br />
I(F)=0 si x=a<br />
<br />
*<br />
A1(x)=a -> I(F)=1<br />
A2(x)=b -> I(F)=1<br />
A3(x)=c -> I(F)=1<br />
A4(x)=d -> I(F)=1<br />
<br />
*<br />
A1(x)=a, A1(y)=a -> I(F)=1<br />
A1(x)=a, A2(y)=b -> I(F)=1<br />
A1(x)=a, A3(y)=c -> I(F)=1<br />
A1(x)=a, A4(y)=d -> I(F)=1<br />
<br />
A2(x)=b, A1(y)=a -> I(F)=1<br />
A2(x)=b, A2(y)=b -> I(F)=1<br />
A2(x)=b, A3(y)=c -> I(F)=1<br />
A2(x)=b, A4(y)=d -> I(F)=1<br />
<br />
A3(x)=c, A1(y)=a -> I(F)=1<br />
A3(x)=c, A2(y)=b -> I(F)=0<br />
A3(x)=c, A3(y)=c -> I(F)=1<br />
A3(x)=c, A4(y)=d -> I(F)=1<br />
<br />
A4(x)=d, A1(y)=a -> I(F)=1<br />
A4(x)=d, A2(y)=b -> I(F)=1<br />
A4(x)=d, A3(y)=c -> I(F)=1<br />
A4(x)=d, A4(y)=d -> I(F)=1<br />
<br />
----<br />
'''Ejercicio 7.'''<br />
<br />
En el lenguaje con igualdad L = { a, f } , siendo f un símbolo de función<br />
de aridad 1 y a una constante, se consideran las siguientes fórmulas:<br />
* F₁ : = ∀ x [ f ( x ) ≠ a ] ,<br />
* F₂ : = ∀ x ∀ y [ f ( x ) = f ( y ) → x = y ] ,<br />
* F₃ : = ∀ x [ x ≠ a → ∃ y [ f ( y ) = x ]] .<br />
<br />
Probar que ninguna de estas fórmulas es consecuencia lógica de las dos restantes.<br />
----<br />
<br />
'''Solución:'''<br />
<br />
joslopjim4<br />
<br />
* Caso 1: F3 no consecuencia lógica de F1 y F2<br />
<br />
U=naturales<br />
I(a)=0<br />
I(f)=x+2<br />
<br />
----<br />
'''Ejercicio 8.''' Dar una fórmula F, tal que todo modelo de F tenga al menos 3 elementos. Generalizarlo a n cualquiera.<br />
<br />
----<br />
<br />
'''Solución:'''</div>
Joslopjim4
https://www.glc.us.es/~jalonso/LMF2018/index.php?title=Relaci%C3%B3n_5&diff=153
Relación 5
2018-03-24T20:13:09Z
<p>Joslopjim4: </p>
<hr />
<div>=== Relación 5: Sintaxis y semántica de la Lógica de primer orden ===<br />
<br />
<br />
----<br />
'''Ejercicio 1.''' Formalizar el siguiente argumento (ejercicio 5 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Todo aquel que entre en el país y no sea un VIP será cacheado por un aduanero.<br />
* Hay un contrabandista que entra en el país y que solo podrá ser cacheado por contrabandistas.<br />
* Ningún contrabandista es un VIP.<br />
* Por lo tanto, algún aduanero es contrabandista.<br />
----<br />
<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*all x ( entra ( x ) & ( -vip ( x ) ) ) -> ( exists y ( aduanero ( y ) & cachea ( y , x ) ) ) (Da error sintáctico)<br />
* exists x (contrabandista(x) & entra(x) & ( all y (cachea(y,x) & contrabandista(y))))<br />
* all x (contrabandista(x) -> (- vip(x)))<br />
*exists x (aduanero(x) & contrabandista(x))<br />
<br />
----<br />
'''Ejercicio 2.''' Formalizar el siguiente argumento (ejercicio 15 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Todos los robots obedecen a los amigos del programador jefe.<br />
* Alvaro es amigo del programador jefe.<br />
* Benito no obedece a Alvaro.<br />
* Benito no es un robot.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*<br />
*amigo(Alvaro)<br />
*-obedece(Benito,Alvaro)<br />
*-robot(Benito)<br />
<br />
----<br />
'''Ejercicio 3.''' Formalizar el siguiente argumento (ejercicio 16 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Hay algún pez x que para cualquier pez y, si el pez x no se come al pez y entonces existe un pez z tal que z es un tiburón o bien z protege al pez y.<br />
* No hay ningún pez que se coma a todos los demás.<br />
* Ningún pez protege a ningún otro.<br />
* Por lo tanto, existe algún tiburón en la pecera.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
----<br />
'''Ejercicio 4.''' Formalizar el siguiente argumento (ejercicio 21 de LPO de APPLI2), verificando la corrección de la solución:<br />
<br />
* Carlos afeita a todos los habitantes de Las Chinas que no se afeitan a sí mismo y sólo a ellos.<br />
* Carlos es un habitante de las Chinas.<br />
* Por lo tanto, Carlos no afeita a nadie.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
<br />
----<br />
'''Ejercicio 5.''' <br />
Sea F la fórmula P(x) → P (a) , donde a es un símbolo de constante. Dar un<br />
ejemplo de una interpretación en la que F sea verdadera. Y un ejemplo de una<br />
interpretación en la que F sea falsa.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*Verdadera<br />
<br />
U={1}<br />
I(P)={1}<br />
I(a)=1<br />
<br />
*Falsa<br />
<br />
U={1,2}<br />
I(P)={1}<br />
I(a)=2<br />
<br />
----<br />
'''Ejercicio 6.''' <br />
Sea L un lenguaje de primer orden con dos símbolos de predicado, P (de aridad<br />
1), Q (de aridad 2) y un símbolo de función, f , de aridad 1. Sea I = ( U, I )<br />
la estructura dada por:<br />
U = { a, b, c, d } ;<br />
I ( P ) = { a, b } ,<br />
I ( Q ) = {( a, b ) , ( b, b ) , ( c, b )} ,<br />
I ( f ) = {( a, b ) , ( b, b ) , ( c, a ) , ( d, c )} .<br />
¿Cuál es el valor de cada una de las siguientes fórmulas en dicha estructura? <br />
* P ( x ) → ∃ yQ ( y, x ) .<br />
*. ∀ xQ ( f ( x ) , x ) .<br />
* Q ( f ( x ) , x ) → Q ( x, x ) .<br />
* Q ( x, y ) → P ( x ) .<br />
----<br />
<br />
'''Solución:'''<br />
<br />
joslopjim4<br />
<br />
*<br />
A1(x)=a -> I(F)=0<br />
A2(x)=b -> I(F)=1<br />
A3(x)=c -> I(F)=1<br />
A4(x)=d -> I(F)=1<br />
<br />
*<br />
I(F)=0 si x=a<br />
<br />
*<br />
A1(x)=a -> I(F)=1<br />
A2(x)=b -> I(F)=1<br />
A3(x)=c -> I(F)=1<br />
A4(x)=d -> I(F)=1<br />
<br />
*<br />
A1(x)=a, A1(y)=a -> I(F)=1<br />
A1(x)=a, A2(y)=b -> I(F)=1<br />
A1(x)=a, A3(y)=c -> I(F)=1<br />
A1(x)=a, A4(y)=d -> I(F)=1<br />
<br />
A2(x)=b, A1(y)=a -> I(F)=1<br />
A2(x)=b, A2(y)=b -> I(F)=1<br />
A2(x)=b, A3(y)=c -> I(F)=1<br />
A2(x)=b, A4(y)=d -> I(F)=1<br />
<br />
A3(x)=c, A1(y)=a -> I(F)=1<br />
A3(x)=c, A2(y)=b -> I(F)=0<br />
A3(x)=c, A3(y)=c -> I(F)=1<br />
A3(x)=c, A4(y)=d -> I(F)=1<br />
<br />
A4(x)=d, A1(y)=a -> I(F)=1<br />
A4(x)=d, A2(y)=b -> I(F)=1<br />
A4(x)=d, A3(y)=c -> I(F)=1<br />
A4(x)=d, A4(y)=d -> I(F)=1<br />
<br />
----<br />
'''Ejercicio 7.'''<br />
<br />
En el lenguaje con igualdad L = { a, f } , siendo f un símbolo de función<br />
de aridad 1 y a una constante, se consideran las siguientes fórmulas:<br />
* F₁ : = ∀ x [ f ( x ) ≠ a ] ,<br />
* F₂ : = ∀ x ∀ y [ f ( x ) = f ( y ) → x = y ] ,<br />
* F₃ : = ∀ x [ x ≠ a → ∃ y [ f ( y ) = x ]] .<br />
<br />
Probar que ninguna de estas fórmulas es consecuencia lógica de las dos restantes.<br />
----<br />
<br />
'''Solución:'''<br />
<br />
joslopjim4<br />
<br />
U=naturales<br />
I(a)=0<br />
I(f)=x+2<br />
<br />
----<br />
'''Ejercicio 8.''' Dar una fórmula F, tal que todo modelo de F tenga al menos 3 elementos. Generalizarlo a n cualquiera.<br />
<br />
----<br />
<br />
'''Solución:'''</div>
Joslopjim4
https://www.glc.us.es/~jalonso/LMF2018/index.php?title=Relaci%C3%B3n_5&diff=152
Relación 5
2018-03-24T20:11:12Z
<p>Joslopjim4: /* Relación 5: Sintaxis y semántica de la Lógica de primer orden */</p>
<hr />
<div>=== Relación 5: Sintaxis y semántica de la Lógica de primer orden ===<br />
<br />
<br />
----<br />
'''Ejercicio 1.''' Formalizar el siguiente argumento (ejercicio 5 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Todo aquel que entre en el país y no sea un VIP será cacheado por un aduanero.<br />
* Hay un contrabandista que entra en el país y que solo podrá ser cacheado por contrabandistas.<br />
* Ningún contrabandista es un VIP.<br />
* Por lo tanto, algún aduanero es contrabandista.<br />
----<br />
<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*all x ( entra ( x ) & ( -vip ( x ) ) ) -> ( exists y ( aduanero ( y ) & cachea ( y , x ) ) ) (Da error sintáctico)<br />
* exists x (contrabandista(x) & entra(x) & ( all y (cachea(y,x) & contrabandista(y))))<br />
* all x (contrabandista(x) -> (- vip(x)))<br />
*exists x (aduanero(x) & contrabandista(x))<br />
<br />
----<br />
'''Ejercicio 2.''' Formalizar el siguiente argumento (ejercicio 15 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Todos los robots obedecen a los amigos del programador jefe.<br />
* Alvaro es amigo del programador jefe.<br />
* Benito no obedece a Alvaro.<br />
* Benito no es un robot.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*<br />
*amigo(Alvaro)<br />
*-obedece(Benito,Alvaro)<br />
*-robot(Benito)<br />
<br />
----<br />
'''Ejercicio 3.''' Formalizar el siguiente argumento (ejercicio 16 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Hay algún pez x que para cualquier pez y, si el pez x no se come al pez y entonces existe un pez z tal que z es un tiburón o bien z protege al pez y.<br />
* No hay ningún pez que se coma a todos los demás.<br />
* Ningún pez protege a ningún otro.<br />
* Por lo tanto, existe algún tiburón en la pecera.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
----<br />
'''Ejercicio 4.''' Formalizar el siguiente argumento (ejercicio 21 de LPO de APPLI2), verificando la corrección de la solución:<br />
<br />
* Carlos afeita a todos los habitantes de Las Chinas que no se afeitan a sí mismo y sólo a ellos.<br />
* Carlos es un habitante de las Chinas.<br />
* Por lo tanto, Carlos no afeita a nadie.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
<br />
----<br />
'''Ejercicio 5.''' <br />
Sea F la fórmula P(x) → P (a) , donde a es un símbolo de constante. Dar un<br />
ejemplo de una interpretación en la que F sea verdadera. Y un ejemplo de una<br />
interpretación en la que F sea falsa.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*Verdadera<br />
<br />
U={1}<br />
I(P)={1}<br />
I(a)=1<br />
<br />
*Falsa<br />
<br />
U={1,2}<br />
I(P)={1}<br />
I(a)=2<br />
<br />
----<br />
'''Ejercicio 6.''' <br />
Sea L un lenguaje de primer orden con dos símbolos de predicado, P (de aridad<br />
1), Q (de aridad 2) y un símbolo de función, f , de aridad 1. Sea I = ( U, I )<br />
la estructura dada por:<br />
U = { a, b, c, d } ;<br />
I ( P ) = { a, b } ,<br />
I ( Q ) = {( a, b ) , ( b, b ) , ( c, b )} ,<br />
I ( f ) = {( a, b ) , ( b, b ) , ( c, a ) , ( d, c )} .<br />
¿Cuál es el valor de cada una de las siguientes fórmulas en dicha estructura? <br />
* P ( x ) → ∃ yQ ( y, x ) .<br />
*. ∀ xQ ( f ( x ) , x ) .<br />
* Q ( f ( x ) , x ) → Q ( x, x ) .<br />
* Q ( x, y ) → P ( x ) .<br />
----<br />
<br />
'''Solución:'''<br />
<br />
joslopjim4<br />
<br />
*<br />
A1(x)=a -> I(F)=0<br />
A2(x)=b -> I(F)=1<br />
A3(x)=c -> I(F)=1<br />
A4(x)=d -> I(F)=1<br />
<br />
*<br />
I(F)=0 si x=a<br />
<br />
*<br />
A1(x)=a -> I(F)=1<br />
A2(x)=b -> I(F)=1<br />
A3(x)=c -> I(F)=1<br />
A4(x)=d -> I(F)=1<br />
<br />
*<br />
A1(x)=a, A1(y)=a -> I(F)=1<br />
A1(x)=a, A2(y)=b -> I(F)=1<br />
A1(x)=a, A3(y)=c -> I(F)=1<br />
A1(x)=a, A4(y)=d -> I(F)=1<br />
<br />
A2(x)=b, A1(y)=a -> I(F)=1<br />
A2(x)=b, A2(y)=b -> I(F)=1<br />
A2(x)=b, A3(y)=c -> I(F)=1<br />
A2(x)=b, A4(y)=d -> I(F)=1<br />
<br />
A3(x)=c, A1(y)=a -> I(F)=1<br />
A3(x)=c, A2(y)=b -> I(F)=0<br />
A3(x)=c, A3(y)=c -> I(F)=1<br />
A3(x)=c, A4(y)=d -> I(F)=1<br />
<br />
A4(x)=d, A1(y)=a -> I(F)=1<br />
A4(x)=d, A2(y)=b -> I(F)=1<br />
A4(x)=d, A3(y)=c -> I(F)=1<br />
A4(x)=d, A4(y)=d -> I(F)=1<br />
<br />
----<br />
'''Ejercicio 7.'''<br />
<br />
En el lenguaje con igualdad L = { a, f } , siendo f un símbolo de función<br />
de aridad 1 y a una constante, se consideran las siguientes fórmulas:<br />
* F₁ : = ∀ x [ f ( x ) ≠ a ] ,<br />
* F₂ : = ∀ x ∀ y [ f ( x ) = f ( y ) → x = y ] ,<br />
* F₃ : = ∀ x [ x ≠ a → ∃ y [ f ( y ) = x ]] .<br />
<br />
Probar que ninguna de estas fórmulas es consecuencia lógica de las dos restantes.<br />
----<br />
<br />
'''Solución:'''<br />
<br />
joslopjim4<br />
<br />
U=naturales<br />
I(a)=0<br />
I(f)=x+2<br />
<br />
----<br />
'''Ejercicio 8.''' Dar una fórmula F, tal que todo modelo de F tenga al menos 3 elementos. Generalizarlo a n cualquiera.<br />
<br />
----<br />
<br />
'''Solución:'''</div>
Joslopjim4
https://www.glc.us.es/~jalonso/LMF2018/index.php?title=Relaci%C3%B3n_5&diff=151
Relación 5
2018-03-24T20:10:41Z
<p>Joslopjim4: </p>
<hr />
<div>=== Relación 5: Sintaxis y semántica de la Lógica de primer orden ===<br />
<br />
<br />
----<br />
'''Ejercicio 1.''' Formalizar el siguiente argumento (ejercicio 5 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Todo aquel que entre en el país y no sea un VIP será cacheado por un aduanero.<br />
* Hay un contrabandista que entra en el país y que solo podrá ser cacheado por contrabandistas.<br />
* Ningún contrabandista es un VIP.<br />
* Por lo tanto, algún aduanero es contrabandista.<br />
----<br />
<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*all x ( entra ( x ) & ( -vip ( x ) ) ) -> ( exists y ( aduanero ( y ) & cachea ( y , x ) ) ) (Da error sintáctico)<br />
* exists x (contrabandista(x) & entra(x) & ( all y (cachea(y,x) & contrabandista(y))))<br />
* all x (contrabandista(x) -> (- vip(x)))<br />
*exists x (aduanero(x) & contrabandista(x))<br />
<br />
----<br />
'''Ejercicio 2.''' Formalizar el siguiente argumento (ejercicio 15 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Todos los robots obedecen a los amigos del programador jefe.<br />
* Alvaro es amigo del programador jefe.<br />
* Benito no obedece a Alvaro.<br />
* Benito no es un robot.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*<br />
*amigo(Alvaro)<br />
*-obedece(Benito,Alvaro)<br />
*-robot(Benito)<br />
<br />
----<br />
'''Ejercicio 3.''' Formalizar el siguiente argumento (ejercicio 16 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Hay algún pez x que para cualquier pez y, si el pez x no se come al pez y entonces existe un pez z tal que z es un tiburón o bien z protege al pez y.<br />
* No hay ningún pez que se coma a todos los demás.<br />
* Ningún pez protege a ningún otro.<br />
* Por lo tanto, existe algún tiburón en la pecera.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
----<br />
'''Ejercicio 4.''' Formalizar el siguiente argumento (ejercicio 21 de LPO de APPLI2), verificando la corrección de la solución:<br />
<br />
* Carlos afeita a todos los habitantes de Las Chinas que no se afeitan a sí mismo y sólo a ellos.<br />
* Carlos es un habitante de las Chinas.<br />
* Por lo tanto, Carlos no afeita a nadie.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
<br />
----<br />
'''Ejercicio 5.''' <br />
Sea F la fórmula P(x) → P (a) , donde a es un símbolo de constante. Dar un<br />
ejemplo de una interpretación en la que F sea verdadera. Y un ejemplo de una<br />
interpretación en la que F sea falsa.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*Verdadera<br />
<br />
U={1}<br />
I(P)={1}<br />
I(a)=1<br />
<br />
*Falsa<br />
<br />
U={1,2}<br />
I(P)={1}<br />
I(a)=2<br />
<br />
----<br />
'''Ejercicio 6.''' <br />
Sea L un lenguaje de primer orden con dos símbolos de predicado, P (de aridad<br />
1), Q (de aridad 2) y un símbolo de función, f , de aridad 1. Sea I = ( U, I )<br />
la estructura dada por:<br />
U = { a, b, c, d } ;<br />
I ( P ) = { a, b } ,<br />
I ( Q ) = {( a, b ) , ( b, b ) , ( c, b )} ,<br />
I ( f ) = {( a, b ) , ( b, b ) , ( c, a ) , ( d, c )} .<br />
¿Cuál es el valor de cada una de las siguientes fórmulas en dicha estructura? <br />
* P ( x ) → ∃ yQ ( y, x ) .<br />
*. ∀ xQ ( f ( x ) , x ) .<br />
* Q ( f ( x ) , x ) → Q ( x, x ) .<br />
* Q ( x, y ) → P ( x ) .<br />
----<br />
<br />
'''Solución:'''<br />
<br />
joslopjim4<br />
<br />
*<br />
A1(x)=a -> I(F)=0<br />
A2(x)=b -> I(F)=1<br />
A3(x)=c -> I(F)=1<br />
A4(x)=d -> I(F)=1<br />
<br />
*<br />
I(F)=0 si x=a<br />
<br />
*<br />
A1(x)=a -> I(F)=1<br />
A2(x)=b -> I(F)=1<br />
A3(x)=c -> I(F)=1<br />
A4(x)=d -> I(F)=1<br />
<br />
*<br />
A1(x)=a, A1(y)=a -> I(F)=1<br />
A1(x)=a, A2(y)=b -> I(F)=1<br />
A1(x)=a, A3(y)=c -> I(F)=1<br />
A1(x)=a, A4(y)=d -> I(F)=1<br />
<br />
A2(x)=b, A1(y)=a -> I(F)=1<br />
A2(x)=b, A2(y)=b -> I(F)=1<br />
A2(x)=b, A3(y)=c -> I(F)=1<br />
A2(x)=b, A4(y)=d -> I(F)=1<br />
<br />
A3(x)=c, A1(y)=a -> I(F)=1<br />
A3(x)=c, A2(y)=b -> I(F)=0<br />
A3(x)=c, A3(y)=c -> I(F)=1<br />
A3(x)=c, A4(y)=d -> I(F)=1<br />
<br />
A4(x)=d, A1(y)=a -> I(F)=1<br />
A4(x)=d, A2(y)=b -> I(F)=1<br />
A4(x)=d, A3(y)=c -> I(F)=1<br />
A4(x)=d, A4(y)=d -> I(F)=1<br />
<br />
----<br />
'''Ejercicio 7.'''<br />
<br />
En el lenguaje con igualdad L = { a, f } , siendo f un símbolo de función<br />
de aridad 1 y a una constante, se consideran las siguientes fórmulas:<br />
* F₁ : = ∀ x [ f ( x ) ≠ a ] ,<br />
* F₂ : = ∀ x ∀ y [ f ( x ) = f ( y ) → x = y ] ,<br />
* F₃ : = ∀ x [ x ≠ a → ∃ y [ f ( y ) = x ]] .<br />
<br />
Probar que ninguna de estas fórmulas es consecuencia lógica de las dos restantes.<br />
----<br />
<br />
'''Solución:'''<br />
<br />
joslopjim4<br />
<br />
U=naturales<br />
I(a)=0<br />
I(f)=x+2<br />
<br />
----<br />
'''Ejercicio 8.''' Dar una fórmula F, tal que todo modelo de F tenga al menos 3 elementos. Generalizarlo a n cualquiera.<br />
<br />
----<br />
<br />
'''Solución:'''</div>
Joslopjim4
https://www.glc.us.es/~jalonso/LMF2018/index.php?title=Relaci%C3%B3n_5&diff=150
Relación 5
2018-03-24T20:09:32Z
<p>Joslopjim4: </p>
<hr />
<div>=== Relación 5: Sintaxis y semántica de la Lógica de primer orden ===<br />
<br />
<br />
----<br />
'''Ejercicio 1.''' Formalizar el siguiente argumento (ejercicio 5 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Todo aquel que entre en el país y no sea un VIP será cacheado por un aduanero.<br />
* Hay un contrabandista que entra en el país y que solo podrá ser cacheado por contrabandistas.<br />
* Ningún contrabandista es un VIP.<br />
* Por lo tanto, algún aduanero es contrabandista.<br />
----<br />
<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*all x ( entra ( x ) & ( -vip ( x ) ) ) -> ( exists y ( aduanero ( y ) & cachea ( y , x ) ) ) (Da error sintáctico)<br />
* exists x (contrabandista(x) & entra(x) & ( all y (cachea(y,x) & contrabandista(y))))<br />
* all x (contrabandista(x) -> (- vip(x)))<br />
*exists x (aduanero(x) & contrabandista(x))<br />
<br />
----<br />
'''Ejercicio 2.''' Formalizar el siguiente argumento (ejercicio 15 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Todos los robots obedecen a los amigos del programador jefe.<br />
* Alvaro es amigo del programador jefe.<br />
* Benito no obedece a Alvaro.<br />
* Benito no es un robot.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*<br />
*amigo(Alvaro)<br />
*-obedece(Benito,Alvaro)<br />
*-robot(Benito)<br />
<br />
----<br />
'''Ejercicio 3.''' Formalizar el siguiente argumento (ejercicio 16 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Hay algún pez x que para cualquier pez y, si el pez x no se come al pez y entonces existe un pez z tal que z es un tiburón o bien z protege al pez y.<br />
* No hay ningún pez que se coma a todos los demás.<br />
* Ningún pez protege a ningún otro.<br />
* Por lo tanto, existe algún tiburón en la pecera.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
----<br />
'''Ejercicio 4.''' Formalizar el siguiente argumento (ejercicio 21 de LPO de APPLI2), verificando la corrección de la solución:<br />
<br />
* Carlos afeita a todos los habitantes de Las Chinas que no se afeitan a sí mismo y sólo a ellos.<br />
* Carlos es un habitante de las Chinas.<br />
* Por lo tanto, Carlos no afeita a nadie.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
<br />
----<br />
'''Ejercicio 5.''' <br />
Sea F la fórmula P(x) → P (a) , donde a es un símbolo de constante. Dar un<br />
ejemplo de una interpretación en la que F sea verdadera. Y un ejemplo de una<br />
interpretación en la que F sea falsa.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*Verdadera<br />
<br />
U={1}<br />
I(P)={1}<br />
I(a)=1<br />
<br />
*Falsa<br />
<br />
U={1,2}<br />
I(P)={1}<br />
I(a)=2<br />
<br />
----<br />
'''Ejercicio 6.''' <br />
Sea L un lenguaje de primer orden con dos símbolos de predicado, P (de aridad<br />
1), Q (de aridad 2) y un símbolo de función, f , de aridad 1. Sea I = ( U, I )<br />
la estructura dada por:<br />
U = { a, b, c, d } ;<br />
I ( P ) = { a, b } ,<br />
I ( Q ) = {( a, b ) , ( b, b ) , ( c, b )} ,<br />
I ( f ) = {( a, b ) , ( b, b ) , ( c, a ) , ( d, c )} .<br />
¿Cuál es el valor de cada una de las siguientes fórmulas en dicha estructura? <br />
* P ( x ) → ∃ yQ ( y, x ) .<br />
*. ∀ xQ ( f ( x ) , x ) .<br />
* Q ( f ( x ) , x ) → Q ( x, x ) .<br />
* Q ( x, y ) → P ( x ) .<br />
----<br />
<br />
'''Solución:'''<br />
<br />
joslopjim4<br />
<br />
*<br />
A1(x)=a -> I(F)=0<br />
A2(x)=b -> I(F)=1<br />
A3(x)=c -> I(F)=1<br />
A4(x)=d -> I(F)=1<br />
<br />
*<br />
I(F)=0 si x=a<br />
<br />
*<br />
A1(x)=a -> I(F)=1<br />
A2(x)=b -> I(F)=1<br />
A3(x)=c -> I(F)=1<br />
A4(x)=d -> I(F)=1<br />
<br />
A1(x)=a, A1(y)=a -> I(F)=1<br />
A1(x)=a, A2(y)=b -> I(F)=1<br />
A1(x)=a, A3(y)=c -> I(F)=1<br />
A1(x)=a, A4(y)=d -> I(F)=1<br />
<br />
A2(x)=b, A1(y)=a -> I(F)=1<br />
A2(x)=b, A2(y)=b -> I(F)=1<br />
A2(x)=b, A3(y)=c -> I(F)=1<br />
A2(x)=b, A4(y)=d -> I(F)=1<br />
<br />
A3(x)=c, A1(y)=a -> I(F)=1<br />
A3(x)=c, A2(y)=b -> I(F)=0<br />
A3(x)=c, A3(y)=c -> I(F)=1<br />
A3(x)=c, A4(y)=d -> I(F)=1<br />
<br />
A4(x)=d, A1(y)=a -> I(F)=1<br />
A4(x)=d, A2(y)=b -> I(F)=1<br />
A4(x)=d, A3(y)=c -> I(F)=1<br />
A4(x)=d, A4(y)=d -> I(F)=1<br />
<br />
----<br />
'''Ejercicio 7.'''<br />
<br />
En el lenguaje con igualdad L = { a, f } , siendo f un símbolo de función<br />
de aridad 1 y a una constante, se consideran las siguientes fórmulas:<br />
* F₁ : = ∀ x [ f ( x ) ≠ a ] ,<br />
* F₂ : = ∀ x ∀ y [ f ( x ) = f ( y ) → x = y ] ,<br />
* F₃ : = ∀ x [ x ≠ a → ∃ y [ f ( y ) = x ]] .<br />
<br />
Probar que ninguna de estas fórmulas es consecuencia lógica de las dos restantes.<br />
----<br />
<br />
'''Solución:'''<br />
<br />
joslopjim4<br />
<br />
U=naturales<br />
I(a)=0<br />
I(f)=x+2<br />
<br />
----<br />
'''Ejercicio 8.''' Dar una fórmula F, tal que todo modelo de F tenga al menos 3 elementos. Generalizarlo a n cualquiera.<br />
<br />
----<br />
<br />
'''Solución:'''</div>
Joslopjim4
https://www.glc.us.es/~jalonso/LMF2018/index.php?title=Relaci%C3%B3n_5&diff=149
Relación 5
2018-03-24T20:08:48Z
<p>Joslopjim4: </p>
<hr />
<div>=== Relación 5: Sintaxis y semántica de la Lógica de primer orden ===<br />
<br />
<br />
----<br />
'''Ejercicio 1.''' Formalizar el siguiente argumento (ejercicio 5 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Todo aquel que entre en el país y no sea un VIP será cacheado por un aduanero.<br />
* Hay un contrabandista que entra en el país y que solo podrá ser cacheado por contrabandistas.<br />
* Ningún contrabandista es un VIP.<br />
* Por lo tanto, algún aduanero es contrabandista.<br />
----<br />
<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*all x ( entra ( x ) & ( -vip ( x ) ) ) -> ( exists y ( aduanero ( y ) & cachea ( y , x ) ) ) (Da error sintáctico)<br />
* exists x (contrabandista(x) & entra(x) & ( all y (cachea(y,x) & contrabandista(y))))<br />
* all x (contrabandista(x) -> (- vip(x)))<br />
*exists x (aduanero(x) & contrabandista(x))<br />
<br />
----<br />
'''Ejercicio 2.''' Formalizar el siguiente argumento (ejercicio 15 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Todos los robots obedecen a los amigos del programador jefe.<br />
* Alvaro es amigo del programador jefe.<br />
* Benito no obedece a Alvaro.<br />
* Benito no es un robot.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*<br />
*amigo(Alvaro)<br />
*-obedece(Benito,Alvaro)<br />
*-robot(Benito)<br />
<br />
----<br />
'''Ejercicio 3.''' Formalizar el siguiente argumento (ejercicio 16 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Hay algún pez x que para cualquier pez y, si el pez x no se come al pez y entonces existe un pez z tal que z es un tiburón o bien z protege al pez y.<br />
* No hay ningún pez que se coma a todos los demás.<br />
* Ningún pez protege a ningún otro.<br />
* Por lo tanto, existe algún tiburón en la pecera.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
----<br />
'''Ejercicio 4.''' Formalizar el siguiente argumento (ejercicio 21 de LPO de APPLI2), verificando la corrección de la solución:<br />
<br />
* Carlos afeita a todos los habitantes de Las Chinas que no se afeitan a sí mismo y sólo a ellos.<br />
* Carlos es un habitante de las Chinas.<br />
* Por lo tanto, Carlos no afeita a nadie.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
<br />
----<br />
'''Ejercicio 5.''' <br />
Sea F la fórmula P(x) → P (a) , donde a es un símbolo de constante. Dar un<br />
ejemplo de una interpretación en la que F sea verdadera. Y un ejemplo de una<br />
interpretación en la que F sea falsa.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*Verdadera<br />
<br />
U={1}<br />
I(P)={1}<br />
I(a)=1<br />
<br />
*Falsa<br />
<br />
U={1,2}<br />
I(P)={1}<br />
I(a)=2<br />
<br />
----<br />
'''Ejercicio 6.''' <br />
Sea L un lenguaje de primer orden con dos símbolos de predicado, P (de aridad<br />
1), Q (de aridad 2) y un símbolo de función, f , de aridad 1. Sea I = ( U, I )<br />
la estructura dada por:<br />
U = { a, b, c, d } ;<br />
I ( P ) = { a, b } ,<br />
I ( Q ) = {( a, b ) , ( b, b ) , ( c, b )} ,<br />
I ( f ) = {( a, b ) , ( b, b ) , ( c, a ) , ( d, c )} .<br />
¿Cuál es el valor de cada una de las siguientes fórmulas en dicha estructura? <br />
* P ( x ) → ∃ yQ ( y, x ) .<br />
*. ∀ xQ ( f ( x ) , x ) .<br />
* Q ( f ( x ) , x ) → Q ( x, x ) .<br />
* Q ( x, y ) → P ( x ) .<br />
----<br />
<br />
'''Solución:'''<br />
<br />
joslopjim4<br />
<br />
*<br />
A1(x)=a -> I(F)=0<br />
A2(x)=b -> I(F)=1<br />
A3(x)=c -> I(F)=1<br />
A4(x)=d -> I(F)=1<br />
<br />
*<br />
I(F)=0 si x=a<br />
<br />
*<br />
A1(x)=a -> I(F)=1<br />
A2(x)=b -> I(F)=1<br />
A3(x)=c -> I(F)=1<br />
A4(x)=d -> I(F)=1<br />
<br />
A1(x)=a, A1(y)=a -> I(F)=1<br />
A1(x)=a, A2(y)=b -> I(F)=1<br />
A1(x)=a, A3(y)=c -> I(F)=1<br />
A1(x)=a, A4(y)=d -> I(F)=1<br />
<br />
A2(x)=b, A1(y)=a -> I(F)=1<br />
A2(x)=b, A2(y)=b -> I(F)=1<br />
A2(x)=b, A3(y)=c -> I(F)=1<br />
A2(x)=b, A4(y)=d -> I(F)=1<br />
<br />
A3(x)=c, A1(y)=a -> I(F)=1<br />
A3(x)=c, A2(y)=b -> I(F)=0<br />
A3(x)=c, A3(y)=c -> I(F)=1<br />
A3(x)=c, A4(y)=d -> I(F)=1<br />
<br />
A4(x)=d, A1(y)=a -> I(F)=1<br />
A4(x)=d, A2(y)=b -> I(F)=1<br />
A4(x)=d, A3(y)=c -> I(F)=1<br />
A4(x)=d, A4(y)=d -> I(F)=1<br />
<br />
----<br />
'''Ejercicio 7.'''<br />
<br />
En el lenguaje con igualdad L = { a, f } , siendo f un símbolo de función<br />
de aridad 1 y a una constante, se consideran las siguientes fórmulas:<br />
* F₁ : = ∀ x [ f ( x ) ≠ a ] ,<br />
* F₂ : = ∀ x ∀ y [ f ( x ) = f ( y ) → x = y ] ,<br />
* F₃ : = ∀ x [ x ≠ a → ∃ y [ f ( y ) = x ]] .<br />
<br />
Probar que ninguna de estas fórmulas es consecuencia lógica de las dos restantes.<br />
----<br />
<br />
'''Solución:'''<br />
<br />
joslopjim4<br />
<br />
U=naturales<br />
I(a)=0<br />
I(f)=x+2<br />
<br />
----<br />
'''Ejercicio 8.''' Dar una fórmula F, tal que todo modelo de F tenga al menos 3 elementos. Generalizarlo a n cualquiera.<br />
<br />
----<br />
<br />
'''Solución:'''</div>
Joslopjim4
https://www.glc.us.es/~jalonso/LMF2018/index.php?title=Relaci%C3%B3n_5&diff=148
Relación 5
2018-03-24T20:07:01Z
<p>Joslopjim4: </p>
<hr />
<div>=== Relación 5: Sintaxis y semántica de la Lógica de primer orden ===<br />
<br />
<br />
----<br />
'''Ejercicio 1.''' Formalizar el siguiente argumento (ejercicio 5 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Todo aquel que entre en el país y no sea un VIP será cacheado por un aduanero.<br />
* Hay un contrabandista que entra en el país y que solo podrá ser cacheado por contrabandistas.<br />
* Ningún contrabandista es un VIP.<br />
* Por lo tanto, algún aduanero es contrabandista.<br />
----<br />
<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*all x ( entra ( x ) & ( -vip ( x ) ) ) -> ( exists y ( aduanero ( y ) & cachea ( y , x ) ) ) (Da error sintáctico)<br />
* exists x (contrabandista(x) & entra(x) & ( all y (cachea(y,x) & contrabandista(y))))<br />
* all x (contrabandista(x) -> (- vip(x)))<br />
*exists x (aduanero(x) & contrabandista(x))<br />
<br />
----<br />
'''Ejercicio 2.''' Formalizar el siguiente argumento (ejercicio 15 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Todos los robots obedecen a los amigos del programador jefe.<br />
* Alvaro es amigo del programador jefe.<br />
* Benito no obedece a Alvaro.<br />
* Benito no es un robot.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*<br />
*amigo(Alvaro)<br />
*-obedece(Benito,Alvaro)<br />
*-robot(Benito)<br />
<br />
----<br />
'''Ejercicio 3.''' Formalizar el siguiente argumento (ejercicio 16 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Hay algún pez x que para cualquier pez y, si el pez x no se come al pez y entonces existe un pez z tal que z es un tiburón o bien z protege al pez y.<br />
* No hay ningún pez que se coma a todos los demás.<br />
* Ningún pez protege a ningún otro.<br />
* Por lo tanto, existe algún tiburón en la pecera.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
----<br />
'''Ejercicio 4.''' Formalizar el siguiente argumento (ejercicio 21 de LPO de APPLI2), verificando la corrección de la solución:<br />
<br />
* Carlos afeita a todos los habitantes de Las Chinas que no se afeitan a sí mismo y sólo a ellos.<br />
* Carlos es un habitante de las Chinas.<br />
* Por lo tanto, Carlos no afeita a nadie.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
<br />
----<br />
'''Ejercicio 5.''' <br />
Sea F la fórmula P(x) → P (a) , donde a es un símbolo de constante. Dar un<br />
ejemplo de una interpretación en la que F sea verdadera. Y un ejemplo de una<br />
interpretación en la que F sea falsa.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*Verdadera<br />
<br />
U={1}<br />
I(P)={1}<br />
I(a)=1<br />
<br />
*Falsa<br />
<br />
U={1,2}<br />
I(P)={1}<br />
I(a)=2<br />
<br />
----<br />
'''Ejercicio 6.''' <br />
Sea L un lenguaje de primer orden con dos símbolos de predicado, P (de aridad<br />
1), Q (de aridad 2) y un símbolo de función, f , de aridad 1. Sea I = ( U, I )<br />
la estructura dada por:<br />
U = { a, b, c, d } ;<br />
I ( P ) = { a, b } ,<br />
I ( Q ) = {( a, b ) , ( b, b ) , ( c, b )} ,<br />
I ( f ) = {( a, b ) , ( b, b ) , ( c, a ) , ( d, c )} .<br />
¿Cuál es el valor de cada una de las siguientes fórmulas en dicha estructura? <br />
* P ( x ) → ∃ yQ ( y, x ) .<br />
*. ∀ xQ ( f ( x ) , x ) .<br />
* Q ( f ( x ) , x ) → Q ( x, x ) .<br />
* Q ( x, y ) → P ( x ) .<br />
----<br />
<br />
'''Solución:'''<br />
<br />
joslopjim4<br />
<br />
*A1(x)=a -> I(F)=0<br />
A2(x)=b -> I(F)=1<br />
A3(x)=c -> I(F)=1<br />
A4(x)=d -> I(F)=1<br />
<br />
*I(F)=0 si x=a<br />
<br />
*A1(x)=a -> I(F)=1<br />
A2(x)=b -> I(F)=1<br />
A3(x)=c -> I(F)=1<br />
A4(x)=d -> I(F)=1<br />
<br />
*A1(x)=a, A1(y)=a -> I(F)=1<br />
A1(x)=a, A2(y)=b -> I(F)=1<br />
A1(x)=a, A3(y)=c -> I(F)=1<br />
A1(x)=a, A4(y)=d -> I(F)=1<br />
<br />
A2(x)=b, A1(y)=a -> I(F)=1<br />
A2(x)=b, A2(y)=b -> I(F)=1<br />
A2(x)=b, A3(y)=c -> I(F)=1<br />
A2(x)=b, A4(y)=d -> I(F)=1<br />
<br />
A3(x)=c, A1(y)=a -> I(F)=1<br />
A3(x)=c, A2(y)=b -> I(F)=0<br />
A3(x)=c, A3(y)=c -> I(F)=1<br />
A3(x)=c, A4(y)=d -> I(F)=1<br />
<br />
A4(x)=d, A1(y)=a -> I(F)=1<br />
A4(x)=d, A2(y)=b -> I(F)=1<br />
A4(x)=d, A3(y)=c -> I(F)=1<br />
A4(x)=d, A4(y)=d -> I(F)=1<br />
<br />
----<br />
'''Ejercicio 7.'''<br />
<br />
En el lenguaje con igualdad L = { a, f } , siendo f un símbolo de función<br />
de aridad 1 y a una constante, se consideran las siguientes fórmulas:<br />
* F₁ : = ∀ x [ f ( x ) ≠ a ] ,<br />
* F₂ : = ∀ x ∀ y [ f ( x ) = f ( y ) → x = y ] ,<br />
* F₃ : = ∀ x [ x ≠ a → ∃ y [ f ( y ) = x ]] .<br />
<br />
Probar que ninguna de estas fórmulas es consecuencia lógica de las dos restantes.<br />
----<br />
<br />
'''Solución:'''<br />
<br />
joslopjim4<br />
<br />
U=naturales<br />
I(a)=0<br />
I(f)=x+2<br />
<br />
----<br />
'''Ejercicio 8.''' Dar una fórmula F, tal que todo modelo de F tenga al menos 3 elementos. Generalizarlo a n cualquiera.<br />
<br />
----<br />
<br />
'''Solución:'''</div>
Joslopjim4
https://www.glc.us.es/~jalonso/LMF2018/index.php?title=Relaci%C3%B3n_5&diff=147
Relación 5
2018-03-24T19:20:59Z
<p>Joslopjim4: </p>
<hr />
<div>=== Relación 5: Sintaxis y semántica de la Lógica de primer orden ===<br />
<br />
<br />
----<br />
'''Ejercicio 1.''' Formalizar el siguiente argumento (ejercicio 5 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Todo aquel que entre en el país y no sea un VIP será cacheado por un aduanero.<br />
* Hay un contrabandista que entra en el país y que solo podrá ser cacheado por contrabandistas.<br />
* Ningún contrabandista es un VIP.<br />
* Por lo tanto, algún aduanero es contrabandista.<br />
----<br />
<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*all x ( entra ( x ) & ( -vip ( x ) ) ) -> ( exists y ( aduanero ( y ) & cachea ( y , x ) ) ) (Da error sintáctico)<br />
* exists x (contrabandista(x) & entra(x) & ( all y (cachea(y,x) & contrabandista(y))))<br />
* all x (contrabandista(x) -> (- vip(x)))<br />
*exists x (aduanero(x) & contrabandista(x))<br />
<br />
----<br />
'''Ejercicio 2.''' Formalizar el siguiente argumento (ejercicio 15 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Todos los robots obedecen a los amigos del programador jefe.<br />
* Alvaro es amigo del programador jefe.<br />
* Benito no obedece a Alvaro.<br />
* Benito no es un robot.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*<br />
*amigo(Alvaro)<br />
*-obedece(Benito,Alvaro)<br />
*-robot(Benito)<br />
<br />
----<br />
'''Ejercicio 3.''' Formalizar el siguiente argumento (ejercicio 16 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Hay algún pez x que para cualquier pez y, si el pez x no se come al pez y entonces existe un pez z tal que z es un tiburón o bien z protege al pez y.<br />
* No hay ningún pez que se coma a todos los demás.<br />
* Ningún pez protege a ningún otro.<br />
* Por lo tanto, existe algún tiburón en la pecera.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
----<br />
'''Ejercicio 4.''' Formalizar el siguiente argumento (ejercicio 21 de LPO de APPLI2), verificando la corrección de la solución:<br />
<br />
* Carlos afeita a todos los habitantes de Las Chinas que no se afeitan a sí mismo y sólo a ellos.<br />
* Carlos es un habitante de las Chinas.<br />
* Por lo tanto, Carlos no afeita a nadie.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
<br />
----<br />
'''Ejercicio 5.''' <br />
Sea F la fórmula P(x) → P (a) , donde a es un símbolo de constante. Dar un<br />
ejemplo de una interpretación en la que F sea verdadera. Y un ejemplo de una<br />
interpretación en la que F sea falsa.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
----<br />
'''Ejercicio 6.''' <br />
Sea L un lenguaje de primer orden con dos símbolos de predicado, P (de aridad<br />
1), Q (de aridad 2) y un símbolo de función, f , de aridad 1. Sea I = ( U, I )<br />
la estructura dada por:<br />
U = { a, b, c, d } ;<br />
I ( P ) = { a, b } ,<br />
I ( Q ) = {( a, b ) , ( b, b ) , ( c, b )} ,<br />
I ( f ) = {( a, b ) , ( b, b ) , ( c, a ) , ( d, c )} .<br />
¿Cuál es el valor de cada una de las siguientes fórmulas en dicha estructura? <br />
* P ( x ) → ∃ yQ ( y, x ) .<br />
*. ∀ xQ ( f ( x ) , x ) .<br />
* Q ( f ( x ) , x ) → Q ( x, x ) .<br />
* Q ( x, y ) → P ( x ) .<br />
----<br />
<br />
'''Solución:'''<br />
<br />
----<br />
'''Ejercicio 7.'''<br />
<br />
En el lenguaje con igualdad L = { a, f } , siendo f un símbolo de función<br />
de aridad 1 y a una constante, se consideran las siguientes fórmulas:<br />
* F₁ : = ∀ x [ f ( x ) ≠ a ] ,<br />
* F₂ : = ∀ x ∀ y [ f ( x ) = f ( y ) → x = y ] ,<br />
* F₃ : = ∀ x [ x ≠ a → ∃ y [ f ( y ) = x ]] .<br />
<br />
Probar que ninguna de estas fórmulas es consecuencia lógica de las dos restantes.<br />
----<br />
<br />
'''Solución:'''<br />
<br />
----<br />
'''Ejercicio 8.''' Dar una fórmula F, tal que todo modelo de F tenga al menos 3 elementos. Generalizarlo a n cualquiera.<br />
<br />
----<br />
<br />
'''Solución:'''</div>
Joslopjim4
https://www.glc.us.es/~jalonso/LMF2018/index.php?title=Relaci%C3%B3n_5&diff=146
Relación 5
2018-03-24T19:00:52Z
<p>Joslopjim4: </p>
<hr />
<div>=== Relación 5: Sintaxis y semántica de la Lógica de primer orden ===<br />
<br />
<br />
----<br />
'''Ejercicio 1.''' Formalizar el siguiente argumento (ejercicio 5 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Todo aquel que entre en el país y no sea un VIP será cacheado por un aduanero.<br />
* Hay un contrabandista que entra en el país y que solo podrá ser cacheado por contrabandistas.<br />
* Ningún contrabandista es un VIP.<br />
* Por lo tanto, algún aduanero es contrabandista.<br />
----<br />
<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*all x ( entra ( x ) & ( -vip ( x ) ) ) -> ( exists y ( aduanero ( y ) & cachea ( y , x ) ) ) (Da error sintáctico)<br />
* exists x (contrabandista(x) & entra(x) & ( all y (cachea(y,x) & contrabandista(y))))<br />
* all x (contrabandista(x) -> (- vip(x)))<br />
*exists x (aduanero(x) & contrabandista(x))<br />
<br />
----<br />
'''Ejercicio 2.''' Formalizar el siguiente argumento (ejercicio 15 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Todos los robots obedecen a los amigos del programador jefe.<br />
* Alvaro es amigo del programador jefe.<br />
* Benito no obedece a Alvaro.<br />
* Benito no es un robot.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
<br />
----<br />
'''Ejercicio 3.''' Formalizar el siguiente argumento (ejercicio 16 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Hay algún pez x que para cualquier pez y, si el pez x no se come al pez y entonces existe un pez z tal que z es un tiburón o bien z protege al pez y.<br />
* No hay ningún pez que se coma a todos los demás.<br />
* Ningún pez protege a ningún otro.<br />
* Por lo tanto, existe algún tiburón en la pecera.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
<br />
<br />
----<br />
'''Ejercicio 4.''' Formalizar el siguiente argumento (ejercicio 21 de LPO de APPLI2), verificando la corrección de la solución:<br />
<br />
* Carlos afeita a todos los habitantes de Las Chinas que no se afeitan a sí mismo y sólo a ellos.<br />
* Carlos es un habitante de las Chinas.<br />
* Por lo tanto, Carlos no afeita a nadie.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
<br />
----<br />
'''Ejercicio 5.''' <br />
Sea F la fórmula P(x) → P (a) , donde a es un símbolo de constante. Dar un<br />
ejemplo de una interpretación en la que F sea verdadera. Y un ejemplo de una<br />
interpretación en la que F sea falsa.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
----<br />
'''Ejercicio 6.''' <br />
Sea L un lenguaje de primer orden con dos símbolos de predicado, P (de aridad<br />
1), Q (de aridad 2) y un símbolo de función, f , de aridad 1. Sea I = ( U, I )<br />
la estructura dada por:<br />
U = { a, b, c, d } ;<br />
I ( P ) = { a, b } ,<br />
I ( Q ) = {( a, b ) , ( b, b ) , ( c, b )} ,<br />
I ( f ) = {( a, b ) , ( b, b ) , ( c, a ) , ( d, c )} .<br />
¿Cuál es el valor de cada una de las siguientes fórmulas en dicha estructura? <br />
* P ( x ) → ∃ yQ ( y, x ) .<br />
*. ∀ xQ ( f ( x ) , x ) .<br />
* Q ( f ( x ) , x ) → Q ( x, x ) .<br />
* Q ( x, y ) → P ( x ) .<br />
----<br />
<br />
'''Solución:'''<br />
<br />
----<br />
'''Ejercicio 7.'''<br />
<br />
En el lenguaje con igualdad L = { a, f } , siendo f un símbolo de función<br />
de aridad 1 y a una constante, se consideran las siguientes fórmulas:<br />
* F₁ : = ∀ x [ f ( x ) ≠ a ] ,<br />
* F₂ : = ∀ x ∀ y [ f ( x ) = f ( y ) → x = y ] ,<br />
* F₃ : = ∀ x [ x ≠ a → ∃ y [ f ( y ) = x ]] .<br />
<br />
Probar que ninguna de estas fórmulas es consecuencia lógica de las dos restantes.<br />
----<br />
<br />
'''Solución:'''<br />
<br />
----<br />
'''Ejercicio 8.''' Dar una fórmula F, tal que todo modelo de F tenga al menos 3 elementos. Generalizarlo a n cualquiera.<br />
<br />
----<br />
<br />
'''Solución:'''</div>
Joslopjim4
https://www.glc.us.es/~jalonso/LMF2018/index.php?title=Relaci%C3%B3n_5&diff=145
Relación 5
2018-03-24T19:00:30Z
<p>Joslopjim4: </p>
<hr />
<div>=== Relación 5: Sintaxis y semántica de la Lógica de primer orden ===<br />
<br />
<br />
----<br />
'''Ejercicio 1.''' Formalizar el siguiente argumento (ejercicio 5 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Todo aquel que entre en el país y no sea un VIP será cacheado por un aduanero.<br />
* Hay un contrabandista que entra en el país y que solo podrá ser cacheado por contrabandistas.<br />
* Ningún contrabandista es un VIP.<br />
* Por lo tanto, algún aduanero es contrabandista.<br />
----<br />
<br />
<br />
'''Solución:''' <br />
<br />
joslopjim4<br />
<br />
*all x ( entra ( x ) & ( -vip ( x ) ) ) -> ( exists y ( aduanero ( y ) & cachea ( y , x ) ) ) (Da error sintáctico)<br />
* exists x (contrabandista(x) & entra(x) & ( all y (cachea(y,x) & contrabandista(y))))<br />
* all x (contrabandista(x) -> (- vip(x)))<br />
*exists x (aduanero(x) & contrabandista(x))<br />
----<br />
'''Ejercicio 2.''' Formalizar el siguiente argumento (ejercicio 15 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Todos los robots obedecen a los amigos del programador jefe.<br />
* Alvaro es amigo del programador jefe.<br />
* Benito no obedece a Alvaro.<br />
* Benito no es un robot.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
<br />
----<br />
'''Ejercicio 3.''' Formalizar el siguiente argumento (ejercicio 16 de LPO de<br />
APPLI2), verificando la corrección de la solución:<br />
<br />
* Hay algún pez x que para cualquier pez y, si el pez x no se come al pez y entonces existe un pez z tal que z es un tiburón o bien z protege al pez y.<br />
* No hay ningún pez que se coma a todos los demás.<br />
* Ningún pez protege a ningún otro.<br />
* Por lo tanto, existe algún tiburón en la pecera.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
<br />
<br />
----<br />
'''Ejercicio 4.''' Formalizar el siguiente argumento (ejercicio 21 de LPO de APPLI2), verificando la corrección de la solución:<br />
<br />
* Carlos afeita a todos los habitantes de Las Chinas que no se afeitan a sí mismo y sólo a ellos.<br />
* Carlos es un habitante de las Chinas.<br />
* Por lo tanto, Carlos no afeita a nadie.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
<br />
----<br />
'''Ejercicio 5.''' <br />
Sea F la fórmula P(x) → P (a) , donde a es un símbolo de constante. Dar un<br />
ejemplo de una interpretación en la que F sea verdadera. Y un ejemplo de una<br />
interpretación en la que F sea falsa.<br />
<br />
----<br />
<br />
'''Solución:''' <br />
<br />
----<br />
'''Ejercicio 6.''' <br />
Sea L un lenguaje de primer orden con dos símbolos de predicado, P (de aridad<br />
1), Q (de aridad 2) y un símbolo de función, f , de aridad 1. Sea I = ( U, I )<br />
la estructura dada por:<br />
U = { a, b, c, d } ;<br />
I ( P ) = { a, b } ,<br />
I ( Q ) = {( a, b ) , ( b, b ) , ( c, b )} ,<br />
I ( f ) = {( a, b ) , ( b, b ) , ( c, a ) , ( d, c )} .<br />
¿Cuál es el valor de cada una de las siguientes fórmulas en dicha estructura? <br />
* P ( x ) → ∃ yQ ( y, x ) .<br />
*. ∀ xQ ( f ( x ) , x ) .<br />
* Q ( f ( x ) , x ) → Q ( x, x ) .<br />
* Q ( x, y ) → P ( x ) .<br />
----<br />
<br />
'''Solución:'''<br />
<br />
----<br />
'''Ejercicio 7.'''<br />
<br />
En el lenguaje con igualdad L = { a, f } , siendo f un símbolo de función<br />
de aridad 1 y a una constante, se consideran las siguientes fórmulas:<br />
* F₁ : = ∀ x [ f ( x ) ≠ a ] ,<br />
* F₂ : = ∀ x ∀ y [ f ( x ) = f ( y ) → x = y ] ,<br />
* F₃ : = ∀ x [ x ≠ a → ∃ y [ f ( y ) = x ]] .<br />
<br />
Probar que ninguna de estas fórmulas es consecuencia lógica de las dos restantes.<br />
----<br />
<br />
'''Solución:'''<br />
<br />
----<br />
'''Ejercicio 8.''' Dar una fórmula F, tal que todo modelo de F tenga al menos 3 elementos. Generalizarlo a n cualquiera.<br />
<br />
----<br />
<br />
'''Solución:'''</div>
Joslopjim4
https://www.glc.us.es/~jalonso/LMF2018/index.php?title=Relaci%C3%B3n_4&diff=144
Relación 4
2018-03-24T12:48:18Z
<p>Joslopjim4: </p>
<hr />
<div><source lang = "isar"><br />
<br />
chapter {* R4: Deducción natural proposicional *}<br />
<br />
theory R4<br />
imports Main <br />
begin<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
El objetivo de esta relación es demostrar cada uno de los ejercicios<br />
usando sólo las reglas básicas de deducción natural de la lógica<br />
proposicional (sin usar el método auto).<br />
<br />
Las reglas básicas de la deducción natural son las siguientes:<br />
· conjI: ⟦P; Q⟧ ⟹ P ∧ Q<br />
· conjunct1: P ∧ Q ⟹ P<br />
· conjunct2: P ∧ Q ⟹ Q <br />
· notnotD: ¬¬ P ⟹ P<br />
· notnotI: P ⟹ ¬¬ P<br />
· mp: ⟦P ⟶ Q; P⟧ ⟹ Q <br />
· mt: ⟦F ⟶ G; ¬G⟧ ⟹ ¬F <br />
· impI: (P ⟹ Q) ⟹ P ⟶ Q<br />
· disjI1: P ⟹ P ∨ Q<br />
· disjI2: Q ⟹ P ∨ Q<br />
· disjE: ⟦P ∨ Q; P ⟹ R; Q ⟹ R⟧ ⟹ R <br />
· FalseE: False ⟹ P<br />
· notE: ⟦¬P; P⟧ ⟹ R<br />
· notI: (P ⟹ False) ⟹ ¬P<br />
· iffI: ⟦P ⟹ Q; Q ⟹ P⟧ ⟹ P = Q<br />
· iffD1: ⟦Q = P; Q⟧ ⟹ P <br />
· iffD2: ⟦P = Q; Q⟧ ⟹ P<br />
· ccontr: (¬P ⟹ False) ⟹ P<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
text {*<br />
Se usarán las reglas notnotI y mt que demostramos a continuación. *}<br />
<br />
lemma notnotI: "P ⟹ ¬¬ P"<br />
by auto<br />
<br />
lemma mt: "⟦F ⟶ G; ¬G⟧ ⟹ ¬F"<br />
by auto<br />
<br />
section {* Implicaciones *}<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 1. Demostrar<br />
p ⟶ q, p ⊢ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_1: josrodjim2 carmarria inmbenber marmedmar3<br />
assumes 1: "p ⟶ q" and<br />
2: "p"<br />
shows "q"<br />
<br />
proof -<br />
show 1: "q" using 1 2 by (rule mp)<br />
qed<br />
<br />
lemma ej_1: joslopjim4<br />
assumes "p --> q"<br />
"p"<br />
shows "q"<br />
proof -<br />
show "q" using assms(1) assms(2) by (rule mp)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 2. Demostrar<br />
p ⟶ q, q ⟶ r, p ⊢ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_2: josrodjim2 carmarria inmbenber marmedmar3<br />
assumes 1: "p ⟶ q" and<br />
2: "q ⟶ r" and<br />
3: "p" <br />
shows "r"<br />
<br />
proof-<br />
have 4: "q" using 1 3 by (rule mp)<br />
show "r" using 2 4 by (rule mp)<br />
qed<br />
<br />
lemma ej_2: joslopjim4<br />
assumes "p ⟶ q"<br />
"q ⟶ r"<br />
"p" <br />
shows "r"<br />
proof -<br />
have "q" using `p⟶q` `p` by (rule mp)<br />
show "r" using `q⟶r` `q` by (rule mp)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 3. Demostrar<br />
p ⟶ (q ⟶ r), p ⟶ q, p ⊢ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_3: josrodjim2 carmarria inmbenber marmedmar3<br />
assumes 1: "p ⟶ (q ⟶ r)" and<br />
2: "p ⟶ q" and<br />
3: "p"<br />
shows "r"<br />
<br />
proof-<br />
have 4: "q" using 2 3 by (rule mp)<br />
have 5: "(q ⟶ r)" using 1 3 by (rule mp)<br />
show "r" using 5 4 by (rule mp)<br />
<br />
qed<br />
<br />
lemma ej_3: joslopjim4<br />
assumes "p ⟶ (q ⟶ r)"<br />
"p ⟶ q"<br />
"p"<br />
shows "r"<br />
proof -<br />
have "q" using `p⟶q` `p` by (rule mp)<br />
have "q⟶r" using `p ⟶ (q ⟶ r)` `p` by (rule mp)<br />
show "r" using `q⟶r` `q` by (rule mp)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 4. Demostrar<br />
p ⟶ q, q ⟶ r ⊢ p ⟶ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_4: josrodjim2<br />
assumes 1: "p ⟶ q" and<br />
2: "q ⟶ r" <br />
shows "p ⟶ r"<br />
<br />
proof-<br />
{ assume 3: "p"<br />
have 4: "q" using 1 3 by (rule mp)<br />
have 5: "r" using 2 4 by (rule mp)}<br />
hence 6: "p ⟶ r" using 3 5 by (rule impI)<br />
show "p⟶r" using 6 by this<br />
<br />
oops<br />
<br />
lemma ejercicio_4: carmarria<br />
assumes 1: "p ⟶ q" and<br />
2: "q ⟶ r" <br />
shows "p ⟶ r"<br />
proof -<br />
{ assume 3: p <br />
have 4: "q" using 1 3 by (rule mp)<br />
have 5: "r" using 2 4 by (rule mp)<br />
}<br />
thus "p ⟶ r" by (rule impI) <br />
qed<br />
lemma ejercicio_4: inmbenber<br />
assumes 1: "p ⟶ q" and<br />
2: "q ⟶ r" <br />
shows "p ⟶ r"<br />
proof -<br />
{ assume 3: "p" <br />
have 4: "q" using 1 3 by (rule mp)<br />
have 5: "r" using 2 4 by (rule mp) }<br />
thus "p ⟶ r" by (rule impI) <br />
qed<br />
<br />
lemma ejercicio_4:marmedmar3<br />
assumes "p ⟶ q"<br />
"q ⟶ r" <br />
shows "p ⟶ r"<br />
proof <br />
assume "p" <br />
with assms(1) have "q" by (rule mp)<br />
with assms(2) show "r" by (rule mp)<br />
qed <br />
<br />
lemma ej_4: joslopjim4<br />
assumes "p ⟶ q"<br />
"q ⟶ r" <br />
shows "p ⟶ r"<br />
proof (rule impI)<br />
assume "p"<br />
have "q" using assms(1) `p` by (rule mp)<br />
show "r" using assms(2) `q` by (rule mp)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 5. Demostrar<br />
p ⟶ (q ⟶ r) ⊢ q ⟶ (p ⟶ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_5: josrodjim2 (NO DA ERROR, PERO SI ALERTA)<br />
assumes "p ⟶ (q ⟶ r)" <br />
shows "q ⟶ (p ⟶ r)"<br />
<br />
proof -<br />
<br />
{ assume 2: "q"<br />
{assume 3: "p"<br />
have 4: "q ⟶r" using 1 3 by (rule mp)<br />
have 5: "r" using 4 2 by (rule mp)}<br />
hence 6: "p⟶r" using 3 5 by (rule impI)}<br />
hence 7: "q ⟶ (p ⟶ r)" using 2 6 by (rule impI)<br />
show "q ⟶ (p ⟶ r)" using 7 by this<br />
<br />
oops<br />
<br />
lemma ejercicio_5: carmarria <br />
assumes 1: "p ⟶ (q ⟶ r)" <br />
shows "q ⟶ (p ⟶ r)"<br />
<br />
proof - <br />
{assume 2: q<br />
{assume 3: p <br />
have 4: "q ⟶ r" using 1 3 by (rule mp)<br />
have 5: "r" using 4 2 by (rule mp)<br />
}<br />
hence " p ⟶ r" by (rule impI)<br />
}<br />
thus "q ⟶ (p ⟶ r)" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_5: inmbenber<br />
assumes 1: "p ⟶ (q ⟶ r)" <br />
shows "q ⟶ (p ⟶ r)"<br />
<br />
proof - <br />
{assume 2: "q"<br />
{assume 3: "p" <br />
have 4: "q ⟶ r" using 1 3 by (rule mp)<br />
have 5: "r" using 4 2 by (rule mp) }<br />
hence " p ⟶ r" by (rule impI) }<br />
thus "q ⟶ (p ⟶ r)" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_5: marmedmar3<br />
assumes "p ⟶ (q ⟶ r)" <br />
shows "q ⟶ (p ⟶ r)"<br />
proof <br />
assume "q"<br />
show "(p ⟶ r)"<br />
proof <br />
assume "p" <br />
with assms(1) have "(q ⟶ r)" by (rule mp)<br />
thus "r" using `q` by (rule mp) <br />
qed <br />
qed<br />
<br />
lemma ej_5: joslopjim4<br />
assumes "p ⟶ (q ⟶ r)" <br />
shows "q ⟶ (p ⟶ r)"<br />
proof (rule impI)<br />
assume "q"<br />
show "p⟶r"<br />
proof (rule impI)<br />
assume "p"<br />
have "q⟶r" using `p ⟶ (q ⟶ r)` `p` by (rule mp)<br />
show "r" using `q⟶r` `q` by (rule mp)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 6. Demostrar<br />
p ⟶ (q ⟶ r) ⊢ (p ⟶ q) ⟶ (p ⟶ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_6:josrodjim2<br />
assumes 1: "p ⟶ (q ⟶ r)" <br />
shows "(p ⟶ q) ⟶ (p ⟶ r)"<br />
<br />
proof- <br />
<br />
{assume 2: "p⟶q" <br />
{assume 3: "p"<br />
have 4: "q⟶r" using 1 3 by (rule mp)<br />
have 5: "q" using 2 3 by (rule mp)<br />
have 6: "r" using 4 5 by (rule mp)}<br />
hence 7: "p⟶r" using 3 6 by (rule impI)}<br />
hence 8: "(p ⟶ q) ⟶ (p ⟶ r)" using 2 7 by (rule impI)<br />
show "(p ⟶ q) ⟶ (p ⟶ r)" using 8 by this<br />
oops<br />
<br />
lemma ejercicio_6: carmarria, inmbenber<br />
assumes 1: "p ⟶ (q ⟶ r)" <br />
shows "(p ⟶ q) ⟶ (p ⟶ r)"<br />
proof -<br />
{assume 2: "p ⟶ q"<br />
{assume 3: "p"<br />
have 4: "q" using 2 3 by (rule mp)<br />
have 5: "q ⟶ r" using 1 3 by (rule mp)<br />
have 6: "r" using 5 4 by (rule mp)<br />
}<br />
hence 7: "p ⟶ r" by (rule impI)<br />
}<br />
thus "(p ⟶ q) ⟶ (p ⟶ r)" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_6:marmedmar3<br />
assumes 1: "p ⟶ (q ⟶ r)" <br />
shows "(p ⟶ q) ⟶ (p ⟶ r)"<br />
proof <br />
assume 2: "(p ⟶ q)" <br />
show "(p ⟶ r)" <br />
proof <br />
assume 3: "p" <br />
have 4: "q" using 2 3 by (rule mp)<br />
have 5: "(q ⟶ r)" using 1 3 by (rule mp) <br />
show "r" using 5 4 by (rule mp) <br />
qed<br />
qed<br />
<br />
lemma ej_6: joslopjim4<br />
assumes "p ⟶ (q ⟶ r)" <br />
shows "(p ⟶ q) ⟶ (p ⟶ r)"<br />
proof (rule impI)<br />
assume "(p⟶q)"<br />
show "p⟶r"<br />
proof (rule impI)<br />
assume "p"<br />
have "q⟶r" using `p ⟶ (q ⟶ r)` `p` by (rule mp)<br />
have "q" using `p⟶q` `p` by (rule mp)<br />
show "r" using `q⟶r` `q` by (rule mp)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 7. Demostrar<br />
p ⊢ q ⟶ p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_7: (NO SE QUE ESTA MAL)<br />
assumes "p" <br />
shows "q ⟶ p"<br />
<br />
proof -<br />
<br />
{ assume 2: "q"<br />
have 3: "p" using 1<br />
}<br />
hence 4: "q⟶p" using 2 3 by (rule impI)<br />
<br />
show "q⟶p" using 4 by this<br />
<br />
oops<br />
<br />
lemma ejercicio_7: carmarria, inmbenber<br />
assumes 1: "p" <br />
shows "q ⟶ p"<br />
proof -<br />
{assume 2: "q"<br />
have 3: "p" using 1 by this}<br />
thus "q ⟶ p" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_7:marmedmar3<br />
assumes "p" <br />
shows "q ⟶ p"<br />
proof <br />
assume "q" <br />
show "p" using assms(1) by this<br />
qed <br />
<br />
lemma ej_7: joslopjim4<br />
assumes "p" <br />
shows "q ⟶ p"<br />
proof (rule impI)<br />
assume "q"<br />
show "p" using assms by this<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 8. Demostrar<br />
⊢ p ⟶ (q ⟶ p)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_8: carmarria<br />
"p ⟶ (q ⟶ p)"<br />
<br />
proof -<br />
{assume 1: p<br />
{assume 2: q<br />
have 3: p using 1 by this<br />
}<br />
hence 4: "q ⟶ p" by (rule impI)<br />
}<br />
thus "p ⟶ (q ⟶ p)" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_8: marmedmar3<br />
"p ⟶ (q ⟶ p)"<br />
proof<br />
assume 1: "p"<br />
show "(q ⟶ p)"<br />
proof <br />
assume 2: "q" <br />
show "p" using 1 by this <br />
qed<br />
qed<br />
<br />
lemma ej_8: joslopjim4<br />
shows "p⟶(q⟶p)"<br />
proof (rule impI)<br />
assume "p"<br />
show "q⟶p"<br />
proof (rule impI)<br />
assume "q"<br />
show "p" using `p` by this<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 9. Demostrar<br />
p ⟶ q ⊢ (q ⟶ r) ⟶ (p ⟶ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_9: carmarria<br />
assumes 1: "p ⟶ q" <br />
shows "(q ⟶ r) ⟶ (p ⟶ r)"<br />
proof -<br />
{assume 2: "q ⟶ r"<br />
{assume 3: "p"<br />
have 4: "q" using 1 3 by (rule mp)<br />
have 5: "r" using 2 4 by (rule mp)<br />
}<br />
hence 6: "p ⟶ r" by (rule impI)<br />
}<br />
thus 7: "(q ⟶ r) ⟶ (p ⟶ r)" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_9: marmedmar3<br />
assumes "p ⟶ q" <br />
shows "(q ⟶ r) ⟶ (p ⟶ r)"<br />
proof <br />
assume "(q ⟶ r)" <br />
show "(p ⟶ r)"<br />
proof <br />
assume "p" <br />
with `(p ⟶ q)` have "q" by (rule mp)<br />
with `(q ⟶ r)` show "r" by (rule mp)<br />
qed<br />
qed<br />
<br />
lemma ej_9: joslopjim4<br />
assumes "p ⟶ q" <br />
shows "(q ⟶ r) ⟶ (p ⟶ r)"<br />
proof (rule impI)<br />
assume "q⟶r"<br />
show "p⟶r"<br />
proof (rule impI)<br />
assume "p"<br />
have "q" using `p⟶q` `p` by (rule mp)<br />
show "r" using `q⟶r` `q` by (rule mp)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 10. Demostrar<br />
p ⟶ (q ⟶ (r ⟶ s)) ⊢ r ⟶ (q ⟶ (p ⟶ s))<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_10: carmarria<br />
assumes 1: "p ⟶ (q ⟶ (r ⟶ s))" <br />
shows "r ⟶ (q ⟶ (p ⟶ s))"<br />
<br />
proof-<br />
{assume 2: r<br />
{assume 3: q<br />
{assume 4: p<br />
have 5: "q ⟶ (r ⟶ s)" using 1 4 by (rule mp)<br />
have 6: "r ⟶ s" using 5 3 by (rule mp)<br />
have 7: s using 6 2 by (rule mp)<br />
}<br />
hence 8: "p ⟶ s" by (rule impI)<br />
}<br />
hence 9: "q ⟶ (p ⟶ s)" by (rule impI)<br />
}<br />
thus "r ⟶ (q ⟶ (p ⟶ s))" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_10: marmedmar3<br />
assumes 1: "p ⟶ (q ⟶ (r ⟶ s))" <br />
shows "r ⟶ (q ⟶ (p ⟶ s))"<br />
proof <br />
assume 2: "r"<br />
show "(q ⟶ (p ⟶ s))"<br />
proof <br />
assume 3: "q" <br />
show "(p ⟶ s)"<br />
proof <br />
assume 4: "p" <br />
have 5: "(q ⟶ (r ⟶ s))" using 1 4 by (rule mp) <br />
have 6: "(r ⟶ s)" using 5 3 by (rule mp)<br />
show "s" using 6 2 by (rule mp) <br />
qed<br />
qed<br />
qed<br />
<br />
lemma ej_10: joslopjim4<br />
assumes "p ⟶ (q ⟶ (r ⟶ s))" <br />
shows "r ⟶ (q ⟶ (p ⟶ s))"<br />
proof (rule impI)<br />
assume "r"<br />
show "q⟶(p⟶s)"<br />
proof (rule impI)<br />
assume "q"<br />
show "p⟶s"<br />
proof (rule impI)<br />
assume "p"<br />
have "q ⟶ (r ⟶ s)" using `p ⟶ (q ⟶ (r ⟶ s))` `p` by (rule mp)<br />
have "r⟶s" using `q ⟶ (r ⟶ s)` `q` by (rule mp)<br />
show "s" using `r-->s` `r` by (rule mp)<br />
qed<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 11. Demostrar<br />
⊢ (p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_11: carmarria<br />
"(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))"<br />
proof - <br />
{assume 1: "p ⟶ (q ⟶ r)"<br />
{assume 2: "p ⟶ q"<br />
{assume 3: p<br />
have 4: q using 2 3 by (rule mp)<br />
have 5: "q ⟶ r" using 1 3 by (rule mp)<br />
have 6: r using 5 4 by (rule mp)<br />
}<br />
hence 7: "p ⟶ r" by (rule impI)<br />
}<br />
hence 8: "(p ⟶ q) ⟶ (p ⟶ r)" by (rule impI)<br />
}<br />
thus "(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_11: marmedmar3<br />
"(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))"<br />
proof <br />
assume 1: "(p ⟶ (q ⟶ r))" <br />
show "((p ⟶ q) ⟶ (p ⟶ r))"<br />
proof <br />
assume 2: "(p ⟶ q)" <br />
show "(p ⟶ r)" <br />
proof <br />
assume 3: "p" <br />
have 4: "(q ⟶ r)" using 1 3 by (rule mp) <br />
have 5: "q" using 2 3 by (rule mp) <br />
show "r" using 4 5 by (rule mp) <br />
qed<br />
qed<br />
qed<br />
<br />
lemma ej_11: joslopjim4<br />
shows "(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))"<br />
proof (rule impI)<br />
assume "p⟶(q⟶r)"<br />
show "(p⟶q)⟶(p⟶r)"<br />
proof (rule impI)<br />
assume "p⟶q"<br />
show "p⟶r"<br />
proof (rule impI)<br />
assume "p"<br />
have "q" using `p⟶q` `p` by (rule mp)<br />
have "q⟶r" using `p⟶(q⟶r)` `p` by (rule mp)<br />
show "r" using `q⟶r` `q` by (rule mp)<br />
qed<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 12. Demostrar<br />
(p ⟶ q) ⟶ r ⊢ p ⟶ (q ⟶ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_12: carmarria<br />
assumes 1: "(p ⟶ q) ⟶ r" <br />
shows "p ⟶ (q ⟶ r)"<br />
proof -<br />
{assume 2: p<br />
{assume 3: q<br />
{assume 4: p<br />
have 5: q using 3 by this<br />
}<br />
hence 6: "p ⟶ q" by (rule impI)<br />
have 7: r using 1 6 by (rule mp)<br />
}<br />
hence 8: "q ⟶ r" by (rule impI)<br />
}<br />
thus "p ⟶ (q ⟶ r)" by (rule impI)<br />
qed<br />
<br />
lemma ej_12: joslopjim4<br />
assumes "(p ⟶ q) ⟶ r" <br />
shows "p ⟶ (q ⟶ r)"<br />
proof (rule impI)<br />
assume "p"<br />
show "q⟶r"<br />
proof (rule impI)<br />
assume "q"<br />
have "p⟶q"<br />
proof (rule impI)<br />
assume "p"<br />
show "q" using `q` by this<br />
qed<br />
show "r" using `(p ⟶ q) ⟶ r` `p⟶q` by (rule mp)<br />
qed<br />
qed<br />
<br />
section {* Conjunciones *}<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 13. Demostrar<br />
p, q ⊢ p ∧ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_13: carmarria marmedmar3<br />
assumes "p"<br />
"q" <br />
shows "p ∧ q"<br />
<br />
proof-<br />
show "p \<and> q" using assms(1) assms(2) by (rule conjI)<br />
qed<br />
<br />
lemma ej_13: joslopjim4<br />
assumes "p"<br />
"q" <br />
shows "p ∧ q"<br />
proof-<br />
show "p∧q" using `p` `q` by (rule conjI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 14. Demostrar<br />
p ∧ q ⊢ p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_14: carmarria marmedmar3<br />
assumes "p ∧ q" <br />
shows "p"<br />
<br />
proof -<br />
show "p" using assms by (rule conjunct1)<br />
qed<br />
<br />
lemma ej_14: joslopjim4<br />
assumes "p ∧ q" <br />
shows "p"<br />
proof -<br />
show "p" using assms by (rule conjunct1)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 15. Demostrar<br />
p ∧ q ⊢ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_15: carmarria marmedmar3<br />
assumes "p ∧ q" <br />
shows "q"<br />
<br />
proof -<br />
show q using assms(1) by (rule conjunct2)<br />
qed<br />
<br />
lemma ej_15: joslopjim4<br />
assumes "p ∧ q" <br />
shows "q"<br />
proof-<br />
show "q" using assms by (rule conjunct2)<br />
qed<br />
<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 16. Demostrar<br />
p ∧ (q ∧ r) ⊢ (p ∧ q) ∧ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_16: carmarria marmedmar3<br />
assumes 1: "p ∧ (q ∧ r)"<br />
shows "(p ∧ q) ∧ r"<br />
proof -<br />
have 2: "q \<and> r" using 1 by (rule conjunct2)<br />
have 3: p using 1 by (rule conjunct1)<br />
have 4: q using 2 by (rule conjunct1)<br />
have 5: r using 2 by (rule conjunct2)<br />
have 6: "p \<and> q" using 3 4 by (rule conjI)<br />
show "(p \<and> q) \<and> r" using 6 5 by (rule conjI)<br />
qed<br />
<br />
lemma ej_16: joslopjim4<br />
assumes "p ∧ (q ∧ r)"<br />
shows "(p ∧ q) ∧ r"<br />
proof-<br />
have "p" using assms by (rule conjunct1)<br />
have "q∧r" using assms by (rule conjunct2)<br />
have "q" using `q∧r` by (rule conjunct1)<br />
have "r" using `q∧r` by (rule conjunct2)<br />
have "p∧q" using `p` `q` by (rule conjI)<br />
show "(p∧q)∧r" using `p∧q` `r` by (rule conjI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 17. Demostrar<br />
(p ∧ q) ∧ r ⊢ p ∧ (q ∧ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_17:<br />
assumes "(p ∧ q) ∧ r" <br />
shows "p ∧ (q ∧ r)"<br />
proof (rule conjI)<br />
have "p ∧ q" using assms ..<br />
thus "p" ..<br />
next<br />
show "q ∧ r"<br />
proof (rule conjI)<br />
have "p ∧ q" using assms ..<br />
thus "q" ..<br />
next<br />
show "r" using assms ..<br />
qed<br />
qed<br />
<br />
lemma ejercicio_17: carmarria marmedmar3<br />
assumes 1:"(p ∧ q) ∧ r" <br />
shows "p ∧ (q ∧ r)"<br />
proof-<br />
have 2:"p \<and> q" using 1 by (rule conjunct1)<br />
have 3: r using 1 by (rule conjunct2)<br />
have 4: p using 2 by (rule conjunct1)<br />
have 5: q using 2 by (rule conjunct2)<br />
have 6: "q \<and> r" using 5 3 by (rule conjI)<br />
show "p \<and> (q \<and> r)" using 4 6 by (rule conjI)<br />
qed<br />
<br />
<br />
lemma ej_17: joslopjim4<br />
assumes "(p ∧ q) ∧ r" <br />
shows "p ∧ (q ∧ r)"<br />
proof-<br />
have "p∧q" using assms by (rule conjunct1)<br />
have "r" using assms by (rule conjunct2)<br />
have "p" using `p∧q` by (rule conjunct1)<br />
have "q" using `p∧q` by (rule conjunct2)<br />
have "q∧r" using `q` `r` by (rule conjI)<br />
show "p∧(q∧r)" using `p` `q∧r` by (rule conjI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 18. Demostrar<br />
p ∧ q ⊢ p ⟶ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_18: carmarria<br />
assumes "p ∧ q" <br />
shows "p ⟶ q"<br />
proof -<br />
{assume p<br />
have q using assms(1) by (rule conjunct2)<br />
}<br />
thus "p ⟶ q" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_18: marmedmar3<br />
assumes "p ∧ q" <br />
shows "p ⟶ q"<br />
proof <br />
assume "p" <br />
show "q" using assms by (rule conjunct2) <br />
qed <br />
<br />
lemma ej_18: joslopjim4<br />
assumes "p ∧ q" <br />
shows "p ⟶ q"<br />
proof (rule impI)<br />
assume "p"<br />
show "q" using assms by (rule conjunct2)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 19. Demostrar<br />
(p ⟶ q) ∧ (p ⟶ r) ⊢ p ⟶ q ∧ r <br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_19: carmarria<br />
assumes 1: "(p ⟶ q) ∧ (p ⟶ r)" <br />
shows "p ⟶ q ∧ r"<br />
proof -<br />
{assume 2: p<br />
have 3: "p ⟶ q" using 1 by (rule conjunct1)<br />
have 4: "p ⟶ r" using 1 by (rule conjunct2)<br />
have 5: q using 3 2 by (rule mp)<br />
have 6: r using 4 2 by (rule mp)<br />
have 7: "q \<and> r" using 5 6 by (rule conjI)<br />
}<br />
thus "p ⟶ q \<and> r" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_19: marmedmar3<br />
assumes 1: "(p ⟶ q) ∧ (p ⟶ r)" <br />
shows "p ⟶ q ∧ r"<br />
proof <br />
have 2: "(p ⟶ q)" using assms by (rule conjunct1)<br />
have 3: "(p ⟶ r)" using assms by (rule conjunct2)<br />
assume 4: "p"<br />
with `(p ⟶ q)` have 5: "q" by (rule mp)<br />
have 6: "r" using 3 4 by (rule mp) <br />
show "q ∧ r" using 5 6 by (rule conjI)<br />
qed <br />
<br />
lemma ej_19: joslopjim4<br />
assumes "(p ⟶ q) ∧ (p ⟶ r)" <br />
shows "p ⟶ q ∧ r"<br />
proof (rule impI)<br />
assume "p"<br />
have "p⟶q" using assms by (rule conjunct1)<br />
have "q" using `p⟶q` `p` by (rule mp)<br />
have "p⟶r" using assms by (rule conjunct2)<br />
have "r" using `p⟶r` `p` by (rule mp)<br />
show "q∧r" using `q` `r` by (rule conjI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 20. Demostrar<br />
p ⟶ q ∧ r ⊢ (p ⟶ q) ∧ (p ⟶ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_20: carmarria<br />
assumes 1: "p ⟶ q ∧ r" <br />
shows "(p ⟶ q) ∧ (p ⟶ r)"<br />
proof -<br />
{assume 2: p<br />
have 3: "q \<and> r" using 1 2 by (rule mp)<br />
have 4: q using 3 by (rule conjunct1)<br />
}<br />
hence 5: "p ⟶ q" by (rule impI)<br />
{assume 6: p<br />
have 7: "q \<and> r" using 1 6 by (rule mp)<br />
have 8: r using 7 by (rule conjunct2)<br />
}<br />
hence 9: "p ⟶ r" by (rule impI)<br />
show "(p ⟶ q) \<and> (p ⟶ r)" using 5 9 by (rule conjI)<br />
qed<br />
<br />
lemma ej_20: joslopjim4<br />
assumes "p ⟶ q ∧ r" <br />
shows "(p ⟶ q) ∧ (p ⟶ r)"<br />
proof-<br />
have "p⟶q"<br />
proof<br />
assume "p"<br />
have "q∧r" using assms `p` by (rule mp)<br />
show "q" using `q∧r` by (rule conjunct1)<br />
qed<br />
have "p⟶r"<br />
proof<br />
assume "p"<br />
have "q∧r" using assms `p` by (rule mp)<br />
show "r" using `q∧r` by (rule conjunct2)<br />
qed<br />
show "(p⟶q)∧(p⟶r)" using `p⟶q` `p⟶r` by (rule conjI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 21. Demostrar<br />
p ⟶ (q ⟶ r) ⊢ p ∧ q ⟶ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_21: carmarria<br />
assumes 1: "p ⟶ (q ⟶ r)" <br />
shows "p ∧ q ⟶ r"<br />
proof -<br />
{assume 2: "p \<and> q"<br />
have 3: p using 2 by (rule conjunct1)<br />
have 4: q using 2 by (rule conjunct2)<br />
have 5: "q ⟶ r" using 1 3 by (rule mp)<br />
have 6: r using 5 4 by (rule mp)<br />
}<br />
thus "p ∧ q ⟶ r" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_21: marmedmar3<br />
assumes "p ⟶ (q ⟶ r)" <br />
shows "p ∧ q ⟶ r"<br />
proof <br />
assume "p ∧ q" <br />
hence "p" by (rule conjunct1) <br />
with assms have "(q ⟶ r)" by (rule mp)<br />
have "q" using `p ∧ q` by (rule conjunct2) <br />
with `(q ⟶ r)` show "r" by (rule mp) <br />
qed <br />
<br />
lemma ej_21: joslopjim4<br />
assumes "p ⟶ (q ⟶ r)" <br />
shows "p ∧ q ⟶ r"<br />
proof (rule impI)<br />
assume "p∧q"<br />
have "p" using `p∧q` by (rule conjunct1)<br />
have "q" using `p∧q` by (rule conjunct2)<br />
have "q⟶r" using assms `p` by (rule mp)<br />
show "r" using `q⟶r` `q` by (rule mp)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 22. Demostrar<br />
p ∧ q ⟶ r ⊢ p ⟶ (q ⟶ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_22: carmarria<br />
assumes "p ∧ q ⟶ r" <br />
shows "p ⟶ (q ⟶ r)"<br />
proof -<br />
{assume p<br />
{assume q<br />
have "p \<and> q" using `p` `q` by (rule conjI)<br />
have r using assms(1) `p \<and> q` by (rule mp)<br />
}<br />
hence "q ⟶ r" by (rule impI)<br />
}<br />
thus "p ⟶ (q ⟶ r)" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_22: marmedmar3<br />
assumes "p ∧ q ⟶ r" <br />
shows "p ⟶ (q ⟶ r)"<br />
proof <br />
assume "p" <br />
show "(q ⟶ r)"<br />
proof <br />
assume "q" <br />
with `p` have "p ∧ q" by (rule conjI) <br />
with assms show "r" by (rule mp) <br />
qed <br />
qed<br />
<br />
lemma ej_22: joslopjim4<br />
assumes "p ∧ q ⟶ r" <br />
shows "p ⟶ (q ⟶ r)"<br />
proof (rule impI)<br />
assume "p"<br />
show "q⟶r"<br />
proof (rule impI)<br />
assume "q"<br />
have "p∧q" using `p` `q` by (rule conjI)<br />
show "r" using assms `p∧q` by (rule mp)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 23. Demostrar<br />
(p ⟶ q) ⟶ r ⊢ p ∧ q ⟶ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_23: carmarria<br />
assumes "(p ⟶ q) ⟶ r" <br />
shows "p ∧ q ⟶ r"<br />
proof -<br />
{assume "p \<and> q" <br />
{assume p<br />
have q using `p \<and> q` by (rule conjunct2)<br />
}<br />
hence "p ⟶ q" by (rule impI)<br />
have r using assms(1) `p ⟶ q` by (rule mp)<br />
}<br />
thus "p \<and> q ⟶ r" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_23: marmedmar3<br />
assumes "(p ⟶ q) ⟶ r" <br />
shows "p ∧ q ⟶ r"<br />
proof <br />
assume "p ∧ q"<br />
have "p" using `p ∧ q` by (rule conjunct1)<br />
have "q" using `p ∧ q` by (rule conjunct2) <br />
hence "(p ⟶ q)" by (rule impI) <br />
with assms show "r" by (rule mp) <br />
qed <br />
<br />
lemma ej_23: joslopjim4<br />
assumes "(p ⟶ q) ⟶ r" <br />
shows "p ∧ q ⟶ r"<br />
proof (rule impI)<br />
assume "p∧q"<br />
have "p⟶q"<br />
proof<br />
assume "p"<br />
show "q" using `p∧q` by (rule conjunct2)<br />
qed<br />
show "r" using assms `p⟶q` by (rule mp)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 24. Demostrar<br />
p ∧ (q ⟶ r) ⊢ (p ⟶ q) ⟶ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_24: carmarria<br />
assumes 1: "p ∧ (q ⟶ r)" <br />
shows "(p ⟶ q) ⟶ r"<br />
proof -<br />
{assume 2: "p ⟶ q" <br />
have 3: p using 1 by (rule conjunct1)<br />
have 4: "q ⟶ r" using 1 by (rule conjunct2)<br />
have 5: q using 2 3 by (rule mp)<br />
have 6: r using 4 5 by (rule mp)<br />
}<br />
thus "(p ⟶ q) ⟶ r" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_24: marmedmar3<br />
assumes "p ∧ (q ⟶ r)" <br />
shows "(p ⟶ q) ⟶ r"<br />
proof <br />
assume "(p ⟶ q)" <br />
have "(q ⟶ r)" using assms by (rule conjunct2)<br />
have "p" using assms by (rule conjunct1)<br />
with `(p ⟶ q)` have "q" by (rule mp) <br />
with `(q ⟶ r)` show "r" by (rule mp)<br />
qed <br />
<br />
lemma ej_24: joslopjim4<br />
assumes "p ∧ (q ⟶ r)" <br />
shows "(p ⟶ q) ⟶ r"<br />
proof (rule impI)<br />
assume "p⟶q"<br />
have "p" using assms by (rule conjunct1)<br />
have "q⟶r" using assms by (rule conjunct2)<br />
have "q" using `p⟶q` `p` by (rule mp)<br />
show "r" using `q⟶r` `q` by (rule mp)<br />
qed<br />
<br />
section {* Disyunciones *}<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 25. Demostrar<br />
p ⊢ p ∨ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_25: carmarria<br />
assumes "p"<br />
shows "p ∨ q"<br />
proof -<br />
show "p | q" using assms(1) by (rule disjI1)<br />
qed<br />
<br />
lemma ej_25: joslopjim4<br />
assumes "p"<br />
shows "p ∨ q"<br />
proof (rule disjI1)<br />
show "p" using assms by this<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 26. Demostrar<br />
q ⊢ p ∨ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_26: carmarria<br />
assumes "q"<br />
shows "p ∨ q"<br />
proof -<br />
show "p | q" using assms(1) by (rule disjI2)<br />
qed<br />
<br />
lemma ej_26: joslopjim4<br />
assumes "q"<br />
shows "p ∨ q"<br />
proof (rule disjI2)<br />
show "q" using assms by this<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 27. Demostrar<br />
p ∨ q ⊢ q ∨ p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_27: carmarria<br />
assumes "p ∨ q"<br />
shows "q ∨ p"<br />
proof -<br />
have "p | q" using assms(1) by this<br />
moreover<br />
{assume p<br />
have "q | p" using `p` by (rule disjI2)<br />
}<br />
moreover<br />
{assume q<br />
have "q | p" using `q` by (rule disjI1)<br />
}<br />
ultimately show "q | p" by (rule disjE)<br />
qed<br />
<br />
lemma ej_27: joslopjim4<br />
assumes "p ∨ q"<br />
shows "q ∨ p"<br />
using assms<br />
proof <br />
assume "p"<br />
then show "q∨p" by(rule disjI2)<br />
next<br />
assume "q"<br />
show "q∨p" using `q` by(rule disjI1)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 28. Demostrar<br />
q ⟶ r ⊢ p ∨ q ⟶ p ∨ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_28: carmarria<br />
assumes "q ⟶ r" <br />
shows "p ∨ q ⟶ p ∨ r"<br />
proof -<br />
{assume "p | q" <br />
moreover<br />
{assume p<br />
have "p | r" using `p` by (rule disjI1)<br />
}<br />
moreover<br />
{assume q<br />
have r using assms(1) `q` by (rule mp)<br />
have "p | r" using `r` by (rule disjI2)<br />
}<br />
ultimately have "p | r" by (rule disjE)<br />
}<br />
thus "p | q ⟶ p | r" by (rule impI)<br />
qed<br />
<br />
lemma ej_28: joslopjim4<br />
assumes "q ⟶ r" <br />
shows "p ∨ q ⟶ p ∨ r"<br />
proof (rule impI)<br />
assume "p∨q" <br />
show "p∨r"<br />
using `p∨q`<br />
proof <br />
assume "p"<br />
show "p∨r" using `p` by (rule disjI1)<br />
next<br />
assume "q"<br />
have "r" using assms `q` by (rule mp)<br />
show "p∨r" using `r` by (rule disjI2)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 29. Demostrar<br />
p ∨ p ⊢ p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_29: carmarria<br />
assumes "p ∨ p"<br />
shows "p"<br />
proof -<br />
have "p | p" using assms(1) by this<br />
moreover <br />
{assume p<br />
}<br />
moreover<br />
{assume p<br />
}<br />
ultimately show p by (rule disjE)<br />
qed<br />
<br />
lemma ej_29: joslopjim4<br />
assumes "p ∨ p"<br />
shows "p"<br />
proof (rule ccontr)<br />
assume "¬p"<br />
show False<br />
using assms<br />
proof<br />
assume "p"<br />
show False using `¬p` `p` by (rule notE)<br />
next<br />
assume "p"<br />
show False using `¬p` `p` by (rule notE)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 30. Demostrar<br />
p ⊢ p ∨ p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_30: carmarria<br />
assumes "p" <br />
shows "p ∨ p"<br />
proof -<br />
show "p | p" using assms(1) by (rule disjI1)<br />
qed<br />
<br />
lemma ej_30: joslopjim4<br />
assumes "p" <br />
shows "p ∨ p"<br />
proof (rule disjI1)<br />
show "p" using assms by this<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 31. Demostrar<br />
p ∨ (q ∨ r) ⊢ (p ∨ q) ∨ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_31: carmarria<br />
assumes 1: "p ∨ (q ∨ r)" <br />
shows "(p ∨ q) ∨ r"<br />
proof -<br />
have "p | (q | r)" using 1 by this<br />
moreover {assume 2: p<br />
have 3: "p | q" using 2 by (rule disjI1)<br />
have 4: "(p | q) | r" using 3 by (rule disjI1)<br />
}<br />
moreover {assume 5: "q | r"<br />
moreover {assume 6: q<br />
have 7: "p | q" using 6 by (rule disjI2)<br />
have 8: "(p | q) | r " using 7 by (rule disjI1)<br />
}<br />
moreover {assume 9: r<br />
have 10: "(p | q) | r " using 9 by (rule disjI2)<br />
}<br />
ultimately have 11: "(p | q) | r " by (rule disjE)<br />
}<br />
ultimately show "(p | q) | r " by (rule disjE)<br />
qed<br />
<br />
lemma ej_31: joslopjim4<br />
assumes "p ∨ (q ∨ r)" <br />
shows "(p ∨ q) ∨ r"<br />
using assms<br />
proof <br />
assume "p"<br />
have "p∨q" using `p` by (rule disjI1)<br />
thus "(p ∨ q) ∨ r" by (rule disjI1)<br />
next<br />
assume "q∨r"<br />
show "(p ∨ q) ∨ r"<br />
using `q∨r`<br />
proof <br />
assume "q"<br />
have "p∨q" using `q` by (rule disjI2)<br />
show "(p ∨ q) ∨ r" using `p∨q` by (rule disjI1)<br />
next<br />
assume "r"<br />
show "(p ∨ q) ∨ r" using `r` by (rule disjI2)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 32. Demostrar<br />
(p ∨ q) ∨ r ⊢ p ∨ (q ∨ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_32: carmarria<br />
assumes 1: "(p ∨ q) ∨ r" <br />
shows "p ∨ (q ∨ r)"<br />
proof -<br />
have "(p | q) | r" using 1 by this<br />
moreover {assume 2: "p | q" <br />
moreover {assume 3: p <br />
have 4: "p | (q | r)" using 3 by (rule disjI1)<br />
}<br />
moreover {assume 5: q<br />
have 6: "q | r" using 5 by (rule disjI1)<br />
have 7: "p | (q | r)" using 6 by (rule disjI2)<br />
}<br />
ultimately have 8: " p | (q | r)" by (rule disjE)<br />
}<br />
moreover {assume 9: r<br />
have 10: "q | r" using 9 by (rule disjI2)<br />
have 11: "p | (q | r)" using 10 by (rule disjI2)<br />
}<br />
ultimately show "p | (q | r)" by (rule disjE)<br />
qed<br />
<br />
lemma ej_32: joslopjim4<br />
assumes "(p ∨ q) ∨ r" <br />
shows "p ∨ (q ∨ r)"<br />
using assms<br />
proof<br />
assume "p∨q"<br />
show "p ∨ (q ∨ r)"<br />
using `p∨q`<br />
proof <br />
assume "p"<br />
show "p ∨ (q ∨ r)" using `p` by (rule disjI1)<br />
next<br />
assume "q"<br />
have "q∨r" using `q` by (rule disjI1)<br />
show "p ∨ (q ∨ r)" using `q∨r` by (rule disjI2)<br />
qed<br />
next<br />
assume "r"<br />
have "q∨r" using `r` by (rule disjI2)<br />
show "p ∨ (q ∨ r)" using `q∨r` by (rule disjI2)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 33. Demostrar<br />
p ∧ (q ∨ r) ⊢ (p ∧ q) ∨ (p ∧ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_33: carmarria<br />
assumes 1: "p ∧ (q ∨ r)" <br />
shows "(p ∧ q) ∨ (p ∧ r)"<br />
proof -<br />
have 2: p using 1 by (rule conjunct1)<br />
have 3: "q | r" using 1 by (rule conjunct2)<br />
moreover {assume 4: q<br />
have 5: "p & q" using 2 4 by (rule conjI)<br />
have 6: "(p & q) | (p & r)" using 5 by (rule disjI1)<br />
}<br />
moreover {assume 7: r<br />
have 8: "p & r" using 2 7 by (rule conjI)<br />
have 9: "(p & q) | (p & r)" using 8 by (rule disjI2)<br />
}<br />
ultimately show "(p & q) | (p & r)" by (rule disjE)<br />
qed<br />
<br />
lemma ej_33: joslopjim4<br />
assumes "p ∧ (q ∨ r)" <br />
shows "(p ∧ q) ∨ (p ∧ r)"<br />
proof-<br />
have "p" using assms by (rule conjunct1)<br />
have "q∨r" using assms by (rule conjunct2)<br />
show "(p ∧ q) ∨ (p ∧ r)"<br />
using `q∨r` <br />
proof (rule disjE)<br />
assume "q"<br />
have "p∧q" using `p` `q` by (rule conjI)<br />
show "(p ∧ q) ∨ (p ∧ r)" using `p∧q` by (rule disjI1)<br />
next<br />
assume "r"<br />
have "p∧r" using `p` `r` by (rule conjI)<br />
show "(p ∧ q) ∨ (p ∧ r)" using `p∧r` by (rule disjI2)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 34. Demostrar<br />
(p ∧ q) ∨ (p ∧ r) ⊢ p ∧ (q ∨ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_34: carmarria<br />
assumes 1: "(p ∧ q) ∨ (p ∧ r)" <br />
shows "p ∧ (q ∨ r)"<br />
proof -<br />
have "(p & q) | (p & r)" using 1 by this<br />
moreover {assume 2: "p & q"<br />
have 3: p using 2 by (rule conjunct1)<br />
have 4: q using 2 by (rule conjunct2)<br />
have 5: "q | r" using 4 by (rule disjI1)<br />
have 6: "p & (q | r)" using 3 5 by (rule conjI)<br />
}<br />
moreover {assume 7: "p & r"<br />
have 8: p using 7 by (rule conjunct1)<br />
have 9: r using 7 by (rule conjunct2)<br />
have 10: "q | r" using 9 by (rule disjI2)<br />
have 11: "p & (q | r)" using 8 10 by (rule conjI)<br />
}<br />
ultimately show "p & (q | r)" by (rule disjE)<br />
qed<br />
<br />
lemma ej_34: joslopjim4<br />
assumes "(p ∧ q) ∨ (p ∧ r)" <br />
shows "p ∧ (q ∨ r)"<br />
using assms<br />
proof<br />
assume "p∧q"<br />
have "p" using `p∧q` by (rule conjunct1)<br />
have "q" using `p∧q` by (rule conjunct2)<br />
have "q∨r" using `q` by (rule disjI1)<br />
show "p∧(q∨r)" using `p` `q∨r` by (rule conjI)<br />
next<br />
assume "p∧r"<br />
have "p" using `p∧r` by (rule conjunct1)<br />
have "r" using `p∧r` by (rule conjunct2)<br />
have "q∨r" using `r` by (rule disjI2)<br />
show "p∧(q∨r)" using `p` `q∨r` by (rule conjI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 35. Demostrar<br />
p ∨ (q ∧ r) ⊢ (p ∨ q) ∧ (p ∨ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_35: carmarria<br />
assumes 1: "p ∨ (q ∧ r)" <br />
shows "(p ∨ q) ∧ (p ∨ r)"<br />
proof -<br />
have "p | (q & r)" using 1 by this<br />
moreover {assume 2: p<br />
have 3: " p | q" using 2 by (rule disjI1)<br />
have 4: " p | r" using 2 by (rule disjI1)<br />
have 5: "(p | q) & (p | r)" using 3 4 by (rule conjI)<br />
}<br />
moreover {assume 6: "q & r" <br />
have 7: q using 6 by (rule conjunct1)<br />
have 8: r using 6 by (rule conjunct2)<br />
have 9: "p | q" using 7 by (rule disjI2)<br />
have 10: "p | r" using 8 by (rule disjI2)<br />
have 11: "(p | q) & (p | r)" using 9 10 by (rule conjI)<br />
}<br />
ultimately show "(p | q) & (p | r)" by (rule disjE)<br />
qed<br />
<br />
lemma ej_35: joslopjim4<br />
assumes "p ∨ (q ∧ r)" <br />
shows "(p ∨ q) ∧ (p ∨ r)"<br />
using assms<br />
proof (rule disjE)<br />
assume "p"<br />
have "p∨q" using `p` by (rule disjI1)<br />
have "p∨r" using `p` by (rule disjI1)<br />
show "(p ∨ q) ∧ (p ∨ r)" using `p∨q` `p∨r` by (rule conjI)<br />
next<br />
assume "q∧r"<br />
have "q" using `q∧r` by (rule conjunct1)<br />
have "r" using `q∧r` by (rule conjunct2)<br />
have "p∨q" using `q` by (rule disjI2)<br />
have "p∨r" using `r` by (rule disjI2)<br />
show "(p ∨ q) ∧ (p ∨ r)" using `p∨q` `p∨r` by (rule conjI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 36. Demostrar<br />
(p ∨ q) ∧ (p ∨ r) ⊢ p ∨ (q ∧ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_36: carmarria<br />
assumes 1: "(p ∨ q) ∧ (p ∨ r)"<br />
shows "p ∨ (q ∧ r)"<br />
proof -<br />
have 2: "p | q" using 1 by (rule conjunct1)<br />
moreover {assume 3: p<br />
have 4: "p | (q & r)" using 3 by (rule disjI1)<br />
}<br />
moreover {assume 5: q<br />
have 6: "p | r" using 1 by (rule conjunct2)<br />
moreover {assume 7: p<br />
have 8: "p | (q & r)" using 7 by (rule disjI1)<br />
}<br />
moreover {assume 9: r<br />
have 10: "q & r" using 5 9 by (rule conjI)<br />
have 11: "p | (q & r)" using 10 by (rule disjI2)<br />
}<br />
ultimately have 12: "p | (q & r)" by (rule disjE)<br />
}<br />
ultimately show "p | (q & r)" by (rule disjE)<br />
qed<br />
<br />
lemma ej_36: joslopjim4<br />
assumes "(p ∨ q) ∧ (p ∨ r)"<br />
shows "p ∨ (q ∧ r)"<br />
proof-<br />
have "p∨q" using assms by (rule conjunct1)<br />
have "p∨r" using assms by (rule conjunct2)<br />
show "p∨(q ∧ r)"<br />
using `p∨r`<br />
proof<br />
assume "p"<br />
show "p∨(q ∧ r)" using `p` by (rule disjI1)<br />
next<br />
assume "r"<br />
show "p∨(q ∧ r)"<br />
using `p∨q`<br />
proof<br />
assume "p"<br />
show "p∨(q ∧ r)" using `p` by (rule disjI1)<br />
next<br />
assume "q"<br />
have "q∧r" using `q` `r` by (rule conjI)<br />
show "p∨(q ∧ r)" using `q∧r` by (rule disjI2)<br />
qed<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 37. Demostrar<br />
(p ⟶ r) ∧ (q ⟶ r) ⊢ p ∨ q ⟶ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_37: carmarria<br />
assumes 1: "(p ⟶ r) ∧ (q ⟶ r)" <br />
shows "p ∨ q ⟶ r"<br />
proof -<br />
have 2: "p ⟶ r" using 1 by (rule conjunct1)<br />
have 3: "q ⟶ r" using 1 by (rule conjunct2)<br />
{assume 4: "p | q"<br />
moreover {assume 5: p<br />
have 6: r using 2 5 by (rule mp)<br />
}<br />
moreover {assume 7: q<br />
have 8: r using 3 7 by (rule mp)<br />
}<br />
ultimately have r by (rule disjE)<br />
}<br />
thus "p | q ⟶ r" by (rule impI)<br />
qed<br />
<br />
lemma ej_37: joslopjim4<br />
assumes "(p ⟶ r) ∧ (q ⟶ r)" <br />
shows "p ∨ q ⟶ r"<br />
proof (rule impI)<br />
assume "p∨q"<br />
have "p⟶r" using assms by (rule conjunct1)<br />
have "q⟶r" using assms by (rule conjunct2)<br />
show "r"<br />
using `p∨q`<br />
proof <br />
assume "p"<br />
show "r" using `p⟶r` `p` by (rule mp)<br />
next<br />
assume "q"<br />
show "r" using `q⟶r` `q` by (rule mp)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 38. Demostrar<br />
p ∨ q ⟶ r ⊢ (p ⟶ r) ∧ (q ⟶ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_38: carmarria<br />
assumes 1: "p ∨ q ⟶ r" <br />
shows "(p ⟶ r) ∧ (q ⟶ r)"<br />
proof -<br />
{assume 2: p<br />
have 3: "p | q" using 2 by (rule disjI1)<br />
have 4: r using 1 3 by (rule mp)<br />
}<br />
hence 5: "p ⟶ r" by (rule impI)<br />
{assume 6: q<br />
have 7: "p | q" using 6 by (rule disjI2)<br />
have 8: r using 1 7 by (rule mp)<br />
}<br />
hence 9: "q ⟶ r" by (rule impI)<br />
show "(p ⟶ r) & (q ⟶ r)" using 5 9 by (rule conjI)<br />
qed<br />
<br />
lemma ej_38: joslopjim4<br />
assumes "p ∨ q ⟶ r" <br />
shows "(p ⟶ r) ∧ (q ⟶ r)"<br />
proof-<br />
have "p⟶r" <br />
proof (rule impI)<br />
assume "p"<br />
have "p∨q" using `p` by (rule disjI1)<br />
show "r" using assms `p∨q` by (rule mp)<br />
qed<br />
have "q⟶r"<br />
proof (rule impI)<br />
assume "q"<br />
have "p∨q" using `q` by (rule disjI2)<br />
show "r" using assms `p∨q` by (rule mp)<br />
qed<br />
show "(p ⟶ r) ∧ (q ⟶ r)" using `p⟶r` `q⟶r` by (rule conjI)<br />
qed<br />
<br />
section {* Negaciones *}<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 39. Demostrar<br />
p ⊢ ¬¬p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_39: carmarria<br />
assumes "p"<br />
shows "¬¬p"<br />
proof -<br />
show "~~p" using assms(1) by (rule notnotI)<br />
qed<br />
<br />
lemma ej_39: joslopjim4<br />
assumes "p"<br />
shows "¬¬p"<br />
proof-<br />
show "¬¬p" using assms by (rule notnotI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 40. Demostrar<br />
¬p ⊢ p ⟶ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_40: carmarria<br />
assumes "¬p" <br />
shows "p ⟶ q"<br />
proof -<br />
{assume p<br />
have "~p" using assms(1) by this<br />
have False using `~p` `p` by (rule notE)<br />
have q using `False` by (rule FalseE)<br />
}<br />
thus "p ⟶ q" by (rule impI)<br />
qed<br />
<br />
lemma ej_40: joslopjim4<br />
assumes "¬p" <br />
shows "p ⟶ q"<br />
proof (rule impI)<br />
assume "p"<br />
show "q" using assms `p` by (rule notE)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 41. Demostrar<br />
p ⟶ q ⊢ ¬q ⟶ ¬p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_41: carmarria<br />
assumes 1: "p ⟶ q"<br />
shows "¬q ⟶ ¬p"<br />
proof -<br />
{assume 2: "~q"<br />
have 3: "~p" using 1 2 by (rule mt)<br />
}<br />
thus 4: "~q ⟶ ~p" by (rule impI)<br />
qed<br />
<br />
lemma ej_41: joslopjim4<br />
assumes "p ⟶ q"<br />
shows "¬q ⟶ ¬p"<br />
proof (rule impI)<br />
assume "¬q"<br />
show "¬p" using assms `¬q` by (rule mt)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 42. Demostrar<br />
p∨q, ¬q ⊢ p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_42: carmarria<br />
assumes 1: "p∨q" and<br />
2: "¬q" <br />
shows "p"<br />
proof -<br />
have " p | q" using 1 by this<br />
moreover {assume 3: p<br />
}<br />
moreover {assume 4: q<br />
have 5: False using 2 4 by (rule notE)<br />
have 6: p using 5 by (rule FalseE)<br />
}<br />
ultimately show p by (rule disjE)<br />
qed<br />
<br />
lemma ej_42: joslopjim4<br />
assumes "p∨q"<br />
"¬q" <br />
shows "p"<br />
using assms(1)<br />
proof <br />
assume "p"<br />
show "p" using `p` by this<br />
next<br />
assume "q"<br />
show "p" using assms(2) `q` by (rule notE)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 43. Demostrar<br />
p ∨ q, ¬p ⊢ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_43: carmarria<br />
assumes 1: "p ∨ q" and<br />
2: "¬p" <br />
shows "q"<br />
proof -<br />
have " p | q" using 1 by this<br />
moreover {assume 3: p<br />
have 4: False using 2 3 by (rule notE)<br />
have 5: q using 4 by (rule FalseE)<br />
}<br />
moreover {assume 6: q<br />
}<br />
ultimately show q by (rule disjE)<br />
qed<br />
<br />
lemma ej_43: joslopjim4<br />
assumes "p ∨ q"<br />
"¬p" <br />
shows "q"<br />
using assms(1)<br />
proof<br />
assume "p"<br />
show "q" using assms(2) `p` by (rule notE)<br />
next<br />
assume "q"<br />
show "q" using `q` by this<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 44. Demostrar<br />
p ∨ q ⊢ ¬(¬p ∧ ¬q)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_44: carmarria<br />
assumes 1: "p ∨ q" <br />
shows "¬(¬p ∧ ¬q)"<br />
proof -<br />
have "p | q" using 1 by this<br />
moreover {assume 2: p<br />
{assume 3:"~p & ~q"<br />
have 4: "~p" using 3 by (rule conjunct1)<br />
have 5: False using 4 2 by (rule notE)<br />
}<br />
hence 6: "~(~p & ~q)" by (rule notI)<br />
}<br />
moreover {assume 7: q<br />
{assume 8: "~p & ~q"<br />
have 9: "~q" using 8 by (rule conjunct2)<br />
have 10: False using 9 7 by (rule notE)<br />
}<br />
hence 11: "~(~p & ~q)" by (rule notI)<br />
}<br />
ultimately show "~(~p & ~q)" by (rule disjE)<br />
qed<br />
<br />
lemma ej_44: joslopjim4<br />
assumes "p∨q"<br />
shows "¬(¬p∧¬q)"<br />
proof (rule notI)<br />
assume "¬p∧¬q"<br />
show False<br />
using assms<br />
proof<br />
assume "p"<br />
have "¬p" using `¬p∧¬q` by (rule conjunct1)<br />
show False using `¬p` `p` by (rule notE)<br />
next<br />
assume "q"<br />
have "¬q" using `¬p∧¬q` by (rule conjunct2)<br />
show False using `¬q` `q` by (rule notE)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 45. Demostrar<br />
p ∧ q ⊢ ¬(¬p ∨ ¬q)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_45: carmarria<br />
assumes 1: "p ∧ q" <br />
shows "¬(¬p ∨ ¬q)"<br />
proof -<br />
{assume 2: "~p | ~q"<br />
moreover {assume 3: "~p"<br />
have 4: p using 1 by (rule conjunct1)<br />
have 5: False using 3 4 by (rule notE)<br />
}<br />
moreover {assume 6: "~q"<br />
have 7: q using 1 by (rule conjunct2)<br />
have 8: False using 6 7 by (rule notE)<br />
}<br />
ultimately have False by (rule disjE)<br />
}<br />
thus "~(~p | ~q)" by (rule notI)<br />
qed<br />
<br />
lemma ej_45: joslopjim4<br />
assumes "p∧q"<br />
shows "¬(¬p∨¬q)"<br />
proof (rule notI)<br />
assume "¬p∨¬q"<br />
then show False<br />
proof (rule disjE)<br />
assume "¬p"<br />
have "p" using assms by (rule conjunct1)<br />
show False using `¬p` `p` by (rule notE)<br />
next<br />
assume "¬q"<br />
have "q" using assms by (rule conjunct2)<br />
show False using `¬q` `q` by (rule notE)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 46. Demostrar<br />
¬(p ∨ q) ⊢ ¬p ∧ ¬q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_46: carmarria<br />
assumes 1: "¬(p ∨ q)" <br />
shows "¬p ∧ ¬q"<br />
proof -<br />
{assume 2: p<br />
have 3: "p | q" using 2 by (rule disjI1)<br />
have 4: False using 1 3 by (rule notE)<br />
}<br />
hence 5: "~p" by (rule notI)<br />
{assume 6: q<br />
have 7: "p | q" using 6 by (rule disjI2)<br />
have 8: False using 1 7 by (rule notE)<br />
}<br />
hence 9: "~q" by (rule notI)<br />
show "~p & ~q" using 5 9 by (rule conjI)<br />
qed<br />
<br />
lemma ej_46: joslopjim4<br />
assumes "¬(p ∨ q)"<br />
shows "¬p ∧ ¬q"<br />
proof (rule conjI)<br />
show "¬p"<br />
proof (rule ccontr)<br />
assume "¬¬p"<br />
have "p" using `¬¬p` by (rule notnotD)<br />
have "p∨q" using `p` by (rule disjI1)<br />
show False using assms `p∨q` by (rule notE)<br />
qed<br />
show "¬q"<br />
proof (rule ccontr)<br />
assume "¬¬q"<br />
have "q" using `¬¬q` by (rule notnotD)<br />
have "p∨q" using `q` by (rule disjI2)<br />
show False using assms `p∨q` by (rule notE)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 47. Demostrar<br />
¬p ∧ ¬q ⊢ ¬(p ∨ q)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_47: carmarria<br />
assumes 1: "¬p ∧ ¬q" <br />
shows "¬(p ∨ q)"<br />
proof -<br />
{assume 2: "p | q"<br />
moreover {assume 3: p<br />
have 4: "~p" using 1 by (rule conjunct1)<br />
have 5: False using 4 3 by (rule notE)<br />
}<br />
moreover {assume 6: q <br />
have 7: "~q" using 1 by (rule conjunct2)<br />
have 8: False using 7 6 by (rule notE)<br />
}<br />
ultimately have 9: False by (rule disjE)<br />
}<br />
thus "~(p | q)" by (rule notI)<br />
qed <br />
<br />
lemma ej_47: joslopjim4<br />
assumes "¬p∧¬q"<br />
shows "¬(p∨q)"<br />
proof (rule notI)<br />
assume "p∨q"<br />
then show False<br />
proof<br />
assume "p"<br />
have "¬p" using assms by (rule conjunct1)<br />
show False using `¬p` `p` by (rule notE)<br />
next<br />
assume "q"<br />
have "¬q" using assms by (rule conjunct2)<br />
show False using `¬q` `q` by (rule notE)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 48. Demostrar<br />
¬p ∨ ¬q ⊢ ¬(p ∧ q)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_48: carmarria<br />
assumes 1: "¬p ∨ ¬q"<br />
shows "¬(p ∧ q)"<br />
proof -<br />
have "~p | ~q" using 1 by this<br />
moreover {assume 2: "~p"<br />
{assume 3: "p & q"<br />
have 4: p using 3 by (rule conjunct1)<br />
have 5: False using 2 4 by (rule notE)<br />
}<br />
hence 6: "~(p & q)" by (rule notI)<br />
}<br />
moreover {assume 7: "~q"<br />
{assume 8: "p & q"<br />
have 9: q using 8 by (rule conjunct2)<br />
have 10: False using 7 9 by (rule notE)<br />
}<br />
hence 11: "~(p & q)" by (rule notI)<br />
}<br />
ultimately show "~(p & q)" by (rule disjE)<br />
qed<br />
<br />
lemma ej_48: joslopjim4<br />
assumes "¬p ∨ ¬q"<br />
shows "¬(p ∧ q)"<br />
proof (rule notI)<br />
assume "p∧q"<br />
have "p" using `p∧q` by (rule conjunct1)<br />
have "q" using `p∧q` by (rule conjunct2)<br />
show False<br />
using assms<br />
proof<br />
assume "¬p"<br />
have "p" using `p∧q` by (rule conjunct1)<br />
show False using `¬p` `p` by (rule notE)<br />
next<br />
assume "¬q"<br />
have "q" using `p∧q` by (rule conjunct2)<br />
show False using `¬q` `q` by (rule notE)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 49. Demostrar<br />
⊢ ¬(p ∧ ¬p)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_49: carmarria<br />
"¬(p ∧ ¬p)"<br />
proof -<br />
{assume 1: "p & ~p"<br />
have 2: p using 1 by (rule conjunct1)<br />
have 3: "~p" using 1 by (rule conjunct2)<br />
have 4: False using 3 2 by (rule notE)<br />
}<br />
thus "~(p & ~p)" by (rule notI)<br />
qed<br />
<br />
lemma ej_49: joslopjim4<br />
shows "¬(p∧¬p)"<br />
proof<br />
assume "p∧¬p"<br />
have "p" using `p∧¬p` by (rule conjunct1)<br />
have "¬p" using `p∧¬p` by (rule conjunct2)<br />
show False using `¬p` `p` by (rule notE)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 50. Demostrar<br />
p ∧ ¬p ⊢ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_50: carmarria<br />
assumes 1: "p ∧ ¬p" <br />
shows "q"<br />
proof -<br />
have 2: p using 1 by (rule conjunct1)<br />
have 3: "~p" using 1 by (rule conjunct2)<br />
have 4: False using 3 2 by (rule notE)<br />
show q using 4 by (rule FalseE)<br />
qed<br />
<br />
lemma ej_50: joslopjim4<br />
assumes "p∧¬p"<br />
shows "q"<br />
proof-<br />
have "p" using `p∧¬p` by (rule conjunct1)<br />
have "¬p" using `p∧¬p` by (rule conjunct2)<br />
have False using `¬p` `p` by (rule notE)<br />
then show "q" by (rule FalseE)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 51. Demostrar<br />
¬¬p ⊢ p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_51: carmarria joslopjim4<br />
assumes "¬¬p"<br />
shows "p"<br />
proof -<br />
show p using assms by (rule notnotD)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 52. Demostrar<br />
⊢ p ∨ ¬p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_52: carmarria<br />
"p ∨ ¬p"<br />
proof -<br />
{assume 1: "~(p | ~p)"<br />
{assume 2: p <br />
have 3: "p | ~p" using 2 by (rule disjI1)<br />
have 4: False using 1 3 by (rule notE)<br />
}<br />
hence 5: "~p" by (rule notI)<br />
have 6: "p | ~p" using 5 by (rule disjI2)<br />
have 7: False using 1 6 by (rule notE)<br />
}<br />
hence 8: "~~(p | ~p)" by (rule notI)<br />
show "p | ~p" using 8 by (rule notnotD)<br />
qed<br />
<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 53. Demostrar<br />
⊢ ((p ⟶ q) ⟶ p) ⟶ p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ej_53: joslopjim4<br />
shows "((p ⟶ q) ⟶ p) ⟶ p"<br />
proof (rule impI)<br />
assume "(p ⟶ q) ⟶ p"<br />
show "p"<br />
proof (rule ccontr)<br />
assume "¬p"<br />
have "¬(p⟶q)" using `(p ⟶ q) ⟶ p` `¬p` by (rule mt)<br />
have "p⟶q"<br />
proof (rule impI)<br />
assume "p"<br />
show "q" using `¬p` `p` by (rule notE)<br />
qed<br />
show False using `¬(p⟶q)` `p⟶q` by (rule notE)<br />
qed<br />
qed<br />
<br />
lemma ejercicio_53: carmarria<br />
"((p ⟶ q) ⟶ p) ⟶ p"<br />
proof -<br />
{assume 1: "(p ⟶ q) ⟶ p" <br />
{assume 2:"~p"<br />
have 3: "~(p ⟶q)" using 1 2 by (rule mt)<br />
{assume 4: p<br />
have 5: False using 2 4 by (rule notE)<br />
have 6: q using 5 by (rule FalseE)<br />
}<br />
hence 7: "p ⟶q" by (rule impI)<br />
have 8: False using 3 7 by (rule notE)<br />
}<br />
hence 9: "~~p" by (rule notI)<br />
have 10: p using 9 by (rule notnotD)<br />
}<br />
thus "((p ⟶ q)⟶p)⟶p" by (rule impI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 54. Demostrar<br />
¬q ⟶ ¬p ⊢ p ⟶ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ej_54: joslopjim4<br />
assumes "¬q ⟶ ¬p"<br />
shows "p ⟶ q"<br />
proof (rule impI)<br />
assume "p"<br />
have "¬¬p" using `p` by (rule notnotI)<br />
have "¬¬q" using assms `¬¬p` by (rule mt)<br />
show "q" using `¬¬q` by (rule notnotD)<br />
qed<br />
<br />
lemma ejercicio_54: carmarria<br />
assumes 1: "¬q ⟶ ¬p"<br />
shows "p ⟶ q"<br />
proof -<br />
{assume 2: p<br />
have 3: "~~p" using 2 by (rule notnotI)<br />
have 4: "~~q" using 1 3 by (rule mt)<br />
have 5: q using 4 by (rule notnotD)<br />
}<br />
thus "p ⟶q" by (rule impI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 55. Demostrar<br />
¬(¬p ∧ ¬q) ⊢ p ∨ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_55: carmarria<br />
assumes 1: "¬(¬p ∧ ¬q)"<br />
shows "p ∨ q"<br />
proof - <br />
{assume 2: "~(p | q)"<br />
{assume 3: "~p"<br />
{assume 4: "~q"<br />
have 5: "~p & ~q" using 3 4 by (rule conjI)<br />
have 6: False using 1 5 by (rule notE)<br />
}<br />
hence 7: "~~q" by (rule notI)<br />
have 8: q using 7 by (rule notnotD)<br />
have 9: "p | q" using 8 by (rule disjI2)<br />
have 10: False using 2 9 by (rule notE)<br />
}<br />
hence 11: "~~p" by (rule notI)<br />
have 12: p using 11 by(rule notnotD)<br />
have 13: "p | q" using 12 by (rule disjI1)<br />
have 14: False using 2 13 by (rule notE)<br />
}<br />
hence 15: "~~(p | q)" by (rule notI)<br />
show "p | q" using 15 by (rule notnotD)<br />
qed<br />
<br />
lemma ej_55: joslopjim4<br />
assumes "¬(¬p ∧ ¬q)"<br />
shows "p ∨ q"<br />
proof-<br />
have "¬p∨p" by (rule excluded_middle)<br />
thus "p∨q" <br />
proof (rule disjE)<br />
assume "¬p"<br />
thus "p∨q"<br />
proof-<br />
have "¬q∨q" by (rule excluded_middle)<br />
thus "p∨q"<br />
proof<br />
assume "¬q"<br />
have "¬p∧¬q" using `¬p` `¬q` by (rule conjI)<br />
have False using assms `¬p∧¬q` by (rule notE)<br />
then show "p∨q" by (rule FalseE)<br />
next<br />
assume "q"<br />
show "p∨q" using `q` by (rule disjI2)<br />
qed<br />
qed<br />
next<br />
assume "p"<br />
show "p∨q" using `p` by (rule disjI1)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 56. Demostrar<br />
¬(¬p ∨ ¬q) ⊢ p ∧ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_56: carmarria<br />
assumes 1: "¬(¬p ∨ ¬q)" <br />
shows "p ∧ q"<br />
proof -<br />
{assume 2: "~p"<br />
have 3: "~p | ~q" using 2 by (rule disjI1)<br />
have 4: False using 1 3 by (rule notE)<br />
}<br />
hence 5: "~~p" by (rule notI)<br />
have 6: p using 5 by (rule notnotD)<br />
{assume 7: "~q"<br />
have 8: "~p | ~q" using 7 by (rule disjI2)<br />
have 9: False using 1 8 by (rule notE)<br />
}<br />
hence 10: "~~q" by (rule notI)<br />
have 11: q using 10 by (rule notnotD)<br />
show 12: " p & q" using 6 11 by (rule conjI)<br />
qed<br />
<br />
lemma ej_56: joslopjim4<br />
assumes "¬(¬p ∨ ¬q)" <br />
shows "p ∧ q"<br />
proof-<br />
have "¬p∨p" by (rule excluded_middle)<br />
thus "p∧q" <br />
proof<br />
assume "¬p"<br />
have "¬p∨¬q" using `¬p` by (rule disjI1)<br />
have False using assms `¬p∨¬q` by (rule notE)<br />
then show "p∧q" by (rule FalseE)<br />
next<br />
assume "p"<br />
have "¬q∨q" by (rule excluded_middle)<br />
thus "p∧q"<br />
proof<br />
assume "q"<br />
show "p∧q" using `p` `q` by (rule conjI)<br />
next<br />
assume "¬q"<br />
have "¬p∨¬q" using `¬q` by (rule disjI2)<br />
have False using assms `¬p∨¬q` by (rule notE)<br />
then show "p∧q" by (rule FalseE)<br />
qed<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 57. Demostrar<br />
¬(p ∧ q) ⊢ ¬p ∨ ¬q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_57: carmarria<br />
assumes 1: "¬(p ∧ q)"<br />
shows "¬p ∨ ¬q"<br />
proof -<br />
{assume 2: "~(~p | ~q)"<br />
{assume 3: p<br />
{assume 4: q<br />
have 5: "p & q" using 3 4 by (rule conjI)<br />
have 6: False using 1 5 by (rule notE)<br />
}<br />
hence 7: "~q" by (rule notI)<br />
have 8: "~p | ~q" using 7 by (rule disjI2)<br />
have 9: False using 2 8 by (rule notE)<br />
}<br />
hence 10: "~p" by (rule notI)<br />
have 11: "~p | ~q" using 10 by (rule disjI1)<br />
have 12: False using 2 11 by (rule notE)<br />
}<br />
hence 13: "~~(~p | ~q)" by (rule notI)<br />
show "~p | ~q" using 13 by (rule notnotD)<br />
qed<br />
<br />
lemma ej_57: joslopjim4<br />
assumes "¬(p ∧ q)"<br />
shows "¬p ∨ ¬q"<br />
proof-<br />
have "¬p∨p" by (rule excluded_middle)<br />
thus "¬p∨¬q"<br />
proof (rule disjE)<br />
assume "¬p"<br />
show "¬p∨¬q" using `¬p` by (rule disjI1)<br />
next<br />
assume "p"<br />
have "¬q∨q" by (rule excluded_middle)<br />
thus "¬p∨¬q"<br />
proof (rule disjE)<br />
assume "¬q"<br />
show "¬p∨¬q" using `¬q` by (rule disjI2)<br />
next<br />
assume "q"<br />
have "p∧q" using `p` `q` by (rule conjI)<br />
have False using assms `p∧q` by (rule notE)<br />
then show "¬p∨¬q" by (rule FalseE)<br />
qed<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 58. Demostrar<br />
⊢ (p ⟶ q) ∨ (q ⟶ p)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ej_58: joslopjim4<br />
shows "(p ⟶ q) ∨ (q ⟶ p)"<br />
proof -<br />
have "¬p∨p" by (rule excluded_middle)<br />
thus "(p ⟶ q) ∨ (q ⟶ p)"<br />
proof (rule disjE)<br />
assume "¬p"<br />
show "(p ⟶ q) ∨ (q ⟶ p)"<br />
proof (rule disjI1)<br />
show "p⟶q"<br />
proof<br />
assume "p"<br />
show "q" using `¬p` `p` by (rule notE)<br />
qed<br />
qed<br />
next<br />
assume "p"<br />
show "(p ⟶ q) ∨ (q ⟶ p)"<br />
proof (rule disjI2)<br />
show "q⟶p"<br />
proof (rule impI)<br />
assume "q"<br />
show "p" using `p` by this<br />
qed<br />
qed<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 59 (EXTRA). Demostrar<br />
p∧¬(q⟶r) ⊢ (p∧q)∧¬r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ej_59: joslopjim4<br />
assumes "p∧¬(q⟶r)"<br />
shows "(p∧q)∧¬r"<br />
proof<br />
show "p∧q"<br />
proof<br />
show "p" using assms by (rule conjunct1)<br />
next<br />
show "q"<br />
proof (rule ccontr)<br />
assume "¬q"<br />
have "¬(q⟶r)" using assms by (rule conjunct2)<br />
have "q⟶r"<br />
proof<br />
assume "q"<br />
show "r" using `¬q` `q` by (rule notE)<br />
qed<br />
show False using `¬(q⟶r)` `q⟶r` by (rule notE)<br />
qed<br />
qed<br />
next<br />
show "¬r"<br />
proof (rule notI)<br />
assume "r"<br />
have "q⟶r"<br />
proof<br />
assume "q" <br />
show "r" using `r` by this<br />
qed<br />
have "¬(q⟶r)" using assms ..<br />
thus False using `q⟶r` ..<br />
qed<br />
qed<br />
<br />
end<br />
<br />
</source></div>
Joslopjim4
https://www.glc.us.es/~jalonso/LMF2018/index.php?title=Relaci%C3%B3n_4&diff=138
Relación 4
2018-03-22T10:44:53Z
<p>Joslopjim4: </p>
<hr />
<div><source lang = "isar"><br />
<br />
chapter {* R4: Deducción natural proposicional *}<br />
<br />
theory R4<br />
imports Main <br />
begin<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
El objetivo de esta relación es demostrar cada uno de los ejercicios<br />
usando sólo las reglas básicas de deducción natural de la lógica<br />
proposicional (sin usar el método auto).<br />
<br />
Las reglas básicas de la deducción natural son las siguientes:<br />
· conjI: ⟦P; Q⟧ ⟹ P ∧ Q<br />
· conjunct1: P ∧ Q ⟹ P<br />
· conjunct2: P ∧ Q ⟹ Q <br />
· notnotD: ¬¬ P ⟹ P<br />
· notnotI: P ⟹ ¬¬ P<br />
· mp: ⟦P ⟶ Q; P⟧ ⟹ Q <br />
· mt: ⟦F ⟶ G; ¬G⟧ ⟹ ¬F <br />
· impI: (P ⟹ Q) ⟹ P ⟶ Q<br />
· disjI1: P ⟹ P ∨ Q<br />
· disjI2: Q ⟹ P ∨ Q<br />
· disjE: ⟦P ∨ Q; P ⟹ R; Q ⟹ R⟧ ⟹ R <br />
· FalseE: False ⟹ P<br />
· notE: ⟦¬P; P⟧ ⟹ R<br />
· notI: (P ⟹ False) ⟹ ¬P<br />
· iffI: ⟦P ⟹ Q; Q ⟹ P⟧ ⟹ P = Q<br />
· iffD1: ⟦Q = P; Q⟧ ⟹ P <br />
· iffD2: ⟦P = Q; Q⟧ ⟹ P<br />
· ccontr: (¬P ⟹ False) ⟹ P<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
text {*<br />
Se usarán las reglas notnotI y mt que demostramos a continuación. *}<br />
<br />
lemma notnotI: "P ⟹ ¬¬ P"<br />
by auto<br />
<br />
lemma mt: "⟦F ⟶ G; ¬G⟧ ⟹ ¬F"<br />
by auto<br />
<br />
section {* Implicaciones *}<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 1. Demostrar<br />
p ⟶ q, p ⊢ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_1: josrodjim2 carmarria inmbenber marmedmar3<br />
assumes 1: "p ⟶ q" and<br />
2: "p"<br />
shows "q"<br />
<br />
proof -<br />
show 1: "q" using 1 2 by (rule mp)<br />
qed<br />
<br />
lemma ej_1: joslopjim4<br />
assumes "p --> q"<br />
"p"<br />
shows "q"<br />
proof -<br />
show "q" using assms(1) assms(2) by (rule mp)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 2. Demostrar<br />
p ⟶ q, q ⟶ r, p ⊢ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_2: josrodjim2 carmarria inmbenber marmedmar3<br />
assumes 1: "p ⟶ q" and<br />
2: "q ⟶ r" and<br />
3: "p" <br />
shows "r"<br />
<br />
proof-<br />
have 4: "q" using 1 3 by (rule mp)<br />
show "r" using 2 4 by (rule mp)<br />
qed<br />
<br />
lemma ej_2: joslopjim4<br />
assumes "p ⟶ q"<br />
"q ⟶ r"<br />
"p" <br />
shows "r"<br />
proof -<br />
have "q" using `p⟶q` `p` by (rule mp)<br />
show "r" using `q⟶r` `q` by (rule mp)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 3. Demostrar<br />
p ⟶ (q ⟶ r), p ⟶ q, p ⊢ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_3: josrodjim2 carmarria inmbenber marmedmar3<br />
assumes 1: "p ⟶ (q ⟶ r)" and<br />
2: "p ⟶ q" and<br />
3: "p"<br />
shows "r"<br />
<br />
proof-<br />
have 4: "q" using 2 3 by (rule mp)<br />
have 5: "(q ⟶ r)" using 1 3 by (rule mp)<br />
show "r" using 5 4 by (rule mp)<br />
<br />
qed<br />
<br />
lemma ej_3: joslopjim4<br />
assumes "p ⟶ (q ⟶ r)"<br />
"p ⟶ q"<br />
"p"<br />
shows "r"<br />
proof -<br />
have "q" using `p⟶q` `p` by (rule mp)<br />
have "q⟶r" using `p ⟶ (q ⟶ r)` `p` by (rule mp)<br />
show "r" using `q⟶r` `q` by (rule mp)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 4. Demostrar<br />
p ⟶ q, q ⟶ r ⊢ p ⟶ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_4: josrodjim2<br />
assumes 1: "p ⟶ q" and<br />
2: "q ⟶ r" <br />
shows "p ⟶ r"<br />
<br />
proof-<br />
{ assume 3: "p"<br />
have 4: "q" using 1 3 by (rule mp)<br />
have 5: "r" using 2 4 by (rule mp)}<br />
hence 6: "p ⟶ r" using 3 5 by (rule impI)<br />
show "p⟶r" using 6 by this<br />
<br />
oops<br />
<br />
lemma ejercicio_4: carmarria<br />
assumes 1: "p ⟶ q" and<br />
2: "q ⟶ r" <br />
shows "p ⟶ r"<br />
proof -<br />
{ assume 3: p <br />
have 4: "q" using 1 3 by (rule mp)<br />
have 5: "r" using 2 4 by (rule mp)<br />
}<br />
thus "p ⟶ r" by (rule impI) <br />
qed<br />
lemma ejercicio_4: inmbenber<br />
assumes 1: "p ⟶ q" and<br />
2: "q ⟶ r" <br />
shows "p ⟶ r"<br />
proof -<br />
{ assume 3: "p" <br />
have 4: "q" using 1 3 by (rule mp)<br />
have 5: "r" using 2 4 by (rule mp) }<br />
thus "p ⟶ r" by (rule impI) <br />
qed<br />
<br />
lemma ejercicio_4:marmedmar3<br />
assumes "p ⟶ q"<br />
"q ⟶ r" <br />
shows "p ⟶ r"<br />
proof <br />
assume "p" <br />
with assms(1) have "q" by (rule mp)<br />
with assms(2) show "r" by (rule mp)<br />
qed <br />
<br />
lemma ej_4: joslopjim4<br />
assumes "p ⟶ q"<br />
"q ⟶ r" <br />
shows "p ⟶ r"<br />
proof (rule impI)<br />
assume "p"<br />
have "q" using assms(1) `p` by (rule mp)<br />
show "r" using assms(2) `q` by (rule mp)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 5. Demostrar<br />
p ⟶ (q ⟶ r) ⊢ q ⟶ (p ⟶ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_5: josrodjim2 (NO DA ERROR, PERO SI ALERTA)<br />
assumes "p ⟶ (q ⟶ r)" <br />
shows "q ⟶ (p ⟶ r)"<br />
<br />
proof -<br />
<br />
{ assume 2: "q"<br />
{assume 3: "p"<br />
have 4: "q ⟶r" using 1 3 by (rule mp)<br />
have 5: "r" using 4 2 by (rule mp)}<br />
hence 6: "p⟶r" using 3 5 by (rule impI)}<br />
hence 7: "q ⟶ (p ⟶ r)" using 2 6 by (rule impI)<br />
show "q ⟶ (p ⟶ r)" using 7 by this<br />
<br />
oops<br />
<br />
lemma ejercicio_5: carmarria <br />
assumes 1: "p ⟶ (q ⟶ r)" <br />
shows "q ⟶ (p ⟶ r)"<br />
<br />
proof - <br />
{assume 2: q<br />
{assume 3: p <br />
have 4: "q ⟶ r" using 1 3 by (rule mp)<br />
have 5: "r" using 4 2 by (rule mp)<br />
}<br />
hence " p ⟶ r" by (rule impI)<br />
}<br />
thus "q ⟶ (p ⟶ r)" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_5: inmbenber<br />
assumes 1: "p ⟶ (q ⟶ r)" <br />
shows "q ⟶ (p ⟶ r)"<br />
<br />
proof - <br />
{assume 2: "q"<br />
{assume 3: "p" <br />
have 4: "q ⟶ r" using 1 3 by (rule mp)<br />
have 5: "r" using 4 2 by (rule mp) }<br />
hence " p ⟶ r" by (rule impI) }<br />
thus "q ⟶ (p ⟶ r)" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_5: marmedmar3<br />
assumes "p ⟶ (q ⟶ r)" <br />
shows "q ⟶ (p ⟶ r)"<br />
proof <br />
assume "q"<br />
show "(p ⟶ r)"<br />
proof <br />
assume "p" <br />
with assms(1) have "(q ⟶ r)" by (rule mp)<br />
thus "r" using `q` by (rule mp) <br />
qed <br />
qed<br />
<br />
lemma ej_5: joslopjim4<br />
assumes "p ⟶ (q ⟶ r)" <br />
shows "q ⟶ (p ⟶ r)"<br />
proof (rule impI)<br />
assume "q"<br />
show "p⟶r"<br />
proof (rule impI)<br />
assume "p"<br />
have "q⟶r" using `p ⟶ (q ⟶ r)` `p` by (rule mp)<br />
show "r" using `q⟶r` `q` by (rule mp)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 6. Demostrar<br />
p ⟶ (q ⟶ r) ⊢ (p ⟶ q) ⟶ (p ⟶ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_6:josrodjim2<br />
assumes 1: "p ⟶ (q ⟶ r)" <br />
shows "(p ⟶ q) ⟶ (p ⟶ r)"<br />
<br />
proof- <br />
<br />
{assume 2: "p⟶q" <br />
{assume 3: "p"<br />
have 4: "q⟶r" using 1 3 by (rule mp)<br />
have 5: "q" using 2 3 by (rule mp)<br />
have 6: "r" using 4 5 by (rule mp)}<br />
hence 7: "p⟶r" using 3 6 by (rule impI)}<br />
hence 8: "(p ⟶ q) ⟶ (p ⟶ r)" using 2 7 by (rule impI)<br />
show "(p ⟶ q) ⟶ (p ⟶ r)" using 8 by this<br />
oops<br />
<br />
lemma ejercicio_6: carmarria, inmbenber<br />
assumes 1: "p ⟶ (q ⟶ r)" <br />
shows "(p ⟶ q) ⟶ (p ⟶ r)"<br />
proof -<br />
{assume 2: "p ⟶ q"<br />
{assume 3: "p"<br />
have 4: "q" using 2 3 by (rule mp)<br />
have 5: "q ⟶ r" using 1 3 by (rule mp)<br />
have 6: "r" using 5 4 by (rule mp)<br />
}<br />
hence 7: "p ⟶ r" by (rule impI)<br />
}<br />
thus "(p ⟶ q) ⟶ (p ⟶ r)" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_6:marmedmar3<br />
assumes 1: "p ⟶ (q ⟶ r)" <br />
shows "(p ⟶ q) ⟶ (p ⟶ r)"<br />
proof <br />
assume 2: "(p ⟶ q)" <br />
show "(p ⟶ r)" <br />
proof <br />
assume 3: "p" <br />
have 4: "q" using 2 3 by (rule mp)<br />
have 5: "(q ⟶ r)" using 1 3 by (rule mp) <br />
show "r" using 5 4 by (rule mp) <br />
qed<br />
qed<br />
<br />
lemma ej_6: joslopjim4<br />
assumes "p ⟶ (q ⟶ r)" <br />
shows "(p ⟶ q) ⟶ (p ⟶ r)"<br />
proof (rule impI)<br />
assume "(p⟶q)"<br />
show "p⟶r"<br />
proof (rule impI)<br />
assume "p"<br />
have "q⟶r" using `p ⟶ (q ⟶ r)` `p` by (rule mp)<br />
have "q" using `p⟶q` `p` by (rule mp)<br />
show "r" using `q⟶r` `q` by (rule mp)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 7. Demostrar<br />
p ⊢ q ⟶ p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_7: (NO SE QUE ESTA MAL)<br />
assumes "p" <br />
shows "q ⟶ p"<br />
<br />
proof -<br />
<br />
{ assume 2: "q"<br />
have 3: "p" using 1<br />
}<br />
hence 4: "q⟶p" using 2 3 by (rule impI)<br />
<br />
show "q⟶p" using 4 by this<br />
<br />
oops<br />
<br />
lemma ejercicio_7: carmarria, inmbenber<br />
assumes 1: "p" <br />
shows "q ⟶ p"<br />
proof -<br />
{assume 2: "q"<br />
have 3: "p" using 1 by this}<br />
thus "q ⟶ p" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_7:marmedmar3<br />
assumes "p" <br />
shows "q ⟶ p"<br />
proof <br />
assume "q" <br />
show "p" using assms(1) by this<br />
qed <br />
<br />
lemma ej_7: joslopjim4<br />
assumes "p" <br />
shows "q ⟶ p"<br />
proof (rule impI)<br />
assume "q"<br />
show "p" using assms by this<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 8. Demostrar<br />
⊢ p ⟶ (q ⟶ p)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_8: carmarria<br />
"p ⟶ (q ⟶ p)"<br />
<br />
proof -<br />
{assume 1: p<br />
{assume 2: q<br />
have 3: p using 1 by this<br />
}<br />
hence 4: "q ⟶ p" by (rule impI)<br />
}<br />
thus "p ⟶ (q ⟶ p)" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_8: marmedmar3<br />
"p ⟶ (q ⟶ p)"<br />
proof<br />
assume 1: "p"<br />
show "(q ⟶ p)"<br />
proof <br />
assume 2: "q" <br />
show "p" using 1 by this <br />
qed<br />
qed<br />
<br />
lemma ej_8: joslopjim4<br />
shows "p⟶(q⟶p)"<br />
proof (rule impI)<br />
assume "p"<br />
show "q⟶p"<br />
proof (rule impI)<br />
assume "q"<br />
show "p" using `p` by this<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 9. Demostrar<br />
p ⟶ q ⊢ (q ⟶ r) ⟶ (p ⟶ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_9: carmarria<br />
assumes 1: "p ⟶ q" <br />
shows "(q ⟶ r) ⟶ (p ⟶ r)"<br />
proof -<br />
{assume 2: "q ⟶ r"<br />
{assume 3: "p"<br />
have 4: "q" using 1 3 by (rule mp)<br />
have 5: "r" using 2 4 by (rule mp)<br />
}<br />
hence 6: "p ⟶ r" by (rule impI)<br />
}<br />
thus 7: "(q ⟶ r) ⟶ (p ⟶ r)" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_9: marmedmar3<br />
assumes "p ⟶ q" <br />
shows "(q ⟶ r) ⟶ (p ⟶ r)"<br />
proof <br />
assume "(q ⟶ r)" <br />
show "(p ⟶ r)"<br />
proof <br />
assume "p" <br />
with `(p ⟶ q)` have "q" by (rule mp)<br />
with `(q ⟶ r)` show "r" by (rule mp)<br />
qed<br />
qed<br />
<br />
lemma ej_9: joslopjim4<br />
assumes "p ⟶ q" <br />
shows "(q ⟶ r) ⟶ (p ⟶ r)"<br />
proof (rule impI)<br />
assume "q⟶r"<br />
show "p⟶r"<br />
proof (rule impI)<br />
assume "p"<br />
have "q" using `p⟶q` `p` by (rule mp)<br />
show "r" using `q⟶r` `q` by (rule mp)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 10. Demostrar<br />
p ⟶ (q ⟶ (r ⟶ s)) ⊢ r ⟶ (q ⟶ (p ⟶ s))<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_10: carmarria<br />
assumes 1: "p ⟶ (q ⟶ (r ⟶ s))" <br />
shows "r ⟶ (q ⟶ (p ⟶ s))"<br />
<br />
proof-<br />
{assume 2: r<br />
{assume 3: q<br />
{assume 4: p<br />
have 5: "q ⟶ (r ⟶ s)" using 1 4 by (rule mp)<br />
have 6: "r ⟶ s" using 5 3 by (rule mp)<br />
have 7: s using 6 2 by (rule mp)<br />
}<br />
hence 8: "p ⟶ s" by (rule impI)<br />
}<br />
hence 9: "q ⟶ (p ⟶ s)" by (rule impI)<br />
}<br />
thus "r ⟶ (q ⟶ (p ⟶ s))" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_10: marmedmar3<br />
assumes 1: "p ⟶ (q ⟶ (r ⟶ s))" <br />
shows "r ⟶ (q ⟶ (p ⟶ s))"<br />
proof <br />
assume 2: "r"<br />
show "(q ⟶ (p ⟶ s))"<br />
proof <br />
assume 3: "q" <br />
show "(p ⟶ s)"<br />
proof <br />
assume 4: "p" <br />
have 5: "(q ⟶ (r ⟶ s))" using 1 4 by (rule mp) <br />
have 6: "(r ⟶ s)" using 5 3 by (rule mp)<br />
show "s" using 6 2 by (rule mp) <br />
qed<br />
qed<br />
qed<br />
<br />
lemma ej_10: joslopjim4<br />
assumes "p ⟶ (q ⟶ (r ⟶ s))" <br />
shows "r ⟶ (q ⟶ (p ⟶ s))"<br />
proof (rule impI)<br />
assume "r"<br />
show "q⟶(p⟶s)"<br />
proof (rule impI)<br />
assume "q"<br />
show "p⟶s"<br />
proof (rule impI)<br />
assume "p"<br />
have "q ⟶ (r ⟶ s)" using `p ⟶ (q ⟶ (r ⟶ s))` `p` by (rule mp)<br />
have "r⟶s" using `q ⟶ (r ⟶ s)` `q` by (rule mp)<br />
show "s" using `r-->s` `r` by (rule mp)<br />
qed<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 11. Demostrar<br />
⊢ (p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_11: carmarria<br />
"(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))"<br />
proof - <br />
{assume 1: "p ⟶ (q ⟶ r)"<br />
{assume 2: "p ⟶ q"<br />
{assume 3: p<br />
have 4: q using 2 3 by (rule mp)<br />
have 5: "q ⟶ r" using 1 3 by (rule mp)<br />
have 6: r using 5 4 by (rule mp)<br />
}<br />
hence 7: "p ⟶ r" by (rule impI)<br />
}<br />
hence 8: "(p ⟶ q) ⟶ (p ⟶ r)" by (rule impI)<br />
}<br />
thus "(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_11: marmedmar3<br />
"(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))"<br />
proof <br />
assume 1: "(p ⟶ (q ⟶ r))" <br />
show "((p ⟶ q) ⟶ (p ⟶ r))"<br />
proof <br />
assume 2: "(p ⟶ q)" <br />
show "(p ⟶ r)" <br />
proof <br />
assume 3: "p" <br />
have 4: "(q ⟶ r)" using 1 3 by (rule mp) <br />
have 5: "q" using 2 3 by (rule mp) <br />
show "r" using 4 5 by (rule mp) <br />
qed<br />
qed<br />
qed<br />
<br />
lemma ej_11: joslopjim4<br />
shows "(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))"<br />
proof (rule impI)<br />
assume "p⟶(q⟶r)"<br />
show "(p⟶q)⟶(p⟶r)"<br />
proof (rule impI)<br />
assume "p⟶q"<br />
show "p⟶r"<br />
proof (rule impI)<br />
assume "p"<br />
have "q" using `p⟶q` `p` by (rule mp)<br />
have "q⟶r" using `p⟶(q⟶r)` `p` by (rule mp)<br />
show "r" using `q⟶r` `q` by (rule mp)<br />
qed<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 12. Demostrar<br />
(p ⟶ q) ⟶ r ⊢ p ⟶ (q ⟶ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_12: carmarria<br />
assumes 1: "(p ⟶ q) ⟶ r" <br />
shows "p ⟶ (q ⟶ r)"<br />
proof -<br />
{assume 2: p<br />
{assume 3: q<br />
{assume 4: p<br />
have 5: q using 3 by this<br />
}<br />
hence 6: "p ⟶ q" by (rule impI)<br />
have 7: r using 1 6 by (rule mp)<br />
}<br />
hence 8: "q ⟶ r" by (rule impI)<br />
}<br />
thus "p ⟶ (q ⟶ r)" by (rule impI)<br />
qed<br />
<br />
lemma ej_12: joslopjim4<br />
assumes "(p ⟶ q) ⟶ r" <br />
shows "p ⟶ (q ⟶ r)"<br />
proof (rule impI)<br />
assume "p"<br />
show "q⟶r"<br />
proof (rule impI)<br />
assume "q"<br />
have "p⟶q"<br />
proof (rule impI)<br />
assume "p"<br />
show "q" using `q` by this<br />
qed<br />
show "r" using `(p ⟶ q) ⟶ r` `p⟶q` by (rule mp)<br />
qed<br />
qed<br />
<br />
section {* Conjunciones *}<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 13. Demostrar<br />
p, q ⊢ p ∧ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_13: carmarria marmedmar3<br />
assumes "p"<br />
"q" <br />
shows "p ∧ q"<br />
<br />
proof-<br />
show "p \<and> q" using assms(1) assms(2) by (rule conjI)<br />
qed<br />
<br />
lemma ej_13: joslopjim4<br />
assumes "p"<br />
"q" <br />
shows "p ∧ q"<br />
proof-<br />
show "p∧q" using `p` `q` by (rule conjI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 14. Demostrar<br />
p ∧ q ⊢ p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_14: carmarria marmedmar3<br />
assumes "p ∧ q" <br />
shows "p"<br />
<br />
proof -<br />
show "p" using assms by (rule conjunct1)<br />
qed<br />
<br />
lemma ej_14: joslopjim4<br />
assumes "p ∧ q" <br />
shows "p"<br />
proof -<br />
show "p" using assms by (rule conjunct1)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 15. Demostrar<br />
p ∧ q ⊢ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_15: carmarria marmedmar3<br />
assumes "p ∧ q" <br />
shows "q"<br />
<br />
proof -<br />
show q using assms(1) by (rule conjunct2)<br />
qed<br />
<br />
lemma ej_15: joslopjim4<br />
assumes "p ∧ q" <br />
shows "q"<br />
proof-<br />
show "q" using assms by (rule conjunct2)<br />
qed<br />
<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 16. Demostrar<br />
p ∧ (q ∧ r) ⊢ (p ∧ q) ∧ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_16: carmarria marmedmar3<br />
assumes 1: "p ∧ (q ∧ r)"<br />
shows "(p ∧ q) ∧ r"<br />
proof -<br />
have 2: "q \<and> r" using 1 by (rule conjunct2)<br />
have 3: p using 1 by (rule conjunct1)<br />
have 4: q using 2 by (rule conjunct1)<br />
have 5: r using 2 by (rule conjunct2)<br />
have 6: "p \<and> q" using 3 4 by (rule conjI)<br />
show "(p \<and> q) \<and> r" using 6 5 by (rule conjI)<br />
qed<br />
<br />
lemma ej_16: joslopjim4<br />
assumes "p ∧ (q ∧ r)"<br />
shows "(p ∧ q) ∧ r"<br />
proof-<br />
have "p" using assms by (rule conjunct1)<br />
have "q∧r" using assms by (rule conjunct2)<br />
have "q" using `q∧r` by (rule conjunct1)<br />
have "r" using `q∧r` by (rule conjunct2)<br />
have "p∧q" using `p` `q` by (rule conjI)<br />
show "(p∧q)∧r" using `p∧q` `r` by (rule conjI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 17. Demostrar<br />
(p ∧ q) ∧ r ⊢ p ∧ (q ∧ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_17:<br />
assumes "(p ∧ q) ∧ r" <br />
shows "p ∧ (q ∧ r)"<br />
proof (rule conjI)<br />
have "p ∧ q" using assms ..<br />
thus "p" ..<br />
next<br />
show "q ∧ r"<br />
proof (rule conjI)<br />
have "p ∧ q" using assms ..<br />
thus "q" ..<br />
next<br />
show "r" using assms ..<br />
qed<br />
qed<br />
<br />
lemma ejercicio_17: carmarria marmedmar3<br />
assumes 1:"(p ∧ q) ∧ r" <br />
shows "p ∧ (q ∧ r)"<br />
proof-<br />
have 2:"p \<and> q" using 1 by (rule conjunct1)<br />
have 3: r using 1 by (rule conjunct2)<br />
have 4: p using 2 by (rule conjunct1)<br />
have 5: q using 2 by (rule conjunct2)<br />
have 6: "q \<and> r" using 5 3 by (rule conjI)<br />
show "p \<and> (q \<and> r)" using 4 6 by (rule conjI)<br />
qed<br />
<br />
<br />
lemma ej_17: joslopjim4<br />
assumes "(p ∧ q) ∧ r" <br />
shows "p ∧ (q ∧ r)"<br />
proof-<br />
have "p∧q" using assms by (rule conjunct1)<br />
have "r" using assms by (rule conjunct2)<br />
have "p" using `p∧q` by (rule conjunct1)<br />
have "q" using `p∧q` by (rule conjunct2)<br />
have "q∧r" using `q` `r` by (rule conjI)<br />
show "p∧(q∧r)" using `p` `q∧r` by (rule conjI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 18. Demostrar<br />
p ∧ q ⊢ p ⟶ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_18: carmarria<br />
assumes "p ∧ q" <br />
shows "p ⟶ q"<br />
proof -<br />
{assume p<br />
have q using assms(1) by (rule conjunct2)<br />
}<br />
thus "p ⟶ q" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_18: marmedmar3<br />
assumes "p ∧ q" <br />
shows "p ⟶ q"<br />
proof <br />
assume "p" <br />
show "q" using assms by (rule conjunct2) <br />
qed <br />
<br />
lemma ej_18: joslopjim4<br />
assumes "p ∧ q" <br />
shows "p ⟶ q"<br />
proof (rule impI)<br />
assume "p"<br />
show "q" using assms by (rule conjunct2)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 19. Demostrar<br />
(p ⟶ q) ∧ (p ⟶ r) ⊢ p ⟶ q ∧ r <br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_19: carmarria<br />
assumes 1: "(p ⟶ q) ∧ (p ⟶ r)" <br />
shows "p ⟶ q ∧ r"<br />
proof -<br />
{assume 2: p<br />
have 3: "p ⟶ q" using 1 by (rule conjunct1)<br />
have 4: "p ⟶ r" using 1 by (rule conjunct2)<br />
have 5: q using 3 2 by (rule mp)<br />
have 6: r using 4 2 by (rule mp)<br />
have 7: "q \<and> r" using 5 6 by (rule conjI)<br />
}<br />
thus "p ⟶ q \<and> r" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_19: marmedmar3<br />
assumes 1: "(p ⟶ q) ∧ (p ⟶ r)" <br />
shows "p ⟶ q ∧ r"<br />
proof <br />
have 2: "(p ⟶ q)" using assms by (rule conjunct1)<br />
have 3: "(p ⟶ r)" using assms by (rule conjunct2)<br />
assume 4: "p"<br />
with `(p ⟶ q)` have 5: "q" by (rule mp)<br />
have 6: "r" using 3 4 by (rule mp) <br />
show "q ∧ r" using 5 6 by (rule conjI)<br />
qed <br />
<br />
lemma ej_19: joslopjim4<br />
assumes "(p ⟶ q) ∧ (p ⟶ r)" <br />
shows "p ⟶ q ∧ r"<br />
proof (rule impI)<br />
assume "p"<br />
have "p⟶q" using assms by (rule conjunct1)<br />
have "q" using `p⟶q` `p` by (rule mp)<br />
have "p⟶r" using assms by (rule conjunct2)<br />
have "r" using `p⟶r` `p` by (rule mp)<br />
show "q∧r" using `q` `r` by (rule conjI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 20. Demostrar<br />
p ⟶ q ∧ r ⊢ (p ⟶ q) ∧ (p ⟶ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_20: carmarria<br />
assumes 1: "p ⟶ q ∧ r" <br />
shows "(p ⟶ q) ∧ (p ⟶ r)"<br />
proof -<br />
{assume 2: p<br />
have 3: "q \<and> r" using 1 2 by (rule mp)<br />
have 4: q using 3 by (rule conjunct1)<br />
}<br />
hence 5: "p ⟶ q" by (rule impI)<br />
{assume 6: p<br />
have 7: "q \<and> r" using 1 6 by (rule mp)<br />
have 8: r using 7 by (rule conjunct2)<br />
}<br />
hence 9: "p ⟶ r" by (rule impI)<br />
show "(p ⟶ q) \<and> (p ⟶ r)" using 5 9 by (rule conjI)<br />
qed<br />
<br />
lemma ej_20: joslopjim4<br />
assumes "p ⟶ q ∧ r" <br />
shows "(p ⟶ q) ∧ (p ⟶ r)"<br />
proof-<br />
have "p⟶q"<br />
proof<br />
assume "p"<br />
have "q∧r" using assms `p` by (rule mp)<br />
show "q" using `q∧r` by (rule conjunct1)<br />
qed<br />
have "p⟶r"<br />
proof<br />
assume "p"<br />
have "q∧r" using assms `p` by (rule mp)<br />
show "r" using `q∧r` by (rule conjunct2)<br />
qed<br />
show "(p⟶q)∧(p⟶r)" using `p⟶q` `p⟶r` by (rule conjI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 21. Demostrar<br />
p ⟶ (q ⟶ r) ⊢ p ∧ q ⟶ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_21: carmarria<br />
assumes 1: "p ⟶ (q ⟶ r)" <br />
shows "p ∧ q ⟶ r"<br />
proof -<br />
{assume 2: "p \<and> q"<br />
have 3: p using 2 by (rule conjunct1)<br />
have 4: q using 2 by (rule conjunct2)<br />
have 5: "q ⟶ r" using 1 3 by (rule mp)<br />
have 6: r using 5 4 by (rule mp)<br />
}<br />
thus "p ∧ q ⟶ r" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_21: marmedmar3<br />
assumes "p ⟶ (q ⟶ r)" <br />
shows "p ∧ q ⟶ r"<br />
proof <br />
assume "p ∧ q" <br />
hence "p" by (rule conjunct1) <br />
with assms have "(q ⟶ r)" by (rule mp)<br />
have "q" using `p ∧ q` by (rule conjunct2) <br />
with `(q ⟶ r)` show "r" by (rule mp) <br />
qed <br />
<br />
lemma ej_21: joslopjim4<br />
assumes "p ⟶ (q ⟶ r)" <br />
shows "p ∧ q ⟶ r"<br />
proof (rule impI)<br />
assume "p∧q"<br />
have "p" using `p∧q` by (rule conjunct1)<br />
have "q" using `p∧q` by (rule conjunct2)<br />
have "q⟶r" using assms `p` by (rule mp)<br />
show "r" using `q⟶r` `q` by (rule mp)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 22. Demostrar<br />
p ∧ q ⟶ r ⊢ p ⟶ (q ⟶ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_22: carmarria<br />
assumes "p ∧ q ⟶ r" <br />
shows "p ⟶ (q ⟶ r)"<br />
proof -<br />
{assume p<br />
{assume q<br />
have "p \<and> q" using `p` `q` by (rule conjI)<br />
have r using assms(1) `p \<and> q` by (rule mp)<br />
}<br />
hence "q ⟶ r" by (rule impI)<br />
}<br />
thus "p ⟶ (q ⟶ r)" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_22: marmedmar3<br />
assumes "p ∧ q ⟶ r" <br />
shows "p ⟶ (q ⟶ r)"<br />
proof <br />
assume "p" <br />
show "(q ⟶ r)"<br />
proof <br />
assume "q" <br />
with `p` have "p ∧ q" by (rule conjI) <br />
with assms show "r" by (rule mp) <br />
qed <br />
qed<br />
<br />
lemma ej_22: joslopjim4<br />
assumes "p ∧ q ⟶ r" <br />
shows "p ⟶ (q ⟶ r)"<br />
proof (rule impI)<br />
assume "p"<br />
show "q⟶r"<br />
proof (rule impI)<br />
assume "q"<br />
have "p∧q" using `p` `q` by (rule conjI)<br />
show "r" using assms `p∧q` by (rule mp)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 23. Demostrar<br />
(p ⟶ q) ⟶ r ⊢ p ∧ q ⟶ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_23: carmarria<br />
assumes "(p ⟶ q) ⟶ r" <br />
shows "p ∧ q ⟶ r"<br />
proof -<br />
{assume "p \<and> q" <br />
{assume p<br />
have q using `p \<and> q` by (rule conjunct2)<br />
}<br />
hence "p ⟶ q" by (rule impI)<br />
have r using assms(1) `p ⟶ q` by (rule mp)<br />
}<br />
thus "p \<and> q ⟶ r" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_23: marmedmar3<br />
assumes "(p ⟶ q) ⟶ r" <br />
shows "p ∧ q ⟶ r"<br />
proof <br />
assume "p ∧ q"<br />
have "p" using `p ∧ q` by (rule conjunct1)<br />
have "q" using `p ∧ q` by (rule conjunct2) <br />
hence "(p ⟶ q)" by (rule impI) <br />
with assms show "r" by (rule mp) <br />
qed <br />
<br />
lemma ej_23: joslopjim4<br />
assumes "(p ⟶ q) ⟶ r" <br />
shows "p ∧ q ⟶ r"<br />
proof (rule impI)<br />
assume "p∧q"<br />
have "p⟶q"<br />
proof<br />
assume "p"<br />
show "q" using `p∧q` by (rule conjunct2)<br />
qed<br />
show "r" using assms `p⟶q` by (rule mp)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 24. Demostrar<br />
p ∧ (q ⟶ r) ⊢ (p ⟶ q) ⟶ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_24: carmarria<br />
assumes 1: "p ∧ (q ⟶ r)" <br />
shows "(p ⟶ q) ⟶ r"<br />
proof -<br />
{assume 2: "p ⟶ q" <br />
have 3: p using 1 by (rule conjunct1)<br />
have 4: "q ⟶ r" using 1 by (rule conjunct2)<br />
have 5: q using 2 3 by (rule mp)<br />
have 6: r using 4 5 by (rule mp)<br />
}<br />
thus "(p ⟶ q) ⟶ r" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_24: marmedmar3<br />
assumes "p ∧ (q ⟶ r)" <br />
shows "(p ⟶ q) ⟶ r"<br />
proof <br />
assume "(p ⟶ q)" <br />
have "(q ⟶ r)" using assms by (rule conjunct2)<br />
have "p" using assms by (rule conjunct1)<br />
with `(p ⟶ q)` have "q" by (rule mp) <br />
with `(q ⟶ r)` show "r" by (rule mp)<br />
qed <br />
<br />
lemma ej_24: joslopjim4<br />
assumes "p ∧ (q ⟶ r)" <br />
shows "(p ⟶ q) ⟶ r"<br />
proof (rule impI)<br />
assume "p⟶q"<br />
have "p" using assms by (rule conjunct1)<br />
have "q⟶r" using assms by (rule conjunct2)<br />
have "q" using `p⟶q` `p` by (rule mp)<br />
show "r" using `q⟶r` `q` by (rule mp)<br />
qed<br />
<br />
section {* Disyunciones *}<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 25. Demostrar<br />
p ⊢ p ∨ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_25: carmarria<br />
assumes "p"<br />
shows "p ∨ q"<br />
proof -<br />
show "p | q" using assms(1) by (rule disjI1)<br />
qed<br />
<br />
lemma ej_25: joslopjim4<br />
assumes "p"<br />
shows "p ∨ q"<br />
proof (rule disjI1)<br />
show "p" using assms by this<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 26. Demostrar<br />
q ⊢ p ∨ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_26: carmarria<br />
assumes "q"<br />
shows "p ∨ q"<br />
proof -<br />
show "p | q" using assms(1) by (rule disjI2)<br />
qed<br />
<br />
lemma ej_26: joslopjim4<br />
assumes "q"<br />
shows "p ∨ q"<br />
proof (rule disjI2)<br />
show "q" using assms by this<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 27. Demostrar<br />
p ∨ q ⊢ q ∨ p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_27: carmarria<br />
assumes "p ∨ q"<br />
shows "q ∨ p"<br />
proof -<br />
have "p | q" using assms(1) by this<br />
moreover<br />
{assume p<br />
have "q | p" using `p` by (rule disjI2)<br />
}<br />
moreover<br />
{assume q<br />
have "q | p" using `q` by (rule disjI1)<br />
}<br />
ultimately show "q | p" by (rule disjE)<br />
qed<br />
<br />
lemma ej_27: joslopjim4<br />
assumes "p ∨ q"<br />
shows "q ∨ p"<br />
using assms<br />
proof <br />
assume "p"<br />
then show "q∨p" by(rule disjI2)<br />
next<br />
assume "q"<br />
show "q∨p" using `q` by(rule disjI1)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 28. Demostrar<br />
q ⟶ r ⊢ p ∨ q ⟶ p ∨ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_28: carmarria<br />
assumes "q ⟶ r" <br />
shows "p ∨ q ⟶ p ∨ r"<br />
proof -<br />
{assume "p | q" <br />
moreover<br />
{assume p<br />
have "p | r" using `p` by (rule disjI1)<br />
}<br />
moreover<br />
{assume q<br />
have r using assms(1) `q` by (rule mp)<br />
have "p | r" using `r` by (rule disjI2)<br />
}<br />
ultimately have "p | r" by (rule disjE)<br />
}<br />
thus "p | q ⟶ p | r" by (rule impI)<br />
qed<br />
<br />
lemma ej_28: joslopjim4<br />
assumes "q ⟶ r" <br />
shows "p ∨ q ⟶ p ∨ r"<br />
proof (rule impI)<br />
assume "p∨q" <br />
show "p∨r"<br />
using `p∨q`<br />
proof <br />
assume "p"<br />
show "p∨r" using `p` by (rule disjI1)<br />
next<br />
assume "q"<br />
have "r" using assms `q` by (rule mp)<br />
show "p∨r" using `r` by (rule disjI2)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 29. Demostrar<br />
p ∨ p ⊢ p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_29: carmarria<br />
assumes "p ∨ p"<br />
shows "p"<br />
proof -<br />
have "p | p" using assms(1) by this<br />
moreover <br />
{assume p<br />
}<br />
moreover<br />
{assume p<br />
}<br />
ultimately show p by (rule disjE)<br />
qed<br />
<br />
lemma ej_29: joslopjim4<br />
assumes "p ∨ p"<br />
shows "p"<br />
proof (rule ccontr)<br />
assume "¬p"<br />
show False<br />
using assms<br />
proof<br />
assume "p"<br />
show False using `¬p` `p` by (rule notE)<br />
next<br />
assume "p"<br />
show False using `¬p` `p` by (rule notE)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 30. Demostrar<br />
p ⊢ p ∨ p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_30: carmarria<br />
assumes "p" <br />
shows "p ∨ p"<br />
proof -<br />
show "p | p" using assms(1) by (rule disjI1)<br />
qed<br />
<br />
lemma ej_30: joslopjim4<br />
assumes "p" <br />
shows "p ∨ p"<br />
proof (rule disjI1)<br />
show "p" using assms by this<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 31. Demostrar<br />
p ∨ (q ∨ r) ⊢ (p ∨ q) ∨ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_31: carmarria<br />
assumes 1: "p ∨ (q ∨ r)" <br />
shows "(p ∨ q) ∨ r"<br />
proof -<br />
have "p | (q | r)" using 1 by this<br />
moreover {assume 2: p<br />
have 3: "p | q" using 2 by (rule disjI1)<br />
have 4: "(p | q) | r" using 3 by (rule disjI1)<br />
}<br />
moreover {assume 5: "q | r"<br />
moreover {assume 6: q<br />
have 7: "p | q" using 6 by (rule disjI2)<br />
have 8: "(p | q) | r " using 7 by (rule disjI1)<br />
}<br />
moreover {assume 9: r<br />
have 10: "(p | q) | r " using 9 by (rule disjI2)<br />
}<br />
ultimately have 11: "(p | q) | r " by (rule disjE)<br />
}<br />
ultimately show "(p | q) | r " by (rule disjE)<br />
qed<br />
<br />
lemma ej_31: joslopjim4<br />
assumes "p ∨ (q ∨ r)" <br />
shows "(p ∨ q) ∨ r"<br />
using assms<br />
proof <br />
assume "p"<br />
have "p∨q" using `p` by (rule disjI1)<br />
thus "(p ∨ q) ∨ r" by (rule disjI1)<br />
next<br />
assume "q∨r"<br />
show "(p ∨ q) ∨ r"<br />
using `q∨r`<br />
proof <br />
assume "q"<br />
have "p∨q" using `q` by (rule disjI2)<br />
show "(p ∨ q) ∨ r" using `p∨q` by (rule disjI1)<br />
next<br />
assume "r"<br />
show "(p ∨ q) ∨ r" using `r` by (rule disjI2)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 32. Demostrar<br />
(p ∨ q) ∨ r ⊢ p ∨ (q ∨ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_32: carmarria<br />
assumes 1: "(p ∨ q) ∨ r" <br />
shows "p ∨ (q ∨ r)"<br />
proof -<br />
have "(p | q) | r" using 1 by this<br />
moreover {assume 2: "p | q" <br />
moreover {assume 3: p <br />
have 4: "p | (q | r)" using 3 by (rule disjI1)<br />
}<br />
moreover {assume 5: q<br />
have 6: "q | r" using 5 by (rule disjI1)<br />
have 7: "p | (q | r)" using 6 by (rule disjI2)<br />
}<br />
ultimately have 8: " p | (q | r)" by (rule disjE)<br />
}<br />
moreover {assume 9: r<br />
have 10: "q | r" using 9 by (rule disjI2)<br />
have 11: "p | (q | r)" using 10 by (rule disjI2)<br />
}<br />
ultimately show "p | (q | r)" by (rule disjE)<br />
qed<br />
<br />
lemma ej_32: joslopjim4<br />
assumes "(p ∨ q) ∨ r" <br />
shows "p ∨ (q ∨ r)"<br />
using assms<br />
proof<br />
assume "p∨q"<br />
show "p ∨ (q ∨ r)"<br />
using `p∨q`<br />
proof <br />
assume "p"<br />
show "p ∨ (q ∨ r)" using `p` by (rule disjI1)<br />
next<br />
assume "q"<br />
have "q∨r" using `q` by (rule disjI1)<br />
show "p ∨ (q ∨ r)" using `q∨r` by (rule disjI2)<br />
qed<br />
next<br />
assume "r"<br />
have "q∨r" using `r` by (rule disjI2)<br />
show "p ∨ (q ∨ r)" using `q∨r` by (rule disjI2)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 33. Demostrar<br />
p ∧ (q ∨ r) ⊢ (p ∧ q) ∨ (p ∧ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_33: carmarria<br />
assumes 1: "p ∧ (q ∨ r)" <br />
shows "(p ∧ q) ∨ (p ∧ r)"<br />
proof -<br />
have 2: p using 1 by (rule conjunct1)<br />
have 3: "q | r" using 1 by (rule conjunct2)<br />
moreover {assume 4: q<br />
have 5: "p & q" using 2 4 by (rule conjI)<br />
have 6: "(p & q) | (p & r)" using 5 by (rule disjI1)<br />
}<br />
moreover {assume 7: r<br />
have 8: "p & r" using 2 7 by (rule conjI)<br />
have 9: "(p & q) | (p & r)" using 8 by (rule disjI2)<br />
}<br />
ultimately show "(p & q) | (p & r)" by (rule disjE)<br />
qed<br />
<br />
lemma ej_33: joslopjim4<br />
assumes "p ∧ (q ∨ r)" <br />
shows "(p ∧ q) ∨ (p ∧ r)"<br />
proof-<br />
have "p" using assms by (rule conjunct1)<br />
have "q∨r" using assms by (rule conjunct2)<br />
show "(p ∧ q) ∨ (p ∧ r)"<br />
using `q∨r` <br />
proof (rule disjE)<br />
assume "q"<br />
have "p∧q" using `p` `q` by (rule conjI)<br />
show "(p ∧ q) ∨ (p ∧ r)" using `p∧q` by (rule disjI1)<br />
next<br />
assume "r"<br />
have "p∧r" using `p` `r` by (rule conjI)<br />
show "(p ∧ q) ∨ (p ∧ r)" using `p∧r` by (rule disjI2)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 34. Demostrar<br />
(p ∧ q) ∨ (p ∧ r) ⊢ p ∧ (q ∨ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_34: carmarria<br />
assumes 1: "(p ∧ q) ∨ (p ∧ r)" <br />
shows "p ∧ (q ∨ r)"<br />
proof -<br />
have "(p & q) | (p & r)" using 1 by this<br />
moreover {assume 2: "p & q"<br />
have 3: p using 2 by (rule conjunct1)<br />
have 4: q using 2 by (rule conjunct2)<br />
have 5: "q | r" using 4 by (rule disjI1)<br />
have 6: "p & (q | r)" using 3 5 by (rule conjI)<br />
}<br />
moreover {assume 7: "p & r"<br />
have 8: p using 7 by (rule conjunct1)<br />
have 9: r using 7 by (rule conjunct2)<br />
have 10: "q | r" using 9 by (rule disjI2)<br />
have 11: "p & (q | r)" using 8 10 by (rule conjI)<br />
}<br />
ultimately show "p & (q | r)" by (rule disjE)<br />
qed<br />
<br />
lemma ej_34: joslopjim4<br />
assumes "(p ∧ q) ∨ (p ∧ r)" <br />
shows "p ∧ (q ∨ r)"<br />
using assms<br />
proof<br />
assume "p∧q"<br />
have "p" using `p∧q` by (rule conjunct1)<br />
have "q" using `p∧q` by (rule conjunct2)<br />
have "q∨r" using `q` by (rule disjI1)<br />
show "p∧(q∨r)" using `p` `q∨r` by (rule conjI)<br />
next<br />
assume "p∧r"<br />
have "p" using `p∧r` by (rule conjunct1)<br />
have "r" using `p∧r` by (rule conjunct2)<br />
have "q∨r" using `r` by (rule disjI2)<br />
show "p∧(q∨r)" using `p` `q∨r` by (rule conjI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 35. Demostrar<br />
p ∨ (q ∧ r) ⊢ (p ∨ q) ∧ (p ∨ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_35: carmarria<br />
assumes 1: "p ∨ (q ∧ r)" <br />
shows "(p ∨ q) ∧ (p ∨ r)"<br />
proof -<br />
have "p | (q & r)" using 1 by this<br />
moreover {assume 2: p<br />
have 3: " p | q" using 2 by (rule disjI1)<br />
have 4: " p | r" using 2 by (rule disjI1)<br />
have 5: "(p | q) & (p | r)" using 3 4 by (rule conjI)<br />
}<br />
moreover {assume 6: "q & r" <br />
have 7: q using 6 by (rule conjunct1)<br />
have 8: r using 6 by (rule conjunct2)<br />
have 9: "p | q" using 7 by (rule disjI2)<br />
have 10: "p | r" using 8 by (rule disjI2)<br />
have 11: "(p | q) & (p | r)" using 9 10 by (rule conjI)<br />
}<br />
ultimately show "(p | q) & (p | r)" by (rule disjE)<br />
qed<br />
<br />
lemma ej_35: joslopjim4<br />
assumes "p ∨ (q ∧ r)" <br />
shows "(p ∨ q) ∧ (p ∨ r)"<br />
using assms<br />
proof (rule disjE)<br />
assume "p"<br />
have "p∨q" using `p` by (rule disjI1)<br />
have "p∨r" using `p` by (rule disjI1)<br />
show "(p ∨ q) ∧ (p ∨ r)" using `p∨q` `p∨r` by (rule conjI)<br />
next<br />
assume "q∧r"<br />
have "q" using `q∧r` by (rule conjunct1)<br />
have "r" using `q∧r` by (rule conjunct2)<br />
have "p∨q" using `q` by (rule disjI2)<br />
have "p∨r" using `r` by (rule disjI2)<br />
show "(p ∨ q) ∧ (p ∨ r)" using `p∨q` `p∨r` by (rule conjI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 36. Demostrar<br />
(p ∨ q) ∧ (p ∨ r) ⊢ p ∨ (q ∧ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_36: carmarria<br />
assumes 1: "(p ∨ q) ∧ (p ∨ r)"<br />
shows "p ∨ (q ∧ r)"<br />
proof -<br />
have 2: "p | q" using 1 by (rule conjunct1)<br />
moreover {assume 3: p<br />
have 4: "p | (q & r)" using 3 by (rule disjI1)<br />
}<br />
moreover {assume 5: q<br />
have 6: "p | r" using 1 by (rule conjunct2)<br />
moreover {assume 7: p<br />
have 8: "p | (q & r)" using 7 by (rule disjI1)<br />
}<br />
moreover {assume 9: r<br />
have 10: "q & r" using 5 9 by (rule conjI)<br />
have 11: "p | (q & r)" using 10 by (rule disjI2)<br />
}<br />
ultimately have 12: "p | (q & r)" by (rule disjE)<br />
}<br />
ultimately show "p | (q & r)" by (rule disjE)<br />
qed<br />
<br />
lemma ej_36: joslopjim4<br />
assumes "(p ∨ q) ∧ (p ∨ r)"<br />
shows "p ∨ (q ∧ r)"<br />
proof-<br />
have "p∨q" using assms by (rule conjunct1)<br />
have "p∨r" using assms by (rule conjunct2)<br />
show "p∨(q ∧ r)"<br />
using `p∨r`<br />
proof<br />
assume "p"<br />
show "p∨(q ∧ r)" using `p` by (rule disjI1)<br />
next<br />
assume "r"<br />
show "p∨(q ∧ r)"<br />
using `p∨q`<br />
proof<br />
assume "p"<br />
show "p∨(q ∧ r)" using `p` by (rule disjI1)<br />
next<br />
assume "q"<br />
have "q∧r" using `q` `r` by (rule conjI)<br />
show "p∨(q ∧ r)" using `q∧r` by (rule disjI2)<br />
qed<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 37. Demostrar<br />
(p ⟶ r) ∧ (q ⟶ r) ⊢ p ∨ q ⟶ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_37: carmarria<br />
assumes 1: "(p ⟶ r) ∧ (q ⟶ r)" <br />
shows "p ∨ q ⟶ r"<br />
proof -<br />
have 2: "p ⟶ r" using 1 by (rule conjunct1)<br />
have 3: "q ⟶ r" using 1 by (rule conjunct2)<br />
{assume 4: "p | q"<br />
moreover {assume 5: p<br />
have 6: r using 2 5 by (rule mp)<br />
}<br />
moreover {assume 7: q<br />
have 8: r using 3 7 by (rule mp)<br />
}<br />
ultimately have r by (rule disjE)<br />
}<br />
thus "p | q ⟶ r" by (rule impI)<br />
qed<br />
<br />
lemma ej_37: joslopjim4<br />
assumes "(p ⟶ r) ∧ (q ⟶ r)" <br />
shows "p ∨ q ⟶ r"<br />
proof (rule impI)<br />
assume "p∨q"<br />
have "p⟶r" using assms by (rule conjunct1)<br />
have "q⟶r" using assms by (rule conjunct2)<br />
show "r"<br />
using `p∨q`<br />
proof <br />
assume "p"<br />
show "r" using `p⟶r` `p` by (rule mp)<br />
next<br />
assume "q"<br />
show "r" using `q⟶r` `q` by (rule mp)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 38. Demostrar<br />
p ∨ q ⟶ r ⊢ (p ⟶ r) ∧ (q ⟶ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_38: carmarria<br />
assumes 1: "p ∨ q ⟶ r" <br />
shows "(p ⟶ r) ∧ (q ⟶ r)"<br />
proof -<br />
{assume 2: p<br />
have 3: "p | q" using 2 by (rule disjI1)<br />
have 4: r using 1 3 by (rule mp)<br />
}<br />
hence 5: "p ⟶ r" by (rule impI)<br />
{assume 6: q<br />
have 7: "p | q" using 6 by (rule disjI2)<br />
have 8: r using 1 7 by (rule mp)<br />
}<br />
hence 9: "q ⟶ r" by (rule impI)<br />
show "(p ⟶ r) & (q ⟶ r)" using 5 9 by (rule conjI)<br />
qed<br />
<br />
lemma ej_38: joslopjim4<br />
assumes "p ∨ q ⟶ r" <br />
shows "(p ⟶ r) ∧ (q ⟶ r)"<br />
proof-<br />
have "p⟶r" <br />
proof (rule impI)<br />
assume "p"<br />
have "p∨q" using `p` by (rule disjI1)<br />
show "r" using assms `p∨q` by (rule mp)<br />
qed<br />
have "q⟶r"<br />
proof (rule impI)<br />
assume "q"<br />
have "p∨q" using `q` by (rule disjI2)<br />
show "r" using assms `p∨q` by (rule mp)<br />
qed<br />
show "(p ⟶ r) ∧ (q ⟶ r)" using `p⟶r` `q⟶r` by (rule conjI)<br />
qed<br />
<br />
section {* Negaciones *}<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 39. Demostrar<br />
p ⊢ ¬¬p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_39: carmarria<br />
assumes "p"<br />
shows "¬¬p"<br />
proof -<br />
show "~~p" using assms(1) by (rule notnotI)<br />
qed<br />
<br />
lemma ej_39: joslopjim4<br />
assumes "p"<br />
shows "¬¬p"<br />
proof-<br />
show "¬¬p" using assms by (rule notnotI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 40. Demostrar<br />
¬p ⊢ p ⟶ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_40: carmarria<br />
assumes "¬p" <br />
shows "p ⟶ q"<br />
proof -<br />
{assume p<br />
have "~p" using assms(1) by this<br />
have False using `~p` `p` by (rule notE)<br />
have q using `False` by (rule FalseE)<br />
}<br />
thus "p ⟶ q" by (rule impI)<br />
qed<br />
<br />
lemma ej_40: joslopjim4<br />
assumes "¬p" <br />
shows "p ⟶ q"<br />
proof (rule impI)<br />
assume "p"<br />
show "q" using assms `p` by (rule notE)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 41. Demostrar<br />
p ⟶ q ⊢ ¬q ⟶ ¬p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_41: carmarria<br />
assumes 1: "p ⟶ q"<br />
shows "¬q ⟶ ¬p"<br />
proof -<br />
{assume 2: "~q"<br />
have 3: "~p" using 1 2 by (rule mt)<br />
}<br />
thus 4: "~q ⟶ ~p" by (rule impI)<br />
qed<br />
<br />
lemma ej_41: joslopjim4<br />
assumes "p ⟶ q"<br />
shows "¬q ⟶ ¬p"<br />
proof (rule impI)<br />
assume "¬q"<br />
show "¬p" using assms `¬q` by (rule mt)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 42. Demostrar<br />
p∨q, ¬q ⊢ p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_42: carmarria<br />
assumes 1: "p∨q" and<br />
2: "¬q" <br />
shows "p"<br />
proof -<br />
have " p | q" using 1 by this<br />
moreover {assume 3: p<br />
}<br />
moreover {assume 4: q<br />
have 5: False using 2 4 by (rule notE)<br />
have 6: p using 5 by (rule FalseE)<br />
}<br />
ultimately show p by (rule disjE)<br />
qed<br />
<br />
lemma ej_42: joslopjim4<br />
assumes "p∨q"<br />
"¬q" <br />
shows "p"<br />
using assms(1)<br />
proof <br />
assume "p"<br />
show "p" using `p` by this<br />
next<br />
assume "q"<br />
show "p" using assms(2) `q` by (rule notE)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 43. Demostrar<br />
p ∨ q, ¬p ⊢ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_43: carmarria<br />
assumes 1: "p ∨ q" and<br />
2: "¬p" <br />
shows "q"<br />
proof -<br />
have " p | q" using 1 by this<br />
moreover {assume 3: p<br />
have 4: False using 2 3 by (rule notE)<br />
have 5: q using 4 by (rule FalseE)<br />
}<br />
moreover {assume 6: q<br />
}<br />
ultimately show q by (rule disjE)<br />
qed<br />
<br />
lemma ej_43: joslopjim4<br />
assumes "p ∨ q"<br />
"¬p" <br />
shows "q"<br />
using assms(1)<br />
proof<br />
assume "p"<br />
show "q" using assms(2) `p` by (rule notE)<br />
next<br />
assume "q"<br />
show "q" using `q` by this<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 44. Demostrar<br />
p ∨ q ⊢ ¬(¬p ∧ ¬q)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_44: carmarria<br />
assumes 1: "p ∨ q" <br />
shows "¬(¬p ∧ ¬q)"<br />
proof -<br />
have "p | q" using 1 by this<br />
moreover {assume 2: p<br />
{assume 3:"~p & ~q"<br />
have 4: "~p" using 3 by (rule conjunct1)<br />
have 5: False using 4 2 by (rule notE)<br />
}<br />
hence 6: "~(~p & ~q)" by (rule notI)<br />
}<br />
moreover {assume 7: q<br />
{assume 8: "~p & ~q"<br />
have 9: "~q" using 8 by (rule conjunct2)<br />
have 10: False using 9 7 by (rule notE)<br />
}<br />
hence 11: "~(~p & ~q)" by (rule notI)<br />
}<br />
ultimately show "~(~p & ~q)" by (rule disjE)<br />
qed<br />
<br />
lemma ej_44: joslopjim4<br />
assumes "p∨q"<br />
shows "¬(¬p∧¬q)"<br />
proof (rule notI)<br />
assume "¬p∧¬q"<br />
show False<br />
using assms<br />
proof<br />
assume "p"<br />
have "¬p" using `¬p∧¬q` by (rule conjunct1)<br />
show False using `¬p` `p` by (rule notE)<br />
next<br />
assume "q"<br />
have "¬q" using `¬p∧¬q` by (rule conjunct2)<br />
show False using `¬q` `q` by (rule notE)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 45. Demostrar<br />
p ∧ q ⊢ ¬(¬p ∨ ¬q)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_45: carmarria<br />
assumes 1: "p ∧ q" <br />
shows "¬(¬p ∨ ¬q)"<br />
proof -<br />
{assume 2: "~p | ~q"<br />
moreover {assume 3: "~p"<br />
have 4: p using 1 by (rule conjunct1)<br />
have 5: False using 3 4 by (rule notE)<br />
}<br />
moreover {assume 6: "~q"<br />
have 7: q using 1 by (rule conjunct2)<br />
have 8: False using 6 7 by (rule notE)<br />
}<br />
ultimately have False by (rule disjE)<br />
}<br />
thus "~(~p | ~q)" by (rule notI)<br />
qed<br />
<br />
lemma ej_45: joslopjim4<br />
assumes "p∧q"<br />
shows "¬(¬p∨¬q)"<br />
proof (rule notI)<br />
assume "¬p∨¬q"<br />
then show False<br />
proof (rule disjE)<br />
assume "¬p"<br />
have "p" using assms by (rule conjunct1)<br />
show False using `¬p` `p` by (rule notE)<br />
next<br />
assume "¬q"<br />
have "q" using assms by (rule conjunct2)<br />
show False using `¬q` `q` by (rule notE)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 46. Demostrar<br />
¬(p ∨ q) ⊢ ¬p ∧ ¬q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_46: carmarria<br />
assumes 1: "¬(p ∨ q)" <br />
shows "¬p ∧ ¬q"<br />
proof -<br />
{assume 2: p<br />
have 3: "p | q" using 2 by (rule disjI1)<br />
have 4: False using 1 3 by (rule notE)<br />
}<br />
hence 5: "~p" by (rule notI)<br />
{assume 6: q<br />
have 7: "p | q" using 6 by (rule disjI2)<br />
have 8: False using 1 7 by (rule notE)<br />
}<br />
hence 9: "~q" by (rule notI)<br />
show "~p & ~q" using 5 9 by (rule conjI)<br />
qed<br />
<br />
lemma ej_46: joslopjim4<br />
assumes "¬(p ∨ q)"<br />
shows "¬p ∧ ¬q"<br />
proof (rule conjI)<br />
show "¬p"<br />
proof (rule ccontr)<br />
assume "¬¬p"<br />
have "p" using `¬¬p` by (rule notnotD)<br />
have "p∨q" using `p` by (rule disjI1)<br />
show False using assms `p∨q` by (rule notE)<br />
qed<br />
show "¬q"<br />
proof (rule ccontr)<br />
assume "¬¬q"<br />
have "q" using `¬¬q` by (rule notnotD)<br />
have "p∨q" using `q` by (rule disjI2)<br />
show False using assms `p∨q` by (rule notE)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 47. Demostrar<br />
¬p ∧ ¬q ⊢ ¬(p ∨ q)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_47: carmarria<br />
assumes 1: "¬p ∧ ¬q" <br />
shows "¬(p ∨ q)"<br />
proof -<br />
{assume 2: "p | q"<br />
moreover {assume 3: p<br />
have 4: "~p" using 1 by (rule conjunct1)<br />
have 5: False using 4 3 by (rule notE)<br />
}<br />
moreover {assume 6: q <br />
have 7: "~q" using 1 by (rule conjunct2)<br />
have 8: False using 7 6 by (rule notE)<br />
}<br />
ultimately have 9: False by (rule disjE)<br />
}<br />
thus "~(p | q)" by (rule notI)<br />
qed <br />
<br />
lemma ej_47: joslopjim4<br />
assumes "¬p∧¬q"<br />
shows "¬(p∨q)"<br />
proof (rule notI)<br />
assume "p∨q"<br />
then show False<br />
proof<br />
assume "p"<br />
have "¬p" using assms by (rule conjunct1)<br />
show False using `¬p` `p` by (rule notE)<br />
next<br />
assume "q"<br />
have "¬q" using assms by (rule conjunct2)<br />
show False using `¬q` `q` by (rule notE)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 48. Demostrar<br />
¬p ∨ ¬q ⊢ ¬(p ∧ q)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_48: carmarria<br />
assumes 1: "¬p ∨ ¬q"<br />
shows "¬(p ∧ q)"<br />
proof -<br />
have "~p | ~q" using 1 by this<br />
moreover {assume 2: "~p"<br />
{assume 3: "p & q"<br />
have 4: p using 3 by (rule conjunct1)<br />
have 5: False using 2 4 by (rule notE)<br />
}<br />
hence 6: "~(p & q)" by (rule notI)<br />
}<br />
moreover {assume 7: "~q"<br />
{assume 8: "p & q"<br />
have 9: q using 8 by (rule conjunct2)<br />
have 10: False using 7 9 by (rule notE)<br />
}<br />
hence 11: "~(p & q)" by (rule notI)<br />
}<br />
ultimately show "~(p & q)" by (rule disjE)<br />
qed<br />
<br />
lemma ej_48: joslopjim4<br />
assumes "¬p ∨ ¬q"<br />
shows "¬(p ∧ q)"<br />
proof (rule notI)<br />
assume "p∧q"<br />
have "p" using `p∧q` by (rule conjunct1)<br />
have "q" using `p∧q` by (rule conjunct2)<br />
show False<br />
using assms<br />
proof<br />
assume "¬p"<br />
have "p" using `p∧q` by (rule conjunct1)<br />
show False using `¬p` `p` by (rule notE)<br />
next<br />
assume "¬q"<br />
have "q" using `p∧q` by (rule conjunct2)<br />
show False using `¬q` `q` by (rule notE)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 49. Demostrar<br />
⊢ ¬(p ∧ ¬p)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_49: carmarria<br />
"¬(p ∧ ¬p)"<br />
proof -<br />
{assume 1: "p & ~p"<br />
have 2: p using 1 by (rule conjunct1)<br />
have 3: "~p" using 1 by (rule conjunct2)<br />
have 4: False using 3 2 by (rule notE)<br />
}<br />
thus "~(p & ~p)" by (rule notI)<br />
qed<br />
<br />
lemma ej_49: joslopjim4<br />
shows "¬(p∧¬p)"<br />
proof<br />
assume "p∧¬p"<br />
have "p" using `p∧¬p` by (rule conjunct1)<br />
have "¬p" using `p∧¬p` by (rule conjunct2)<br />
show False using `¬p` `p` by (rule notE)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 50. Demostrar<br />
p ∧ ¬p ⊢ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_50: carmarria<br />
assumes 1: "p ∧ ¬p" <br />
shows "q"<br />
proof -<br />
have 2: p using 1 by (rule conjunct1)<br />
have 3: "~p" using 1 by (rule conjunct2)<br />
have 4: False using 3 2 by (rule notE)<br />
show q using 4 by (rule FalseE)<br />
qed<br />
<br />
lemma ej_50: joslopjim4<br />
assumes "p∧¬p"<br />
shows "q"<br />
proof-<br />
have "p" using `p∧¬p` by (rule conjunct1)<br />
have "¬p" using `p∧¬p` by (rule conjunct2)<br />
have False using `¬p` `p` by (rule notE)<br />
then show "q" by (rule FalseE)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 51. Demostrar<br />
¬¬p ⊢ p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_51: carmarria joslopjim4<br />
assumes "¬¬p"<br />
shows "p"<br />
proof -<br />
show p using assms by (rule notnotD)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 52. Demostrar<br />
⊢ p ∨ ¬p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_52: carmarria<br />
"p ∨ ¬p"<br />
proof -<br />
{assume 1: "~(p | ~p)"<br />
{assume 2: p <br />
have 3: "p | ~p" using 2 by (rule disjI1)<br />
have 4: False using 1 3 by (rule notE)<br />
}<br />
hence 5: "~p" by (rule notI)<br />
have 6: "p | ~p" using 5 by (rule disjI2)<br />
have 7: False using 1 6 by (rule notE)<br />
}<br />
hence 8: "~~(p | ~p)" by (rule notI)<br />
show "p | ~p" using 8 by (rule notnotD)<br />
qed<br />
<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 53. Demostrar<br />
⊢ ((p ⟶ q) ⟶ p) ⟶ p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ej_53: joslopjim4<br />
shows "((p ⟶ q) ⟶ p) ⟶ p"<br />
proof (rule impI)<br />
assume "(p ⟶ q) ⟶ p"<br />
show "p"<br />
proof (rule ccontr)<br />
assume "¬p"<br />
have "¬(p⟶q)" using `(p ⟶ q) ⟶ p` `¬p` by (rule mt)<br />
have "p⟶q"<br />
proof (rule impI)<br />
assume "p"<br />
show "q" using `¬p` `p` by (rule notE)<br />
qed<br />
show False using `¬(p⟶q)` `p⟶q` by (rule notE)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 54. Demostrar<br />
¬q ⟶ ¬p ⊢ p ⟶ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ej_54: joslopjim4<br />
assumes "¬q ⟶ ¬p"<br />
shows "p ⟶ q"<br />
proof (rule impI)<br />
assume "p"<br />
have "¬¬p" using `p` by (rule notnotI)<br />
have "¬¬q" using assms `¬¬p` by (rule mt)<br />
show "q" using `¬¬q` by (rule notnotD)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 55. Demostrar<br />
¬(¬p ∧ ¬q) ⊢ p ∨ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_55:<br />
assumes "¬(¬p ∧ ¬q)"<br />
shows "p ∨ q"<br />
oops<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 56. Demostrar<br />
¬(¬p ∨ ¬q) ⊢ p ∧ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_56:<br />
assumes "¬(¬p ∨ ¬q)" <br />
shows "p ∧ q"<br />
oops<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 57. Demostrar<br />
¬(p ∧ q) ⊢ ¬p ∨ ¬q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_57:<br />
assumes "¬(p ∧ q)"<br />
shows "¬p ∨ ¬q"<br />
oops<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 58. Demostrar<br />
⊢ (p ⟶ q) ∨ (q ⟶ p)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_58:<br />
"(p ⟶ q) ∨ (q ⟶ p)"<br />
oops<br />
<br />
end<br />
<br />
</source></div>
Joslopjim4
https://www.glc.us.es/~jalonso/LMF2018/index.php?title=Relaci%C3%B3n_1&diff=137
Relación 1
2018-03-21T22:59:16Z
<p>Joslopjim4: /* Relación 1: Sintaxis y semántica de la lógica proposicional */</p>
<hr />
<div>=== Relación 1: Sintaxis y semántica de la lógica proposicional ===<br />
<br />
----<br />
'''Ejercicio 1.''' Formalizar el siguiente argumento <br />
<blockquote><br />
''Siempre que un número x es divisible por 10, acaba en 0. El número x no acaba en 0. Por lo tanto, x no es divisible por 10.''<br />
</blockquote> <br />
Usando los símbolos D: el número es divisible por 10 y C: el número acaba en cero.<br />
----<br />
<br />
'''Solución:''' <br />
josrodjim2, jescammor1, inmbenber,josgarfer12,antnavcue,sarolizap,manberdel1,joslopjim4<br />
<br />
D -> C. ¬C |= ¬D<br />
<br />
----<br />
'''Ejercicio 2.''' Formalizar el siguiente argumento <br />
<blockquote><br />
''Si la válvula está abierta o la monitorización está preparada, entonces se envía una señal de reconocimiento y un mensaje de funcionamiento al controlador del ordenador. Si se envía un mensaje de funcionamiento al controlador del ordenador o el sistema está en estado normal, entonces se aceptan las órdenes del operador. Por lo tanto, si la válvula está abierta, entonces se aceptan las órdenes del operador.'' <br />
</blockquote><br />
Usando los símbolos V: La válvula está abierta, P: La monitorización está preparada, R: Envía una señal de reconocimiento, F: Envía un mensaje de funcionamiento, A: Se aceptan órdenes del operador y N: El sistema está en estado normal.<br />
----<br />
<br />
'''Solución:'''<br />
josrodjim2, jescammor1,josgarfer12,antnavcue,sarolizap,manberdel1<br />
<br />
(V v P) -> (R ∧ F). (F v N) -> A. V |= A<br />
<br />
inmbenber, joslopjim4<br />
<br />
(V v P) -> (R ^ F), (F v N) -> A |= (V -> A)<br />
----<br />
'''Ejercicio 3.''' Formalizar el siguiente argumento <br />
<blockquote><br />
''Cuando tanto la temperatura como la presión atmosférica permanecen contantes, no llueve. La temperatura permanece constante. Por lo tanto, en caso de que llueva, la presión atmosférica no permanece constante.'' <br />
</blockquote><br />
Usando los símbolos T: La temperatura permanece constante, P: La presión atmosférica permanece constante y L: Llueve<br />
----<br />
<br />
'''Solución:'''<br />
josrodjim2, jescammor1, inmbenber,,josgarfer12,antnavcue,sarolizap,manberdel1,joslopjim4<br />
<br />
(T ∧ P) -> ¬L. T |= (L -> ¬P)<br />
----<br />
'''Ejercicio 4.''' Formalizar el siguiente argumento <br />
<blockquote><br />
''En cierto experimento, cuando hemos empleado un fármaco A, el paciente ha mejorado considerablemente en el caso, y sólo en el caso, en que no se haya empleado también un fármaco B. Además, o se ha empleado el fármaco A o se ha empleado el fármaco B. En consecuencia, podemos afirmar que si no hemos empleado el fármaco B, el paciente ha mejorado considerablemente.'' <br />
</blockquote><br />
Usando los símbolos A: Hemos empleado el fármaco A, B: Hemos empleado el fármaco B y M: El paciente ha mejorado notablemente.<br />
----<br />
<br />
'''Solución:'''<br />
josrodjim2,sarolizap<br />
<br />
(A ∧ ¬B) -> M. A v B. ¬B |= M<br />
<br />
jescammor1<br />
<br />
(A ∧ ¬B) -> M, (A v B) |= (¬B -> M)<br />
<br />
inmbenber,manberdel1<br />
<br />
A -> (M <-> ¬B), A v B |= (¬B -> M)<br />
<br />
josgarfer12,joslopjim4<br />
(A ∧ ¬B) <-> M, A v B-> ¬B|=M<br />
<br />
antnavcue<br />
<br />
A -> (M <-> ¬B) , A v B-> ¬B|=M<br />
----<br />
'''Ejercicio 5.''' Definir por recursión sobre fórmulas las siguientes funciones<br />
* nv(F) que calcula el número variables proposicionales que ocurren en la fórmula F. Por ejemplo,<br />
: nv(p → p ∨ q) = 3.<br />
* prof(F) que calcula la profundidad del árbol de análisis de la fórmula F. Por ejemplo,<br />
: prof(p → p ∨ q) = 2.<br />
<br />
Demostrar por inducción, que para toda fórmula F,<br />
: nv(F) ≤ 2^prof(F)<br />
----<br />
<br />
'''Solución:'''<br />
josrodjim2, jescammor1, inmbenber,sarolizap,joslopjim4<br />
<br />
*nv(F)= 1 si F atomica, nv(G) si F es ¬G, nv(G)+nv(H) si F es (G*H)<br />
<br />
* prof(F)= 0 si F atomica, 1+prof(G) si F es ¬G, 1+max{prof(G),prof(H)} si F es (G*H)<br />
<br />
* Caso base: F atomica. 1 = nv(F) <= 2^prof(F) = 2^0 = 1. Se cumple.<br />
<br />
* Supongo que se cumple para F y G.<br />
<br />
nv(¬F) = nv(F) <= 2^prof(F) <= 2^(1+prof(F)) = 2^prof(¬F)<br />
<br />
nv(F*G) = nv(F) + nv(G) <= 2^prof(F) + 2^prof(G) <= 2^(1 + max{prof(F),prof(G)}) = 2^prof(F*G)<br />
<br />
----<br />
'''Ejercicio 6.''' ¿Existe un conjunto S de tres fórmulas tal que de todos los subconjuntos de S sólo uno es consistente?<br />
----<br />
<br />
'''Solución:'''<br />
josrodjim2, jescammor1<br />
<br />
Sea S = { F1= ¬(P v ¬P), F2= P ∧ ¬P, F3= P}<br />
<br />
F3 es sastifacible I(P) = 1.<br />
<br />
Pero cualquiera de sus subformulas incluida ella misma, son inconsistente por construccion o usando el ejercicio 9 apartado primero.<br />
<br />
inmbenber<br />
<br />
Sí, basta tomar S un conjunto que tenga una fórmula satisfacible y las otras insatisfacibles, así solo el subconjunto formado por la fórmula satisfacible es consistente. Pongamos un ejemplo: <br />
<br />
S = {F,G,H} donde F = p <-> q, G = p ^ ¬p, H = (p <-> q)^(p -> ¬q)^p. Se comprueba que F es satisfacible (un modelo es I(p)=I(q)=1) y que G y H son insatisfacibles. Se ve que de todos los subconjuntos posibles de S, el único que es consistente es S1={F}<br />
<br />
joslopjim4<br />
<br />
Basta tomar F=p v ¬p, G= q v ¬q, H= r v ¬r<br />
<br />
----<br />
'''Ejercicio 7.''' ¿Es cierto que si F → G y F son satisfacibles, entonces G es satisfacible? Si es cierto, dar una explicación. Si no es cierto, dar un contraejemplo.<br />
----<br />
<br />
'''Solución:'''<br />
josrodjim2, jescammor1<br />
<br />
Si F -> G es satisfacible y F es satisfacible para el mismo modelo, entonces existe I(¬F v G) = 1, I(F)=1.<br />
<br />
Que I(¬F v G) = 1 implica que I(¬F) = 0 o bien I(G)=1, como ya tenemos I(F) = 1. La unica posibilidad es que I(G) = 1. Por lo tanto G es satisfacible.<br />
<br />
inmbenber, joslopjim4<br />
<br />
G no es necesariamente satisfacible, ya que si es una contradicción (falsa en todas sus interpretaciones), existen interpretaciones para las que F->G es cierta (cuando I(F)=0). Por ejemplo, F = ¬p y G = p ^ ¬p. Al hacer su tabla de verdad, notamos que F y F->G son satisfacibles, en cambio, G es insatisfacible (es una contradicción).<br />
<br />
Si F y F->G son safistacibles a la vez, es decir, existe una interpretación I tal que I(F)=I(F->G)=1, entonces sí tendríamos que G es satisfacible, ya que sabiendo que I(F)=I(F->G)=1, la única opción que tenemos es que I(G)=1.<br />
<br />
----<br />
'''Ejercicio 8.''' Demostrar o refutar las siguientes proposiciones:<br />
<br />
# Si F es una fórmula satisfacible, entonces todas las subfórmulas de F son satisfacibles.<br />
# Existen fórmulas válidas tales que todas sus subfórmulas son válidas.<br />
----<br />
<br />
'''Solución:'''<br />
<br />
javdelcru, jescammor1<br />
<br />
* 1) Falso. (p -> ¬p) -> p es satisfacible, pero la subformula p -> ¬p no lo es<br />
* 2) Falso. Todas las formulas tienen al menos una subformula que es atómica y no es válida ya que puede ser falsa.<br />
<br />
josrodjim2, inmbenber,joslopjim4<br />
<br />
* 1) Falso. (P ∧ ¬P) es insatisfacible, cual es una subformula de la tautologia ¬(P ∧ ¬P).<br />
<br />
----<br />
'''Ejercicio 9.'''<br />
Sean S y T conjuntos de fórmulas. Demostrar o refutar las siguientes<br />
afirmaciones: <br />
* Si S es consistente y T es inconsistente, entonces S ∪ T es inconsistente.<br />
* Si S es consistente y T es inconsistente, entonces S ∩ T es consistente.<br />
* Si S es consistente y T es inconsistente, entonces S ∪ T es inconsistente.<br />
* Si S es consistente y T es inconsistente, entonces S ∩ T es consistente.<br />
----<br />
<br />
'''Solución:'''<br />
josrodjim2, inmbenber, joslopjim4<br />
<br />
* 1. Por reducción al absurdo, supongamos que S v T es consistente.<br />
Significa que existe I modelo para todas sus fórmulas, pero el subconjunto de formulas T es inconsistente, por lo que no tiene modelo. Contradcción.<br />
<br />
* 2. S ∩ T puede ser vacío cual es consistente.<br />
S ∩ T si no es vacío es S o un subconjunto suyo el cual es consistente por hipótesis.<br />
<br />
inmbenber<br />
<br />
* 1. Sean S={F1, F2,..., Fn} consistente y T={G1, G2,..., Gm} inconsistente. Por ser S consistente, existe I modelo de S, entonces I(F1^F2^...^Fn)=1. Como T es inconsistente, entonces no tiene ningún modelo, es decir, I(G1^G2^...^Gm)=0 para cualquier interpretación I. Se tiene entonces que (F1^...^Fn^G1^...^Gm)=(F1^...^Fn)^(G1^...^Gm). Por tanto no existe modelo para (F1^...^Fn)^(G1^...^Gm). Lo que implica que S ∪ T es inconsistente.<br />
<br />
<br />
----<br />
'''Ejercicio 10.'''<br />
Da un ejemplo de tres fórmulas F₁ , F₂ , y F₃ tales que F₁ ∧ F₂ ∧ F₃ sea<br />
insatisfactible y donde cualquier conjunción de todas ellas menos una sea<br />
satisfactible. Generalízalo a n fórmulas.<br />
----<br />
<br />
'''Solución:''' <br />
<br />
javdelcru, josrodjim2, jescammor1, joslopjim4<br />
<br />
F_1 = p; F_2 = q; F_3 = ¬(p ∧ q)<br />
F_1 ∧ F_2 ∧ F_3 es insatisfacible pero dos a dos si son satisfacible.<br />
<br />
Generalizamos a n formulas <br />
F_1 = p_1, F_2=p_2 ... F_(n-1) = p_(n-1), F_n = ¬(p_1 ∧ p_2 ∧ .... ∧ p_(n-1))<br />
Igual que en el caso de 3 formulas, la conjunción de todas en insatisfacible ya que F_n es verdadera solo si las anteriores son falsas y las conjunciones de todas menos una (k) son satisfacibles con la interpretación I(p_t)= 1 para todos los t distintos de k y I(p_k) = 0.<br />
<br />
inmbenber<br />
<br />
<math>F_{1}</math> = ¬p, <math>F_{2}</math> = ¬q, <math>F_{3}</math> = p v q<br />
<br />
Generalizamos a n formulas:<br />
<math>F_{1}</math> = ¬<math>p_{1}</math>, <math>F_{2}</math> = ¬<math>p_{2}</math>,..., <math>F_{n-1}</math> = ¬<math>p_{n-1}</math>, <math>F_{n}</math> = <math>p_{1}</math> v <math>p_{2}</math> v ... v <math>p_{n-1}</math><br />
<br />
<br />
<br />
----<br />
'''Ejercicio 11.'''<br />
* Probar que la fórmula (((p → q) → p) → p) es una tautología <br />
* Si definimos recursivamente A(0) = (p → q) y A(n+1) = (A(n) → p), ¿para qué valores de n es A es una tautología?<br />
----<br />
<br />
'''Solución:'''<br />
josrodjim2<br />
<br />
* 1. Se puede escribir la formula como ¬(¬(¬P v Q) v P) v P<br />
<br />
Aplico tabla de verdad<br />
<br />
Para I(P)=1 y I(Q) cualquiera, se ve que siempre es verdad.<br />
<br />
Veamos los demas casos.<br />
<br />
I(P)=0,I(Q)=0 se tiene ¬(¬(1 v 0)) v 0 ----> ¬(¬1) v 0 ---->1 v 0 ---->1 es verdad<br />
<br />
I(P)=0,I(Q)=1 se tiene ¬(¬(1 v 1)) v 0 ----> ¬(¬1) v 0 ---->1 v 0 ---->1 es verdad<br />
<br />
Entonces la formula es tautologia.<br />
<br />
* 2. Para n >= 2 A(n)=p, teniendose ¬P v P cual es tautologia.<br />
<br />
jescammor1<br />
<br />
* 2. Estudiando los casos básicos podemos generalizar fácilmente. Como ya se ha dicho si I(P)=1 siempre es válida de n=1 en adelante así que nos ocupamos del caso I(P)=0. En A(1) esta interpretación nos da 1->0 independientemente del valor de Q (es decir, no es tautología). Por otro lado, A(2) nos da 0->1 en la misma interpretación, lo que nos da la tautología.<br />
<br />
Está claro que si seguimos el proceso tenemos que A(3) no es tautología (1->0), A(4) si (0->1),... Luego es tautología para los n pares.<br />
<br />
inmbenber, joslopjim4<br />
<br />
* 1. Haciendo su tabla de verdad tenemos que (((p → q) → p) → p) es una tautología, ya que para cualquier interpretación se tiene que I(F) = 1 con F = (((p → q) → p) → p)<br />
* 2. A partir de su tabla de verdad, es fácil comprobar que es tautología para n par</div>
Joslopjim4
https://www.glc.us.es/~jalonso/LMF2018/index.php?title=Relaci%C3%B3n_4&diff=136
Relación 4
2018-03-21T22:52:36Z
<p>Joslopjim4: </p>
<hr />
<div><source lang = "isar"><br />
<br />
chapter {* R4: Deducción natural proposicional *}<br />
<br />
theory R4<br />
imports Main <br />
begin<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
El objetivo de esta relación es demostrar cada uno de los ejercicios<br />
usando sólo las reglas básicas de deducción natural de la lógica<br />
proposicional (sin usar el método auto).<br />
<br />
Las reglas básicas de la deducción natural son las siguientes:<br />
· conjI: ⟦P; Q⟧ ⟹ P ∧ Q<br />
· conjunct1: P ∧ Q ⟹ P<br />
· conjunct2: P ∧ Q ⟹ Q <br />
· notnotD: ¬¬ P ⟹ P<br />
· notnotI: P ⟹ ¬¬ P<br />
· mp: ⟦P ⟶ Q; P⟧ ⟹ Q <br />
· mt: ⟦F ⟶ G; ¬G⟧ ⟹ ¬F <br />
· impI: (P ⟹ Q) ⟹ P ⟶ Q<br />
· disjI1: P ⟹ P ∨ Q<br />
· disjI2: Q ⟹ P ∨ Q<br />
· disjE: ⟦P ∨ Q; P ⟹ R; Q ⟹ R⟧ ⟹ R <br />
· FalseE: False ⟹ P<br />
· notE: ⟦¬P; P⟧ ⟹ R<br />
· notI: (P ⟹ False) ⟹ ¬P<br />
· iffI: ⟦P ⟹ Q; Q ⟹ P⟧ ⟹ P = Q<br />
· iffD1: ⟦Q = P; Q⟧ ⟹ P <br />
· iffD2: ⟦P = Q; Q⟧ ⟹ P<br />
· ccontr: (¬P ⟹ False) ⟹ P<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
text {*<br />
Se usarán las reglas notnotI y mt que demostramos a continuación. *}<br />
<br />
lemma notnotI: "P ⟹ ¬¬ P"<br />
by auto<br />
<br />
lemma mt: "⟦F ⟶ G; ¬G⟧ ⟹ ¬F"<br />
by auto<br />
<br />
section {* Implicaciones *}<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 1. Demostrar<br />
p ⟶ q, p ⊢ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_1: josrodjim2 carmarria inmbenber marmedmar3<br />
assumes 1: "p ⟶ q" and<br />
2: "p"<br />
shows "q"<br />
<br />
proof -<br />
show 1: "q" using 1 2 by (rule mp)<br />
qed<br />
<br />
lemma ej_1: joslopjim4<br />
assumes "p --> q"<br />
"p"<br />
shows "q"<br />
proof -<br />
show "q" using assms(1) assms(2) by (rule mp)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 2. Demostrar<br />
p ⟶ q, q ⟶ r, p ⊢ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_2: josrodjim2 carmarria inmbenber marmedmar3<br />
assumes 1: "p ⟶ q" and<br />
2: "q ⟶ r" and<br />
3: "p" <br />
shows "r"<br />
<br />
proof-<br />
have 4: "q" using 1 3 by (rule mp)<br />
show "r" using 2 4 by (rule mp)<br />
qed<br />
<br />
lemma ej_2: joslopjim4<br />
assumes "p ⟶ q"<br />
"q ⟶ r"<br />
"p" <br />
shows "r"<br />
proof -<br />
have "q" using `p⟶q` `p` by (rule mp)<br />
show "r" using `q⟶r` `q` by (rule mp)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 3. Demostrar<br />
p ⟶ (q ⟶ r), p ⟶ q, p ⊢ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_3: josrodjim2 carmarria inmbenber marmedmar3<br />
assumes 1: "p ⟶ (q ⟶ r)" and<br />
2: "p ⟶ q" and<br />
3: "p"<br />
shows "r"<br />
<br />
proof-<br />
have 4: "q" using 2 3 by (rule mp)<br />
have 5: "(q ⟶ r)" using 1 3 by (rule mp)<br />
show "r" using 5 4 by (rule mp)<br />
<br />
qed<br />
<br />
lemma ej_3: joslopjim4<br />
assumes "p ⟶ (q ⟶ r)"<br />
"p ⟶ q"<br />
"p"<br />
shows "r"<br />
proof -<br />
have "q" using `p⟶q` `p` by (rule mp)<br />
have "q⟶r" using `p ⟶ (q ⟶ r)` `p` by (rule mp)<br />
show "r" using `q⟶r` `q` by (rule mp)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 4. Demostrar<br />
p ⟶ q, q ⟶ r ⊢ p ⟶ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_4: josrodjim2<br />
assumes 1: "p ⟶ q" and<br />
2: "q ⟶ r" <br />
shows "p ⟶ r"<br />
<br />
proof-<br />
{ assume 3: "p"<br />
have 4: "q" using 1 3 by (rule mp)<br />
have 5: "r" using 2 4 by (rule mp)}<br />
hence 6: "p ⟶ r" using 3 5 by (rule impI)<br />
show "p⟶r" using 6 by this<br />
<br />
oops<br />
<br />
lemma ejercicio_4: carmarria<br />
assumes 1: "p ⟶ q" and<br />
2: "q ⟶ r" <br />
shows "p ⟶ r"<br />
proof -<br />
{ assume 3: p <br />
have 4: "q" using 1 3 by (rule mp)<br />
have 5: "r" using 2 4 by (rule mp)<br />
}<br />
thus "p ⟶ r" by (rule impI) <br />
qed<br />
lemma ejercicio_4: inmbenber<br />
assumes 1: "p ⟶ q" and<br />
2: "q ⟶ r" <br />
shows "p ⟶ r"<br />
proof -<br />
{ assume 3: "p" <br />
have 4: "q" using 1 3 by (rule mp)<br />
have 5: "r" using 2 4 by (rule mp) }<br />
thus "p ⟶ r" by (rule impI) <br />
qed<br />
<br />
lemma ejercicio_4:marmedmar3<br />
assumes "p ⟶ q"<br />
"q ⟶ r" <br />
shows "p ⟶ r"<br />
proof <br />
assume "p" <br />
with assms(1) have "q" by (rule mp)<br />
with assms(2) show "r" by (rule mp)<br />
qed <br />
<br />
lemma ej_4: joslopjim4<br />
assumes "p ⟶ q"<br />
"q ⟶ r" <br />
shows "p ⟶ r"<br />
proof (rule impI)<br />
assume "p"<br />
have "q" using assms(1) `p` by (rule mp)<br />
show "r" using assms(2) `q` by (rule mp)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 5. Demostrar<br />
p ⟶ (q ⟶ r) ⊢ q ⟶ (p ⟶ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_5: josrodjim2 (NO DA ERROR, PERO SI ALERTA)<br />
assumes "p ⟶ (q ⟶ r)" <br />
shows "q ⟶ (p ⟶ r)"<br />
<br />
proof -<br />
<br />
{ assume 2: "q"<br />
{assume 3: "p"<br />
have 4: "q ⟶r" using 1 3 by (rule mp)<br />
have 5: "r" using 4 2 by (rule mp)}<br />
hence 6: "p⟶r" using 3 5 by (rule impI)}<br />
hence 7: "q ⟶ (p ⟶ r)" using 2 6 by (rule impI)<br />
show "q ⟶ (p ⟶ r)" using 7 by this<br />
<br />
oops<br />
<br />
lemma ejercicio_5: carmarria <br />
assumes 1: "p ⟶ (q ⟶ r)" <br />
shows "q ⟶ (p ⟶ r)"<br />
<br />
proof - <br />
{assume 2: q<br />
{assume 3: p <br />
have 4: "q ⟶ r" using 1 3 by (rule mp)<br />
have 5: "r" using 4 2 by (rule mp)<br />
}<br />
hence " p ⟶ r" by (rule impI)<br />
}<br />
thus "q ⟶ (p ⟶ r)" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_5: inmbenber<br />
assumes 1: "p ⟶ (q ⟶ r)" <br />
shows "q ⟶ (p ⟶ r)"<br />
<br />
proof - <br />
{assume 2: "q"<br />
{assume 3: "p" <br />
have 4: "q ⟶ r" using 1 3 by (rule mp)<br />
have 5: "r" using 4 2 by (rule mp) }<br />
hence " p ⟶ r" by (rule impI) }<br />
thus "q ⟶ (p ⟶ r)" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_5: marmedmar3<br />
assumes "p ⟶ (q ⟶ r)" <br />
shows "q ⟶ (p ⟶ r)"<br />
proof <br />
assume "q"<br />
show "(p ⟶ r)"<br />
proof <br />
assume "p" <br />
with assms(1) have "(q ⟶ r)" by (rule mp)<br />
thus "r" using `q` by (rule mp) <br />
qed <br />
qed<br />
<br />
lemma ej_5: joslopjim4<br />
assumes "p ⟶ (q ⟶ r)" <br />
shows "q ⟶ (p ⟶ r)"<br />
proof (rule impI)<br />
assume "q"<br />
show "p⟶r"<br />
proof (rule impI)<br />
assume "p"<br />
have "q⟶r" using `p ⟶ (q ⟶ r)` `p` by (rule mp)<br />
show "r" using `q⟶r` `q` by (rule mp)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 6. Demostrar<br />
p ⟶ (q ⟶ r) ⊢ (p ⟶ q) ⟶ (p ⟶ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_6:josrodjim2<br />
assumes 1: "p ⟶ (q ⟶ r)" <br />
shows "(p ⟶ q) ⟶ (p ⟶ r)"<br />
<br />
proof- <br />
<br />
{assume 2: "p⟶q" <br />
{assume 3: "p"<br />
have 4: "q⟶r" using 1 3 by (rule mp)<br />
have 5: "q" using 2 3 by (rule mp)<br />
have 6: "r" using 4 5 by (rule mp)}<br />
hence 7: "p⟶r" using 3 6 by (rule impI)}<br />
hence 8: "(p ⟶ q) ⟶ (p ⟶ r)" using 2 7 by (rule impI)<br />
show "(p ⟶ q) ⟶ (p ⟶ r)" using 8 by this<br />
oops<br />
<br />
lemma ejercicio_6: carmarria, inmbenber<br />
assumes 1: "p ⟶ (q ⟶ r)" <br />
shows "(p ⟶ q) ⟶ (p ⟶ r)"<br />
proof -<br />
{assume 2: "p ⟶ q"<br />
{assume 3: "p"<br />
have 4: "q" using 2 3 by (rule mp)<br />
have 5: "q ⟶ r" using 1 3 by (rule mp)<br />
have 6: "r" using 5 4 by (rule mp)<br />
}<br />
hence 7: "p ⟶ r" by (rule impI)<br />
}<br />
thus "(p ⟶ q) ⟶ (p ⟶ r)" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_6:marmedmar3<br />
assumes 1: "p ⟶ (q ⟶ r)" <br />
shows "(p ⟶ q) ⟶ (p ⟶ r)"<br />
proof <br />
assume 2: "(p ⟶ q)" <br />
show "(p ⟶ r)" <br />
proof <br />
assume 3: "p" <br />
have 4: "q" using 2 3 by (rule mp)<br />
have 5: "(q ⟶ r)" using 1 3 by (rule mp) <br />
show "r" using 5 4 by (rule mp) <br />
qed<br />
qed<br />
<br />
lemma ej_6: joslopjim4<br />
assumes "p ⟶ (q ⟶ r)" <br />
shows "(p ⟶ q) ⟶ (p ⟶ r)"<br />
proof (rule impI)<br />
assume "(p⟶q)"<br />
show "p⟶r"<br />
proof (rule impI)<br />
assume "p"<br />
have "q⟶r" using `p ⟶ (q ⟶ r)` `p` by (rule mp)<br />
have "q" using `p⟶q` `p` by (rule mp)<br />
show "r" using `q⟶r` `q` by (rule mp)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 7. Demostrar<br />
p ⊢ q ⟶ p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_7: (NO SE QUE ESTA MAL)<br />
assumes "p" <br />
shows "q ⟶ p"<br />
<br />
proof -<br />
<br />
{ assume 2: "q"<br />
have 3: "p" using 1<br />
}<br />
hence 4: "q⟶p" using 2 3 by (rule impI)<br />
<br />
show "q⟶p" using 4 by this<br />
<br />
oops<br />
<br />
lemma ejercicio_7: carmarria, inmbenber<br />
assumes 1: "p" <br />
shows "q ⟶ p"<br />
proof -<br />
{assume 2: "q"<br />
have 3: "p" using 1 by this}<br />
thus "q ⟶ p" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_7:marmedmar3<br />
assumes "p" <br />
shows "q ⟶ p"<br />
proof <br />
assume "q" <br />
show "p" using assms(1) by this<br />
qed <br />
<br />
lemma ej_7: joslopjim4<br />
assumes "p" <br />
shows "q ⟶ p"<br />
proof (rule impI)<br />
assume "q"<br />
show "p" using assms by this<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 8. Demostrar<br />
⊢ p ⟶ (q ⟶ p)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_8: carmarria<br />
"p ⟶ (q ⟶ p)"<br />
<br />
proof -<br />
{assume 1: p<br />
{assume 2: q<br />
have 3: p using 1 by this<br />
}<br />
hence 4: "q ⟶ p" by (rule impI)<br />
}<br />
thus "p ⟶ (q ⟶ p)" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_8: marmedmar3<br />
"p ⟶ (q ⟶ p)"<br />
proof<br />
assume 1: "p"<br />
show "(q ⟶ p)"<br />
proof <br />
assume 2: "q" <br />
show "p" using 1 by this <br />
qed<br />
qed<br />
<br />
lemma ej_8: joslopjim4<br />
shows "p⟶(q⟶p)"<br />
proof (rule impI)<br />
assume "p"<br />
show "q⟶p"<br />
proof (rule impI)<br />
assume "q"<br />
show "p" using `p` by this<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 9. Demostrar<br />
p ⟶ q ⊢ (q ⟶ r) ⟶ (p ⟶ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_9: carmarria<br />
assumes 1: "p ⟶ q" <br />
shows "(q ⟶ r) ⟶ (p ⟶ r)"<br />
proof -<br />
{assume 2: "q ⟶ r"<br />
{assume 3: "p"<br />
have 4: "q" using 1 3 by (rule mp)<br />
have 5: "r" using 2 4 by (rule mp)<br />
}<br />
hence 6: "p ⟶ r" by (rule impI)<br />
}<br />
thus 7: "(q ⟶ r) ⟶ (p ⟶ r)" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_9: marmedmar3<br />
assumes "p ⟶ q" <br />
shows "(q ⟶ r) ⟶ (p ⟶ r)"<br />
proof <br />
assume "(q ⟶ r)" <br />
show "(p ⟶ r)"<br />
proof <br />
assume "p" <br />
with `(p ⟶ q)` have "q" by (rule mp)<br />
with `(q ⟶ r)` show "r" by (rule mp)<br />
qed<br />
qed<br />
<br />
lemma ej_9: joslopjim4<br />
assumes "p ⟶ q" <br />
shows "(q ⟶ r) ⟶ (p ⟶ r)"<br />
proof (rule impI)<br />
assume "q⟶r"<br />
show "p⟶r"<br />
proof (rule impI)<br />
assume "p"<br />
have "q" using `p⟶q` `p` by (rule mp)<br />
show "r" using `q⟶r` `q` by (rule mp)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 10. Demostrar<br />
p ⟶ (q ⟶ (r ⟶ s)) ⊢ r ⟶ (q ⟶ (p ⟶ s))<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_10: carmarria<br />
assumes 1: "p ⟶ (q ⟶ (r ⟶ s))" <br />
shows "r ⟶ (q ⟶ (p ⟶ s))"<br />
<br />
proof-<br />
{assume 2: r<br />
{assume 3: q<br />
{assume 4: p<br />
have 5: "q ⟶ (r ⟶ s)" using 1 4 by (rule mp)<br />
have 6: "r ⟶ s" using 5 3 by (rule mp)<br />
have 7: s using 6 2 by (rule mp)<br />
}<br />
hence 8: "p ⟶ s" by (rule impI)<br />
}<br />
hence 9: "q ⟶ (p ⟶ s)" by (rule impI)<br />
}<br />
thus "r ⟶ (q ⟶ (p ⟶ s))" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_10: marmedmar3<br />
assumes 1: "p ⟶ (q ⟶ (r ⟶ s))" <br />
shows "r ⟶ (q ⟶ (p ⟶ s))"<br />
proof <br />
assume 2: "r"<br />
show "(q ⟶ (p ⟶ s))"<br />
proof <br />
assume 3: "q" <br />
show "(p ⟶ s)"<br />
proof <br />
assume 4: "p" <br />
have 5: "(q ⟶ (r ⟶ s))" using 1 4 by (rule mp) <br />
have 6: "(r ⟶ s)" using 5 3 by (rule mp)<br />
show "s" using 6 2 by (rule mp) <br />
qed<br />
qed<br />
qed<br />
<br />
lemma ej_10: joslopjim4<br />
assumes "p ⟶ (q ⟶ (r ⟶ s))" <br />
shows "r ⟶ (q ⟶ (p ⟶ s))"<br />
proof (rule impI)<br />
assume "r"<br />
show "q⟶(p⟶s)"<br />
proof (rule impI)<br />
assume "q"<br />
show "p⟶s"<br />
proof (rule impI)<br />
assume "p"<br />
have "q ⟶ (r ⟶ s)" using `p ⟶ (q ⟶ (r ⟶ s))` `p` by (rule mp)<br />
have "r⟶s" using `q ⟶ (r ⟶ s)` `q` by (rule mp)<br />
show "s" using `r-->s` `r` by (rule mp)<br />
qed<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 11. Demostrar<br />
⊢ (p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_11: carmarria<br />
"(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))"<br />
proof - <br />
{assume 1: "p ⟶ (q ⟶ r)"<br />
{assume 2: "p ⟶ q"<br />
{assume 3: p<br />
have 4: q using 2 3 by (rule mp)<br />
have 5: "q ⟶ r" using 1 3 by (rule mp)<br />
have 6: r using 5 4 by (rule mp)<br />
}<br />
hence 7: "p ⟶ r" by (rule impI)<br />
}<br />
hence 8: "(p ⟶ q) ⟶ (p ⟶ r)" by (rule impI)<br />
}<br />
thus "(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_11: marmedmar3<br />
"(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))"<br />
proof <br />
assume 1: "(p ⟶ (q ⟶ r))" <br />
show "((p ⟶ q) ⟶ (p ⟶ r))"<br />
proof <br />
assume 2: "(p ⟶ q)" <br />
show "(p ⟶ r)" <br />
proof <br />
assume 3: "p" <br />
have 4: "(q ⟶ r)" using 1 3 by (rule mp) <br />
have 5: "q" using 2 3 by (rule mp) <br />
show "r" using 4 5 by (rule mp) <br />
qed<br />
qed<br />
qed<br />
<br />
lemma ej_11: joslopjim4<br />
shows "(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))"<br />
proof (rule impI)<br />
assume "p⟶(q⟶r)"<br />
show "(p⟶q)⟶(p⟶r)"<br />
proof (rule impI)<br />
assume "p⟶q"<br />
show "p⟶r"<br />
proof (rule impI)<br />
assume "p"<br />
have "q" using `p⟶q` `p` by (rule mp)<br />
have "q⟶r" using `p⟶(q⟶r)` `p` by (rule mp)<br />
show "r" using `q⟶r` `q` by (rule mp)<br />
qed<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 12. Demostrar<br />
(p ⟶ q) ⟶ r ⊢ p ⟶ (q ⟶ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_12: carmarria<br />
assumes 1: "(p ⟶ q) ⟶ r" <br />
shows "p ⟶ (q ⟶ r)"<br />
proof -<br />
{assume 2: p<br />
{assume 3: q<br />
{assume 4: p<br />
have 5: q using 3 by this<br />
}<br />
hence 6: "p ⟶ q" by (rule impI)<br />
have 7: r using 1 6 by (rule mp)<br />
}<br />
hence 8: "q ⟶ r" by (rule impI)<br />
}<br />
thus "p ⟶ (q ⟶ r)" by (rule impI)<br />
qed<br />
<br />
lemma ej_12: joslopjim4<br />
assumes "(p ⟶ q) ⟶ r" <br />
shows "p ⟶ (q ⟶ r)"<br />
proof (rule impI)<br />
assume "p"<br />
show "q⟶r"<br />
proof (rule impI)<br />
assume "q"<br />
have "p⟶q"<br />
proof (rule impI)<br />
assume "p"<br />
show "q" using `q` by this<br />
qed<br />
show "r" using `(p ⟶ q) ⟶ r` `p⟶q` by (rule mp)<br />
qed<br />
qed<br />
<br />
section {* Conjunciones *}<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 13. Demostrar<br />
p, q ⊢ p ∧ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_13: carmarria marmedmar3<br />
assumes "p"<br />
"q" <br />
shows "p ∧ q"<br />
<br />
proof-<br />
show "p \<and> q" using assms(1) assms(2) by (rule conjI)<br />
qed<br />
<br />
lemma ej_13: joslopjim4<br />
assumes "p"<br />
"q" <br />
shows "p ∧ q"<br />
proof-<br />
show "p∧q" using `p` `q` by (rule conjI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 14. Demostrar<br />
p ∧ q ⊢ p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_14: carmarria marmedmar3<br />
assumes "p ∧ q" <br />
shows "p"<br />
<br />
proof -<br />
show "p" using assms by (rule conjunct1)<br />
qed<br />
<br />
lemma ej_14: joslopjim4<br />
assumes "p ∧ q" <br />
shows "p"<br />
proof -<br />
show "p" using assms by (rule conjunct1)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 15. Demostrar<br />
p ∧ q ⊢ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_15: carmarria marmedmar3<br />
assumes "p ∧ q" <br />
shows "q"<br />
<br />
proof -<br />
show q using assms(1) by (rule conjunct2)<br />
qed<br />
<br />
lemma ej_15: joslopjim4<br />
assumes "p ∧ q" <br />
shows "q"<br />
proof-<br />
show "q" using assms by (rule conjunct2)<br />
qed<br />
<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 16. Demostrar<br />
p ∧ (q ∧ r) ⊢ (p ∧ q) ∧ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_16: carmarria marmedmar3<br />
assumes 1: "p ∧ (q ∧ r)"<br />
shows "(p ∧ q) ∧ r"<br />
proof -<br />
have 2: "q \<and> r" using 1 by (rule conjunct2)<br />
have 3: p using 1 by (rule conjunct1)<br />
have 4: q using 2 by (rule conjunct1)<br />
have 5: r using 2 by (rule conjunct2)<br />
have 6: "p \<and> q" using 3 4 by (rule conjI)<br />
show "(p \<and> q) \<and> r" using 6 5 by (rule conjI)<br />
qed<br />
<br />
lemma ej_16: joslopjim4<br />
assumes "p ∧ (q ∧ r)"<br />
shows "(p ∧ q) ∧ r"<br />
proof-<br />
have "p" using assms by (rule conjunct1)<br />
have "q∧r" using assms by (rule conjunct2)<br />
have "q" using `q∧r` by (rule conjunct1)<br />
have "r" using `q∧r` by (rule conjunct2)<br />
have "p∧q" using `p` `q` by (rule conjI)<br />
show "(p∧q)∧r" using `p∧q` `r` by (rule conjI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 17. Demostrar<br />
(p ∧ q) ∧ r ⊢ p ∧ (q ∧ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_17:<br />
assumes "(p ∧ q) ∧ r" <br />
shows "p ∧ (q ∧ r)"<br />
proof (rule conjI)<br />
have "p ∧ q" using assms ..<br />
thus "p" ..<br />
next<br />
show "q ∧ r"<br />
proof (rule conjI)<br />
have "p ∧ q" using assms ..<br />
thus "q" ..<br />
next<br />
show "r" using assms ..<br />
qed<br />
qed<br />
<br />
lemma ejercicio_17: carmarria marmedmar3<br />
assumes 1:"(p ∧ q) ∧ r" <br />
shows "p ∧ (q ∧ r)"<br />
proof-<br />
have 2:"p \<and> q" using 1 by (rule conjunct1)<br />
have 3: r using 1 by (rule conjunct2)<br />
have 4: p using 2 by (rule conjunct1)<br />
have 5: q using 2 by (rule conjunct2)<br />
have 6: "q \<and> r" using 5 3 by (rule conjI)<br />
show "p \<and> (q \<and> r)" using 4 6 by (rule conjI)<br />
qed<br />
<br />
<br />
lemma ej_17: joslopjim4<br />
assumes "(p ∧ q) ∧ r" <br />
shows "p ∧ (q ∧ r)"<br />
proof-<br />
have "p∧q" using assms by (rule conjunct1)<br />
have "r" using assms by (rule conjunct2)<br />
have "p" using `p∧q` by (rule conjunct1)<br />
have "q" using `p∧q` by (rule conjunct2)<br />
have "q∧r" using `q` `r` by (rule conjI)<br />
show "p∧(q∧r)" using `p` `q∧r` by (rule conjI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 18. Demostrar<br />
p ∧ q ⊢ p ⟶ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_18: carmarria<br />
assumes "p ∧ q" <br />
shows "p ⟶ q"<br />
proof -<br />
{assume p<br />
have q using assms(1) by (rule conjunct2)<br />
}<br />
thus "p ⟶ q" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_18: marmedmar3<br />
assumes "p ∧ q" <br />
shows "p ⟶ q"<br />
proof <br />
assume "p" <br />
show "q" using assms by (rule conjunct2) <br />
qed <br />
<br />
lemma ej_18: joslopjim4<br />
assumes "p ∧ q" <br />
shows "p ⟶ q"<br />
proof (rule impI)<br />
assume "p"<br />
show "q" using assms by (rule conjunct2)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 19. Demostrar<br />
(p ⟶ q) ∧ (p ⟶ r) ⊢ p ⟶ q ∧ r <br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_19: carmarria<br />
assumes 1: "(p ⟶ q) ∧ (p ⟶ r)" <br />
shows "p ⟶ q ∧ r"<br />
proof -<br />
{assume 2: p<br />
have 3: "p ⟶ q" using 1 by (rule conjunct1)<br />
have 4: "p ⟶ r" using 1 by (rule conjunct2)<br />
have 5: q using 3 2 by (rule mp)<br />
have 6: r using 4 2 by (rule mp)<br />
have 7: "q \<and> r" using 5 6 by (rule conjI)<br />
}<br />
thus "p ⟶ q \<and> r" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_19: marmedmar3<br />
assumes 1: "(p ⟶ q) ∧ (p ⟶ r)" <br />
shows "p ⟶ q ∧ r"<br />
proof <br />
have 2: "(p ⟶ q)" using assms by (rule conjunct1)<br />
have 3: "(p ⟶ r)" using assms by (rule conjunct2)<br />
assume 4: "p"<br />
with `(p ⟶ q)` have 5: "q" by (rule mp)<br />
have 6: "r" using 3 4 by (rule mp) <br />
show "q ∧ r" using 5 6 by (rule conjI)<br />
qed <br />
<br />
lemma ej_19: joslopjim4<br />
assumes "(p ⟶ q) ∧ (p ⟶ r)" <br />
shows "p ⟶ q ∧ r"<br />
proof (rule impI)<br />
assume "p"<br />
have "p⟶q" using assms by (rule conjunct1)<br />
have "q" using `p⟶q` `p` by (rule mp)<br />
have "p⟶r" using assms by (rule conjunct2)<br />
have "r" using `p⟶r` `p` by (rule mp)<br />
show "q∧r" using `q` `r` by (rule conjI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 20. Demostrar<br />
p ⟶ q ∧ r ⊢ (p ⟶ q) ∧ (p ⟶ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_20: carmarria<br />
assumes 1: "p ⟶ q ∧ r" <br />
shows "(p ⟶ q) ∧ (p ⟶ r)"<br />
proof -<br />
{assume 2: p<br />
have 3: "q \<and> r" using 1 2 by (rule mp)<br />
have 4: q using 3 by (rule conjunct1)<br />
}<br />
hence 5: "p ⟶ q" by (rule impI)<br />
{assume 6: p<br />
have 7: "q \<and> r" using 1 6 by (rule mp)<br />
have 8: r using 7 by (rule conjunct2)<br />
}<br />
hence 9: "p ⟶ r" by (rule impI)<br />
show "(p ⟶ q) \<and> (p ⟶ r)" using 5 9 by (rule conjI)<br />
qed<br />
<br />
lemma ej_20: joslopjim4<br />
assumes "p ⟶ q ∧ r" <br />
shows "(p ⟶ q) ∧ (p ⟶ r)"<br />
proof-<br />
have "p⟶q"<br />
proof<br />
assume "p"<br />
have "q∧r" using assms `p` by (rule mp)<br />
show "q" using `q∧r` by (rule conjunct1)<br />
qed<br />
have "p⟶r"<br />
proof<br />
assume "p"<br />
have "q∧r" using assms `p` by (rule mp)<br />
show "r" using `q∧r` by (rule conjunct2)<br />
qed<br />
show "(p⟶q)∧(p⟶r)" using `p⟶q` `p⟶r` by (rule conjI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 21. Demostrar<br />
p ⟶ (q ⟶ r) ⊢ p ∧ q ⟶ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_21: carmarria<br />
assumes 1: "p ⟶ (q ⟶ r)" <br />
shows "p ∧ q ⟶ r"<br />
proof -<br />
{assume 2: "p \<and> q"<br />
have 3: p using 2 by (rule conjunct1)<br />
have 4: q using 2 by (rule conjunct2)<br />
have 5: "q ⟶ r" using 1 3 by (rule mp)<br />
have 6: r using 5 4 by (rule mp)<br />
}<br />
thus "p ∧ q ⟶ r" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_21: marmedmar3<br />
assumes "p ⟶ (q ⟶ r)" <br />
shows "p ∧ q ⟶ r"<br />
proof <br />
assume "p ∧ q" <br />
hence "p" by (rule conjunct1) <br />
with assms have "(q ⟶ r)" by (rule mp)<br />
have "q" using `p ∧ q` by (rule conjunct2) <br />
with `(q ⟶ r)` show "r" by (rule mp) <br />
qed <br />
<br />
lemma ej_21: joslopjim4<br />
assumes "p ⟶ (q ⟶ r)" <br />
shows "p ∧ q ⟶ r"<br />
proof (rule impI)<br />
assume "p∧q"<br />
have "p" using `p∧q` by (rule conjunct1)<br />
have "q" using `p∧q` by (rule conjunct2)<br />
have "q⟶r" using assms `p` by (rule mp)<br />
show "r" using `q⟶r` `q` by (rule mp)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 22. Demostrar<br />
p ∧ q ⟶ r ⊢ p ⟶ (q ⟶ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_22: carmarria<br />
assumes "p ∧ q ⟶ r" <br />
shows "p ⟶ (q ⟶ r)"<br />
proof -<br />
{assume p<br />
{assume q<br />
have "p \<and> q" using `p` `q` by (rule conjI)<br />
have r using assms(1) `p \<and> q` by (rule mp)<br />
}<br />
hence "q ⟶ r" by (rule impI)<br />
}<br />
thus "p ⟶ (q ⟶ r)" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_22: marmedmar3<br />
assumes "p ∧ q ⟶ r" <br />
shows "p ⟶ (q ⟶ r)"<br />
proof <br />
assume "p" <br />
show "(q ⟶ r)"<br />
proof <br />
assume "q" <br />
with `p` have "p ∧ q" by (rule conjI) <br />
with assms show "r" by (rule mp) <br />
qed <br />
qed<br />
<br />
lemma ej_22: joslopjim4<br />
assumes "p ∧ q ⟶ r" <br />
shows "p ⟶ (q ⟶ r)"<br />
proof (rule impI)<br />
assume "p"<br />
show "q⟶r"<br />
proof (rule impI)<br />
assume "q"<br />
have "p∧q" using `p` `q` by (rule conjI)<br />
show "r" using assms `p∧q` by (rule mp)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 23. Demostrar<br />
(p ⟶ q) ⟶ r ⊢ p ∧ q ⟶ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_23: carmarria<br />
assumes "(p ⟶ q) ⟶ r" <br />
shows "p ∧ q ⟶ r"<br />
proof -<br />
{assume "p \<and> q" <br />
{assume p<br />
have q using `p \<and> q` by (rule conjunct2)<br />
}<br />
hence "p ⟶ q" by (rule impI)<br />
have r using assms(1) `p ⟶ q` by (rule mp)<br />
}<br />
thus "p \<and> q ⟶ r" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_23: marmedmar3<br />
assumes "(p ⟶ q) ⟶ r" <br />
shows "p ∧ q ⟶ r"<br />
proof <br />
assume "p ∧ q"<br />
have "p" using `p ∧ q` by (rule conjunct1)<br />
have "q" using `p ∧ q` by (rule conjunct2) <br />
hence "(p ⟶ q)" by (rule impI) <br />
with assms show "r" by (rule mp) <br />
qed <br />
<br />
lemma ej_23: joslopjim4<br />
assumes "(p ⟶ q) ⟶ r" <br />
shows "p ∧ q ⟶ r"<br />
proof (rule impI)<br />
assume "p∧q"<br />
have "p⟶q"<br />
proof<br />
assume "p"<br />
show "q" using `p∧q` by (rule conjunct2)<br />
qed<br />
show "r" using assms `p⟶q` by (rule mp)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 24. Demostrar<br />
p ∧ (q ⟶ r) ⊢ (p ⟶ q) ⟶ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_24: carmarria<br />
assumes 1: "p ∧ (q ⟶ r)" <br />
shows "(p ⟶ q) ⟶ r"<br />
proof -<br />
{assume 2: "p ⟶ q" <br />
have 3: p using 1 by (rule conjunct1)<br />
have 4: "q ⟶ r" using 1 by (rule conjunct2)<br />
have 5: q using 2 3 by (rule mp)<br />
have 6: r using 4 5 by (rule mp)<br />
}<br />
thus "(p ⟶ q) ⟶ r" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_24: marmedmar3<br />
assumes "p ∧ (q ⟶ r)" <br />
shows "(p ⟶ q) ⟶ r"<br />
proof <br />
assume "(p ⟶ q)" <br />
have "(q ⟶ r)" using assms by (rule conjunct2)<br />
have "p" using assms by (rule conjunct1)<br />
with `(p ⟶ q)` have "q" by (rule mp) <br />
with `(q ⟶ r)` show "r" by (rule mp)<br />
qed <br />
<br />
lemma ej_24: joslopjim4<br />
assumes "p ∧ (q ⟶ r)" <br />
shows "(p ⟶ q) ⟶ r"<br />
proof (rule impI)<br />
assume "p⟶q"<br />
have "p" using assms by (rule conjunct1)<br />
have "q⟶r" using assms by (rule conjunct2)<br />
have "q" using `p⟶q` `p` by (rule mp)<br />
show "r" using `q⟶r` `q` by (rule mp)<br />
qed<br />
<br />
section {* Disyunciones *}<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 25. Demostrar<br />
p ⊢ p ∨ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_25: carmarria<br />
assumes "p"<br />
shows "p ∨ q"<br />
proof -<br />
show "p | q" using assms(1) by (rule disjI1)<br />
qed<br />
<br />
lemma ej_25: joslopjim4<br />
assumes "p"<br />
shows "p ∨ q"<br />
proof (rule disjI1)<br />
show "p" using assms by this<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 26. Demostrar<br />
q ⊢ p ∨ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_26: carmarria<br />
assumes "q"<br />
shows "p ∨ q"<br />
proof -<br />
show "p | q" using assms(1) by (rule disjI2)<br />
qed<br />
<br />
lemma ej_26: joslopjim4<br />
assumes "q"<br />
shows "p ∨ q"<br />
proof (rule disjI2)<br />
show "q" using assms by this<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 27. Demostrar<br />
p ∨ q ⊢ q ∨ p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_27: carmarria<br />
assumes "p ∨ q"<br />
shows "q ∨ p"<br />
proof -<br />
have "p | q" using assms(1) by this<br />
moreover<br />
{assume p<br />
have "q | p" using `p` by (rule disjI2)<br />
}<br />
moreover<br />
{assume q<br />
have "q | p" using `q` by (rule disjI1)<br />
}<br />
ultimately show "q | p" by (rule disjE)<br />
qed<br />
<br />
lemma ej_27: joslopjim4<br />
assumes "p ∨ q"<br />
shows "q ∨ p"<br />
using assms<br />
proof <br />
assume "p"<br />
then show "q∨p" by(rule disjI2)<br />
next<br />
assume "q"<br />
show "q∨p" using `q` by(rule disjI1)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 28. Demostrar<br />
q ⟶ r ⊢ p ∨ q ⟶ p ∨ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_28: carmarria<br />
assumes "q ⟶ r" <br />
shows "p ∨ q ⟶ p ∨ r"<br />
proof -<br />
{assume "p | q" <br />
moreover<br />
{assume p<br />
have "p | r" using `p` by (rule disjI1)<br />
}<br />
moreover<br />
{assume q<br />
have r using assms(1) `q` by (rule mp)<br />
have "p | r" using `r` by (rule disjI2)<br />
}<br />
ultimately have "p | r" by (rule disjE)<br />
}<br />
thus "p | q ⟶ p | r" by (rule impI)<br />
qed<br />
<br />
lemma ej_28: joslopjim4<br />
assumes "q ⟶ r" <br />
shows "p ∨ q ⟶ p ∨ r"<br />
proof (rule impI)<br />
assume "p∨q" <br />
show "p∨r"<br />
using `p∨q`<br />
proof <br />
assume "p"<br />
show "p∨r" using `p` by (rule disjI1)<br />
next<br />
assume "q"<br />
have "r" using assms `q` by (rule mp)<br />
show "p∨r" using `r` by (rule disjI2)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 29. Demostrar<br />
p ∨ p ⊢ p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_29: carmarria<br />
assumes "p ∨ p"<br />
shows "p"<br />
proof -<br />
have "p | p" using assms(1) by this<br />
moreover <br />
{assume p<br />
}<br />
moreover<br />
{assume p<br />
}<br />
ultimately show p by (rule disjE)<br />
qed<br />
<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 30. Demostrar<br />
p ⊢ p ∨ p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_30: carmarria<br />
assumes "p" <br />
shows "p ∨ p"<br />
proof -<br />
show "p | p" using assms(1) by (rule disjI1)<br />
qed<br />
<br />
lemma ej_30: joslopjim4<br />
assumes "p" <br />
shows "p ∨ p"<br />
proof (rule disjI1)<br />
show "p" using assms by this<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 31. Demostrar<br />
p ∨ (q ∨ r) ⊢ (p ∨ q) ∨ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_31: carmarria<br />
assumes 1: "p ∨ (q ∨ r)" <br />
shows "(p ∨ q) ∨ r"<br />
proof -<br />
have "p | (q | r)" using 1 by this<br />
moreover {assume 2: p<br />
have 3: "p | q" using 2 by (rule disjI1)<br />
have 4: "(p | q) | r" using 3 by (rule disjI1)<br />
}<br />
moreover {assume 5: "q | r"<br />
moreover {assume 6: q<br />
have 7: "p | q" using 6 by (rule disjI2)<br />
have 8: "(p | q) | r " using 7 by (rule disjI1)<br />
}<br />
moreover {assume 9: r<br />
have 10: "(p | q) | r " using 9 by (rule disjI2)<br />
}<br />
ultimately have 11: "(p | q) | r " by (rule disjE)<br />
}<br />
ultimately show "(p | q) | r " by (rule disjE)<br />
qed<br />
<br />
lemma ej_31: joslopjim4<br />
assumes "p ∨ (q ∨ r)" <br />
shows "(p ∨ q) ∨ r"<br />
using assms<br />
proof <br />
assume "p"<br />
have "p∨q" using `p` by (rule disjI1)<br />
thus "(p ∨ q) ∨ r" by (rule disjI1)<br />
next<br />
assume "q∨r"<br />
show "(p ∨ q) ∨ r"<br />
using `q∨r`<br />
proof <br />
assume "q"<br />
have "p∨q" using `q` by (rule disjI2)<br />
show "(p ∨ q) ∨ r" using `p∨q` by (rule disjI1)<br />
next<br />
assume "r"<br />
show "(p ∨ q) ∨ r" using `r` by (rule disjI2)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 32. Demostrar<br />
(p ∨ q) ∨ r ⊢ p ∨ (q ∨ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_32: carmarria<br />
assumes 1: "(p ∨ q) ∨ r" <br />
shows "p ∨ (q ∨ r)"<br />
proof -<br />
have "(p | q) | r" using 1 by this<br />
moreover {assume 2: "p | q" <br />
moreover {assume 3: p <br />
have 4: "p | (q | r)" using 3 by (rule disjI1)<br />
}<br />
moreover {assume 5: q<br />
have 6: "q | r" using 5 by (rule disjI1)<br />
have 7: "p | (q | r)" using 6 by (rule disjI2)<br />
}<br />
ultimately have 8: " p | (q | r)" by (rule disjE)<br />
}<br />
moreover {assume 9: r<br />
have 10: "q | r" using 9 by (rule disjI2)<br />
have 11: "p | (q | r)" using 10 by (rule disjI2)<br />
}<br />
ultimately show "p | (q | r)" by (rule disjE)<br />
qed<br />
<br />
lemma ej_32: joslopjim4<br />
assumes "(p ∨ q) ∨ r" <br />
shows "p ∨ (q ∨ r)"<br />
using assms<br />
proof<br />
assume "p∨q"<br />
show "p ∨ (q ∨ r)"<br />
using `p∨q`<br />
proof <br />
assume "p"<br />
show "p ∨ (q ∨ r)" using `p` by (rule disjI1)<br />
next<br />
assume "q"<br />
have "q∨r" using `q` by (rule disjI1)<br />
show "p ∨ (q ∨ r)" using `q∨r` by (rule disjI2)<br />
qed<br />
next<br />
assume "r"<br />
have "q∨r" using `r` by (rule disjI2)<br />
show "p ∨ (q ∨ r)" using `q∨r` by (rule disjI2)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 33. Demostrar<br />
p ∧ (q ∨ r) ⊢ (p ∧ q) ∨ (p ∧ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_33: carmarria<br />
assumes 1: "p ∧ (q ∨ r)" <br />
shows "(p ∧ q) ∨ (p ∧ r)"<br />
proof -<br />
have 2: p using 1 by (rule conjunct1)<br />
have 3: "q | r" using 1 by (rule conjunct2)<br />
moreover {assume 4: q<br />
have 5: "p & q" using 2 4 by (rule conjI)<br />
have 6: "(p & q) | (p & r)" using 5 by (rule disjI1)<br />
}<br />
moreover {assume 7: r<br />
have 8: "p & r" using 2 7 by (rule conjI)<br />
have 9: "(p & q) | (p & r)" using 8 by (rule disjI2)<br />
}<br />
ultimately show "(p & q) | (p & r)" by (rule disjE)<br />
qed<br />
<br />
lemma ej_33: joslopjim4<br />
assumes "p ∧ (q ∨ r)" <br />
shows "(p ∧ q) ∨ (p ∧ r)"<br />
proof-<br />
have "p" using assms by (rule conjunct1)<br />
have "q∨r" using assms by (rule conjunct2)<br />
show "(p ∧ q) ∨ (p ∧ r)"<br />
using `q∨r` <br />
proof (rule disjE)<br />
assume "q"<br />
have "p∧q" using `p` `q` by (rule conjI)<br />
show "(p ∧ q) ∨ (p ∧ r)" using `p∧q` by (rule disjI1)<br />
next<br />
assume "r"<br />
have "p∧r" using `p` `r` by (rule conjI)<br />
show "(p ∧ q) ∨ (p ∧ r)" using `p∧r` by (rule disjI2)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 34. Demostrar<br />
(p ∧ q) ∨ (p ∧ r) ⊢ p ∧ (q ∨ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_34: carmarria<br />
assumes 1: "(p ∧ q) ∨ (p ∧ r)" <br />
shows "p ∧ (q ∨ r)"<br />
proof -<br />
have "(p & q) | (p & r)" using 1 by this<br />
moreover {assume 2: "p & q"<br />
have 3: p using 2 by (rule conjunct1)<br />
have 4: q using 2 by (rule conjunct2)<br />
have 5: "q | r" using 4 by (rule disjI1)<br />
have 6: "p & (q | r)" using 3 5 by (rule conjI)<br />
}<br />
moreover {assume 7: "p & r"<br />
have 8: p using 7 by (rule conjunct1)<br />
have 9: r using 7 by (rule conjunct2)<br />
have 10: "q | r" using 9 by (rule disjI2)<br />
have 11: "p & (q | r)" using 8 10 by (rule conjI)<br />
}<br />
ultimately show "p & (q | r)" by (rule disjE)<br />
qed<br />
<br />
lemma ej_34: joslopjim4<br />
assumes "(p ∧ q) ∨ (p ∧ r)" <br />
shows "p ∧ (q ∨ r)"<br />
using assms<br />
proof<br />
assume "p∧q"<br />
have "p" using `p∧q` by (rule conjunct1)<br />
have "q" using `p∧q` by (rule conjunct2)<br />
have "q∨r" using `q` by (rule disjI1)<br />
show "p∧(q∨r)" using `p` `q∨r` by (rule conjI)<br />
next<br />
assume "p∧r"<br />
have "p" using `p∧r` by (rule conjunct1)<br />
have "r" using `p∧r` by (rule conjunct2)<br />
have "q∨r" using `r` by (rule disjI2)<br />
show "p∧(q∨r)" using `p` `q∨r` by (rule conjI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 35. Demostrar<br />
p ∨ (q ∧ r) ⊢ (p ∨ q) ∧ (p ∨ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_35: carmarria<br />
assumes 1: "p ∨ (q ∧ r)" <br />
shows "(p ∨ q) ∧ (p ∨ r)"<br />
proof -<br />
have "p | (q & r)" using 1 by this<br />
moreover {assume 2: p<br />
have 3: " p | q" using 2 by (rule disjI1)<br />
have 4: " p | r" using 2 by (rule disjI1)<br />
have 5: "(p | q) & (p | r)" using 3 4 by (rule conjI)<br />
}<br />
moreover {assume 6: "q & r" <br />
have 7: q using 6 by (rule conjunct1)<br />
have 8: r using 6 by (rule conjunct2)<br />
have 9: "p | q" using 7 by (rule disjI2)<br />
have 10: "p | r" using 8 by (rule disjI2)<br />
have 11: "(p | q) & (p | r)" using 9 10 by (rule conjI)<br />
}<br />
ultimately show "(p | q) & (p | r)" by (rule disjE)<br />
qed<br />
<br />
lemma ej_35: joslopjim4<br />
assumes "p ∨ (q ∧ r)" <br />
shows "(p ∨ q) ∧ (p ∨ r)"<br />
using assms<br />
proof (rule disjE)<br />
assume "p"<br />
have "p∨q" using `p` by (rule disjI1)<br />
have "p∨r" using `p` by (rule disjI1)<br />
show "(p ∨ q) ∧ (p ∨ r)" using `p∨q` `p∨r` by (rule conjI)<br />
next<br />
assume "q∧r"<br />
have "q" using `q∧r` by (rule conjunct1)<br />
have "r" using `q∧r` by (rule conjunct2)<br />
have "p∨q" using `q` by (rule disjI2)<br />
have "p∨r" using `r` by (rule disjI2)<br />
show "(p ∨ q) ∧ (p ∨ r)" using `p∨q` `p∨r` by (rule conjI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 36. Demostrar<br />
(p ∨ q) ∧ (p ∨ r) ⊢ p ∨ (q ∧ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_36: carmarria<br />
assumes 1: "(p ∨ q) ∧ (p ∨ r)"<br />
shows "p ∨ (q ∧ r)"<br />
proof -<br />
have 2: "p | q" using 1 by (rule conjunct1)<br />
moreover {assume 3: p<br />
have 4: "p | (q & r)" using 3 by (rule disjI1)<br />
}<br />
moreover {assume 5: q<br />
have 6: "p | r" using 1 by (rule conjunct2)<br />
moreover {assume 7: p<br />
have 8: "p | (q & r)" using 7 by (rule disjI1)<br />
}<br />
moreover {assume 9: r<br />
have 10: "q & r" using 5 9 by (rule conjI)<br />
have 11: "p | (q & r)" using 10 by (rule disjI2)<br />
}<br />
ultimately have 12: "p | (q & r)" by (rule disjE)<br />
}<br />
ultimately show "p | (q & r)" by (rule disjE)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 37. Demostrar<br />
(p ⟶ r) ∧ (q ⟶ r) ⊢ p ∨ q ⟶ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_37: carmarria<br />
assumes 1: "(p ⟶ r) ∧ (q ⟶ r)" <br />
shows "p ∨ q ⟶ r"<br />
proof -<br />
have 2: "p ⟶ r" using 1 by (rule conjunct1)<br />
have 3: "q ⟶ r" using 1 by (rule conjunct2)<br />
{assume 4: "p | q"<br />
moreover {assume 5: p<br />
have 6: r using 2 5 by (rule mp)<br />
}<br />
moreover {assume 7: q<br />
have 8: r using 3 7 by (rule mp)<br />
}<br />
ultimately have r by (rule disjE)<br />
}<br />
thus "p | q ⟶ r" by (rule impI)<br />
qed<br />
<br />
lemma ej_37: joslopjim4<br />
assumes "(p ⟶ r) ∧ (q ⟶ r)" <br />
shows "p ∨ q ⟶ r"<br />
proof (rule impI)<br />
assume "p∨q"<br />
have "p⟶r" using assms by (rule conjunct1)<br />
have "q⟶r" using assms by (rule conjunct2)<br />
show "r"<br />
using `p∨q`<br />
proof <br />
assume "p"<br />
show "r" using `p⟶r` `p` by (rule mp)<br />
next<br />
assume "q"<br />
show "r" using `q⟶r` `q` by (rule mp)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 38. Demostrar<br />
p ∨ q ⟶ r ⊢ (p ⟶ r) ∧ (q ⟶ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_38: carmarria<br />
assumes 1: "p ∨ q ⟶ r" <br />
shows "(p ⟶ r) ∧ (q ⟶ r)"<br />
proof -<br />
{assume 2: p<br />
have 3: "p | q" using 2 by (rule disjI1)<br />
have 4: r using 1 3 by (rule mp)<br />
}<br />
hence 5: "p ⟶ r" by (rule impI)<br />
{assume 6: q<br />
have 7: "p | q" using 6 by (rule disjI2)<br />
have 8: r using 1 7 by (rule mp)<br />
}<br />
hence 9: "q ⟶ r" by (rule impI)<br />
show "(p ⟶ r) & (q ⟶ r)" using 5 9 by (rule conjI)<br />
qed<br />
<br />
lemma ej_38: joslopjim4<br />
assumes "p ∨ q ⟶ r" <br />
shows "(p ⟶ r) ∧ (q ⟶ r)"<br />
proof-<br />
have "p⟶r" <br />
proof (rule impI)<br />
assume "p"<br />
have "p∨q" using `p` by (rule disjI1)<br />
show "r" using assms `p∨q` by (rule mp)<br />
qed<br />
have "q⟶r"<br />
proof (rule impI)<br />
assume "q"<br />
have "p∨q" using `q` by (rule disjI2)<br />
show "r" using assms `p∨q` by (rule mp)<br />
qed<br />
show "(p ⟶ r) ∧ (q ⟶ r)" using `p⟶r` `q⟶r` by (rule conjI)<br />
qed<br />
<br />
section {* Negaciones *}<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 39. Demostrar<br />
p ⊢ ¬¬p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_39: carmarria<br />
assumes "p"<br />
shows "¬¬p"<br />
proof -<br />
show "~~p" using assms(1) by (rule notnotI)<br />
qed<br />
<br />
lemma ej_39: joslopjim4<br />
assumes "p"<br />
shows "¬¬p"<br />
proof-<br />
show "¬¬p" using assms by (rule notnotI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 40. Demostrar<br />
¬p ⊢ p ⟶ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_40: carmarria<br />
assumes "¬p" <br />
shows "p ⟶ q"<br />
proof -<br />
{assume p<br />
have "~p" using assms(1) by this<br />
have False using `~p` `p` by (rule notE)<br />
have q using `False` by (rule FalseE)<br />
}<br />
thus "p ⟶ q" by (rule impI)<br />
qed<br />
<br />
lemma ej_40: joslopjim4<br />
assumes "¬p" <br />
shows "p ⟶ q"<br />
proof (rule impI)<br />
assume "p"<br />
show "q" using assms `p` by (rule notE)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 41. Demostrar<br />
p ⟶ q ⊢ ¬q ⟶ ¬p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_41: carmarria<br />
assumes 1: "p ⟶ q"<br />
shows "¬q ⟶ ¬p"<br />
proof -<br />
{assume 2: "~q"<br />
have 3: "~p" using 1 2 by (rule mt)<br />
}<br />
thus 4: "~q ⟶ ~p" by (rule impI)<br />
qed<br />
<br />
lemma ej_41: joslopjim4<br />
assumes "p ⟶ q"<br />
shows "¬q ⟶ ¬p"<br />
proof (rule impI)<br />
assume "¬q"<br />
show "¬p" using assms `¬q` by (rule mt)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 42. Demostrar<br />
p∨q, ¬q ⊢ p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_42: carmarria<br />
assumes 1: "p∨q" and<br />
2: "¬q" <br />
shows "p"<br />
proof -<br />
have " p | q" using 1 by this<br />
moreover {assume 3: p<br />
}<br />
moreover {assume 4: q<br />
have 5: False using 2 4 by (rule notE)<br />
have 6: p using 5 by (rule FalseE)<br />
}<br />
ultimately show p by (rule disjE)<br />
qed<br />
<br />
lemma ej_42: joslopjim4<br />
assumes "p∨q"<br />
"¬q" <br />
shows "p"<br />
using assms(1)<br />
proof <br />
assume "p"<br />
show "p" using `p` by this<br />
next<br />
assume "q"<br />
show "p" using assms(2) `q` by (rule notE)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 43. Demostrar<br />
p ∨ q, ¬p ⊢ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_43: carmarria<br />
assumes 1: "p ∨ q" and<br />
2: "¬p" <br />
shows "q"<br />
proof -<br />
have " p | q" using 1 by this<br />
moreover {assume 3: p<br />
have 4: False using 2 3 by (rule notE)<br />
have 5: q using 4 by (rule FalseE)<br />
}<br />
moreover {assume 6: q<br />
}<br />
ultimately show q by (rule disjE)<br />
qed<br />
<br />
lemma ej_43: joslopjim4<br />
assumes "p ∨ q"<br />
"¬p" <br />
shows "q"<br />
using assms(1)<br />
proof<br />
assume "p"<br />
show "q" using assms(2) `p` by (rule notE)<br />
next<br />
assume "q"<br />
show "q" using `q` by this<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 44. Demostrar<br />
p ∨ q ⊢ ¬(¬p ∧ ¬q)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_44: carmarria<br />
assumes 1: "p ∨ q" <br />
shows "¬(¬p ∧ ¬q)"<br />
proof -<br />
have "p | q" using 1 by this<br />
moreover {assume 2: p<br />
{assume 3:"~p & ~q"<br />
have 4: "~p" using 3 by (rule conjunct1)<br />
have 5: False using 4 2 by (rule notE)<br />
}<br />
hence 6: "~(~p & ~q)" by (rule notI)<br />
}<br />
moreover {assume 7: q<br />
{assume 8: "~p & ~q"<br />
have 9: "~q" using 8 by (rule conjunct2)<br />
have 10: False using 9 7 by (rule notE)<br />
}<br />
hence 11: "~(~p & ~q)" by (rule notI)<br />
}<br />
ultimately show "~(~p & ~q)" by (rule disjE)<br />
qed<br />
<br />
lemma ej_44: joslopjim4<br />
assumes "p∨q"<br />
shows "¬(¬p∧¬q)"<br />
proof (rule notI)<br />
assume "¬p∧¬q"<br />
show False<br />
using assms<br />
proof<br />
assume "p"<br />
have "¬p" using `¬p∧¬q` by (rule conjunct1)<br />
show False using `¬p` `p` by (rule notE)<br />
next<br />
assume "q"<br />
have "¬q" using `¬p∧¬q` by (rule conjunct2)<br />
show False using `¬q` `q` by (rule notE)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 45. Demostrar<br />
p ∧ q ⊢ ¬(¬p ∨ ¬q)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_45: carmarria<br />
assumes 1: "p ∧ q" <br />
shows "¬(¬p ∨ ¬q)"<br />
proof -<br />
{assume 2: "~p | ~q"<br />
moreover {assume 3: "~p"<br />
have 4: p using 1 by (rule conjunct1)<br />
have 5: False using 3 4 by (rule notE)<br />
}<br />
moreover {assume 6: "~q"<br />
have 7: q using 1 by (rule conjunct2)<br />
have 8: False using 6 7 by (rule notE)<br />
}<br />
ultimately have False by (rule disjE)<br />
}<br />
thus "~(~p | ~q)" by (rule notI)<br />
qed<br />
<br />
lemma ej_45: joslopjim4<br />
assumes "p∧q"<br />
shows "¬(¬p∨¬q)"<br />
proof (rule notI)<br />
assume "¬p∨¬q"<br />
then show False<br />
proof (rule disjE)<br />
assume "¬p"<br />
have "p" using assms by (rule conjunct1)<br />
show False using `¬p` `p` by (rule notE)<br />
next<br />
assume "¬q"<br />
have "q" using assms by (rule conjunct2)<br />
show False using `¬q` `q` by (rule notE)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 46. Demostrar<br />
¬(p ∨ q) ⊢ ¬p ∧ ¬q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_46: carmarria<br />
assumes 1: "¬(p ∨ q)" <br />
shows "¬p ∧ ¬q"<br />
proof -<br />
{assume 2: p<br />
have 3: "p | q" using 2 by (rule disjI1)<br />
have 4: False using 1 3 by (rule notE)<br />
}<br />
hence 5: "~p" by (rule notI)<br />
{assume 6: q<br />
have 7: "p | q" using 6 by (rule disjI2)<br />
have 8: False using 1 7 by (rule notE)<br />
}<br />
hence 9: "~q" by (rule notI)<br />
show "~p & ~q" using 5 9 by (rule conjI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 47. Demostrar<br />
¬p ∧ ¬q ⊢ ¬(p ∨ q)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_47: carmarria<br />
assumes 1: "¬p ∧ ¬q" <br />
shows "¬(p ∨ q)"<br />
proof -<br />
{assume 2: "p | q"<br />
moreover {assume 3: p<br />
have 4: "~p" using 1 by (rule conjunct1)<br />
have 5: False using 4 3 by (rule notE)<br />
}<br />
moreover {assume 6: q <br />
have 7: "~q" using 1 by (rule conjunct2)<br />
have 8: False using 7 6 by (rule notE)<br />
}<br />
ultimately have 9: False by (rule disjE)<br />
}<br />
thus "~(p | q)" by (rule notI)<br />
qed <br />
<br />
lemma ej_47: joslopjim4<br />
assumes "¬p∧¬q"<br />
shows "¬(p∨q)"<br />
proof (rule notI)<br />
assume "p∨q"<br />
then show False<br />
proof<br />
assume "p"<br />
have "¬p" using assms by (rule conjunct1)<br />
show False using `¬p` `p` by (rule notE)<br />
next<br />
assume "q"<br />
have "¬q" using assms by (rule conjunct2)<br />
show False using `¬q` `q` by (rule notE)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 48. Demostrar<br />
¬p ∨ ¬q ⊢ ¬(p ∧ q)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_48: carmarria<br />
assumes 1: "¬p ∨ ¬q"<br />
shows "¬(p ∧ q)"<br />
proof -<br />
have "~p | ~q" using 1 by this<br />
moreover {assume 2: "~p"<br />
{assume 3: "p & q"<br />
have 4: p using 3 by (rule conjunct1)<br />
have 5: False using 2 4 by (rule notE)<br />
}<br />
hence 6: "~(p & q)" by (rule notI)<br />
}<br />
moreover {assume 7: "~q"<br />
{assume 8: "p & q"<br />
have 9: q using 8 by (rule conjunct2)<br />
have 10: False using 7 9 by (rule notE)<br />
}<br />
hence 11: "~(p & q)" by (rule notI)<br />
}<br />
ultimately show "~(p & q)" by (rule disjE)<br />
qed<br />
<br />
lemma ej_48: joslopjim4<br />
assumes "¬p ∨ ¬q"<br />
shows "¬(p ∧ q)"<br />
proof (rule notI)<br />
assume "p∧q"<br />
have "p" using `p∧q` by (rule conjunct1)<br />
have "q" using `p∧q` by (rule conjunct2)<br />
show False<br />
using assms<br />
proof<br />
assume "¬p"<br />
have "p" using `p∧q` by (rule conjunct1)<br />
show False using `¬p` `p` by (rule notE)<br />
next<br />
assume "¬q"<br />
have "q" using `p∧q` by (rule conjunct2)<br />
show False using `¬q` `q` by (rule notE)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 49. Demostrar<br />
⊢ ¬(p ∧ ¬p)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_49: carmarria<br />
"¬(p ∧ ¬p)"<br />
proof -<br />
{assume 1: "p & ~p"<br />
have 2: p using 1 by (rule conjunct1)<br />
have 3: "~p" using 1 by (rule conjunct2)<br />
have 4: False using 3 2 by (rule notE)<br />
}<br />
thus "~(p & ~p)" by (rule notI)<br />
qed<br />
<br />
lemma ej_49: joslopjim4<br />
shows "¬(p∧¬p)"<br />
proof<br />
assume "p∧¬p"<br />
have "p" using `p∧¬p` by (rule conjunct1)<br />
have "¬p" using `p∧¬p` by (rule conjunct2)<br />
show False using `¬p` `p` by (rule notE)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 50. Demostrar<br />
p ∧ ¬p ⊢ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_50: carmarria<br />
assumes 1: "p ∧ ¬p" <br />
shows "q"<br />
proof -<br />
have 2: p using 1 by (rule conjunct1)<br />
have 3: "~p" using 1 by (rule conjunct2)<br />
have 4: False using 3 2 by (rule notE)<br />
show q using 4 by (rule FalseE)<br />
qed<br />
<br />
lemma ej_50: joslopjim4<br />
assumes "p∧¬p"<br />
shows "q"<br />
proof-<br />
have "p" using `p∧¬p` by (rule conjunct1)<br />
have "¬p" using `p∧¬p` by (rule conjunct2)<br />
have False using `¬p` `p` by (rule notE)<br />
then show "q" by (rule FalseE)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 51. Demostrar<br />
¬¬p ⊢ p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_51: carmarria joslopjim4<br />
assumes "¬¬p"<br />
shows "p"<br />
proof -<br />
show p using assms by (rule notnotD)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 52. Demostrar<br />
⊢ p ∨ ¬p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_52: carmarria<br />
"p ∨ ¬p"<br />
proof -<br />
{assume 1: "~(p | ~p)"<br />
{assume 2: p <br />
have 3: "p | ~p" using 2 by (rule disjI1)<br />
have 4: False using 1 3 by (rule notE)<br />
}<br />
hence 5: "~p" by (rule notI)<br />
have 6: "p | ~p" using 5 by (rule disjI2)<br />
have 7: False using 1 6 by (rule notE)<br />
}<br />
hence 8: "~~(p | ~p)" by (rule notI)<br />
show "p | ~p" using 8 by (rule notnotD)<br />
qed<br />
<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 53. Demostrar<br />
⊢ ((p ⟶ q) ⟶ p) ⟶ p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ej_53: joslopjim4<br />
shows "((p ⟶ q) ⟶ p) ⟶ p"<br />
proof (rule impI)<br />
assume "(p ⟶ q) ⟶ p"<br />
show "p"<br />
proof (rule ccontr)<br />
assume "¬p"<br />
have "¬(p⟶q)" using `(p ⟶ q) ⟶ p` `¬p` by (rule mt)<br />
have "p⟶q"<br />
proof (rule impI)<br />
assume "p"<br />
show "q" using `¬p` `p` by (rule notE)<br />
qed<br />
show False using `¬(p⟶q)` `p⟶q` by (rule notE)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 54. Demostrar<br />
¬q ⟶ ¬p ⊢ p ⟶ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ej_54: joslopjim4<br />
assumes "¬q ⟶ ¬p"<br />
shows "p ⟶ q"<br />
proof (rule impI)<br />
assume "p"<br />
have "¬¬p" using `p` by (rule notnotI)<br />
have "¬¬q" using assms `¬¬p` by (rule mt)<br />
show "q" using `¬¬q` by (rule notnotD)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 55. Demostrar<br />
¬(¬p ∧ ¬q) ⊢ p ∨ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_55:<br />
assumes "¬(¬p ∧ ¬q)"<br />
shows "p ∨ q"<br />
oops<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 56. Demostrar<br />
¬(¬p ∨ ¬q) ⊢ p ∧ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_56:<br />
assumes "¬(¬p ∨ ¬q)" <br />
shows "p ∧ q"<br />
oops<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 57. Demostrar<br />
¬(p ∧ q) ⊢ ¬p ∨ ¬q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_57:<br />
assumes "¬(p ∧ q)"<br />
shows "¬p ∨ ¬q"<br />
oops<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 58. Demostrar<br />
⊢ (p ⟶ q) ∨ (q ⟶ p)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_58:<br />
"(p ⟶ q) ∨ (q ⟶ p)"<br />
oops<br />
<br />
end<br />
<br />
</source></div>
Joslopjim4
https://www.glc.us.es/~jalonso/LMF2018/index.php?title=Relaci%C3%B3n_4&diff=135
Relación 4
2018-03-21T20:41:01Z
<p>Joslopjim4: </p>
<hr />
<div><source lang = "isar"><br />
<br />
chapter {* R4: Deducción natural proposicional *}<br />
<br />
theory R4<br />
imports Main <br />
begin<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
El objetivo de esta relación es demostrar cada uno de los ejercicios<br />
usando sólo las reglas básicas de deducción natural de la lógica<br />
proposicional (sin usar el método auto).<br />
<br />
Las reglas básicas de la deducción natural son las siguientes:<br />
· conjI: ⟦P; Q⟧ ⟹ P ∧ Q<br />
· conjunct1: P ∧ Q ⟹ P<br />
· conjunct2: P ∧ Q ⟹ Q <br />
· notnotD: ¬¬ P ⟹ P<br />
· notnotI: P ⟹ ¬¬ P<br />
· mp: ⟦P ⟶ Q; P⟧ ⟹ Q <br />
· mt: ⟦F ⟶ G; ¬G⟧ ⟹ ¬F <br />
· impI: (P ⟹ Q) ⟹ P ⟶ Q<br />
· disjI1: P ⟹ P ∨ Q<br />
· disjI2: Q ⟹ P ∨ Q<br />
· disjE: ⟦P ∨ Q; P ⟹ R; Q ⟹ R⟧ ⟹ R <br />
· FalseE: False ⟹ P<br />
· notE: ⟦¬P; P⟧ ⟹ R<br />
· notI: (P ⟹ False) ⟹ ¬P<br />
· iffI: ⟦P ⟹ Q; Q ⟹ P⟧ ⟹ P = Q<br />
· iffD1: ⟦Q = P; Q⟧ ⟹ P <br />
· iffD2: ⟦P = Q; Q⟧ ⟹ P<br />
· ccontr: (¬P ⟹ False) ⟹ P<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
text {*<br />
Se usarán las reglas notnotI y mt que demostramos a continuación. *}<br />
<br />
lemma notnotI: "P ⟹ ¬¬ P"<br />
by auto<br />
<br />
lemma mt: "⟦F ⟶ G; ¬G⟧ ⟹ ¬F"<br />
by auto<br />
<br />
section {* Implicaciones *}<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 1. Demostrar<br />
p ⟶ q, p ⊢ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_1: josrodjim2 carmarria inmbenber marmedmar3<br />
assumes 1: "p ⟶ q" and<br />
2: "p"<br />
shows "q"<br />
<br />
proof -<br />
show 1: "q" using 1 2 by (rule mp)<br />
qed<br />
<br />
lemma ej_1: joslopjim4<br />
assumes "p --> q"<br />
"p"<br />
shows "q"<br />
proof -<br />
show "q" using assms(1) assms(2) by (rule mp)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 2. Demostrar<br />
p ⟶ q, q ⟶ r, p ⊢ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_2: josrodjim2 carmarria inmbenber marmedmar3<br />
assumes 1: "p ⟶ q" and<br />
2: "q ⟶ r" and<br />
3: "p" <br />
shows "r"<br />
<br />
proof-<br />
have 4: "q" using 1 3 by (rule mp)<br />
show "r" using 2 4 by (rule mp)<br />
qed<br />
<br />
lemma ej_2: joslopjim4<br />
assumes "p ⟶ q"<br />
"q ⟶ r"<br />
"p" <br />
shows "r"<br />
proof -<br />
have "q" using `p⟶q` `p` by (rule mp)<br />
show "r" using `q⟶r` `q` by (rule mp)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 3. Demostrar<br />
p ⟶ (q ⟶ r), p ⟶ q, p ⊢ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_3: josrodjim2 carmarria inmbenber marmedmar3<br />
assumes 1: "p ⟶ (q ⟶ r)" and<br />
2: "p ⟶ q" and<br />
3: "p"<br />
shows "r"<br />
<br />
proof-<br />
have 4: "q" using 2 3 by (rule mp)<br />
have 5: "(q ⟶ r)" using 1 3 by (rule mp)<br />
show "r" using 5 4 by (rule mp)<br />
<br />
qed<br />
<br />
lemma ej_3: joslopjim4<br />
assumes "p ⟶ (q ⟶ r)"<br />
"p ⟶ q"<br />
"p"<br />
shows "r"<br />
proof -<br />
have "q" using `p⟶q` `p` by (rule mp)<br />
have "q⟶r" using `p ⟶ (q ⟶ r)` `p` by (rule mp)<br />
show "r" using `q⟶r` `q` by (rule mp)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 4. Demostrar<br />
p ⟶ q, q ⟶ r ⊢ p ⟶ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_4: josrodjim2<br />
assumes 1: "p ⟶ q" and<br />
2: "q ⟶ r" <br />
shows "p ⟶ r"<br />
<br />
proof-<br />
{ assume 3: "p"<br />
have 4: "q" using 1 3 by (rule mp)<br />
have 5: "r" using 2 4 by (rule mp)}<br />
hence 6: "p ⟶ r" using 3 5 by (rule impI)<br />
show "p⟶r" using 6 by this<br />
<br />
oops<br />
<br />
lemma ejercicio_4: carmarria<br />
assumes 1: "p ⟶ q" and<br />
2: "q ⟶ r" <br />
shows "p ⟶ r"<br />
proof -<br />
{ assume 3: p <br />
have 4: "q" using 1 3 by (rule mp)<br />
have 5: "r" using 2 4 by (rule mp)<br />
}<br />
thus "p ⟶ r" by (rule impI) <br />
qed<br />
lemma ejercicio_4: inmbenber<br />
assumes 1: "p ⟶ q" and<br />
2: "q ⟶ r" <br />
shows "p ⟶ r"<br />
proof -<br />
{ assume 3: "p" <br />
have 4: "q" using 1 3 by (rule mp)<br />
have 5: "r" using 2 4 by (rule mp) }<br />
thus "p ⟶ r" by (rule impI) <br />
qed<br />
<br />
lemma ejercicio_4:marmedmar3<br />
assumes "p ⟶ q"<br />
"q ⟶ r" <br />
shows "p ⟶ r"<br />
proof <br />
assume "p" <br />
with assms(1) have "q" by (rule mp)<br />
with assms(2) show "r" by (rule mp)<br />
qed <br />
<br />
lemma ej_4: joslopjim4<br />
assumes "p ⟶ q"<br />
"q ⟶ r" <br />
shows "p ⟶ r"<br />
proof (rule impI)<br />
assume "p"<br />
have "q" using assms(1) `p` by (rule mp)<br />
show "r" using assms(2) `q` by (rule mp)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 5. Demostrar<br />
p ⟶ (q ⟶ r) ⊢ q ⟶ (p ⟶ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_5: josrodjim2 (NO DA ERROR, PERO SI ALERTA)<br />
assumes "p ⟶ (q ⟶ r)" <br />
shows "q ⟶ (p ⟶ r)"<br />
<br />
proof -<br />
<br />
{ assume 2: "q"<br />
{assume 3: "p"<br />
have 4: "q ⟶r" using 1 3 by (rule mp)<br />
have 5: "r" using 4 2 by (rule mp)}<br />
hence 6: "p⟶r" using 3 5 by (rule impI)}<br />
hence 7: "q ⟶ (p ⟶ r)" using 2 6 by (rule impI)<br />
show "q ⟶ (p ⟶ r)" using 7 by this<br />
<br />
oops<br />
<br />
lemma ejercicio_5: carmarria <br />
assumes 1: "p ⟶ (q ⟶ r)" <br />
shows "q ⟶ (p ⟶ r)"<br />
<br />
proof - <br />
{assume 2: q<br />
{assume 3: p <br />
have 4: "q ⟶ r" using 1 3 by (rule mp)<br />
have 5: "r" using 4 2 by (rule mp)<br />
}<br />
hence " p ⟶ r" by (rule impI)<br />
}<br />
thus "q ⟶ (p ⟶ r)" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_5: inmbenber<br />
assumes 1: "p ⟶ (q ⟶ r)" <br />
shows "q ⟶ (p ⟶ r)"<br />
<br />
proof - <br />
{assume 2: "q"<br />
{assume 3: "p" <br />
have 4: "q ⟶ r" using 1 3 by (rule mp)<br />
have 5: "r" using 4 2 by (rule mp) }<br />
hence " p ⟶ r" by (rule impI) }<br />
thus "q ⟶ (p ⟶ r)" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_5: marmedmar3<br />
assumes "p ⟶ (q ⟶ r)" <br />
shows "q ⟶ (p ⟶ r)"<br />
proof <br />
assume "q"<br />
show "(p ⟶ r)"<br />
proof <br />
assume "p" <br />
with assms(1) have "(q ⟶ r)" by (rule mp)<br />
thus "r" using `q` by (rule mp) <br />
qed <br />
qed<br />
<br />
lemma ej_5: joslopjim4<br />
assumes "p ⟶ (q ⟶ r)" <br />
shows "q ⟶ (p ⟶ r)"<br />
proof (rule impI)<br />
assume "q"<br />
show "p⟶r"<br />
proof (rule impI)<br />
assume "p"<br />
have "q⟶r" using `p ⟶ (q ⟶ r)` `p` by (rule mp)<br />
show "r" using `q⟶r` `q` by (rule mp)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 6. Demostrar<br />
p ⟶ (q ⟶ r) ⊢ (p ⟶ q) ⟶ (p ⟶ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_6:josrodjim2<br />
assumes 1: "p ⟶ (q ⟶ r)" <br />
shows "(p ⟶ q) ⟶ (p ⟶ r)"<br />
<br />
proof- <br />
<br />
{assume 2: "p⟶q" <br />
{assume 3: "p"<br />
have 4: "q⟶r" using 1 3 by (rule mp)<br />
have 5: "q" using 2 3 by (rule mp)<br />
have 6: "r" using 4 5 by (rule mp)}<br />
hence 7: "p⟶r" using 3 6 by (rule impI)}<br />
hence 8: "(p ⟶ q) ⟶ (p ⟶ r)" using 2 7 by (rule impI)<br />
show "(p ⟶ q) ⟶ (p ⟶ r)" using 8 by this<br />
oops<br />
<br />
lemma ejercicio_6: carmarria, inmbenber<br />
assumes 1: "p ⟶ (q ⟶ r)" <br />
shows "(p ⟶ q) ⟶ (p ⟶ r)"<br />
proof -<br />
{assume 2: "p ⟶ q"<br />
{assume 3: "p"<br />
have 4: "q" using 2 3 by (rule mp)<br />
have 5: "q ⟶ r" using 1 3 by (rule mp)<br />
have 6: "r" using 5 4 by (rule mp)<br />
}<br />
hence 7: "p ⟶ r" by (rule impI)<br />
}<br />
thus "(p ⟶ q) ⟶ (p ⟶ r)" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_6:marmedmar3<br />
assumes 1: "p ⟶ (q ⟶ r)" <br />
shows "(p ⟶ q) ⟶ (p ⟶ r)"<br />
proof <br />
assume 2: "(p ⟶ q)" <br />
show "(p ⟶ r)" <br />
proof <br />
assume 3: "p" <br />
have 4: "q" using 2 3 by (rule mp)<br />
have 5: "(q ⟶ r)" using 1 3 by (rule mp) <br />
show "r" using 5 4 by (rule mp) <br />
qed<br />
qed<br />
<br />
lemma ej_6: joslopjim4<br />
assumes "p ⟶ (q ⟶ r)" <br />
shows "(p ⟶ q) ⟶ (p ⟶ r)"<br />
proof (rule impI)<br />
assume "(p⟶q)"<br />
show "p⟶r"<br />
proof (rule impI)<br />
assume "p"<br />
have "q⟶r" using `p ⟶ (q ⟶ r)` `p` by (rule mp)<br />
have "q" using `p⟶q` `p` by (rule mp)<br />
show "r" using `q⟶r` `q` by (rule mp)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 7. Demostrar<br />
p ⊢ q ⟶ p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_7: (NO SE QUE ESTA MAL)<br />
assumes "p" <br />
shows "q ⟶ p"<br />
<br />
proof -<br />
<br />
{ assume 2: "q"<br />
have 3: "p" using 1<br />
}<br />
hence 4: "q⟶p" using 2 3 by (rule impI)<br />
<br />
show "q⟶p" using 4 by this<br />
<br />
oops<br />
<br />
lemma ejercicio_7: carmarria, inmbenber<br />
assumes 1: "p" <br />
shows "q ⟶ p"<br />
proof -<br />
{assume 2: "q"<br />
have 3: "p" using 1 by this}<br />
thus "q ⟶ p" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_7:marmedmar3<br />
assumes "p" <br />
shows "q ⟶ p"<br />
proof <br />
assume "q" <br />
show "p" using assms(1) by this<br />
qed <br />
<br />
lemma ej_7: joslopjim4<br />
assumes "p" <br />
shows "q ⟶ p"<br />
proof (rule impI)<br />
assume "q"<br />
show "p" using assms by this<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 8. Demostrar<br />
⊢ p ⟶ (q ⟶ p)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_8: carmarria<br />
"p ⟶ (q ⟶ p)"<br />
<br />
proof -<br />
{assume 1: p<br />
{assume 2: q<br />
have 3: p using 1 by this<br />
}<br />
hence 4: "q ⟶ p" by (rule impI)<br />
}<br />
thus "p ⟶ (q ⟶ p)" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_8: marmedmar3<br />
"p ⟶ (q ⟶ p)"<br />
proof<br />
assume 1: "p"<br />
show "(q ⟶ p)"<br />
proof <br />
assume 2: "q" <br />
show "p" using 1 by this <br />
qed<br />
qed<br />
<br />
lemma ej_8: joslopjim4<br />
shows "p⟶(q⟶p)"<br />
proof (rule impI)<br />
assume "p"<br />
show "q⟶p"<br />
proof (rule impI)<br />
assume "q"<br />
show "p" using `p` by this<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 9. Demostrar<br />
p ⟶ q ⊢ (q ⟶ r) ⟶ (p ⟶ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_9: carmarria<br />
assumes 1: "p ⟶ q" <br />
shows "(q ⟶ r) ⟶ (p ⟶ r)"<br />
proof -<br />
{assume 2: "q ⟶ r"<br />
{assume 3: "p"<br />
have 4: "q" using 1 3 by (rule mp)<br />
have 5: "r" using 2 4 by (rule mp)<br />
}<br />
hence 6: "p ⟶ r" by (rule impI)<br />
}<br />
thus 7: "(q ⟶ r) ⟶ (p ⟶ r)" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_9: marmedmar3<br />
assumes "p ⟶ q" <br />
shows "(q ⟶ r) ⟶ (p ⟶ r)"<br />
proof <br />
assume "(q ⟶ r)" <br />
show "(p ⟶ r)"<br />
proof <br />
assume "p" <br />
with `(p ⟶ q)` have "q" by (rule mp)<br />
with `(q ⟶ r)` show "r" by (rule mp)<br />
qed<br />
qed<br />
<br />
lemma ej_9: joslopjim4<br />
assumes "p ⟶ q" <br />
shows "(q ⟶ r) ⟶ (p ⟶ r)"<br />
proof (rule impI)<br />
assume "q⟶r"<br />
show "p⟶r"<br />
proof (rule impI)<br />
assume "p"<br />
have "q" using `p⟶q` `p` by (rule mp)<br />
show "r" using `q⟶r` `q` by (rule mp)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 10. Demostrar<br />
p ⟶ (q ⟶ (r ⟶ s)) ⊢ r ⟶ (q ⟶ (p ⟶ s))<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_10: carmarria<br />
assumes 1: "p ⟶ (q ⟶ (r ⟶ s))" <br />
shows "r ⟶ (q ⟶ (p ⟶ s))"<br />
<br />
proof-<br />
{assume 2: r<br />
{assume 3: q<br />
{assume 4: p<br />
have 5: "q ⟶ (r ⟶ s)" using 1 4 by (rule mp)<br />
have 6: "r ⟶ s" using 5 3 by (rule mp)<br />
have 7: s using 6 2 by (rule mp)<br />
}<br />
hence 8: "p ⟶ s" by (rule impI)<br />
}<br />
hence 9: "q ⟶ (p ⟶ s)" by (rule impI)<br />
}<br />
thus "r ⟶ (q ⟶ (p ⟶ s))" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_10: marmedmar3<br />
assumes 1: "p ⟶ (q ⟶ (r ⟶ s))" <br />
shows "r ⟶ (q ⟶ (p ⟶ s))"<br />
proof <br />
assume 2: "r"<br />
show "(q ⟶ (p ⟶ s))"<br />
proof <br />
assume 3: "q" <br />
show "(p ⟶ s)"<br />
proof <br />
assume 4: "p" <br />
have 5: "(q ⟶ (r ⟶ s))" using 1 4 by (rule mp) <br />
have 6: "(r ⟶ s)" using 5 3 by (rule mp)<br />
show "s" using 6 2 by (rule mp) <br />
qed<br />
qed<br />
qed<br />
<br />
lemma ej_10: joslopjim4<br />
assumes "p ⟶ (q ⟶ (r ⟶ s))" <br />
shows "r ⟶ (q ⟶ (p ⟶ s))"<br />
proof (rule impI)<br />
assume "r"<br />
show "q⟶(p⟶s)"<br />
proof (rule impI)<br />
assume "q"<br />
show "p⟶s"<br />
proof (rule impI)<br />
assume "p"<br />
have "q ⟶ (r ⟶ s)" using `p ⟶ (q ⟶ (r ⟶ s))` `p` by (rule mp)<br />
have "r⟶s" using `q ⟶ (r ⟶ s)` `q` by (rule mp)<br />
show "s" using `r-->s` `r` by (rule mp)<br />
qed<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 11. Demostrar<br />
⊢ (p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_11: carmarria<br />
"(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))"<br />
proof - <br />
{assume 1: "p ⟶ (q ⟶ r)"<br />
{assume 2: "p ⟶ q"<br />
{assume 3: p<br />
have 4: q using 2 3 by (rule mp)<br />
have 5: "q ⟶ r" using 1 3 by (rule mp)<br />
have 6: r using 5 4 by (rule mp)<br />
}<br />
hence 7: "p ⟶ r" by (rule impI)<br />
}<br />
hence 8: "(p ⟶ q) ⟶ (p ⟶ r)" by (rule impI)<br />
}<br />
thus "(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_11: marmedmar3<br />
"(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))"<br />
proof <br />
assume 1: "(p ⟶ (q ⟶ r))" <br />
show "((p ⟶ q) ⟶ (p ⟶ r))"<br />
proof <br />
assume 2: "(p ⟶ q)" <br />
show "(p ⟶ r)" <br />
proof <br />
assume 3: "p" <br />
have 4: "(q ⟶ r)" using 1 3 by (rule mp) <br />
have 5: "q" using 2 3 by (rule mp) <br />
show "r" using 4 5 by (rule mp) <br />
qed<br />
qed<br />
qed<br />
<br />
lemma ej_11: joslopjim4<br />
shows "(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))"<br />
proof (rule impI)<br />
assume "p⟶(q⟶r)"<br />
show "(p⟶q)⟶(p⟶r)"<br />
proof (rule impI)<br />
assume "p⟶q"<br />
show "p⟶r"<br />
proof (rule impI)<br />
assume "p"<br />
have "q" using `p⟶q` `p` by (rule mp)<br />
have "q⟶r" using `p⟶(q⟶r)` `p` by (rule mp)<br />
show "r" using `q⟶r` `q` by (rule mp)<br />
qed<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 12. Demostrar<br />
(p ⟶ q) ⟶ r ⊢ p ⟶ (q ⟶ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_12: carmarria<br />
assumes 1: "(p ⟶ q) ⟶ r" <br />
shows "p ⟶ (q ⟶ r)"<br />
proof -<br />
{assume 2: p<br />
{assume 3: q<br />
{assume 4: p<br />
have 5: q using 3 by this<br />
}<br />
hence 6: "p ⟶ q" by (rule impI)<br />
have 7: r using 1 6 by (rule mp)<br />
}<br />
hence 8: "q ⟶ r" by (rule impI)<br />
}<br />
thus "p ⟶ (q ⟶ r)" by (rule impI)<br />
qed<br />
<br />
lemma ej_12: joslopjim4<br />
assumes "(p ⟶ q) ⟶ r" <br />
shows "p ⟶ (q ⟶ r)"<br />
proof (rule impI)<br />
assume "p"<br />
show "q⟶r"<br />
proof (rule impI)<br />
assume "q"<br />
have "p⟶q"<br />
proof (rule impI)<br />
assume "p"<br />
show "q" using `q` by this<br />
qed<br />
show "r" using `(p ⟶ q) ⟶ r` `p⟶q` by (rule mp)<br />
qed<br />
qed<br />
<br />
section {* Conjunciones *}<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 13. Demostrar<br />
p, q ⊢ p ∧ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_13: carmarria marmedmar3<br />
assumes "p"<br />
"q" <br />
shows "p ∧ q"<br />
<br />
proof-<br />
show "p \<and> q" using assms(1) assms(2) by (rule conjI)<br />
qed<br />
<br />
lemma ej_13: joslopjim4<br />
assumes "p"<br />
"q" <br />
shows "p ∧ q"<br />
proof-<br />
show "p∧q" using `p` `q` by (rule conjI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 14. Demostrar<br />
p ∧ q ⊢ p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_14: carmarria marmedmar3<br />
assumes "p ∧ q" <br />
shows "p"<br />
<br />
proof -<br />
show "p" using assms by (rule conjunct1)<br />
qed<br />
<br />
lemma ej_14: joslopjim4<br />
assumes "p ∧ q" <br />
shows "p"<br />
proof -<br />
show "p" using assms by (rule conjunct1)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 15. Demostrar<br />
p ∧ q ⊢ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_15: carmarria marmedmar3<br />
assumes "p ∧ q" <br />
shows "q"<br />
<br />
proof -<br />
show q using assms(1) by (rule conjunct2)<br />
qed<br />
<br />
lemma ej_15: joslopjim4<br />
assumes "p ∧ q" <br />
shows "q"<br />
proof-<br />
show "q" using assms by (rule conjunct2)<br />
qed<br />
<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 16. Demostrar<br />
p ∧ (q ∧ r) ⊢ (p ∧ q) ∧ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_16: carmarria marmedmar3<br />
assumes 1: "p ∧ (q ∧ r)"<br />
shows "(p ∧ q) ∧ r"<br />
proof -<br />
have 2: "q \<and> r" using 1 by (rule conjunct2)<br />
have 3: p using 1 by (rule conjunct1)<br />
have 4: q using 2 by (rule conjunct1)<br />
have 5: r using 2 by (rule conjunct2)<br />
have 6: "p \<and> q" using 3 4 by (rule conjI)<br />
show "(p \<and> q) \<and> r" using 6 5 by (rule conjI)<br />
qed<br />
<br />
lemma ej_16: joslopjim4<br />
assumes "p ∧ (q ∧ r)"<br />
shows "(p ∧ q) ∧ r"<br />
proof-<br />
have "p" using assms by (rule conjunct1)<br />
have "q∧r" using assms by (rule conjunct2)<br />
have "q" using `q∧r` by (rule conjunct1)<br />
have "r" using `q∧r` by (rule conjunct2)<br />
have "p∧q" using `p` `q` by (rule conjI)<br />
show "(p∧q)∧r" using `p∧q` `r` by (rule conjI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 17. Demostrar<br />
(p ∧ q) ∧ r ⊢ p ∧ (q ∧ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_17:<br />
assumes "(p ∧ q) ∧ r" <br />
shows "p ∧ (q ∧ r)"<br />
proof (rule conjI)<br />
have "p ∧ q" using assms ..<br />
thus "p" ..<br />
next<br />
show "q ∧ r"<br />
proof (rule conjI)<br />
have "p ∧ q" using assms ..<br />
thus "q" ..<br />
next<br />
show "r" using assms ..<br />
qed<br />
qed<br />
<br />
lemma ejercicio_17: carmarria marmedmar3<br />
assumes 1:"(p ∧ q) ∧ r" <br />
shows "p ∧ (q ∧ r)"<br />
proof-<br />
have 2:"p \<and> q" using 1 by (rule conjunct1)<br />
have 3: r using 1 by (rule conjunct2)<br />
have 4: p using 2 by (rule conjunct1)<br />
have 5: q using 2 by (rule conjunct2)<br />
have 6: "q \<and> r" using 5 3 by (rule conjI)<br />
show "p \<and> (q \<and> r)" using 4 6 by (rule conjI)<br />
qed<br />
<br />
<br />
lemma ej_17: joslopjim4<br />
assumes "(p ∧ q) ∧ r" <br />
shows "p ∧ (q ∧ r)"<br />
proof-<br />
have "p∧q" using assms by (rule conjunct1)<br />
have "r" using assms by (rule conjunct2)<br />
have "p" using `p∧q` by (rule conjunct1)<br />
have "q" using `p∧q` by (rule conjunct2)<br />
have "q∧r" using `q` `r` by (rule conjI)<br />
show "p∧(q∧r)" using `p` `q∧r` by (rule conjI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 18. Demostrar<br />
p ∧ q ⊢ p ⟶ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_18: carmarria<br />
assumes "p ∧ q" <br />
shows "p ⟶ q"<br />
proof -<br />
{assume p<br />
have q using assms(1) by (rule conjunct2)<br />
}<br />
thus "p ⟶ q" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_18: marmedmar3<br />
assumes "p ∧ q" <br />
shows "p ⟶ q"<br />
proof <br />
assume "p" <br />
show "q" using assms by (rule conjunct2) <br />
qed <br />
<br />
lemma ej_18: joslopjim4<br />
assumes "p ∧ q" <br />
shows "p ⟶ q"<br />
proof (rule impI)<br />
assume "p"<br />
show "q" using assms by (rule conjunct2)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 19. Demostrar<br />
(p ⟶ q) ∧ (p ⟶ r) ⊢ p ⟶ q ∧ r <br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_19: carmarria<br />
assumes 1: "(p ⟶ q) ∧ (p ⟶ r)" <br />
shows "p ⟶ q ∧ r"<br />
proof -<br />
{assume 2: p<br />
have 3: "p ⟶ q" using 1 by (rule conjunct1)<br />
have 4: "p ⟶ r" using 1 by (rule conjunct2)<br />
have 5: q using 3 2 by (rule mp)<br />
have 6: r using 4 2 by (rule mp)<br />
have 7: "q \<and> r" using 5 6 by (rule conjI)<br />
}<br />
thus "p ⟶ q \<and> r" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_19: marmedmar3<br />
assumes 1: "(p ⟶ q) ∧ (p ⟶ r)" <br />
shows "p ⟶ q ∧ r"<br />
proof <br />
have 2: "(p ⟶ q)" using assms by (rule conjunct1)<br />
have 3: "(p ⟶ r)" using assms by (rule conjunct2)<br />
assume 4: "p"<br />
with `(p ⟶ q)` have 5: "q" by (rule mp)<br />
have 6: "r" using 3 4 by (rule mp) <br />
show "q ∧ r" using 5 6 by (rule conjI)<br />
qed <br />
<br />
lemma ej_19: joslopjim4<br />
assumes "(p ⟶ q) ∧ (p ⟶ r)" <br />
shows "p ⟶ q ∧ r"<br />
proof (rule impI)<br />
assume "p"<br />
have "p⟶q" using assms by (rule conjunct1)<br />
have "q" using `p⟶q` `p` by (rule mp)<br />
have "p⟶r" using assms by (rule conjunct2)<br />
have "r" using `p⟶r` `p` by (rule mp)<br />
show "q∧r" using `q` `r` by (rule conjI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 20. Demostrar<br />
p ⟶ q ∧ r ⊢ (p ⟶ q) ∧ (p ⟶ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_20: carmarria<br />
assumes 1: "p ⟶ q ∧ r" <br />
shows "(p ⟶ q) ∧ (p ⟶ r)"<br />
proof -<br />
{assume 2: p<br />
have 3: "q \<and> r" using 1 2 by (rule mp)<br />
have 4: q using 3 by (rule conjunct1)<br />
}<br />
hence 5: "p ⟶ q" by (rule impI)<br />
{assume 6: p<br />
have 7: "q \<and> r" using 1 6 by (rule mp)<br />
have 8: r using 7 by (rule conjunct2)<br />
}<br />
hence 9: "p ⟶ r" by (rule impI)<br />
show "(p ⟶ q) \<and> (p ⟶ r)" using 5 9 by (rule conjI)<br />
qed<br />
<br />
lemma ej_20: joslopjim4<br />
assumes "p ⟶ q ∧ r" <br />
shows "(p ⟶ q) ∧ (p ⟶ r)"<br />
proof-<br />
have "p⟶q"<br />
proof<br />
assume "p"<br />
have "q∧r" using assms `p` by (rule mp)<br />
show "q" using `q∧r` by (rule conjunct1)<br />
qed<br />
have "p⟶r"<br />
proof<br />
assume "p"<br />
have "q∧r" using assms `p` by (rule mp)<br />
show "r" using `q∧r` by (rule conjunct2)<br />
qed<br />
show "(p⟶q)∧(p⟶r)" using `p⟶q` `p⟶r` by (rule conjI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 21. Demostrar<br />
p ⟶ (q ⟶ r) ⊢ p ∧ q ⟶ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_21: carmarria<br />
assumes 1: "p ⟶ (q ⟶ r)" <br />
shows "p ∧ q ⟶ r"<br />
proof -<br />
{assume 2: "p \<and> q"<br />
have 3: p using 2 by (rule conjunct1)<br />
have 4: q using 2 by (rule conjunct2)<br />
have 5: "q ⟶ r" using 1 3 by (rule mp)<br />
have 6: r using 5 4 by (rule mp)<br />
}<br />
thus "p ∧ q ⟶ r" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_21: marmedmar3<br />
assumes "p ⟶ (q ⟶ r)" <br />
shows "p ∧ q ⟶ r"<br />
proof <br />
assume "p ∧ q" <br />
hence "p" by (rule conjunct1) <br />
with assms have "(q ⟶ r)" by (rule mp)<br />
have "q" using `p ∧ q` by (rule conjunct2) <br />
with `(q ⟶ r)` show "r" by (rule mp) <br />
qed <br />
<br />
lemma ej_21: joslopjim4<br />
assumes "p ⟶ (q ⟶ r)" <br />
shows "p ∧ q ⟶ r"<br />
proof (rule impI)<br />
assume "p∧q"<br />
have "p" using `p∧q` by (rule conjunct1)<br />
have "q" using `p∧q` by (rule conjunct2)<br />
have "q⟶r" using assms `p` by (rule mp)<br />
show "r" using `q⟶r` `q` by (rule mp)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 22. Demostrar<br />
p ∧ q ⟶ r ⊢ p ⟶ (q ⟶ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_22: carmarria<br />
assumes "p ∧ q ⟶ r" <br />
shows "p ⟶ (q ⟶ r)"<br />
proof -<br />
{assume p<br />
{assume q<br />
have "p \<and> q" using `p` `q` by (rule conjI)<br />
have r using assms(1) `p \<and> q` by (rule mp)<br />
}<br />
hence "q ⟶ r" by (rule impI)<br />
}<br />
thus "p ⟶ (q ⟶ r)" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_22: marmedmar3<br />
assumes "p ∧ q ⟶ r" <br />
shows "p ⟶ (q ⟶ r)"<br />
proof <br />
assume "p" <br />
show "(q ⟶ r)"<br />
proof <br />
assume "q" <br />
with `p` have "p ∧ q" by (rule conjI) <br />
with assms show "r" by (rule mp) <br />
qed <br />
qed<br />
<br />
lemma ej_22: joslopjim4<br />
assumes "p ∧ q ⟶ r" <br />
shows "p ⟶ (q ⟶ r)"<br />
proof (rule impI)<br />
assume "p"<br />
show "q⟶r"<br />
proof (rule impI)<br />
assume "q"<br />
have "p∧q" using `p` `q` by (rule conjI)<br />
show "r" using assms `p∧q` by (rule mp)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 23. Demostrar<br />
(p ⟶ q) ⟶ r ⊢ p ∧ q ⟶ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_23: carmarria<br />
assumes "(p ⟶ q) ⟶ r" <br />
shows "p ∧ q ⟶ r"<br />
proof -<br />
{assume "p \<and> q" <br />
{assume p<br />
have q using `p \<and> q` by (rule conjunct2)<br />
}<br />
hence "p ⟶ q" by (rule impI)<br />
have r using assms(1) `p ⟶ q` by (rule mp)<br />
}<br />
thus "p \<and> q ⟶ r" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_23: marmedmar3<br />
assumes "(p ⟶ q) ⟶ r" <br />
shows "p ∧ q ⟶ r"<br />
proof <br />
assume "p ∧ q"<br />
have "p" using `p ∧ q` by (rule conjunct1)<br />
have "q" using `p ∧ q` by (rule conjunct2) <br />
hence "(p ⟶ q)" by (rule impI) <br />
with assms show "r" by (rule mp) <br />
qed <br />
<br />
lemma ej_23: joslopjim4<br />
assumes "(p ⟶ q) ⟶ r" <br />
shows "p ∧ q ⟶ r"<br />
proof (rule impI)<br />
assume "p∧q"<br />
have "p⟶q"<br />
proof<br />
assume "p"<br />
show "q" using `p∧q` by (rule conjunct2)<br />
qed<br />
show "r" using assms `p⟶q` by (rule mp)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 24. Demostrar<br />
p ∧ (q ⟶ r) ⊢ (p ⟶ q) ⟶ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_24: carmarria<br />
assumes 1: "p ∧ (q ⟶ r)" <br />
shows "(p ⟶ q) ⟶ r"<br />
proof -<br />
{assume 2: "p ⟶ q" <br />
have 3: p using 1 by (rule conjunct1)<br />
have 4: "q ⟶ r" using 1 by (rule conjunct2)<br />
have 5: q using 2 3 by (rule mp)<br />
have 6: r using 4 5 by (rule mp)<br />
}<br />
thus "(p ⟶ q) ⟶ r" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_24: marmedmar3<br />
assumes "p ∧ (q ⟶ r)" <br />
shows "(p ⟶ q) ⟶ r"<br />
proof <br />
assume "(p ⟶ q)" <br />
have "(q ⟶ r)" using assms by (rule conjunct2)<br />
have "p" using assms by (rule conjunct1)<br />
with `(p ⟶ q)` have "q" by (rule mp) <br />
with `(q ⟶ r)` show "r" by (rule mp)<br />
qed <br />
<br />
lemma ej_24: joslopjim4<br />
assumes "p ∧ (q ⟶ r)" <br />
shows "(p ⟶ q) ⟶ r"<br />
proof (rule impI)<br />
assume "p⟶q"<br />
have "p" using assms by (rule conjunct1)<br />
have "q⟶r" using assms by (rule conjunct2)<br />
have "q" using `p⟶q` `p` by (rule mp)<br />
show "r" using `q⟶r` `q` by (rule mp)<br />
qed<br />
<br />
section {* Disyunciones *}<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 25. Demostrar<br />
p ⊢ p ∨ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_25: carmarria<br />
assumes "p"<br />
shows "p ∨ q"<br />
proof -<br />
show "p | q" using assms(1) by (rule disjI1)<br />
qed<br />
<br />
lemma ej_25: joslopjim4<br />
assumes "p"<br />
shows "p ∨ q"<br />
proof (rule disjI1)<br />
show "p" using assms by this<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 26. Demostrar<br />
q ⊢ p ∨ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_26: carmarria<br />
assumes "q"<br />
shows "p ∨ q"<br />
proof -<br />
show "p | q" using assms(1) by (rule disjI2)<br />
qed<br />
<br />
lemma ej_26: joslopjim4<br />
assumes "q"<br />
shows "p ∨ q"<br />
proof (rule disjI2)<br />
show "q" using assms by this<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 27. Demostrar<br />
p ∨ q ⊢ q ∨ p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_27: carmarria<br />
assumes "p ∨ q"<br />
shows "q ∨ p"<br />
proof -<br />
have "p | q" using assms(1) by this<br />
moreover<br />
{assume p<br />
have "q | p" using `p` by (rule disjI2)<br />
}<br />
moreover<br />
{assume q<br />
have "q | p" using `q` by (rule disjI1)<br />
}<br />
ultimately show "q | p" by (rule disjE)<br />
qed<br />
<br />
lemma ej_27: joslopjim4<br />
assumes "p ∨ q"<br />
shows "q ∨ p"<br />
using assms<br />
proof <br />
assume "p"<br />
then show "q∨p" by(rule disjI2)<br />
next<br />
assume "q"<br />
show "q∨p" using `q` by(rule disjI1)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 28. Demostrar<br />
q ⟶ r ⊢ p ∨ q ⟶ p ∨ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_28: carmarria<br />
assumes "q ⟶ r" <br />
shows "p ∨ q ⟶ p ∨ r"<br />
proof -<br />
{assume "p | q" <br />
moreover<br />
{assume p<br />
have "p | r" using `p` by (rule disjI1)<br />
}<br />
moreover<br />
{assume q<br />
have r using assms(1) `q` by (rule mp)<br />
have "p | r" using `r` by (rule disjI2)<br />
}<br />
ultimately have "p | r" by (rule disjE)<br />
}<br />
thus "p | q ⟶ p | r" by (rule impI)<br />
qed<br />
<br />
lemma ej_28: joslopjim4<br />
assumes "q ⟶ r" <br />
shows "p ∨ q ⟶ p ∨ r"<br />
proof (rule impI)<br />
assume "p∨q" <br />
show "p∨r"<br />
using `p∨q`<br />
proof <br />
assume "p"<br />
show "p∨r" using `p` by (rule disjI1)<br />
next<br />
assume "q"<br />
have "r" using assms `q` by (rule mp)<br />
show "p∨r" using `r` by (rule disjI2)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 29. Demostrar<br />
p ∨ p ⊢ p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_29: carmarria<br />
assumes "p ∨ p"<br />
shows "p"<br />
proof -<br />
have "p | p" using assms(1) by this<br />
moreover <br />
{assume p<br />
}<br />
moreover<br />
{assume p<br />
}<br />
ultimately show p by (rule disjE)<br />
qed<br />
<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 30. Demostrar<br />
p ⊢ p ∨ p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_30: carmarria<br />
assumes "p" <br />
shows "p ∨ p"<br />
proof -<br />
show "p | p" using assms(1) by (rule disjI1)<br />
qed<br />
<br />
lemma ej_30: joslopjim4<br />
assumes "p" <br />
shows "p ∨ p"<br />
proof (rule disjI1)<br />
show "p" using assms by this<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 31. Demostrar<br />
p ∨ (q ∨ r) ⊢ (p ∨ q) ∨ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_31: carmarria<br />
assumes 1: "p ∨ (q ∨ r)" <br />
shows "(p ∨ q) ∨ r"<br />
proof -<br />
have "p | (q | r)" using 1 by this<br />
moreover {assume 2: p<br />
have 3: "p | q" using 2 by (rule disjI1)<br />
have 4: "(p | q) | r" using 3 by (rule disjI1)<br />
}<br />
moreover {assume 5: "q | r"<br />
moreover {assume 6: q<br />
have 7: "p | q" using 6 by (rule disjI2)<br />
have 8: "(p | q) | r " using 7 by (rule disjI1)<br />
}<br />
moreover {assume 9: r<br />
have 10: "(p | q) | r " using 9 by (rule disjI2)<br />
}<br />
ultimately have 11: "(p | q) | r " by (rule disjE)<br />
}<br />
ultimately show "(p | q) | r " by (rule disjE)<br />
qed<br />
<br />
lemma ej_31: joslopjim4<br />
assumes "p ∨ (q ∨ r)" <br />
shows "(p ∨ q) ∨ r"<br />
using assms<br />
proof <br />
assume "p"<br />
have "p∨q" using `p` by (rule disjI1)<br />
thus "(p ∨ q) ∨ r" by (rule disjI1)<br />
next<br />
assume "q∨r"<br />
show "(p ∨ q) ∨ r"<br />
using `q∨r`<br />
proof <br />
assume "q"<br />
have "p∨q" using `q` by (rule disjI2)<br />
show "(p ∨ q) ∨ r" using `p∨q` by (rule disjI1)<br />
next<br />
assume "r"<br />
show "(p ∨ q) ∨ r" using `r` by (rule disjI2)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 32. Demostrar<br />
(p ∨ q) ∨ r ⊢ p ∨ (q ∨ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_32: carmarria<br />
assumes 1: "(p ∨ q) ∨ r" <br />
shows "p ∨ (q ∨ r)"<br />
proof -<br />
have "(p | q) | r" using 1 by this<br />
moreover {assume 2: "p | q" <br />
moreover {assume 3: p <br />
have 4: "p | (q | r)" using 3 by (rule disjI1)<br />
}<br />
moreover {assume 5: q<br />
have 6: "q | r" using 5 by (rule disjI1)<br />
have 7: "p | (q | r)" using 6 by (rule disjI2)<br />
}<br />
ultimately have 8: " p | (q | r)" by (rule disjE)<br />
}<br />
moreover {assume 9: r<br />
have 10: "q | r" using 9 by (rule disjI2)<br />
have 11: "p | (q | r)" using 10 by (rule disjI2)<br />
}<br />
ultimately show "p | (q | r)" by (rule disjE)<br />
qed<br />
<br />
lemma ej_32: joslopjim4<br />
assumes "(p ∨ q) ∨ r" <br />
shows "p ∨ (q ∨ r)"<br />
using assms<br />
proof<br />
assume "p∨q"<br />
show "p ∨ (q ∨ r)"<br />
using `p∨q`<br />
proof <br />
assume "p"<br />
show "p ∨ (q ∨ r)" using `p` by (rule disjI1)<br />
next<br />
assume "q"<br />
have "q∨r" using `q` by (rule disjI1)<br />
show "p ∨ (q ∨ r)" using `q∨r` by (rule disjI2)<br />
qed<br />
next<br />
assume "r"<br />
have "q∨r" using `r` by (rule disjI2)<br />
show "p ∨ (q ∨ r)" using `q∨r` by (rule disjI2)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 33. Demostrar<br />
p ∧ (q ∨ r) ⊢ (p ∧ q) ∨ (p ∧ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_33: carmarria<br />
assumes 1: "p ∧ (q ∨ r)" <br />
shows "(p ∧ q) ∨ (p ∧ r)"<br />
proof -<br />
have 2: p using 1 by (rule conjunct1)<br />
have 3: "q | r" using 1 by (rule conjunct2)<br />
moreover {assume 4: q<br />
have 5: "p & q" using 2 4 by (rule conjI)<br />
have 6: "(p & q) | (p & r)" using 5 by (rule disjI1)<br />
}<br />
moreover {assume 7: r<br />
have 8: "p & r" using 2 7 by (rule conjI)<br />
have 9: "(p & q) | (p & r)" using 8 by (rule disjI2)<br />
}<br />
ultimately show "(p & q) | (p & r)" by (rule disjE)<br />
qed<br />
<br />
lemma ej_33: joslopjim4<br />
assumes "p ∧ (q ∨ r)" <br />
shows "(p ∧ q) ∨ (p ∧ r)"<br />
proof-<br />
have "p" using assms by (rule conjunct1)<br />
have "q∨r" using assms by (rule conjunct2)<br />
show "(p ∧ q) ∨ (p ∧ r)"<br />
using `q∨r` <br />
proof (rule disjE)<br />
assume "q"<br />
have "p∧q" using `p` `q` by (rule conjI)<br />
show "(p ∧ q) ∨ (p ∧ r)" using `p∧q` by (rule disjI1)<br />
next<br />
assume "r"<br />
have "p∧r" using `p` `r` by (rule conjI)<br />
show "(p ∧ q) ∨ (p ∧ r)" using `p∧r` by (rule disjI2)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 34. Demostrar<br />
(p ∧ q) ∨ (p ∧ r) ⊢ p ∧ (q ∨ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_34: carmarria<br />
assumes 1: "(p ∧ q) ∨ (p ∧ r)" <br />
shows "p ∧ (q ∨ r)"<br />
proof -<br />
have "(p & q) | (p & r)" using 1 by this<br />
moreover {assume 2: "p & q"<br />
have 3: p using 2 by (rule conjunct1)<br />
have 4: q using 2 by (rule conjunct2)<br />
have 5: "q | r" using 4 by (rule disjI1)<br />
have 6: "p & (q | r)" using 3 5 by (rule conjI)<br />
}<br />
moreover {assume 7: "p & r"<br />
have 8: p using 7 by (rule conjunct1)<br />
have 9: r using 7 by (rule conjunct2)<br />
have 10: "q | r" using 9 by (rule disjI2)<br />
have 11: "p & (q | r)" using 8 10 by (rule conjI)<br />
}<br />
ultimately show "p & (q | r)" by (rule disjE)<br />
qed<br />
<br />
lemma ej_34: joslopjim4<br />
assumes "(p ∧ q) ∨ (p ∧ r)" <br />
shows "p ∧ (q ∨ r)"<br />
using assms<br />
proof<br />
assume "p∧q"<br />
have "p" using `p∧q` by (rule conjunct1)<br />
have "q" using `p∧q` by (rule conjunct2)<br />
have "q∨r" using `q` by (rule disjI1)<br />
show "p∧(q∨r)" using `p` `q∨r` by (rule conjI)<br />
next<br />
assume "p∧r"<br />
have "p" using `p∧r` by (rule conjunct1)<br />
have "r" using `p∧r` by (rule conjunct2)<br />
have "q∨r" using `r` by (rule disjI2)<br />
show "p∧(q∨r)" using `p` `q∨r` by (rule conjI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 35. Demostrar<br />
p ∨ (q ∧ r) ⊢ (p ∨ q) ∧ (p ∨ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_35: carmarria<br />
assumes 1: "p ∨ (q ∧ r)" <br />
shows "(p ∨ q) ∧ (p ∨ r)"<br />
proof -<br />
have "p | (q & r)" using 1 by this<br />
moreover {assume 2: p<br />
have 3: " p | q" using 2 by (rule disjI1)<br />
have 4: " p | r" using 2 by (rule disjI1)<br />
have 5: "(p | q) & (p | r)" using 3 4 by (rule conjI)<br />
}<br />
moreover {assume 6: "q & r" <br />
have 7: q using 6 by (rule conjunct1)<br />
have 8: r using 6 by (rule conjunct2)<br />
have 9: "p | q" using 7 by (rule disjI2)<br />
have 10: "p | r" using 8 by (rule disjI2)<br />
have 11: "(p | q) & (p | r)" using 9 10 by (rule conjI)<br />
}<br />
ultimately show "(p | q) & (p | r)" by (rule disjE)<br />
qed<br />
<br />
lemma ej_35: joslopjim4<br />
assumes "p ∨ (q ∧ r)" <br />
shows "(p ∨ q) ∧ (p ∨ r)"<br />
using assms<br />
proof (rule disjE)<br />
assume "p"<br />
have "p∨q" using `p` by (rule disjI1)<br />
have "p∨r" using `p` by (rule disjI1)<br />
show "(p ∨ q) ∧ (p ∨ r)" using `p∨q` `p∨r` by (rule conjI)<br />
next<br />
assume "q∧r"<br />
have "q" using `q∧r` by (rule conjunct1)<br />
have "r" using `q∧r` by (rule conjunct2)<br />
have "p∨q" using `q` by (rule disjI2)<br />
have "p∨r" using `r` by (rule disjI2)<br />
show "(p ∨ q) ∧ (p ∨ r)" using `p∨q` `p∨r` by (rule conjI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 36. Demostrar<br />
(p ∨ q) ∧ (p ∨ r) ⊢ p ∨ (q ∧ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_36: carmarria<br />
assumes 1: "(p ∨ q) ∧ (p ∨ r)"<br />
shows "p ∨ (q ∧ r)"<br />
proof -<br />
have 2: "p | q" using 1 by (rule conjunct1)<br />
moreover {assume 3: p<br />
have 4: "p | (q & r)" using 3 by (rule disjI1)<br />
}<br />
moreover {assume 5: q<br />
have 6: "p | r" using 1 by (rule conjunct2)<br />
moreover {assume 7: p<br />
have 8: "p | (q & r)" using 7 by (rule disjI1)<br />
}<br />
moreover {assume 9: r<br />
have 10: "q & r" using 5 9 by (rule conjI)<br />
have 11: "p | (q & r)" using 10 by (rule disjI2)<br />
}<br />
ultimately have 12: "p | (q & r)" by (rule disjE)<br />
}<br />
ultimately show "p | (q & r)" by (rule disjE)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 37. Demostrar<br />
(p ⟶ r) ∧ (q ⟶ r) ⊢ p ∨ q ⟶ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_37: carmarria<br />
assumes 1: "(p ⟶ r) ∧ (q ⟶ r)" <br />
shows "p ∨ q ⟶ r"<br />
proof -<br />
have 2: "p ⟶ r" using 1 by (rule conjunct1)<br />
have 3: "q ⟶ r" using 1 by (rule conjunct2)<br />
{assume 4: "p | q"<br />
moreover {assume 5: p<br />
have 6: r using 2 5 by (rule mp)<br />
}<br />
moreover {assume 7: q<br />
have 8: r using 3 7 by (rule mp)<br />
}<br />
ultimately have r by (rule disjE)<br />
}<br />
thus "p | q ⟶ r" by (rule impI)<br />
qed<br />
<br />
lemma ej_37: joslopjim4<br />
assumes "(p ⟶ r) ∧ (q ⟶ r)" <br />
shows "p ∨ q ⟶ r"<br />
proof (rule impI)<br />
assume "p∨q"<br />
have "p⟶r" using assms by (rule conjunct1)<br />
have "q⟶r" using assms by (rule conjunct2)<br />
show "r"<br />
using `p∨q`<br />
proof <br />
assume "p"<br />
show "r" using `p⟶r` `p` by (rule mp)<br />
next<br />
assume "q"<br />
show "r" using `q⟶r` `q` by (rule mp)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 38. Demostrar<br />
p ∨ q ⟶ r ⊢ (p ⟶ r) ∧ (q ⟶ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_38: carmarria<br />
assumes 1: "p ∨ q ⟶ r" <br />
shows "(p ⟶ r) ∧ (q ⟶ r)"<br />
proof -<br />
{assume 2: p<br />
have 3: "p | q" using 2 by (rule disjI1)<br />
have 4: r using 1 3 by (rule mp)<br />
}<br />
hence 5: "p ⟶ r" by (rule impI)<br />
{assume 6: q<br />
have 7: "p | q" using 6 by (rule disjI2)<br />
have 8: r using 1 7 by (rule mp)<br />
}<br />
hence 9: "q ⟶ r" by (rule impI)<br />
show "(p ⟶ r) & (q ⟶ r)" using 5 9 by (rule conjI)<br />
qed<br />
<br />
lemma ej_38: joslopjim4<br />
assumes "p ∨ q ⟶ r" <br />
shows "(p ⟶ r) ∧ (q ⟶ r)"<br />
proof-<br />
have "p⟶r" <br />
proof (rule impI)<br />
assume "p"<br />
have "p∨q" using `p` by (rule disjI1)<br />
show "r" using assms `p∨q` by (rule mp)<br />
qed<br />
have "q⟶r"<br />
proof (rule impI)<br />
assume "q"<br />
have "p∨q" using `q` by (rule disjI2)<br />
show "r" using assms `p∨q` by (rule mp)<br />
qed<br />
show "(p ⟶ r) ∧ (q ⟶ r)" using `p⟶r` `q⟶r` by (rule conjI)<br />
qed<br />
<br />
section {* Negaciones *}<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 39. Demostrar<br />
p ⊢ ¬¬p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_39: carmarria<br />
assumes "p"<br />
shows "¬¬p"<br />
proof -<br />
show "~~p" using assms(1) by (rule notnotI)<br />
qed<br />
<br />
lemma ej_39: joslopjim4<br />
assumes "p"<br />
shows "¬¬p"<br />
proof-<br />
show "¬¬p" using assms by (rule notnotI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 40. Demostrar<br />
¬p ⊢ p ⟶ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_40: carmarria<br />
assumes "¬p" <br />
shows "p ⟶ q"<br />
proof -<br />
{assume p<br />
have "~p" using assms(1) by this<br />
have False using `~p` `p` by (rule notE)<br />
have q using `False` by (rule FalseE)<br />
}<br />
thus "p ⟶ q" by (rule impI)<br />
qed<br />
<br />
lemma ej_40: joslopjim4<br />
assumes "¬p" <br />
shows "p ⟶ q"<br />
proof (rule impI)<br />
assume "p"<br />
show "q" using assms `p` by (rule notE)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 41. Demostrar<br />
p ⟶ q ⊢ ¬q ⟶ ¬p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_41: carmarria<br />
assumes 1: "p ⟶ q"<br />
shows "¬q ⟶ ¬p"<br />
proof -<br />
{assume 2: "~q"<br />
have 3: "~p" using 1 2 by (rule mt)<br />
}<br />
thus 4: "~q ⟶ ~p" by (rule impI)<br />
qed<br />
<br />
lemma ej_41: joslopjim4<br />
assumes "p ⟶ q"<br />
shows "¬q ⟶ ¬p"<br />
proof (rule impI)<br />
assume "¬q"<br />
show "¬p" using assms `¬q` by (rule mt)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 42. Demostrar<br />
p∨q, ¬q ⊢ p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_42: carmarria<br />
assumes 1: "p∨q" and<br />
2: "¬q" <br />
shows "p"<br />
proof -<br />
have " p | q" using 1 by this<br />
moreover {assume 3: p<br />
}<br />
moreover {assume 4: q<br />
have 5: False using 2 4 by (rule notE)<br />
have 6: p using 5 by (rule FalseE)<br />
}<br />
ultimately show p by (rule disjE)<br />
qed<br />
<br />
lemma ej_42: joslopjim4<br />
assumes "p∨q"<br />
"¬q" <br />
shows "p"<br />
using assms(1)<br />
proof <br />
assume "p"<br />
show "p" using `p` by this<br />
next<br />
assume "q"<br />
show "p" using assms(2) `q` by (rule notE)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 43. Demostrar<br />
p ∨ q, ¬p ⊢ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_43: carmarria<br />
assumes 1: "p ∨ q" and<br />
2: "¬p" <br />
shows "q"<br />
proof -<br />
have " p | q" using 1 by this<br />
moreover {assume 3: p<br />
have 4: False using 2 3 by (rule notE)<br />
have 5: q using 4 by (rule FalseE)<br />
}<br />
moreover {assume 6: q<br />
}<br />
ultimately show q by (rule disjE)<br />
qed<br />
<br />
lemma ej_43: joslopjim4<br />
assumes "p ∨ q"<br />
"¬p" <br />
shows "q"<br />
using assms(1)<br />
proof<br />
assume "p"<br />
show "q" using assms(2) `p` by (rule notE)<br />
next<br />
assume "q"<br />
show "q" using `q` by this<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 44. Demostrar<br />
p ∨ q ⊢ ¬(¬p ∧ ¬q)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_44: carmarria<br />
assumes 1: "p ∨ q" <br />
shows "¬(¬p ∧ ¬q)"<br />
proof -<br />
have "p | q" using 1 by this<br />
moreover {assume 2: p<br />
{assume 3:"~p & ~q"<br />
have 4: "~p" using 3 by (rule conjunct1)<br />
have 5: False using 4 2 by (rule notE)<br />
}<br />
hence 6: "~(~p & ~q)" by (rule notI)<br />
}<br />
moreover {assume 7: q<br />
{assume 8: "~p & ~q"<br />
have 9: "~q" using 8 by (rule conjunct2)<br />
have 10: False using 9 7 by (rule notE)<br />
}<br />
hence 11: "~(~p & ~q)" by (rule notI)<br />
}<br />
ultimately show "~(~p & ~q)" by (rule disjE)<br />
qed<br />
<br />
lemma ej_44: joslopjim4<br />
assumes "p∨q"<br />
shows "¬(¬p∧¬q)"<br />
proof (rule notI)<br />
assume "¬p∧¬q"<br />
show False<br />
using assms<br />
proof<br />
assume "p"<br />
have "¬p" using `¬p∧¬q` by (rule conjunct1)<br />
show False using `¬p` `p` by (rule notE)<br />
next<br />
assume "q"<br />
have "¬q" using `¬p∧¬q` by (rule conjunct2)<br />
show False using `¬q` `q` by (rule notE)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 45. Demostrar<br />
p ∧ q ⊢ ¬(¬p ∨ ¬q)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_45: carmarria<br />
assumes 1: "p ∧ q" <br />
shows "¬(¬p ∨ ¬q)"<br />
proof -<br />
{assume 2: "~p | ~q"<br />
moreover {assume 3: "~p"<br />
have 4: p using 1 by (rule conjunct1)<br />
have 5: False using 3 4 by (rule notE)<br />
}<br />
moreover {assume 6: "~q"<br />
have 7: q using 1 by (rule conjunct2)<br />
have 8: False using 6 7 by (rule notE)<br />
}<br />
ultimately have False by (rule disjE)<br />
}<br />
thus "~(~p | ~q)" by (rule notI)<br />
qed<br />
<br />
lemma ej_45: joslopjim4<br />
assumes "p∧q"<br />
shows "¬(¬p∨¬q)"<br />
proof (rule notI)<br />
assume "¬p∨¬q"<br />
then show False<br />
proof (rule disjE)<br />
assume "¬p"<br />
have "p" using assms by (rule conjunct1)<br />
show False using `¬p` `p` by (rule notE)<br />
next<br />
assume "¬q"<br />
have "q" using assms by (rule conjunct2)<br />
show False using `¬q` `q` by (rule notE)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 46. Demostrar<br />
¬(p ∨ q) ⊢ ¬p ∧ ¬q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_46: carmarria<br />
assumes 1: "¬(p ∨ q)" <br />
shows "¬p ∧ ¬q"<br />
proof -<br />
{assume 2: p<br />
have 3: "p | q" using 2 by (rule disjI1)<br />
have 4: False using 1 3 by (rule notE)<br />
}<br />
hence 5: "~p" by (rule notI)<br />
{assume 6: q<br />
have 7: "p | q" using 6 by (rule disjI2)<br />
have 8: False using 1 7 by (rule notE)<br />
}<br />
hence 9: "~q" by (rule notI)<br />
show "~p & ~q" using 5 9 by (rule conjI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 47. Demostrar<br />
¬p ∧ ¬q ⊢ ¬(p ∨ q)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_47: carmarria<br />
assumes 1: "¬p ∧ ¬q" <br />
shows "¬(p ∨ q)"<br />
proof -<br />
{assume 2: "p | q"<br />
moreover {assume 3: p<br />
have 4: "~p" using 1 by (rule conjunct1)<br />
have 5: False using 4 3 by (rule notE)<br />
}<br />
moreover {assume 6: q <br />
have 7: "~q" using 1 by (rule conjunct2)<br />
have 8: False using 7 6 by (rule notE)<br />
}<br />
ultimately have 9: False by (rule disjE)<br />
}<br />
thus "~(p | q)" by (rule notI)<br />
qed <br />
<br />
lemma ej_47: joslopjim4<br />
assumes "¬p∧¬q"<br />
shows "¬(p∨q)"<br />
proof (rule notI)<br />
assume "p∨q"<br />
then show False<br />
proof<br />
assume "p"<br />
have "¬p" using assms by (rule conjunct1)<br />
show False using `¬p` `p` by (rule notE)<br />
next<br />
assume "q"<br />
have "¬q" using assms by (rule conjunct2)<br />
show False using `¬q` `q` by (rule notE)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 48. Demostrar<br />
¬p ∨ ¬q ⊢ ¬(p ∧ q)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_48: carmarria<br />
assumes 1: "¬p ∨ ¬q"<br />
shows "¬(p ∧ q)"<br />
proof -<br />
have "~p | ~q" using 1 by this<br />
moreover {assume 2: "~p"<br />
{assume 3: "p & q"<br />
have 4: p using 3 by (rule conjunct1)<br />
have 5: False using 2 4 by (rule notE)<br />
}<br />
hence 6: "~(p & q)" by (rule notI)<br />
}<br />
moreover {assume 7: "~q"<br />
{assume 8: "p & q"<br />
have 9: q using 8 by (rule conjunct2)<br />
have 10: False using 7 9 by (rule notE)<br />
}<br />
hence 11: "~(p & q)" by (rule notI)<br />
}<br />
ultimately show "~(p & q)" by (rule disjE)<br />
qed<br />
<br />
lemma ej_48: joslopjim4<br />
assumes "¬p ∨ ¬q"<br />
shows "¬(p ∧ q)"<br />
proof (rule notI)<br />
assume "p∧q"<br />
have "p" using `p∧q` by (rule conjunct1)<br />
have "q" using `p∧q` by (rule conjunct2)<br />
show False<br />
using assms<br />
proof<br />
assume "¬p"<br />
have "p" using `p∧q` by (rule conjunct1)<br />
show False using `¬p` `p` by (rule notE)<br />
next<br />
assume "¬q"<br />
have "q" using `p∧q` by (rule conjunct2)<br />
show False using `¬q` `q` by (rule notE)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 49. Demostrar<br />
⊢ ¬(p ∧ ¬p)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_49: carmarria<br />
"¬(p ∧ ¬p)"<br />
proof -<br />
{assume 1: "p & ~p"<br />
have 2: p using 1 by (rule conjunct1)<br />
have 3: "~p" using 1 by (rule conjunct2)<br />
have 4: False using 3 2 by (rule notE)<br />
}<br />
thus "~(p & ~p)" by (rule notI)<br />
qed<br />
<br />
lemma ej_49: joslopjim4<br />
shows "¬(p∧¬p)"<br />
proof<br />
assume "p∧¬p"<br />
have "p" using `p∧¬p` by (rule conjunct1)<br />
have "¬p" using `p∧¬p` by (rule conjunct2)<br />
show False using `¬p` `p` by (rule notE)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 50. Demostrar<br />
p ∧ ¬p ⊢ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_50: carmarria<br />
assumes 1: "p ∧ ¬p" <br />
shows "q"<br />
proof -<br />
have 2: p using 1 by (rule conjunct1)<br />
have 3: "~p" using 1 by (rule conjunct2)<br />
have 4: False using 3 2 by (rule notE)<br />
show q using 4 by (rule FalseE)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 51. Demostrar<br />
¬¬p ⊢ p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_51: carmarria<br />
assumes "¬¬p"<br />
shows "p"<br />
proof -<br />
show p using assms(1) by (rule notnotD)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 52. Demostrar<br />
⊢ p ∨ ¬p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_52: carmarria<br />
"p ∨ ¬p"<br />
proof -<br />
{assume 1: "~(p | ~p)"<br />
{assume 2: p <br />
have 3: "p | ~p" using 2 by (rule disjI1)<br />
have 4: False using 1 3 by (rule notE)<br />
}<br />
hence 5: "~p" by (rule notI)<br />
have 6: "p | ~p" using 5 by (rule disjI2)<br />
have 7: False using 1 6 by (rule notE)<br />
}<br />
hence 8: "~~(p | ~p)" by (rule notI)<br />
show "p | ~p" using 8 by (rule notnotD)<br />
qed<br />
<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 53. Demostrar<br />
⊢ ((p ⟶ q) ⟶ p) ⟶ p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_53:<br />
"((p ⟶ q) ⟶ p) ⟶ p"<br />
oops<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 54. Demostrar<br />
¬q ⟶ ¬p ⊢ p ⟶ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_54:<br />
assumes "¬q ⟶ ¬p"<br />
shows "p ⟶ q"<br />
oops<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 55. Demostrar<br />
¬(¬p ∧ ¬q) ⊢ p ∨ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_55:<br />
assumes "¬(¬p ∧ ¬q)"<br />
shows "p ∨ q"<br />
oops<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 56. Demostrar<br />
¬(¬p ∨ ¬q) ⊢ p ∧ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_56:<br />
assumes "¬(¬p ∨ ¬q)" <br />
shows "p ∧ q"<br />
oops<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 57. Demostrar<br />
¬(p ∧ q) ⊢ ¬p ∨ ¬q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_57:<br />
assumes "¬(p ∧ q)"<br />
shows "¬p ∨ ¬q"<br />
oops<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 58. Demostrar<br />
⊢ (p ⟶ q) ∨ (q ⟶ p)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_58:<br />
"(p ⟶ q) ∨ (q ⟶ p)"<br />
oops<br />
<br />
end<br />
<br />
</source></div>
Joslopjim4
https://www.glc.us.es/~jalonso/LMF2018/index.php?title=Relaci%C3%B3n_4&diff=134
Relación 4
2018-03-21T20:36:23Z
<p>Joslopjim4: </p>
<hr />
<div><source lang = "isar"><br />
<br />
chapter {* R4: Deducción natural proposicional *}<br />
<br />
theory R4<br />
imports Main <br />
begin<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
El objetivo de esta relación es demostrar cada uno de los ejercicios<br />
usando sólo las reglas básicas de deducción natural de la lógica<br />
proposicional (sin usar el método auto).<br />
<br />
Las reglas básicas de la deducción natural son las siguientes:<br />
· conjI: ⟦P; Q⟧ ⟹ P ∧ Q<br />
· conjunct1: P ∧ Q ⟹ P<br />
· conjunct2: P ∧ Q ⟹ Q <br />
· notnotD: ¬¬ P ⟹ P<br />
· notnotI: P ⟹ ¬¬ P<br />
· mp: ⟦P ⟶ Q; P⟧ ⟹ Q <br />
· mt: ⟦F ⟶ G; ¬G⟧ ⟹ ¬F <br />
· impI: (P ⟹ Q) ⟹ P ⟶ Q<br />
· disjI1: P ⟹ P ∨ Q<br />
· disjI2: Q ⟹ P ∨ Q<br />
· disjE: ⟦P ∨ Q; P ⟹ R; Q ⟹ R⟧ ⟹ R <br />
· FalseE: False ⟹ P<br />
· notE: ⟦¬P; P⟧ ⟹ R<br />
· notI: (P ⟹ False) ⟹ ¬P<br />
· iffI: ⟦P ⟹ Q; Q ⟹ P⟧ ⟹ P = Q<br />
· iffD1: ⟦Q = P; Q⟧ ⟹ P <br />
· iffD2: ⟦P = Q; Q⟧ ⟹ P<br />
· ccontr: (¬P ⟹ False) ⟹ P<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
text {*<br />
Se usarán las reglas notnotI y mt que demostramos a continuación. *}<br />
<br />
lemma notnotI: "P ⟹ ¬¬ P"<br />
by auto<br />
<br />
lemma mt: "⟦F ⟶ G; ¬G⟧ ⟹ ¬F"<br />
by auto<br />
<br />
section {* Implicaciones *}<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 1. Demostrar<br />
p ⟶ q, p ⊢ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_1: josrodjim2 carmarria inmbenber marmedmar3<br />
assumes 1: "p ⟶ q" and<br />
2: "p"<br />
shows "q"<br />
<br />
proof -<br />
show 1: "q" using 1 2 by (rule mp)<br />
qed<br />
<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 2. Demostrar<br />
p ⟶ q, q ⟶ r, p ⊢ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_2: josrodjim2 carmarria inmbenber marmedmar3<br />
assumes 1: "p ⟶ q" and<br />
2: "q ⟶ r" and<br />
3: "p" <br />
shows "r"<br />
<br />
proof-<br />
have 4: "q" using 1 3 by (rule mp)<br />
show "r" using 2 4 by (rule mp)<br />
qed<br />
<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 3. Demostrar<br />
p ⟶ (q ⟶ r), p ⟶ q, p ⊢ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_3: josrodjim2 carmarria inmbenber marmedmar3<br />
assumes 1: "p ⟶ (q ⟶ r)" and<br />
2: "p ⟶ q" and<br />
3: "p"<br />
shows "r"<br />
<br />
proof-<br />
have 4: "q" using 2 3 by (rule mp)<br />
have 5: "(q ⟶ r)" using 1 3 by (rule mp)<br />
show "r" using 5 4 by (rule mp)<br />
<br />
qed<br />
<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 4. Demostrar<br />
p ⟶ q, q ⟶ r ⊢ p ⟶ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_4: josrodjim2<br />
assumes 1: "p ⟶ q" and<br />
2: "q ⟶ r" <br />
shows "p ⟶ r"<br />
<br />
proof-<br />
{ assume 3: "p"<br />
have 4: "q" using 1 3 by (rule mp)<br />
have 5: "r" using 2 4 by (rule mp)}<br />
hence 6: "p ⟶ r" using 3 5 by (rule impI)<br />
show "p⟶r" using 6 by this<br />
<br />
oops<br />
<br />
lemma ejercicio_4: carmarria<br />
assumes 1: "p ⟶ q" and<br />
2: "q ⟶ r" <br />
shows "p ⟶ r"<br />
proof -<br />
{ assume 3: p <br />
have 4: "q" using 1 3 by (rule mp)<br />
have 5: "r" using 2 4 by (rule mp)<br />
}<br />
thus "p ⟶ r" by (rule impI) <br />
qed<br />
lemma ejercicio_4: inmbenber<br />
assumes 1: "p ⟶ q" and<br />
2: "q ⟶ r" <br />
shows "p ⟶ r"<br />
proof -<br />
{ assume 3: "p" <br />
have 4: "q" using 1 3 by (rule mp)<br />
have 5: "r" using 2 4 by (rule mp) }<br />
thus "p ⟶ r" by (rule impI) <br />
qed<br />
<br />
lemma ejercicio_4:marmedmar3<br />
assumes "p ⟶ q"<br />
"q ⟶ r" <br />
shows "p ⟶ r"<br />
proof <br />
assume "p" <br />
with assms(1) have "q" by (rule mp)<br />
with assms(2) show "r" by (rule mp)<br />
qed <br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 5. Demostrar<br />
p ⟶ (q ⟶ r) ⊢ q ⟶ (p ⟶ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_5: josrodjim2 (NO DA ERROR, PERO SI ALERTA)<br />
assumes "p ⟶ (q ⟶ r)" <br />
shows "q ⟶ (p ⟶ r)"<br />
<br />
proof -<br />
<br />
{ assume 2: "q"<br />
{assume 3: "p"<br />
have 4: "q ⟶r" using 1 3 by (rule mp)<br />
have 5: "r" using 4 2 by (rule mp)}<br />
hence 6: "p⟶r" using 3 5 by (rule impI)}<br />
hence 7: "q ⟶ (p ⟶ r)" using 2 6 by (rule impI)<br />
show "q ⟶ (p ⟶ r)" using 7 by this<br />
<br />
oops<br />
<br />
lemma ejercicio_5: carmarria <br />
assumes 1: "p ⟶ (q ⟶ r)" <br />
shows "q ⟶ (p ⟶ r)"<br />
<br />
proof - <br />
{assume 2: q<br />
{assume 3: p <br />
have 4: "q ⟶ r" using 1 3 by (rule mp)<br />
have 5: "r" using 4 2 by (rule mp)<br />
}<br />
hence " p ⟶ r" by (rule impI)<br />
}<br />
thus "q ⟶ (p ⟶ r)" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_5: inmbenber<br />
assumes 1: "p ⟶ (q ⟶ r)" <br />
shows "q ⟶ (p ⟶ r)"<br />
<br />
proof - <br />
{assume 2: "q"<br />
{assume 3: "p" <br />
have 4: "q ⟶ r" using 1 3 by (rule mp)<br />
have 5: "r" using 4 2 by (rule mp) }<br />
hence " p ⟶ r" by (rule impI) }<br />
thus "q ⟶ (p ⟶ r)" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_5: marmedmar3<br />
assumes "p ⟶ (q ⟶ r)" <br />
shows "q ⟶ (p ⟶ r)"<br />
proof <br />
assume "q"<br />
show "(p ⟶ r)"<br />
proof <br />
assume "p" <br />
with assms(1) have "(q ⟶ r)" by (rule mp)<br />
thus "r" using `q` by (rule mp) <br />
qed <br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 6. Demostrar<br />
p ⟶ (q ⟶ r) ⊢ (p ⟶ q) ⟶ (p ⟶ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_6:josrodjim2<br />
assumes 1: "p ⟶ (q ⟶ r)" <br />
shows "(p ⟶ q) ⟶ (p ⟶ r)"<br />
<br />
proof- <br />
<br />
{assume 2: "p⟶q" <br />
{assume 3: "p"<br />
have 4: "q⟶r" using 1 3 by (rule mp)<br />
have 5: "q" using 2 3 by (rule mp)<br />
have 6: "r" using 4 5 by (rule mp)}<br />
hence 7: "p⟶r" using 3 6 by (rule impI)}<br />
hence 8: "(p ⟶ q) ⟶ (p ⟶ r)" using 2 7 by (rule impI)<br />
show "(p ⟶ q) ⟶ (p ⟶ r)" using 8 by this<br />
oops<br />
<br />
lemma ejercicio_6: carmarria, inmbenber<br />
assumes 1: "p ⟶ (q ⟶ r)" <br />
shows "(p ⟶ q) ⟶ (p ⟶ r)"<br />
proof -<br />
{assume 2: "p ⟶ q"<br />
{assume 3: "p"<br />
have 4: "q" using 2 3 by (rule mp)<br />
have 5: "q ⟶ r" using 1 3 by (rule mp)<br />
have 6: "r" using 5 4 by (rule mp)<br />
}<br />
hence 7: "p ⟶ r" by (rule impI)<br />
}<br />
thus "(p ⟶ q) ⟶ (p ⟶ r)" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_6:marmedmar3<br />
assumes 1: "p ⟶ (q ⟶ r)" <br />
shows "(p ⟶ q) ⟶ (p ⟶ r)"<br />
proof <br />
assume 2: "(p ⟶ q)" <br />
show "(p ⟶ r)" <br />
proof <br />
assume 3: "p" <br />
have 4: "q" using 2 3 by (rule mp)<br />
have 5: "(q ⟶ r)" using 1 3 by (rule mp) <br />
show "r" using 5 4 by (rule mp) <br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 7. Demostrar<br />
p ⊢ q ⟶ p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_7: (NO SE QUE ESTA MAL)<br />
assumes "p" <br />
shows "q ⟶ p"<br />
<br />
proof -<br />
<br />
{ assume 2: "q"<br />
have 3: "p" using 1<br />
}<br />
hence 4: "q⟶p" using 2 3 by (rule impI)<br />
<br />
show "q⟶p" using 4 by this<br />
<br />
oops<br />
<br />
lemma ejercicio_7: carmarria, inmbenber<br />
assumes 1: "p" <br />
shows "q ⟶ p"<br />
proof -<br />
{assume 2: "q"<br />
have 3: "p" using 1 by this}<br />
thus "q ⟶ p" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_7:marmedmar3<br />
assumes "p" <br />
shows "q ⟶ p"<br />
proof <br />
assume "q" <br />
show "p" using assms(1) by this<br />
qed <br />
<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 8. Demostrar<br />
⊢ p ⟶ (q ⟶ p)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_8: carmarria<br />
"p ⟶ (q ⟶ p)"<br />
<br />
proof -<br />
{assume 1: p<br />
{assume 2: q<br />
have 3: p using 1 by this<br />
}<br />
hence 4: "q ⟶ p" by (rule impI)<br />
}<br />
thus "p ⟶ (q ⟶ p)" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_8: marmedmar3<br />
"p ⟶ (q ⟶ p)"<br />
proof<br />
assume 1: "p"<br />
show "(q ⟶ p)"<br />
proof <br />
assume 2: "q" <br />
show "p" using 1 by this <br />
qed<br />
qed<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 9. Demostrar<br />
p ⟶ q ⊢ (q ⟶ r) ⟶ (p ⟶ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_9: carmarria<br />
assumes 1: "p ⟶ q" <br />
shows "(q ⟶ r) ⟶ (p ⟶ r)"<br />
proof -<br />
{assume 2: "q ⟶ r"<br />
{assume 3: "p"<br />
have 4: "q" using 1 3 by (rule mp)<br />
have 5: "r" using 2 4 by (rule mp)<br />
}<br />
hence 6: "p ⟶ r" by (rule impI)<br />
}<br />
thus 7: "(q ⟶ r) ⟶ (p ⟶ r)" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_9: marmedmar3<br />
assumes "p ⟶ q" <br />
shows "(q ⟶ r) ⟶ (p ⟶ r)"<br />
proof <br />
assume "(q ⟶ r)" <br />
show "(p ⟶ r)"<br />
proof <br />
assume "p" <br />
with `(p ⟶ q)` have "q" by (rule mp)<br />
with `(q ⟶ r)` show "r" by (rule mp)<br />
qed<br />
qed<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 10. Demostrar<br />
p ⟶ (q ⟶ (r ⟶ s)) ⊢ r ⟶ (q ⟶ (p ⟶ s))<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_10: carmarria<br />
assumes 1: "p ⟶ (q ⟶ (r ⟶ s))" <br />
shows "r ⟶ (q ⟶ (p ⟶ s))"<br />
<br />
proof-<br />
{assume 2: r<br />
{assume 3: q<br />
{assume 4: p<br />
have 5: "q ⟶ (r ⟶ s)" using 1 4 by (rule mp)<br />
have 6: "r ⟶ s" using 5 3 by (rule mp)<br />
have 7: s using 6 2 by (rule mp)<br />
}<br />
hence 8: "p ⟶ s" by (rule impI)<br />
}<br />
hence 9: "q ⟶ (p ⟶ s)" by (rule impI)<br />
}<br />
thus "r ⟶ (q ⟶ (p ⟶ s))" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_10: marmedmar3<br />
assumes 1: "p ⟶ (q ⟶ (r ⟶ s))" <br />
shows "r ⟶ (q ⟶ (p ⟶ s))"<br />
proof <br />
assume 2: "r"<br />
show "(q ⟶ (p ⟶ s))"<br />
proof <br />
assume 3: "q" <br />
show "(p ⟶ s)"<br />
proof <br />
assume 4: "p" <br />
have 5: "(q ⟶ (r ⟶ s))" using 1 4 by (rule mp) <br />
have 6: "(r ⟶ s)" using 5 3 by (rule mp)<br />
show "s" using 6 2 by (rule mp) <br />
qed<br />
qed<br />
qed<br />
<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 11. Demostrar<br />
⊢ (p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_11: carmarria<br />
"(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))"<br />
proof - <br />
{assume 1: "p ⟶ (q ⟶ r)"<br />
{assume 2: "p ⟶ q"<br />
{assume 3: p<br />
have 4: q using 2 3 by (rule mp)<br />
have 5: "q ⟶ r" using 1 3 by (rule mp)<br />
have 6: r using 5 4 by (rule mp)<br />
}<br />
hence 7: "p ⟶ r" by (rule impI)<br />
}<br />
hence 8: "(p ⟶ q) ⟶ (p ⟶ r)" by (rule impI)<br />
}<br />
thus "(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_11: marmedmar3<br />
"(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))"<br />
proof <br />
assume 1: "(p ⟶ (q ⟶ r))" <br />
show "((p ⟶ q) ⟶ (p ⟶ r))"<br />
proof <br />
assume 2: "(p ⟶ q)" <br />
show "(p ⟶ r)" <br />
proof <br />
assume 3: "p" <br />
have 4: "(q ⟶ r)" using 1 3 by (rule mp) <br />
have 5: "q" using 2 3 by (rule mp) <br />
show "r" using 4 5 by (rule mp) <br />
qed<br />
qed<br />
qed<br />
<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 12. Demostrar<br />
(p ⟶ q) ⟶ r ⊢ p ⟶ (q ⟶ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_12: carmarria<br />
assumes 1: "(p ⟶ q) ⟶ r" <br />
shows "p ⟶ (q ⟶ r)"<br />
proof -<br />
{assume 2: p<br />
{assume 3: q<br />
{assume 4: p<br />
have 5: q using 3 by this<br />
}<br />
hence 6: "p ⟶ q" by (rule impI)<br />
have 7: r using 1 6 by (rule mp)<br />
}<br />
hence 8: "q ⟶ r" by (rule impI)<br />
}<br />
thus "p ⟶ (q ⟶ r)" by (rule impI)<br />
qed<br />
<br />
section {* Conjunciones *}<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 13. Demostrar<br />
p, q ⊢ p ∧ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_13: carmarria marmedmar3<br />
assumes "p"<br />
"q" <br />
shows "p ∧ q"<br />
<br />
proof-<br />
show "p \<and> q" using assms(1) assms(2) by (rule conjI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 14. Demostrar<br />
p ∧ q ⊢ p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_14: carmarria marmedmar3<br />
assumes "p ∧ q" <br />
shows "p"<br />
<br />
proof -<br />
show "p" using assms by (rule conjunct1)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 15. Demostrar<br />
p ∧ q ⊢ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_15: carmarria marmedmar3<br />
assumes "p ∧ q" <br />
shows "q"<br />
<br />
proof -<br />
show q using assms(1) by (rule conjunct2)<br />
qed<br />
<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 16. Demostrar<br />
p ∧ (q ∧ r) ⊢ (p ∧ q) ∧ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_16: carmarria marmedmar3<br />
assumes 1: "p ∧ (q ∧ r)"<br />
shows "(p ∧ q) ∧ r"<br />
proof -<br />
have 2: "q \<and> r" using 1 by (rule conjunct2)<br />
have 3: p using 1 by (rule conjunct1)<br />
have 4: q using 2 by (rule conjunct1)<br />
have 5: r using 2 by (rule conjunct2)<br />
have 6: "p \<and> q" using 3 4 by (rule conjI)<br />
show "(p \<and> q) \<and> r" using 6 5 by (rule conjI)<br />
qed<br />
<br />
lemma ej_16: joslopjim4<br />
assumes "p ∧ (q ∧ r)"<br />
shows "(p ∧ q) ∧ r"<br />
proof-<br />
have "p" using assms by (rule conjunct1)<br />
have "q∧r" using assms by (rule conjunct2)<br />
have "q" using `q∧r` by (rule conjunct1)<br />
have "r" using `q∧r` by (rule conjunct2)<br />
have "p∧q" using `p` `q` by (rule conjI)<br />
show "(p∧q)∧r" using `p∧q` `r` by (rule conjI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 17. Demostrar<br />
(p ∧ q) ∧ r ⊢ p ∧ (q ∧ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_17:<br />
assumes "(p ∧ q) ∧ r" <br />
shows "p ∧ (q ∧ r)"<br />
proof (rule conjI)<br />
have "p ∧ q" using assms ..<br />
thus "p" ..<br />
next<br />
show "q ∧ r"<br />
proof (rule conjI)<br />
have "p ∧ q" using assms ..<br />
thus "q" ..<br />
next<br />
show "r" using assms ..<br />
qed<br />
qed<br />
<br />
lemma ejercicio_17: carmarria marmedmar3<br />
assumes 1:"(p ∧ q) ∧ r" <br />
shows "p ∧ (q ∧ r)"<br />
proof-<br />
have 2:"p \<and> q" using 1 by (rule conjunct1)<br />
have 3: r using 1 by (rule conjunct2)<br />
have 4: p using 2 by (rule conjunct1)<br />
have 5: q using 2 by (rule conjunct2)<br />
have 6: "q \<and> r" using 5 3 by (rule conjI)<br />
show "p \<and> (q \<and> r)" using 4 6 by (rule conjI)<br />
qed<br />
<br />
<br />
lemma ej_17: joslopjim4<br />
assumes "(p ∧ q) ∧ r" <br />
shows "p ∧ (q ∧ r)"<br />
proof-<br />
have "p∧q" using assms by (rule conjunct1)<br />
have "r" using assms by (rule conjunct2)<br />
have "p" using `p∧q` by (rule conjunct1)<br />
have "q" using `p∧q` by (rule conjunct2)<br />
have "q∧r" using `q` `r` by (rule conjI)<br />
show "p∧(q∧r)" using `p` `q∧r` by (rule conjI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 18. Demostrar<br />
p ∧ q ⊢ p ⟶ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_18: carmarria<br />
assumes "p ∧ q" <br />
shows "p ⟶ q"<br />
proof -<br />
{assume p<br />
have q using assms(1) by (rule conjunct2)<br />
}<br />
thus "p ⟶ q" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_18: marmedmar3<br />
assumes "p ∧ q" <br />
shows "p ⟶ q"<br />
proof <br />
assume "p" <br />
show "q" using assms by (rule conjunct2) <br />
qed <br />
<br />
lemma ej_18: joslopjim4<br />
assumes "p ∧ q" <br />
shows "p ⟶ q"<br />
proof (rule impI)<br />
assume "p"<br />
show "q" using assms by (rule conjunct2)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 19. Demostrar<br />
(p ⟶ q) ∧ (p ⟶ r) ⊢ p ⟶ q ∧ r <br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_19: carmarria<br />
assumes 1: "(p ⟶ q) ∧ (p ⟶ r)" <br />
shows "p ⟶ q ∧ r"<br />
proof -<br />
{assume 2: p<br />
have 3: "p ⟶ q" using 1 by (rule conjunct1)<br />
have 4: "p ⟶ r" using 1 by (rule conjunct2)<br />
have 5: q using 3 2 by (rule mp)<br />
have 6: r using 4 2 by (rule mp)<br />
have 7: "q \<and> r" using 5 6 by (rule conjI)<br />
}<br />
thus "p ⟶ q \<and> r" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_19: marmedmar3<br />
assumes 1: "(p ⟶ q) ∧ (p ⟶ r)" <br />
shows "p ⟶ q ∧ r"<br />
proof <br />
have 2: "(p ⟶ q)" using assms by (rule conjunct1)<br />
have 3: "(p ⟶ r)" using assms by (rule conjunct2)<br />
assume 4: "p"<br />
with `(p ⟶ q)` have 5: "q" by (rule mp)<br />
have 6: "r" using 3 4 by (rule mp) <br />
show "q ∧ r" using 5 6 by (rule conjI)<br />
qed <br />
<br />
lemma ej_19: joslopjim4<br />
assumes "(p ⟶ q) ∧ (p ⟶ r)" <br />
shows "p ⟶ q ∧ r"<br />
proof (rule impI)<br />
assume "p"<br />
have "p⟶q" using assms by (rule conjunct1)<br />
have "q" using `p⟶q` `p` by (rule mp)<br />
have "p⟶r" using assms by (rule conjunct2)<br />
have "r" using `p⟶r` `p` by (rule mp)<br />
show "q∧r" using `q` `r` by (rule conjI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 20. Demostrar<br />
p ⟶ q ∧ r ⊢ (p ⟶ q) ∧ (p ⟶ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_20: carmarria<br />
assumes 1: "p ⟶ q ∧ r" <br />
shows "(p ⟶ q) ∧ (p ⟶ r)"<br />
proof -<br />
{assume 2: p<br />
have 3: "q \<and> r" using 1 2 by (rule mp)<br />
have 4: q using 3 by (rule conjunct1)<br />
}<br />
hence 5: "p ⟶ q" by (rule impI)<br />
{assume 6: p<br />
have 7: "q \<and> r" using 1 6 by (rule mp)<br />
have 8: r using 7 by (rule conjunct2)<br />
}<br />
hence 9: "p ⟶ r" by (rule impI)<br />
show "(p ⟶ q) \<and> (p ⟶ r)" using 5 9 by (rule conjI)<br />
qed<br />
<br />
lemma ej_20: joslopjim4<br />
assumes "p ⟶ q ∧ r" <br />
shows "(p ⟶ q) ∧ (p ⟶ r)"<br />
proof-<br />
have "p⟶q"<br />
proof<br />
assume "p"<br />
have "q∧r" using assms `p` by (rule mp)<br />
show "q" using `q∧r` by (rule conjunct1)<br />
qed<br />
have "p⟶r"<br />
proof<br />
assume "p"<br />
have "q∧r" using assms `p` by (rule mp)<br />
show "r" using `q∧r` by (rule conjunct2)<br />
qed<br />
show "(p⟶q)∧(p⟶r)" using `p⟶q` `p⟶r` by (rule conjI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 21. Demostrar<br />
p ⟶ (q ⟶ r) ⊢ p ∧ q ⟶ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_21: carmarria<br />
assumes 1: "p ⟶ (q ⟶ r)" <br />
shows "p ∧ q ⟶ r"<br />
proof -<br />
{assume 2: "p \<and> q"<br />
have 3: p using 2 by (rule conjunct1)<br />
have 4: q using 2 by (rule conjunct2)<br />
have 5: "q ⟶ r" using 1 3 by (rule mp)<br />
have 6: r using 5 4 by (rule mp)<br />
}<br />
thus "p ∧ q ⟶ r" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_21: marmedmar3<br />
assumes "p ⟶ (q ⟶ r)" <br />
shows "p ∧ q ⟶ r"<br />
proof <br />
assume "p ∧ q" <br />
hence "p" by (rule conjunct1) <br />
with assms have "(q ⟶ r)" by (rule mp)<br />
have "q" using `p ∧ q` by (rule conjunct2) <br />
with `(q ⟶ r)` show "r" by (rule mp) <br />
qed <br />
<br />
lemma ej_21: joslopjim4<br />
assumes "p ⟶ (q ⟶ r)" <br />
shows "p ∧ q ⟶ r"<br />
proof (rule impI)<br />
assume "p∧q"<br />
have "p" using `p∧q` by (rule conjunct1)<br />
have "q" using `p∧q` by (rule conjunct2)<br />
have "q⟶r" using assms `p` by (rule mp)<br />
show "r" using `q⟶r` `q` by (rule mp)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 22. Demostrar<br />
p ∧ q ⟶ r ⊢ p ⟶ (q ⟶ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_22: carmarria<br />
assumes "p ∧ q ⟶ r" <br />
shows "p ⟶ (q ⟶ r)"<br />
proof -<br />
{assume p<br />
{assume q<br />
have "p \<and> q" using `p` `q` by (rule conjI)<br />
have r using assms(1) `p \<and> q` by (rule mp)<br />
}<br />
hence "q ⟶ r" by (rule impI)<br />
}<br />
thus "p ⟶ (q ⟶ r)" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_22: marmedmar3<br />
assumes "p ∧ q ⟶ r" <br />
shows "p ⟶ (q ⟶ r)"<br />
proof <br />
assume "p" <br />
show "(q ⟶ r)"<br />
proof <br />
assume "q" <br />
with `p` have "p ∧ q" by (rule conjI) <br />
with assms show "r" by (rule mp) <br />
qed <br />
qed<br />
<br />
lemma ej_22: joslopjim4<br />
assumes "p ∧ q ⟶ r" <br />
shows "p ⟶ (q ⟶ r)"<br />
proof (rule impI)<br />
assume "p"<br />
show "q⟶r"<br />
proof (rule impI)<br />
assume "q"<br />
have "p∧q" using `p` `q` by (rule conjI)<br />
show "r" using assms `p∧q` by (rule mp)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 23. Demostrar<br />
(p ⟶ q) ⟶ r ⊢ p ∧ q ⟶ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_23: carmarria<br />
assumes "(p ⟶ q) ⟶ r" <br />
shows "p ∧ q ⟶ r"<br />
proof -<br />
{assume "p \<and> q" <br />
{assume p<br />
have q using `p \<and> q` by (rule conjunct2)<br />
}<br />
hence "p ⟶ q" by (rule impI)<br />
have r using assms(1) `p ⟶ q` by (rule mp)<br />
}<br />
thus "p \<and> q ⟶ r" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_23: marmedmar3<br />
assumes "(p ⟶ q) ⟶ r" <br />
shows "p ∧ q ⟶ r"<br />
proof <br />
assume "p ∧ q"<br />
have "p" using `p ∧ q` by (rule conjunct1)<br />
have "q" using `p ∧ q` by (rule conjunct2) <br />
hence "(p ⟶ q)" by (rule impI) <br />
with assms show "r" by (rule mp) <br />
qed <br />
<br />
lemma ej_23: joslopjim4<br />
assumes "(p ⟶ q) ⟶ r" <br />
shows "p ∧ q ⟶ r"<br />
proof (rule impI)<br />
assume "p∧q"<br />
have "p⟶q"<br />
proof<br />
assume "p"<br />
show "q" using `p∧q` by (rule conjunct2)<br />
qed<br />
show "r" using assms `p⟶q` by (rule mp)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 24. Demostrar<br />
p ∧ (q ⟶ r) ⊢ (p ⟶ q) ⟶ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_24: carmarria<br />
assumes 1: "p ∧ (q ⟶ r)" <br />
shows "(p ⟶ q) ⟶ r"<br />
proof -<br />
{assume 2: "p ⟶ q" <br />
have 3: p using 1 by (rule conjunct1)<br />
have 4: "q ⟶ r" using 1 by (rule conjunct2)<br />
have 5: q using 2 3 by (rule mp)<br />
have 6: r using 4 5 by (rule mp)<br />
}<br />
thus "(p ⟶ q) ⟶ r" by (rule impI)<br />
qed<br />
<br />
lemma ejercicio_24: marmedmar3<br />
assumes "p ∧ (q ⟶ r)" <br />
shows "(p ⟶ q) ⟶ r"<br />
proof <br />
assume "(p ⟶ q)" <br />
have "(q ⟶ r)" using assms by (rule conjunct2)<br />
have "p" using assms by (rule conjunct1)<br />
with `(p ⟶ q)` have "q" by (rule mp) <br />
with `(q ⟶ r)` show "r" by (rule mp)<br />
qed <br />
<br />
lemma ej_24: joslopjim4<br />
assumes "p ∧ (q ⟶ r)" <br />
shows "(p ⟶ q) ⟶ r"<br />
proof (rule impI)<br />
assume "p⟶q"<br />
have "p" using assms by (rule conjunct1)<br />
have "q⟶r" using assms by (rule conjunct2)<br />
have "q" using `p⟶q` `p` by (rule mp)<br />
show "r" using `q⟶r` `q` by (rule mp)<br />
qed<br />
<br />
section {* Disyunciones *}<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 25. Demostrar<br />
p ⊢ p ∨ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_25: carmarria<br />
assumes "p"<br />
shows "p ∨ q"<br />
proof -<br />
show "p | q" using assms(1) by (rule disjI1)<br />
qed<br />
<br />
lemma ej_25: joslopjim4<br />
assumes "p"<br />
shows "p ∨ q"<br />
proof (rule disjI1)<br />
show "p" using assms by this<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 26. Demostrar<br />
q ⊢ p ∨ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_26: carmarria<br />
assumes "q"<br />
shows "p ∨ q"<br />
proof -<br />
show "p | q" using assms(1) by (rule disjI2)<br />
qed<br />
<br />
lemma ej_26: joslopjim4<br />
assumes "q"<br />
shows "p ∨ q"<br />
proof (rule disjI2)<br />
show "q" using assms by this<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 27. Demostrar<br />
p ∨ q ⊢ q ∨ p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_27: carmarria<br />
assumes "p ∨ q"<br />
shows "q ∨ p"<br />
proof -<br />
have "p | q" using assms(1) by this<br />
moreover<br />
{assume p<br />
have "q | p" using `p` by (rule disjI2)<br />
}<br />
moreover<br />
{assume q<br />
have "q | p" using `q` by (rule disjI1)<br />
}<br />
ultimately show "q | p" by (rule disjE)<br />
qed<br />
<br />
lemma ej_27: joslopjim4<br />
assumes "p ∨ q"<br />
shows "q ∨ p"<br />
using assms<br />
proof <br />
assume "p"<br />
then show "q∨p" by(rule disjI2)<br />
next<br />
assume "q"<br />
show "q∨p" using `q` by(rule disjI1)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 28. Demostrar<br />
q ⟶ r ⊢ p ∨ q ⟶ p ∨ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_28: carmarria<br />
assumes "q ⟶ r" <br />
shows "p ∨ q ⟶ p ∨ r"<br />
proof -<br />
{assume "p | q" <br />
moreover<br />
{assume p<br />
have "p | r" using `p` by (rule disjI1)<br />
}<br />
moreover<br />
{assume q<br />
have r using assms(1) `q` by (rule mp)<br />
have "p | r" using `r` by (rule disjI2)<br />
}<br />
ultimately have "p | r" by (rule disjE)<br />
}<br />
thus "p | q ⟶ p | r" by (rule impI)<br />
qed<br />
<br />
lemma ej_28: joslopjim4<br />
assumes "q ⟶ r" <br />
shows "p ∨ q ⟶ p ∨ r"<br />
proof (rule impI)<br />
assume "p∨q" <br />
show "p∨r"<br />
using `p∨q`<br />
proof <br />
assume "p"<br />
show "p∨r" using `p` by (rule disjI1)<br />
next<br />
assume "q"<br />
have "r" using assms `q` by (rule mp)<br />
show "p∨r" using `r` by (rule disjI2)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 29. Demostrar<br />
p ∨ p ⊢ p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_29: carmarria<br />
assumes "p ∨ p"<br />
shows "p"<br />
proof -<br />
have "p | p" using assms(1) by this<br />
moreover <br />
{assume p<br />
}<br />
moreover<br />
{assume p<br />
}<br />
ultimately show p by (rule disjE)<br />
qed<br />
<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 30. Demostrar<br />
p ⊢ p ∨ p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_30: carmarria<br />
assumes "p" <br />
shows "p ∨ p"<br />
proof -<br />
show "p | p" using assms(1) by (rule disjI1)<br />
qed<br />
<br />
lemma ej_30: joslopjim4<br />
assumes "p" <br />
shows "p ∨ p"<br />
proof (rule disjI1)<br />
show "p" using assms by this<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 31. Demostrar<br />
p ∨ (q ∨ r) ⊢ (p ∨ q) ∨ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_31: carmarria<br />
assumes 1: "p ∨ (q ∨ r)" <br />
shows "(p ∨ q) ∨ r"<br />
proof -<br />
have "p | (q | r)" using 1 by this<br />
moreover {assume 2: p<br />
have 3: "p | q" using 2 by (rule disjI1)<br />
have 4: "(p | q) | r" using 3 by (rule disjI1)<br />
}<br />
moreover {assume 5: "q | r"<br />
moreover {assume 6: q<br />
have 7: "p | q" using 6 by (rule disjI2)<br />
have 8: "(p | q) | r " using 7 by (rule disjI1)<br />
}<br />
moreover {assume 9: r<br />
have 10: "(p | q) | r " using 9 by (rule disjI2)<br />
}<br />
ultimately have 11: "(p | q) | r " by (rule disjE)<br />
}<br />
ultimately show "(p | q) | r " by (rule disjE)<br />
qed<br />
<br />
lemma ej_31: joslopjim4<br />
assumes "p ∨ (q ∨ r)" <br />
shows "(p ∨ q) ∨ r"<br />
using assms<br />
proof <br />
assume "p"<br />
have "p∨q" using `p` by (rule disjI1)<br />
thus "(p ∨ q) ∨ r" by (rule disjI1)<br />
next<br />
assume "q∨r"<br />
show "(p ∨ q) ∨ r"<br />
using `q∨r`<br />
proof <br />
assume "q"<br />
have "p∨q" using `q` by (rule disjI2)<br />
show "(p ∨ q) ∨ r" using `p∨q` by (rule disjI1)<br />
next<br />
assume "r"<br />
show "(p ∨ q) ∨ r" using `r` by (rule disjI2)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 32. Demostrar<br />
(p ∨ q) ∨ r ⊢ p ∨ (q ∨ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_32: carmarria<br />
assumes 1: "(p ∨ q) ∨ r" <br />
shows "p ∨ (q ∨ r)"<br />
proof -<br />
have "(p | q) | r" using 1 by this<br />
moreover {assume 2: "p | q" <br />
moreover {assume 3: p <br />
have 4: "p | (q | r)" using 3 by (rule disjI1)<br />
}<br />
moreover {assume 5: q<br />
have 6: "q | r" using 5 by (rule disjI1)<br />
have 7: "p | (q | r)" using 6 by (rule disjI2)<br />
}<br />
ultimately have 8: " p | (q | r)" by (rule disjE)<br />
}<br />
moreover {assume 9: r<br />
have 10: "q | r" using 9 by (rule disjI2)<br />
have 11: "p | (q | r)" using 10 by (rule disjI2)<br />
}<br />
ultimately show "p | (q | r)" by (rule disjE)<br />
qed<br />
<br />
lemma ej_32: joslopjim4<br />
assumes "(p ∨ q) ∨ r" <br />
shows "p ∨ (q ∨ r)"<br />
using assms<br />
proof<br />
assume "p∨q"<br />
show "p ∨ (q ∨ r)"<br />
using `p∨q`<br />
proof <br />
assume "p"<br />
show "p ∨ (q ∨ r)" using `p` by (rule disjI1)<br />
next<br />
assume "q"<br />
have "q∨r" using `q` by (rule disjI1)<br />
show "p ∨ (q ∨ r)" using `q∨r` by (rule disjI2)<br />
qed<br />
next<br />
assume "r"<br />
have "q∨r" using `r` by (rule disjI2)<br />
show "p ∨ (q ∨ r)" using `q∨r` by (rule disjI2)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 33. Demostrar<br />
p ∧ (q ∨ r) ⊢ (p ∧ q) ∨ (p ∧ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_33: carmarria<br />
assumes 1: "p ∧ (q ∨ r)" <br />
shows "(p ∧ q) ∨ (p ∧ r)"<br />
proof -<br />
have 2: p using 1 by (rule conjunct1)<br />
have 3: "q | r" using 1 by (rule conjunct2)<br />
moreover {assume 4: q<br />
have 5: "p & q" using 2 4 by (rule conjI)<br />
have 6: "(p & q) | (p & r)" using 5 by (rule disjI1)<br />
}<br />
moreover {assume 7: r<br />
have 8: "p & r" using 2 7 by (rule conjI)<br />
have 9: "(p & q) | (p & r)" using 8 by (rule disjI2)<br />
}<br />
ultimately show "(p & q) | (p & r)" by (rule disjE)<br />
qed<br />
<br />
lemma ej_33: joslopjim4<br />
assumes "p ∧ (q ∨ r)" <br />
shows "(p ∧ q) ∨ (p ∧ r)"<br />
proof-<br />
have "p" using assms by (rule conjunct1)<br />
have "q∨r" using assms by (rule conjunct2)<br />
show "(p ∧ q) ∨ (p ∧ r)"<br />
using `q∨r` <br />
proof (rule disjE)<br />
assume "q"<br />
have "p∧q" using `p` `q` by (rule conjI)<br />
show "(p ∧ q) ∨ (p ∧ r)" using `p∧q` by (rule disjI1)<br />
next<br />
assume "r"<br />
have "p∧r" using `p` `r` by (rule conjI)<br />
show "(p ∧ q) ∨ (p ∧ r)" using `p∧r` by (rule disjI2)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 34. Demostrar<br />
(p ∧ q) ∨ (p ∧ r) ⊢ p ∧ (q ∨ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_34: carmarria<br />
assumes 1: "(p ∧ q) ∨ (p ∧ r)" <br />
shows "p ∧ (q ∨ r)"<br />
proof -<br />
have "(p & q) | (p & r)" using 1 by this<br />
moreover {assume 2: "p & q"<br />
have 3: p using 2 by (rule conjunct1)<br />
have 4: q using 2 by (rule conjunct2)<br />
have 5: "q | r" using 4 by (rule disjI1)<br />
have 6: "p & (q | r)" using 3 5 by (rule conjI)<br />
}<br />
moreover {assume 7: "p & r"<br />
have 8: p using 7 by (rule conjunct1)<br />
have 9: r using 7 by (rule conjunct2)<br />
have 10: "q | r" using 9 by (rule disjI2)<br />
have 11: "p & (q | r)" using 8 10 by (rule conjI)<br />
}<br />
ultimately show "p & (q | r)" by (rule disjE)<br />
qed<br />
<br />
lemma ej_34: joslopjim4<br />
assumes "(p ∧ q) ∨ (p ∧ r)" <br />
shows "p ∧ (q ∨ r)"<br />
using assms<br />
proof<br />
assume "p∧q"<br />
have "p" using `p∧q` by (rule conjunct1)<br />
have "q" using `p∧q` by (rule conjunct2)<br />
have "q∨r" using `q` by (rule disjI1)<br />
show "p∧(q∨r)" using `p` `q∨r` by (rule conjI)<br />
next<br />
assume "p∧r"<br />
have "p" using `p∧r` by (rule conjunct1)<br />
have "r" using `p∧r` by (rule conjunct2)<br />
have "q∨r" using `r` by (rule disjI2)<br />
show "p∧(q∨r)" using `p` `q∨r` by (rule conjI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 35. Demostrar<br />
p ∨ (q ∧ r) ⊢ (p ∨ q) ∧ (p ∨ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_35: carmarria<br />
assumes 1: "p ∨ (q ∧ r)" <br />
shows "(p ∨ q) ∧ (p ∨ r)"<br />
proof -<br />
have "p | (q & r)" using 1 by this<br />
moreover {assume 2: p<br />
have 3: " p | q" using 2 by (rule disjI1)<br />
have 4: " p | r" using 2 by (rule disjI1)<br />
have 5: "(p | q) & (p | r)" using 3 4 by (rule conjI)<br />
}<br />
moreover {assume 6: "q & r" <br />
have 7: q using 6 by (rule conjunct1)<br />
have 8: r using 6 by (rule conjunct2)<br />
have 9: "p | q" using 7 by (rule disjI2)<br />
have 10: "p | r" using 8 by (rule disjI2)<br />
have 11: "(p | q) & (p | r)" using 9 10 by (rule conjI)<br />
}<br />
ultimately show "(p | q) & (p | r)" by (rule disjE)<br />
qed<br />
<br />
lemma ej_35: joslopjim4<br />
assumes "p ∨ (q ∧ r)" <br />
shows "(p ∨ q) ∧ (p ∨ r)"<br />
using assms<br />
proof (rule disjE)<br />
assume "p"<br />
have "p∨q" using `p` by (rule disjI1)<br />
have "p∨r" using `p` by (rule disjI1)<br />
show "(p ∨ q) ∧ (p ∨ r)" using `p∨q` `p∨r` by (rule conjI)<br />
next<br />
assume "q∧r"<br />
have "q" using `q∧r` by (rule conjunct1)<br />
have "r" using `q∧r` by (rule conjunct2)<br />
have "p∨q" using `q` by (rule disjI2)<br />
have "p∨r" using `r` by (rule disjI2)<br />
show "(p ∨ q) ∧ (p ∨ r)" using `p∨q` `p∨r` by (rule conjI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 36. Demostrar<br />
(p ∨ q) ∧ (p ∨ r) ⊢ p ∨ (q ∧ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_36: carmarria<br />
assumes 1: "(p ∨ q) ∧ (p ∨ r)"<br />
shows "p ∨ (q ∧ r)"<br />
proof -<br />
have 2: "p | q" using 1 by (rule conjunct1)<br />
moreover {assume 3: p<br />
have 4: "p | (q & r)" using 3 by (rule disjI1)<br />
}<br />
moreover {assume 5: q<br />
have 6: "p | r" using 1 by (rule conjunct2)<br />
moreover {assume 7: p<br />
have 8: "p | (q & r)" using 7 by (rule disjI1)<br />
}<br />
moreover {assume 9: r<br />
have 10: "q & r" using 5 9 by (rule conjI)<br />
have 11: "p | (q & r)" using 10 by (rule disjI2)<br />
}<br />
ultimately have 12: "p | (q & r)" by (rule disjE)<br />
}<br />
ultimately show "p | (q & r)" by (rule disjE)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 37. Demostrar<br />
(p ⟶ r) ∧ (q ⟶ r) ⊢ p ∨ q ⟶ r<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_37: carmarria<br />
assumes 1: "(p ⟶ r) ∧ (q ⟶ r)" <br />
shows "p ∨ q ⟶ r"<br />
proof -<br />
have 2: "p ⟶ r" using 1 by (rule conjunct1)<br />
have 3: "q ⟶ r" using 1 by (rule conjunct2)<br />
{assume 4: "p | q"<br />
moreover {assume 5: p<br />
have 6: r using 2 5 by (rule mp)<br />
}<br />
moreover {assume 7: q<br />
have 8: r using 3 7 by (rule mp)<br />
}<br />
ultimately have r by (rule disjE)<br />
}<br />
thus "p | q ⟶ r" by (rule impI)<br />
qed<br />
<br />
lemma ej_37: joslopjim4<br />
assumes "(p ⟶ r) ∧ (q ⟶ r)" <br />
shows "p ∨ q ⟶ r"<br />
proof (rule impI)<br />
assume "p∨q"<br />
have "p⟶r" using assms by (rule conjunct1)<br />
have "q⟶r" using assms by (rule conjunct2)<br />
show "r"<br />
using `p∨q`<br />
proof <br />
assume "p"<br />
show "r" using `p⟶r` `p` by (rule mp)<br />
next<br />
assume "q"<br />
show "r" using `q⟶r` `q` by (rule mp)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 38. Demostrar<br />
p ∨ q ⟶ r ⊢ (p ⟶ r) ∧ (q ⟶ r)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_38: carmarria<br />
assumes 1: "p ∨ q ⟶ r" <br />
shows "(p ⟶ r) ∧ (q ⟶ r)"<br />
proof -<br />
{assume 2: p<br />
have 3: "p | q" using 2 by (rule disjI1)<br />
have 4: r using 1 3 by (rule mp)<br />
}<br />
hence 5: "p ⟶ r" by (rule impI)<br />
{assume 6: q<br />
have 7: "p | q" using 6 by (rule disjI2)<br />
have 8: r using 1 7 by (rule mp)<br />
}<br />
hence 9: "q ⟶ r" by (rule impI)<br />
show "(p ⟶ r) & (q ⟶ r)" using 5 9 by (rule conjI)<br />
qed<br />
<br />
lemma ej_38: joslopjim4<br />
assumes "p ∨ q ⟶ r" <br />
shows "(p ⟶ r) ∧ (q ⟶ r)"<br />
proof-<br />
have "p⟶r" <br />
proof (rule impI)<br />
assume "p"<br />
have "p∨q" using `p` by (rule disjI1)<br />
show "r" using assms `p∨q` by (rule mp)<br />
qed<br />
have "q⟶r"<br />
proof (rule impI)<br />
assume "q"<br />
have "p∨q" using `q` by (rule disjI2)<br />
show "r" using assms `p∨q` by (rule mp)<br />
qed<br />
show "(p ⟶ r) ∧ (q ⟶ r)" using `p⟶r` `q⟶r` by (rule conjI)<br />
qed<br />
<br />
section {* Negaciones *}<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 39. Demostrar<br />
p ⊢ ¬¬p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_39: carmarria<br />
assumes "p"<br />
shows "¬¬p"<br />
proof -<br />
show "~~p" using assms(1) by (rule notnotI)<br />
qed<br />
<br />
lemma ej_39: joslopjim4<br />
assumes "p"<br />
shows "¬¬p"<br />
proof-<br />
show "¬¬p" using assms by (rule notnotI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 40. Demostrar<br />
¬p ⊢ p ⟶ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_40: carmarria<br />
assumes "¬p" <br />
shows "p ⟶ q"<br />
proof -<br />
{assume p<br />
have "~p" using assms(1) by this<br />
have False using `~p` `p` by (rule notE)<br />
have q using `False` by (rule FalseE)<br />
}<br />
thus "p ⟶ q" by (rule impI)<br />
qed<br />
<br />
lemma ej_40: joslopjim4<br />
assumes "¬p" <br />
shows "p ⟶ q"<br />
proof (rule impI)<br />
assume "p"<br />
show "q" using assms `p` by (rule notE)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 41. Demostrar<br />
p ⟶ q ⊢ ¬q ⟶ ¬p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_41: carmarria<br />
assumes 1: "p ⟶ q"<br />
shows "¬q ⟶ ¬p"<br />
proof -<br />
{assume 2: "~q"<br />
have 3: "~p" using 1 2 by (rule mt)<br />
}<br />
thus 4: "~q ⟶ ~p" by (rule impI)<br />
qed<br />
<br />
lemma ej_41: joslopjim4<br />
assumes "p ⟶ q"<br />
shows "¬q ⟶ ¬p"<br />
proof (rule impI)<br />
assume "¬q"<br />
show "¬p" using assms `¬q` by (rule mt)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 42. Demostrar<br />
p∨q, ¬q ⊢ p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_42: carmarria<br />
assumes 1: "p∨q" and<br />
2: "¬q" <br />
shows "p"<br />
proof -<br />
have " p | q" using 1 by this<br />
moreover {assume 3: p<br />
}<br />
moreover {assume 4: q<br />
have 5: False using 2 4 by (rule notE)<br />
have 6: p using 5 by (rule FalseE)<br />
}<br />
ultimately show p by (rule disjE)<br />
qed<br />
<br />
lemma ej_42: joslopjim4<br />
assumes "p∨q"<br />
"¬q" <br />
shows "p"<br />
using assms(1)<br />
proof <br />
assume "p"<br />
show "p" using `p` by this<br />
next<br />
assume "q"<br />
show "p" using assms(2) `q` by (rule notE)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 43. Demostrar<br />
p ∨ q, ¬p ⊢ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_43: carmarria<br />
assumes 1: "p ∨ q" and<br />
2: "¬p" <br />
shows "q"<br />
proof -<br />
have " p | q" using 1 by this<br />
moreover {assume 3: p<br />
have 4: False using 2 3 by (rule notE)<br />
have 5: q using 4 by (rule FalseE)<br />
}<br />
moreover {assume 6: q<br />
}<br />
ultimately show q by (rule disjE)<br />
qed<br />
<br />
lemma ej_43: joslopjim4<br />
assumes "p ∨ q"<br />
"¬p" <br />
shows "q"<br />
using assms(1)<br />
proof<br />
assume "p"<br />
show "q" using assms(2) `p` by (rule notE)<br />
next<br />
assume "q"<br />
show "q" using `q` by this<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 44. Demostrar<br />
p ∨ q ⊢ ¬(¬p ∧ ¬q)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_44: carmarria<br />
assumes 1: "p ∨ q" <br />
shows "¬(¬p ∧ ¬q)"<br />
proof -<br />
have "p | q" using 1 by this<br />
moreover {assume 2: p<br />
{assume 3:"~p & ~q"<br />
have 4: "~p" using 3 by (rule conjunct1)<br />
have 5: False using 4 2 by (rule notE)<br />
}<br />
hence 6: "~(~p & ~q)" by (rule notI)<br />
}<br />
moreover {assume 7: q<br />
{assume 8: "~p & ~q"<br />
have 9: "~q" using 8 by (rule conjunct2)<br />
have 10: False using 9 7 by (rule notE)<br />
}<br />
hence 11: "~(~p & ~q)" by (rule notI)<br />
}<br />
ultimately show "~(~p & ~q)" by (rule disjE)<br />
qed<br />
<br />
lemma ej_44: joslopjim4<br />
assumes "p∨q"<br />
shows "¬(¬p∧¬q)"<br />
proof (rule notI)<br />
assume "¬p∧¬q"<br />
show False<br />
using assms<br />
proof<br />
assume "p"<br />
have "¬p" using `¬p∧¬q` by (rule conjunct1)<br />
show False using `¬p` `p` by (rule notE)<br />
next<br />
assume "q"<br />
have "¬q" using `¬p∧¬q` by (rule conjunct2)<br />
show False using `¬q` `q` by (rule notE)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 45. Demostrar<br />
p ∧ q ⊢ ¬(¬p ∨ ¬q)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_45: carmarria<br />
assumes 1: "p ∧ q" <br />
shows "¬(¬p ∨ ¬q)"<br />
proof -<br />
{assume 2: "~p | ~q"<br />
moreover {assume 3: "~p"<br />
have 4: p using 1 by (rule conjunct1)<br />
have 5: False using 3 4 by (rule notE)<br />
}<br />
moreover {assume 6: "~q"<br />
have 7: q using 1 by (rule conjunct2)<br />
have 8: False using 6 7 by (rule notE)<br />
}<br />
ultimately have False by (rule disjE)<br />
}<br />
thus "~(~p | ~q)" by (rule notI)<br />
qed<br />
<br />
lemma ej_45: joslopjim4<br />
assumes "p∧q"<br />
shows "¬(¬p∨¬q)"<br />
proof (rule notI)<br />
assume "¬p∨¬q"<br />
then show False<br />
proof (rule disjE)<br />
assume "¬p"<br />
have "p" using assms by (rule conjunct1)<br />
show False using `¬p` `p` by (rule notE)<br />
next<br />
assume "¬q"<br />
have "q" using assms by (rule conjunct2)<br />
show False using `¬q` `q` by (rule notE)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 46. Demostrar<br />
¬(p ∨ q) ⊢ ¬p ∧ ¬q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_46: carmarria<br />
assumes 1: "¬(p ∨ q)" <br />
shows "¬p ∧ ¬q"<br />
proof -<br />
{assume 2: p<br />
have 3: "p | q" using 2 by (rule disjI1)<br />
have 4: False using 1 3 by (rule notE)<br />
}<br />
hence 5: "~p" by (rule notI)<br />
{assume 6: q<br />
have 7: "p | q" using 6 by (rule disjI2)<br />
have 8: False using 1 7 by (rule notE)<br />
}<br />
hence 9: "~q" by (rule notI)<br />
show "~p & ~q" using 5 9 by (rule conjI)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 47. Demostrar<br />
¬p ∧ ¬q ⊢ ¬(p ∨ q)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_47: carmarria<br />
assumes 1: "¬p ∧ ¬q" <br />
shows "¬(p ∨ q)"<br />
proof -<br />
{assume 2: "p | q"<br />
moreover {assume 3: p<br />
have 4: "~p" using 1 by (rule conjunct1)<br />
have 5: False using 4 3 by (rule notE)<br />
}<br />
moreover {assume 6: q <br />
have 7: "~q" using 1 by (rule conjunct2)<br />
have 8: False using 7 6 by (rule notE)<br />
}<br />
ultimately have 9: False by (rule disjE)<br />
}<br />
thus "~(p | q)" by (rule notI)<br />
qed <br />
<br />
lemma ej_47: joslopjim4<br />
assumes "¬p∧¬q"<br />
shows "¬(p∨q)"<br />
proof (rule notI)<br />
assume "p∨q"<br />
then show False<br />
proof<br />
assume "p"<br />
have "¬p" using assms by (rule conjunct1)<br />
show False using `¬p` `p` by (rule notE)<br />
next<br />
assume "q"<br />
have "¬q" using assms by (rule conjunct2)<br />
show False using `¬q` `q` by (rule notE)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 48. Demostrar<br />
¬p ∨ ¬q ⊢ ¬(p ∧ q)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_48: carmarria<br />
assumes 1: "¬p ∨ ¬q"<br />
shows "¬(p ∧ q)"<br />
proof -<br />
have "~p | ~q" using 1 by this<br />
moreover {assume 2: "~p"<br />
{assume 3: "p & q"<br />
have 4: p using 3 by (rule conjunct1)<br />
have 5: False using 2 4 by (rule notE)<br />
}<br />
hence 6: "~(p & q)" by (rule notI)<br />
}<br />
moreover {assume 7: "~q"<br />
{assume 8: "p & q"<br />
have 9: q using 8 by (rule conjunct2)<br />
have 10: False using 7 9 by (rule notE)<br />
}<br />
hence 11: "~(p & q)" by (rule notI)<br />
}<br />
ultimately show "~(p & q)" by (rule disjE)<br />
qed<br />
<br />
lemma ej_48: joslopjim4<br />
assumes "¬p ∨ ¬q"<br />
shows "¬(p ∧ q)"<br />
proof (rule notI)<br />
assume "p∧q"<br />
have "p" using `p∧q` by (rule conjunct1)<br />
have "q" using `p∧q` by (rule conjunct2)<br />
show False<br />
using assms<br />
proof<br />
assume "¬p"<br />
have "p" using `p∧q` by (rule conjunct1)<br />
show False using `¬p` `p` by (rule notE)<br />
next<br />
assume "¬q"<br />
have "q" using `p∧q` by (rule conjunct2)<br />
show False using `¬q` `q` by (rule notE)<br />
qed<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 49. Demostrar<br />
⊢ ¬(p ∧ ¬p)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_49: carmarria<br />
"¬(p ∧ ¬p)"<br />
proof -<br />
{assume 1: "p & ~p"<br />
have 2: p using 1 by (rule conjunct1)<br />
have 3: "~p" using 1 by (rule conjunct2)<br />
have 4: False using 3 2 by (rule notE)<br />
}<br />
thus "~(p & ~p)" by (rule notI)<br />
qed<br />
<br />
lemma ej_49: joslopjim4<br />
shows "¬(p∧¬p)"<br />
proof<br />
assume "p∧¬p"<br />
have "p" using `p∧¬p` by (rule conjunct1)<br />
have "¬p" using `p∧¬p` by (rule conjunct2)<br />
show False using `¬p` `p` by (rule notE)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 50. Demostrar<br />
p ∧ ¬p ⊢ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_50: carmarria<br />
assumes 1: "p ∧ ¬p" <br />
shows "q"<br />
proof -<br />
have 2: p using 1 by (rule conjunct1)<br />
have 3: "~p" using 1 by (rule conjunct2)<br />
have 4: False using 3 2 by (rule notE)<br />
show q using 4 by (rule FalseE)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 51. Demostrar<br />
¬¬p ⊢ p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_51: carmarria<br />
assumes "¬¬p"<br />
shows "p"<br />
proof -<br />
show p using assms(1) by (rule notnotD)<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 52. Demostrar<br />
⊢ p ∨ ¬p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_52: carmarria<br />
"p ∨ ¬p"<br />
proof -<br />
{assume 1: "~(p | ~p)"<br />
{assume 2: p <br />
have 3: "p | ~p" using 2 by (rule disjI1)<br />
have 4: False using 1 3 by (rule notE)<br />
}<br />
hence 5: "~p" by (rule notI)<br />
have 6: "p | ~p" using 5 by (rule disjI2)<br />
have 7: False using 1 6 by (rule notE)<br />
}<br />
hence 8: "~~(p | ~p)" by (rule notI)<br />
show "p | ~p" using 8 by (rule notnotD)<br />
qed<br />
<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 53. Demostrar<br />
⊢ ((p ⟶ q) ⟶ p) ⟶ p<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_53:<br />
"((p ⟶ q) ⟶ p) ⟶ p"<br />
oops<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 54. Demostrar<br />
¬q ⟶ ¬p ⊢ p ⟶ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_54:<br />
assumes "¬q ⟶ ¬p"<br />
shows "p ⟶ q"<br />
oops<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 55. Demostrar<br />
¬(¬p ∧ ¬q) ⊢ p ∨ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_55:<br />
assumes "¬(¬p ∧ ¬q)"<br />
shows "p ∨ q"<br />
oops<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 56. Demostrar<br />
¬(¬p ∨ ¬q) ⊢ p ∧ q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_56:<br />
assumes "¬(¬p ∨ ¬q)" <br />
shows "p ∧ q"<br />
oops<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 57. Demostrar<br />
¬(p ∧ q) ⊢ ¬p ∨ ¬q<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_57:<br />
assumes "¬(p ∧ q)"<br />
shows "¬p ∨ ¬q"<br />
oops<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 58. Demostrar<br />
⊢ (p ⟶ q) ∨ (q ⟶ p)<br />
------------------------------------------------------------------ *}<br />
<br />
lemma ejercicio_58:<br />
"(p ⟶ q) ∨ (q ⟶ p)"<br />
oops<br />
<br />
end<br />
<br />
</source></div>
Joslopjim4