LMF2015: Ejercicios de deducción natural en lógica proposicional con Isabelle/HOL
En la clase de hoy del curso Lógica matemática y fundamentos se han comentado soluciones de los ejercicios de deducción natural en lógica proposicional con Isabelle/HOL.
Para cada uno de los ejercicios se ha presentado distintas demostraciones: desde la detallada (que sea parecida a la mostrada en las transparencias) hasta la automática.
La teoría con la relación de ejercicios y sus soluciones es la siguiente
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header {* R3: Deducción natural en lógica proposicional con Isabelle/HOL *} theory R3 imports Main begin text {* --------------------------------------------------------------------- El objetivo de esta relación es lemas usando sólo las reglas básicas de deducción natural de la lógica proposicional. Los ejercicios son los de la asignatura de "Lógica informática" que se encuentran en http://goo.gl/yrPLn Las reglas básicas de la deducción natural son las siguientes: · conjI: ⟦P; Q⟧ ⟹ P ∧ Q · conjunct1: P ∧ Q ⟹ P · conjunct2: P ∧ Q ⟹ Q · notnotD: ¬¬ P ⟹ P · notnotI: P ⟹ ¬¬ P · mp: ⟦P ⟶ Q; P⟧ ⟹ Q · mt: ⟦F ⟶ G; ¬G⟧ ⟹ ¬F · impI: (P ⟹ Q) ⟹ P ⟶ Q · disjI1: P ⟹ P ∨ Q · disjI2: Q ⟹ P ∨ Q · disjE: ⟦P ∨ Q; P ⟹ R; Q ⟹ R⟧ ⟹ R · FalseE: False ⟹ P · notE: ⟦¬P; P⟧ ⟹ R · notI: (P ⟹ False) ⟹ ¬P · iffI: ⟦P ⟹ Q; Q ⟹ P⟧ ⟹ P = Q · iffD1: ⟦Q = P; Q⟧ ⟹ P · iffD2: ⟦P = Q; Q⟧ ⟹ P · ccontr: (¬P ⟹ False) ⟹ P · excluded_middle: ¬P ∨ P --------------------------------------------------------------------- *} text {* Se usarán las reglas notnotI y mt que demostramos a continuación. *} lemma notnotI: "P ⟹ ¬¬ P" by auto lemma mt: "⟦F ⟶ G; ¬G⟧ ⟹ ¬F" by auto section {* Implicaciones *} text {* --------------------------------------------------------------- Ejercicio 1. Demostrar p ⟶ q, p ⊢ q ------------------------------------------------------------------ *} -- "La demostración detallada es" lemma ejercicio_1_1: assumes 1: "p ⟶ q" and 2: "p" shows "q" proof - show "q" using 1 2 by (rule mp) qed -- "La demostración estructurada es" lemma ejercicio_1_2: assumes "p ⟶ q" "p" shows "q" proof - show "q" using assms .. qed -- "La demostración automática es" lemma ejercicio_1_3: assumes "p ⟶ q" "p" shows "q" using assms by auto text {* --------------------------------------------------------------- Ejercicio 2. Demostrar p ⟶ q, q ⟶ r, p ⊢ r ------------------------------------------------------------------ *} -- "La demostración detallada es" lemma ejercicio_2_1: assumes 1: "p ⟶ q" and 2: "q ⟶ r" and 3: "p" shows "r" proof - have 4: "q" using 1 3 by (rule mp) show "r" using 2 4 by (rule mp) qed -- "La demostración estructurada es" lemma ejercicio_2_2: assumes "p ⟶ q" "q ⟶ r" "p" shows "r" proof - have "q" using assms(1,3) .. show "r" using assms(2) `q` .. qed -- "La demostración automática es" lemma ejercicio_2_3: assumes "p ⟶ q" "q ⟶ r" "p" shows "r" using assms by auto text {* --------------------------------------------------------------- Ejercicio 3. Demostrar p ⟶ (q ⟶ r), p ⟶ q, p ⊢ r ------------------------------------------------------------------ *} -- "La demostración detallada es" lemma ejercicio_3_1: assumes 1: "p ⟶ (q ⟶ r)" and 2: "p ⟶ q" and 3: "p" shows "r" proof - have 4: "q" using 2 3 by (rule mp) have 5: "q ⟶ r" using 1 3 by (rule mp) show "r" using 5 4 by (rule mp) qed -- "La demostración estructurada es" lemma ejercicio_3_2: assumes "p ⟶ (q ⟶ r)" "p ⟶ q" "p" shows "r" proof - have "q" using assms(2,3) .. have "q ⟶ r" using assms(1,3) .. thus "r" using `q` .. qed -- "La demostración automática es" lemma ejercicio_3_3: assumes "p ⟶ (q ⟶ r)" "p ⟶ q" "p" shows "r" using assms by auto text {* --------------------------------------------------------------- Ejercicio 4. Demostrar p ⟶ q, q ⟶ r ⊢ p ⟶ r ------------------------------------------------------------------ *} -- "La demostración detallada es" lemma ejercicio_4_1: assumes 1: "p ⟶ q" and 2: "q ⟶ r" shows "p ⟶ r" proof (rule impI) assume 3: "p" have 4: "q" using 1 3 by (rule mp) show "r" using 2 4 by (rule mp) qed -- "La demostración estructurada es" lemma ejercicio_4_2: assumes "p ⟶ q" "q ⟶ r" shows "p ⟶ r" proof assume "p" with assms(1) have "q" .. with assms(2) show "r" .. qed -- "La demostración automática es" lemma ejercicio_4_3: assumes "p ⟶ q" "q ⟶ r" shows "p ⟶ r" using assms by auto text {* --------------------------------------------------------------- Ejercicio 5. Demostrar p ⟶ (q ⟶ r) ⊢ q ⟶ (p ⟶ r) ------------------------------------------------------------------ *} -- "La demostración detallada es" lemma ejercicio_5_1: assumes 1: "p ⟶ (q ⟶ r)" shows "q ⟶ (p ⟶ r)" proof (rule impI) assume 2: "q" show "p ⟶ r" proof (rule impI) assume 3: "p" have "q ⟶ r" using 1 3 by (rule mp) thus "r" using 2 by (rule mp) qed qed -- "La demostración estructurada es" lemma ejercicio_5_2: assumes "p ⟶ (q ⟶ r)" shows "q ⟶ (p ⟶ r)" proof assume "q" show "p ⟶ r" proof assume "p" with assms(1) have "q ⟶ r" .. thus "r" using `q` .. qed qed -- "La demostración automática es" lemma ejercicio_5_3: assumes "p ⟶ (q ⟶ r)" shows "q ⟶ (p ⟶ r)" using assms by auto text {* --------------------------------------------------------------- Ejercicio 6. Demostrar p ⟶ (q ⟶ r) ⊢ (p ⟶ q) ⟶ (p ⟶ r) ------------------------------------------------------------------ *} -- "La demostración detallada es" lemma ejercicio_6_1: assumes 1: "p ⟶ (q ⟶ r)" shows "(p ⟶ q) ⟶ (p ⟶ r)" proof (rule impI) assume 2: "p ⟶ q" show "p ⟶ r" proof (rule impI) assume 3: "p" have 4: "q" using 2 3 by (rule mp) have 5: "q ⟶ r" using 1 3 by (rule mp) show "r" using 5 4 by (rule mp) qed qed -- "La demostración estructurada es" lemma ejercicio_6_2: assumes "p ⟶ (q ⟶ r)" shows "(p ⟶ q) ⟶ (p ⟶ r)" proof assume "p ⟶ q" show "p ⟶ r" proof assume "p" with `p ⟶ q` have "q" .. have "q ⟶ r" using assms(1) `p` .. thus "r" using `q` .. qed qed -- "La demostración automática es" lemma ejercicio_6_3: assumes "p ⟶ (q ⟶ r)" shows "(p ⟶ q) ⟶ (p ⟶ r)" using assms by auto text {* --------------------------------------------------------------- Ejercicio 7. Demostrar p ⊢ q ⟶ p ------------------------------------------------------------------ *} -- "La demostración detallada es" lemma ejercicio_7_1: assumes 1: "p" shows "q ⟶ p" proof (rule impI) assume 2: "q" show "p" using 1 by this qed -- "La demostración estructurada es" lemma ejercicio_7_2: assumes "p" shows "q ⟶ p" proof assume "q" show "p" using assms(1) . qed -- "La demostración automática es" lemma ejercicio_7_3: assumes "p" shows "q ⟶ p" using assms by auto text {* --------------------------------------------------------------- Ejercicio 8. Demostrar ⊢ p ⟶ (q ⟶ p) ------------------------------------------------------------------ *} -- "La demostración detallada es" lemma ejercicio_8_1: "p ⟶ (q ⟶ p)" proof (rule impI) assume 1: "p" show "q ⟶ p" proof (rule impI) assume 2: "q" show "p" using 1 by this qed qed -- "La demostración estructurada es" lemma ejercicio_8_2: "p ⟶ (q ⟶ p)" proof assume "p" show "q ⟶ p" proof assume "q" show "p" using `p` . qed qed -- "La demostración automática es" lemma ejercicio_8_3: "p ⟶ (q ⟶ p)" by auto text {* --------------------------------------------------------------- Ejercicio 9. Demostrar p ⟶ q ⊢ (q ⟶ r) ⟶ (p ⟶ r) ------------------------------------------------------------------ *} -- "La demostración detallada es" lemma ejercicio_9_1: assumes 1: "p ⟶ q" shows "(q ⟶ r) ⟶ (p ⟶ r)" proof (rule impI) assume 2: "q ⟶ r" show "p ⟶ r" proof (rule impI) assume 3: "p" have 4: "q" using 1 3 by (rule mp) show "r" using 2 4 by (rule mp) qed qed -- "La demostración estructurada es" lemma ejercicio_9_2: assumes "p ⟶ q" shows "(q ⟶ r) ⟶ (p ⟶ r)" proof assume "q ⟶ r" show "p ⟶ r" proof assume "p" with assms(1) have "q" .. with `q ⟶ r` show "r" .. qed qed -- "La demostración automática es" lemma ejercicio_9_3: assumes "p ⟶ q" shows "(q ⟶ r) ⟶ (p ⟶ r)" using assms by auto text {* --------------------------------------------------------------- Ejercicio 10. Demostrar p ⟶ (q ⟶ (r ⟶ s)) ⊢ r ⟶ (q ⟶ (p ⟶ s)) ------------------------------------------------------------------ *} -- "La demostración detallada es" lemma ejercicio_10_1: assumes 1: "p ⟶ (q ⟶ (r ⟶ s))" shows "r ⟶ (q ⟶ (p ⟶ s))" proof - { assume 2: "r" { assume 3: "q" { assume 4: "p" have 5: "q ⟶ (r ⟶ s)" using 1 4 by (rule mp) have 6: "r ⟶ s" using 5 3 by (rule mp) have 7: "s" using 6 2 by (rule mp) } hence 8: "p ⟶ s" by (rule impI) } hence 9: "q ⟶ (p ⟶ s)" by (rule impI) } thus 10: "r ⟶ (q ⟶ (p ⟶ s))" by (rule impI) qed -- "Una variante de la demostración anterior es" lemma ejercicio_10_1b: assumes 1: "p ⟶ (q ⟶ (r ⟶ s))" shows "r ⟶ (q ⟶ (p ⟶ s))" proof (rule impI) assume 2: "r" show "q ⟶ (p ⟶ s)" proof (rule impI) assume 3: "q" show "p ⟶ s" proof (rule impI) assume 4: "p" have 5: "q ⟶ (r ⟶ s)" using 1 4 by (rule mp) have 6: "r ⟶ s" using 5 3 by (rule mp) show "s" using 6 2 by (rule mp) qed qed qed -- "La demostración estructurada es" lemma ejercicio_10_2: assumes "p ⟶ (q ⟶ (r ⟶ s))" shows "r ⟶ (q ⟶ (p ⟶ s))" proof assume "r" show "q ⟶ (p ⟶ s)" proof assume "q" show "p ⟶ s" proof assume "p" with assms(1) have "q ⟶ (r ⟶ s)" .. hence "r ⟶ s" using `q` .. thus "s" using `r` .. qed qed qed -- "La demostración automática es" lemma ejercicio_10_3: assumes "p ⟶ (q ⟶ (r ⟶ s))" shows "r ⟶ (q ⟶ (p ⟶ s))" using assms by auto text {* --------------------------------------------------------------- Ejercicio 11. Demostrar ⊢ (p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r)) ------------------------------------------------------------------ *} -- "La demostración detallada es" lemma ejercicio_11_1: "(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))" proof - { assume 1: "p ⟶ (q ⟶ r)" { assume 2: "p ⟶ q" { assume 3: "p" have 4: "q" using 2 3 by (rule mp) have 5: "q ⟶ r" using 1 3 by (rule mp) have 6: "r" using 5 4 by (rule mp) } hence 7: "p ⟶ r" by (rule impI) } hence 8: "(p ⟶ q) ⟶ (p ⟶ r)" by (rule impI) } thus "(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))" by (rule impI) qed -- "La demostración estructurada es" lemma ejercicio_11_2: "(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))" proof assume "p ⟶ (q ⟶ r)" { assume "p ⟶ q" { assume "p" with `p ⟶ q` have "q" .. have "q ⟶ r" using `p ⟶ (q ⟶ r)` `p` .. hence "r" using `q` .. } hence "p ⟶ r" .. } thus "(p ⟶ q) ⟶ (p ⟶ r)" .. qed -- "La demostración automática es" lemma ejercicio_11_3: "(p ⟶ (q ⟶ r)) ⟶ ((p ⟶ q) ⟶ (p ⟶ r))" by auto text {* --------------------------------------------------------------- Ejercicio 12. Demostrar (p ⟶ q) ⟶ r ⊢ p ⟶ (q ⟶ r) ------------------------------------------------------------------ *} -- "La demostración detallada es" lemma ejercicio_12_1: assumes 1: "(p ⟶ q) ⟶ r" shows "p ⟶ (q ⟶ r)" proof - { assume 2: "p" { assume 3: "q" { assume 4: "p" have 5: "q" using 3 by this } hence 6: "p ⟶ q" by (rule impI) have 7: "r" using 1 6 by (rule mp) } hence 8: "q ⟶ r" by (rule impI) } thus 9: "p ⟶ (q ⟶ r)" by (rule impI) qed -- "La demostración estructurada es" lemma ejercicio_12_2: assumes "(p ⟶ q) ⟶ r" shows "p ⟶ (q ⟶ r)" proof assume "p" { assume "q" { assume "p" have "q" using `q` . } hence "p ⟶ q" .. with assms(1) have "r" .. } thus "q ⟶ r" .. qed -- "La demostración automática es" lemma ejercicio_12_3: assumes "(p ⟶ q) ⟶ r" shows "p ⟶ (q ⟶ r)" using assms by auto section {* Conjunciones *} text {* --------------------------------------------------------------- Ejercicio 13. Demostrar p, q ⊢ p ∧ q ------------------------------------------------------------------ *} -- "La demostración detallada es" lemma ejercicio_13_1: assumes 1: "p" and 2: "q" shows "p ∧ q" proof - show "p ∧ q" using assms by (rule conjI) qed -- "La demostración estructurada es" lemma ejercicio_13_2: assumes "p" "q" shows "p ∧ q" proof show "p" using assms(1) . next show "q" using assms(2) . qed -- "La demostración automática es" lemma ejercicio_13_3: assumes "p" "q" shows "p ∧ q" using assms by auto text {* --------------------------------------------------------------- Ejercicio 14. Demostrar p ∧ q ⊢ p ------------------------------------------------------------------ *} -- "La demostración detallada es" lemma ejercicio_14_1: assumes 1: "p ∧ q" shows "p" proof - show "p" using 1 by (rule conjunct1) qed -- "La demostración estructurada es" lemma ejercicio_14_2: assumes "p ∧ q" shows "p" proof - show "p" using assms .. qed -- "La demostración automática es" lemma ejercicio_14_3: assumes "p ∧ q" shows "p" using assms by auto text {* --------------------------------------------------------------- Ejercicio 15. Demostrar p ∧ q ⊢ q ------------------------------------------------------------------ *} -- "La demostración detallada es" lemma ejercicio_15_1: assumes 1: "p ∧ q" shows "q" proof - show "q" using 1 by (rule conjunct2) qed -- "La demostración estructurada es" lemma ejercicio_15_2: assumes "p ∧ q" shows "q" proof - show "q" using assms .. qed -- "La demostración automática es" lemma ejercicio_15_3: assumes "p ∧ q" shows "q" using assms by auto text {* --------------------------------------------------------------- Ejercicio 16. Demostrar p ∧ (q ∧ r) ⊢ (p ∧ q) ∧ r ------------------------------------------------------------------ *} -- "La demostración detallada es" lemma ejercicio_16_1: assumes 1: "p ∧ (q ∧ r)" shows "(p ∧ q) ∧ r" proof - have 2: "p" using 1 by (rule conjunct1) have 3: "q ∧ r" using 1 by (rule conjunct2) have 4: "q" using 3 by (rule conjunct1) have 5: "p ∧ q" using 2 4 by (rule conjI) have 6: "r" using 3 by (rule conjunct2) show 7: "(p ∧ q) ∧ r" using 5 6 by (rule conjI) qed -- "La demostración estructurada es" lemma ejercicio_16_2: assumes "p ∧ (q ∧ r)" shows "(p ∧ q) ∧ r" proof show "p ∧ q" proof show "p" using assms .. next have "q ∧ r" using assms .. thus "q" .. qed next have "q ∧ r" using assms .. thus "r" .. qed -- "La demostración automática es" lemma ejercicio_16_3: assumes "p ∧ (q ∧ r)" shows "(p ∧ q) ∧ r" using assms by auto text {* --------------------------------------------------------------- Ejercicio 17. Demostrar (p ∧ q) ∧ r ⊢ p ∧ (q ∧ r) ------------------------------------------------------------------ *} -- "La demostración detallada es" lemma ejercicio_17_1: assumes 1: "(p ∧ q) ∧ r" shows "p ∧ (q ∧ r)" proof - have 2: "p ∧ q" using 1 by (rule conjunct1) have 3: "p" using 2 by (rule conjunct1) have 4: "q" using 2 by (rule conjunct2) have 5: "r" using 1 by (rule conjunct2) have 6: "q ∧ r" using 4 5 by (rule conjI) show 7: "p ∧ (q ∧ r)" using 3 6 by (rule conjI) qed -- "La demostración estructurada es" lemma ejercicio_17_2: assumes "(p ∧ q) ∧ r" shows "p ∧ (q ∧ r)" proof have "p ∧ q" using assms .. thus "p" .. next show "q ∧ r" proof have "p ∧ q" using assms .. thus "q" .. next show "r" using assms .. qed qed -- "La demostración automática es" lemma ejercicio_17_3: assumes "(p ∧ q) ∧ r" shows "p ∧ (q ∧ r)" using assms by auto text {* --------------------------------------------------------------- Ejercicio 18. Demostrar p ∧ q ⊢ p ⟶ q ------------------------------------------------------------------ *} -- "La demostración detallada es" lemma ejercicio_18_1: assumes 1: "p ∧ q" shows "p ⟶ q" proof (rule impI) assume 2: "p" show "q" using 1 by (rule conjunct2) qed -- "La demostración estructurada es" lemma ejercicio_18_2: assumes "p ∧ q" shows "p ⟶ q" proof assume "p" show "q" using assms .. qed -- "La demostración automática es" lemma ejercicio_18_3: assumes "p ∧ q" shows "p ⟶ q" using assms by auto text {* --------------------------------------------------------------- Ejercicio 19. Demostrar (p ⟶ q) ∧ (p ⟶ r) ⊢ p ⟶ q ∧ r ------------------------------------------------------------------ *} -- "La demostración detallada es" lemma ejercicio_19_1: assumes 1: "(p ⟶ q) ∧ (p ⟶ r)" shows "p ⟶ q ∧ r" proof (rule impI) assume 2: "p" have 3: "p ⟶ q" using 1 by (rule conjunct1) have 4: "q" using 3 2 by (rule mp) have 5: "p ⟶ r" using 1 by (rule conjunct2) have 6: "r" using 5 2 by (rule mp) show 7: "q ∧ r" using 4 6 by (rule conjI) qed -- "La demostración estructurada es" lemma ejercicio_19_2: assumes "(p ⟶ q) ∧ (p ⟶ r)" shows "p ⟶ q ∧ r" proof assume "p" show "q ∧ r" proof have "p ⟶ q" using assms .. thus "q" using `p` .. next have "p ⟶ r" using assms .. thus "r" using `p` .. qed qed -- "La demostración automática es" lemma ejercicio_19_3: assumes "(p ⟶ q) ∧ (p ⟶ r)" shows "p ⟶ q ∧ r" using assms by auto text {* --------------------------------------------------------------- Ejercicio 20. Demostrar p ⟶ q ∧ r ⊢ (p ⟶ q) ∧ (p ⟶ r) ------------------------------------------------------------------ *} -- "La demostración detallada es" lemma ejercicio_20_1: assumes 1: "p ⟶ q ∧ r" shows "(p ⟶ q) ∧ (p ⟶ r)" proof (rule conjI) show "p ⟶ q" proof (rule impI) assume 2: "p" have "q ∧ r" using 1 2 by (rule mp) thus "q" by (rule conjunct1) qed next show "p ⟶ r" proof (rule impI) assume 3: "p" have "q ∧ r" using 1 3 by (rule mp) thus "r" by (rule conjunct2) qed qed -- "La demostración estructurada es" lemma ejercicio_20_2: assumes "p ⟶ q ∧ r" shows "(p ⟶ q) ∧ (p ⟶ r)" proof show "p ⟶ q" proof assume "p" have "q ∧ r" using assms `p` .. thus "q" .. qed next show "p ⟶ r" proof assume "p" have "q ∧ r" using assms `p` .. thus "r" .. qed qed -- "La demostración automática es" lemma ejercicio_20_3: assumes "p ⟶ q ∧ r" shows "(p ⟶ q) ∧ (p ⟶ r)" using assms by auto text {* --------------------------------------------------------------- Ejercicio 21. Demostrar p ⟶ (q ⟶ r) ⊢ p ∧ q ⟶ r ------------------------------------------------------------------ *} -- "La demostración detallada es" lemma ejercicio_21_1: assumes 1: "p ⟶ (q ⟶ r)" shows "p ∧ q ⟶ r" proof (rule impI) assume 2: "p ∧ q" have 3: "p" using 2 by (rule conjunct1) have 4: "q" using 2 by (rule conjunct2) have 5: "q ⟶ r" using 1 3 by (rule mp) show 6: "r" using 5 4 by (rule mp) qed -- "La demostración estructurada es" lemma ejercicio_21_2: assumes "p ⟶ (q ⟶ r)" shows "p ∧ q ⟶ r" proof assume "p ∧ q" hence "p" .. have "q" using `p ∧ q` .. have "q ⟶ r" using assms(1) `p` .. thus "r" using `q` .. qed -- "La demostración automática es" lemma ejercicio_21_3: assumes "p ⟶ (q ⟶ r)" shows "p ∧ q ⟶ r" using assms by auto text {* --------------------------------------------------------------- Ejercicio 22. Demostrar p ∧ q ⟶ r ⊢ p ⟶ (q ⟶ r) ------------------------------------------------------------------ *} -- "La demostración detallada es" lemma ejercicio_22_1: assumes 1: "p ∧ q ⟶ r" shows "p ⟶ (q ⟶ r)" proof (rule impI) assume 2: "p" { assume 3: "q" have 4: "p ∧ q" using 2 3 by (rule conjI) have "r" using 1 4 by (rule mp) } thus "q ⟶ r" by (rule impI) qed -- "La demostración estructurada es" lemma ejercicio_22_2: assumes "p ∧ q ⟶ r" shows "p ⟶ (q ⟶ r)" proof assume "p" show "q ⟶ r" proof assume "q" with `p` have "p ∧ q" .. with assms(1) show "r" .. qed qed -- "La demostración automática es" lemma ejercicio_22_3: assumes "p ∧ q ⟶ r" shows "p ⟶ (q ⟶ r)" using assms by auto text {* --------------------------------------------------------------- Ejercicio 23. Demostrar (p ⟶ q) ⟶ r ⊢ p ∧ q ⟶ r ------------------------------------------------------------------ *} -- "La demostración detallada es" lemma ejercicio_23_1: assumes 1: "(p ⟶ q) ⟶ r" shows "p ∧ q ⟶ r" proof (rule impI) assume 2: "p ∧ q" have 3: "p ⟶ q" proof (rule impI) assume 4: "p" show "q" using 2 by (rule conjunct2) qed show "r" using 1 3 by (rule mp) qed -- "La demostración estructurada es" lemma ejercicio_23_2: assumes "(p ⟶ q) ⟶ r" shows "p ∧ q ⟶ r" proof assume "p ∧ q" have "p ⟶ q" proof assume "p" show "q" using `p ∧ q` .. qed with assms show "r" .. qed -- "La demostración automática es" lemma ejercicio_23_3: assumes "(p ⟶ q) ⟶ r" shows "p ∧ q ⟶ r" using assms by auto text {* --------------------------------------------------------------- Ejercicio 24. Demostrar p ∧ (q ⟶ r) ⊢ (p ⟶ q) ⟶ r ------------------------------------------------------------------ *} -- "La demostración detallada es" lemma ejercicio_24_1: assumes 1: "p ∧ (q ⟶ r)" shows "(p ⟶ q) ⟶ r" proof (rule impI) assume 2: "p ⟶ q" have 3: "p" using 1 by (rule conjunct1) have 4: "q" using 2 3 by (rule mp) have 5: "q ⟶ r" using 1 by (rule conjunct2) thus "r" using 4 by (rule mp) qed -- "La demostración estructurada es" lemma ejercicio_24_2: assumes "p ∧ (q ⟶ r)" shows "(p ⟶ q) ⟶ r" proof assume "p ⟶ q" have "p" using assms .. with `p ⟶ q` have "q" .. have "q ⟶ r" using assms .. thus "r" using `q` .. qed -- "La demostración automática es" lemma ejercicio_24_3: assumes "p ∧ (q ⟶ r)" shows "(p ⟶ q) ⟶ r" using assms by auto section {* Disyunciones *} text {* --------------------------------------------------------------- Ejercicio 25. Demostrar p ⊢ p ∨ q ------------------------------------------------------------------ *} -- "La demostración detallada es" lemma ejercicio_25_1: assumes 1: "p" shows "p ∨ q" proof - show "p ∨ q" using 1 by (rule disjI1) qed -- "La demostración estructurada es" lemma ejercicio_25_2: assumes "p" shows "p ∨ q" proof - show "p ∨ q" using assms .. qed -- "La demostración estructurada es" lemma ejercicio_25_3: assumes "p" shows "p ∨ q" using assms by auto text {* --------------------------------------------------------------- Ejercicio 26. Demostrar q ⊢ p ∨ q ------------------------------------------------------------------ *} -- "La demostración detallada es" lemma ejercicio_26_1: assumes 1: "q" shows "p ∨ q" proof - show "p ∨ q" using 1 by (rule disjI2) qed -- "La demostración estructurada es" lemma ejercicio_26_2: assumes "q" shows "p ∨ q" proof - show "p ∨ q" using assms .. qed -- "La demostración estructurada es" lemma ejercicio_26_3: assumes "q" shows "p ∨ q" using assms by auto text {* --------------------------------------------------------------- Ejercicio 27. Demostrar p ∨ q ⊢ q ∨ p ------------------------------------------------------------------ *} -- "La demostración detallada es" lemma ejercicio_27_1: assumes 1: "p ∨ q" shows "q ∨ p" using 1 proof (rule disjE) assume "p" thus "q ∨ p" by (rule disjI2) next assume "q" thus "q ∨ p" by (rule disjI1) qed -- "La demostración estructurada es" lemma ejercicio_27_2: assumes "p ∨ q" shows "q ∨ p" using assms proof assume "p" thus "q ∨ p" .. next assume "q" thus "q ∨ p" .. qed -- "La demostración estructurada es" lemma ejercicio_27_3: assumes "p ∨ q" shows "q ∨ p" using assms by auto text {* --------------------------------------------------------------- Ejercicio 28. Demostrar q ⟶ r ⊢ p ∨ q ⟶ p ∨ r ------------------------------------------------------------------ *} -- "La demostración detallada es" lemma ejercicio_28_1: assumes 1: "q ⟶ r" shows "p ∨ q ⟶ p ∨ r" proof (rule impI) assume "p ∨ q" thus "p ∨ r" proof (rule disjE) assume "p" thus "p ∨ r" by (rule disjI1) next assume "q" have "r" using assms `q` by (rule mp) thus "p ∨ r" by (rule disjI2) qed qed -- "La demostración estructurada es" lemma ejercicio_28_2: assumes "q ⟶ r" shows "p ∨ q ⟶ p ∨ r" proof assume "p ∨ q" thus "p ∨ r" proof assume "p" thus "p ∨ r" .. next assume "q" have "r" using assms `q` .. thus "p ∨ r" .. qed qed -- "La demostración automática es" lemma ejercicio_28_3: assumes "q ⟶ r" shows "p ∨ q ⟶ p ∨ r" using assms by auto text {* --------------------------------------------------------------- Ejercicio 29. Demostrar p ∨ p ⊢ p ------------------------------------------------------------------ *} -- "La demostración detallada es" lemma ejercicio_29_1: assumes 1: "p ∨ p" shows "p" using 1 proof (rule disjE) assume "p" thus "p" by this next assume "p" thus "p" by this qed -- "La demostración estructurada es" lemma ejercicio_29_2: assumes "p ∨ p" shows "p" using assms proof assume "p" thus "p" . next assume "p" thus "p" . qed -- "La demostración estructurada es" lemma ejercicio_29_3: assumes "p ∨ p" shows "p" using assms by auto text {* --------------------------------------------------------------- Ejercicio 30. Demostrar p ⊢ p ∨ p ------------------------------------------------------------------ *} -- "La demostración detallada es" lemma ejercicio_30_1: assumes 1: "p" shows "p ∨ p" proof - show "p ∨ p" using 1 by (rule disjI1) qed -- "La demostración estructurada es" lemma ejercicio_30_2: assumes "p" shows "p ∨ p" proof - show "p ∨ p" using assms .. qed -- "La demostración estructurada es" lemma ejercicio_30_3: assumes "p" shows "p ∨ p" using assms by auto text {* --------------------------------------------------------------- Ejercicio 31. Demostrar p ∨ (q ∨ r) ⊢ (p ∨ q) ∨ r ------------------------------------------------------------------ *} -- "La demostración detallada es" lemma ejercicio_31_1: assumes 1: "p ∨ (q ∨ r)" shows "(p ∨ q) ∨ r" using 1 proof (rule disjE) assume "p" hence "p ∨ q" by (rule disjI1) thus "(p ∨ q) ∨ r" by (rule disjI1) next assume "q ∨ r" thus "(p ∨ q) ∨ r" proof (rule disjE) assume "q" hence "p ∨ q" by (rule disjI2) thus "(p ∨ q) ∨ r" by (rule disjI1) next assume "r" thus "(p ∨ q) ∨ r" by (rule disjI2) qed qed -- "La demostración estructurada es" lemma ejercicio_31_2: assumes "p ∨ (q ∨ r)" shows "(p ∨ q) ∨ r" using assms proof assume "p" hence "p ∨ q" .. thus "(p ∨ q) ∨ r" .. next assume "q ∨ r" thus "(p ∨ q) ∨ r" proof assume "q" hence "p ∨ q" .. thus "(p ∨ q) ∨ r" .. next assume "r" thus "(p ∨ q) ∨ r" .. qed qed -- "La demostración estructurada es" lemma ejercicio_31_3: assumes "p ∨ (q ∨ r)" shows "(p ∨ q) ∨ r" using assms by auto text {* --------------------------------------------------------------- Ejercicio 32. Demostrar (p ∨ q) ∨ r ⊢ p ∨ (q ∨ r) ------------------------------------------------------------------ *} -- "La demostración detallada es" lemma ejercicio_32_1: assumes 1: "(p ∨ q) ∨ r" shows "p ∨ (q ∨ r)" using 1 proof (rule disjE) assume "p ∨ q" thus "p ∨ (q ∨ r)" proof (rule disjE) assume "p" thus "p ∨ (q ∨ r)" by (rule disjI1) next assume "q" hence "q ∨ r" by (rule disjI1) thus "p ∨ (q ∨ r)" by (rule disjI2) qed next assume "r" hence "q ∨ r" by (rule disjI2) thus "p ∨ (q ∨ r)" by (rule disjI2) qed -- "La demostración estructurada es" lemma ejercicio_32_2: assumes "(p ∨ q) ∨ r" shows "p ∨ (q ∨ r)" using assms proof assume "p ∨ q" thus "p ∨ (q ∨ r)" proof assume "p" thus "p ∨ (q ∨ r)" .. next assume "q" hence "q ∨ r" .. thus "p ∨ (q ∨ r)" .. qed next assume "r" hence "q ∨ r" .. thus "p ∨ (q ∨ r)" .. qed -- "La demostración estructurada es" lemma ejercicio_32_3: assumes "(p ∨ q) ∨ r" shows "p ∨ (q ∨ r)" using assms by auto text {* --------------------------------------------------------------- Ejercicio 33. Demostrar p ∧ (q ∨ r) ⊢ (p ∧ q) ∨ (p ∧ r) ------------------------------------------------------------------ *} -- "La demostración detallada es" lemma ejercicio_33_1: assumes 1: "p ∧ (q ∨ r)" shows "(p ∧ q) ∨ (p ∧ r)" proof - have 2: "p" using 1 by (rule conjunct1) have 3: "q ∨ r" using 1 by (rule conjunct2) thus "(p ∧ q) ∨ (p ∧ r)" proof (rule disjE) assume 4: "q" have "p ∧ q" using 2 4 by (rule conjI) thus "(p ∧ q) ∨ (p ∧ r)" by (rule disjI1) next assume 5: "r" have "q ∨ r" using 1 by (rule conjunct2) thus "(p ∧ q) ∨ (p ∧ r)" proof (rule disjE) assume "q" have "p ∧ q" using `p` `q` by (rule conjI) thus "(p ∧ q) ∨ (p ∧ r)" by (rule disjI1) next assume "r" have "p ∧ r" using `p` `r` by (rule conjI) thus "(p ∧ q) ∨ (p ∧ r)" by (rule disjI2) qed qed qed -- "La demostración estructurada es" lemma ejercicio_33_2: assumes "p ∧ (q ∨ r)" shows "(p ∧ q) ∨ (p ∧ r)" proof - have "p" using assms .. have "q ∨ r" using assms .. thus "(p ∧ q) ∨ (p ∧ r)" proof assume "q" with `p` have "p ∧ q" .. thus "(p ∧ q) ∨ (p ∧ r)" .. next assume "r" hence "q ∨ r" .. thus "(p ∧ q) ∨ (p ∧ r)" proof assume "q" with `p` have "p ∧ q" .. thus "(p ∧ q) ∨ (p ∧ r)" .. next assume "r" with `p` have "p ∧ r" .. thus "(p ∧ q) ∨ (p ∧ r)" .. qed qed qed -- "La demostración automática es" lemma ejercicio_33_3: assumes "p ∧ (q ∨ r)" shows "(p ∧ q) ∨ (p ∧ r)" using assms by auto text {* --------------------------------------------------------------- Ejercicio 34. Demostrar (p ∧ q) ∨ (p ∧ r) ⊢ p ∧ (q ∨ r) ------------------------------------------------------------------ *} -- "La demostración detallada es" lemma ejercicio_34_1: assumes "(p ∧ q) ∨ (p ∧ r)" shows "p ∧ (q ∨ r)" using assms proof (rule disjE) assume 1: "p ∧ q" show "p ∧ (q ∨ r)" proof (rule conjI) show "p" using 1 by (rule conjunct1) next have "q" using 1 by (rule conjunct2) thus "q ∨ r" by (rule disjI1) qed next assume 2: "p ∧ r" hence 3: "p" by (rule conjunct1) have "r" using 2 by (rule conjunct2) hence 4: "q ∨ r" by (rule disjI2) show "p ∧ (q ∨ r)" using 3 4 by (rule conjI) qed -- "La demostración estructurada es" lemma ejercicio_34_2: assumes "(p ∧ q) ∨ (p ∧ r)" shows "p ∧ (q ∨ r)" using assms proof assume "p ∧ q" show "p ∧ (q ∨ r)" proof show "p" using `p ∧ q` .. next have "q" using `p ∧ q` .. thus "q ∨ r" .. qed next assume "p ∧ r" hence "p" .. have "r" using `p ∧ r` .. hence "q ∨ r" .. with `p` show "p ∧ (q ∨ r)" .. qed -- "La demostración estructurada es" lemma ejercicio_34_3: assumes "(p ∧ q) ∨ (p ∧ r)" shows "p ∧ (q ∨ r)" using assms by auto text {* --------------------------------------------------------------- Ejercicio 35. Demostrar p ∨ (q ∧ r) ⊢ (p ∨ q) ∧ (p ∨ r) ------------------------------------------------------------------ *} -- "La demostración estructurada es" lemma ejercicio_35_1: assumes "p ∨ (q ∧ r)" shows "(p ∨ q) ∧ (p ∨ r)" using assms proof assume "p" show "(p ∨ q) ∧ (p ∨ r)" proof show "p ∨ q" using `p` .. next show "p ∨ r" using `p` .. qed next assume "q ∧ r" show "(p ∨ q) ∧ (p ∨ r)" proof have "q" using `q ∧ r` .. thus "p ∨ q" .. next have "r" using `q ∧ r` .. thus "p ∨ r" .. qed qed -- "La demostración detallada es" lemma ejercicio_35_2: assumes "p ∨ (q ∧ r)" shows "(p ∨ q) ∧ (p ∨ r)" using assms proof (rule disjE) assume "p" show "(p ∨ q) ∧ (p ∨ r)" proof (rule conjI) show "p ∨ q" using `p` by (rule disjI1) next show "p ∨ r" using `p` by (rule disjI1) qed next assume "q ∧ r" show "(p ∨ q) ∧ (p ∨ r)" proof (rule conjI) have "q" using `q ∧ r` .. thus "p ∨ q" by (rule disjI2) next have "r" using `q ∧ r` by (rule conjunct2) thus "p ∨ r" by (rule disjI2) qed qed -- "La demostración automática es" lemma ejercicio_35_3: assumes "p ∨ (q ∧ r)" shows "(p ∨ q) ∧ (p ∨ r)" using assms by auto text {* --------------------------------------------------------------- Ejercicio 36. Demostrar (p ∨ q) ∧ (p ∨ r) ⊢ p ∨ (q ∧ r) ------------------------------------------------------------------ *} -- "La demostración estructurada es" lemma ejercicio_36_1: assumes "(p ∨ q) ∧ (p ∨ r)" shows "p ∨ (q ∧ r)" proof - have "p ∨ q" using assms .. thus "p ∨ (q ∧ r)" proof assume "p" thus "p ∨ (q ∧ r)" .. next assume "q" have "p ∨ r" using assms .. thus "p ∨ (q ∧ r)" proof assume "p" thus "p ∨ (q ∧ r)" .. next assume "r" with `q` have "q ∧ r" .. thus "p ∨ (q ∧ r)" .. qed qed qed -- "La demostración detallada es" lemma ejercicio_36_2: assumes "(p ∨ q) ∧ (p ∨ r)" shows "p ∨ (q ∧ r)" proof - have "p ∨ q" using assms by (rule conjunct1) thus "p ∨ (q ∧ r)" proof (rule disjE) assume "p" thus "p ∨ (q ∧ r)" by (rule disjI1) next assume "q" have "p ∨ r" using assms by (rule conjunct2) thus "p ∨ (q ∧ r)" proof (rule disjE) assume "p" thus "p ∨ (q ∧ r)" by (rule disjI1) next assume "r" have "q ∧ r" using `q` `r` by (rule conjI) thus "p ∨ (q ∧ r)" by (rule disjI2) qed qed qed -- "La demostración automática es" lemma ejercicio_36_3: assumes "(p ∨ q) ∧ (p ∨ r)" shows "p ∨ (q ∧ r)" using assms by auto text {* --------------------------------------------------------------- Ejercicio 37. Demostrar (p ⟶ r) ∧ (q ⟶ r) ⊢ p ∨ q ⟶ r ------------------------------------------------------------------ *} -- "La demostración estructurada es" lemma ejercicio_37_1: assumes "(p ⟶ r) ∧ (q ⟶ r)" shows "p ∨ q ⟶ r" proof assume "p ∨ q" thus "r" proof assume "p" have "p ⟶ r" using assms .. thus "r" using `p` .. next assume "q" have "q ⟶ r" using assms .. thus "r" using `q` .. qed qed -- "La demostración detallada es" lemma ejercicio_37_2: assumes "(p ⟶ r) ∧ (q ⟶ r)" shows "p ∨ q ⟶ r" proof (rule impI) assume "p ∨ q" thus "r" proof (rule disjE) assume "p" have "p ⟶ r" using assms by (rule conjunct1) thus "r" using `p` by (rule mp) next assume "q" have "q ⟶ r" using assms by (rule conjunct2) thus "r" using `q` by (rule mp) qed qed -- "La demostración automática es" lemma ejercicio_37_3: assumes "(p ⟶ r) ∧ (q ⟶ r)" shows "p ∨ q ⟶ r" using assms by auto text {* --------------------------------------------------------------- Ejercicio 38. Demostrar p ∨ q ⟶ r ⊢ (p ⟶ r) ∧ (q ⟶ r) ------------------------------------------------------------------ *} -- "La demostración estructurada es" lemma ejercicio_38_1: assumes "p ∨ q ⟶ r" shows "(p ⟶ r) ∧ (q ⟶ r)" proof show "p ⟶ r" proof assume "p" hence "p ∨ q" .. with assms show "r" .. qed next show "q ⟶ r" proof assume "q" hence "p ∨ q" .. with assms show "r" .. qed qed -- "La demostración detallada es" lemma ejercicio_38_2: assumes "p ∨ q ⟶ r" shows "(p ⟶ r) ∧ (q ⟶ r)" proof (rule conjI) show "p ⟶ r" proof (rule impI) assume "p" hence "p ∨ q" by (rule disjI1) show "r" using assms `p ∨ q` by (rule mp) qed next show "q ⟶ r" proof (rule impI) assume "q" hence "p ∨ q" by (rule disjI2) show "r" using assms `p ∨ q` by (rule mp) qed qed -- "La demostración automática es" lemma ejercicio_38_3: assumes "p ∨ q ⟶ r" shows "(p ⟶ r) ∧ (q ⟶ r)" using assms by auto section {* Negación *} text {* --------------------------------------------------------------- Ejercicio 39. Demostrar p ⊢ ¬¬p ------------------------------------------------------------------ *} -- "La demostración detallada es" lemma ejercicio_39_1: assumes "p" shows "¬¬p" proof - show "¬¬p" using assms by (rule notnotI) qed -- "La demostración automática es" lemma ejercicio_39_2: assumes "p" shows "¬¬p" using assms by auto text {* --------------------------------------------------------------- Ejercicio 40. Demostrar ¬p ⊢ p ⟶ q ------------------------------------------------------------------ *} -- "La demostración estructurada es" lemma ejercicio_40_1: assumes "¬p" shows "p ⟶ q" proof assume "p" with assms(1) show "q" .. qed -- "La demostración detallada es" lemma ejercicio_40_2: assumes "¬p" shows "p ⟶ q" proof (rule impI) assume "p" show "q" using assms(1) `p` by (rule notE) qed -- "La demostración automática es" lemma ejercicio_40_3: assumes "¬p" shows "p ⟶ q" using assms by auto text {* --------------------------------------------------------------- Ejercicio 41. Demostrar p ⟶ q ⊢ ¬q ⟶ ¬p ------------------------------------------------------------------ *} -- "La demostración estructurada es" lemma ejercicio_41_1: assumes "p ⟶ q" shows "¬q ⟶ ¬p" proof assume "¬q" with assms(1) show "¬p" by (rule mt) qed -- "La demostración detallada es" lemma ejercicio_41_2: assumes "p ⟶ q" shows "¬q ⟶ ¬p" proof (rule impI) assume "¬q" show "¬p" using assms(1) `¬q` by (rule mt) qed -- "La demostración automática es" lemma ejercicio_41_3: assumes "p ⟶ q" shows "¬q ⟶ ¬p" using assms by auto text {* --------------------------------------------------------------- Ejercicio 42. Demostrar p ∨ q, ¬q ⊢ p ------------------------------------------------------------------ *} -- "La demostración estructurada es" lemma ejercicio_42_1: assumes "p ∨ q" "¬q" shows "p" using assms(1) proof assume "p" thus "p" . next assume "q" with assms(2) show "p" .. qed -- "La demostración detallada es" lemma ejercicio_42_2: assumes "p ∨ q" "¬q" shows "p" using assms(1) proof (rule disjE) assume "p" thus "p" by this next assume "q" show "p" using assms(2) `q` by (rule notE) qed -- "La demostración automática es" lemma ejercicio_42_3: assumes "p ∨ q" "¬q" shows "p" using assms by auto text {* --------------------------------------------------------------- Ejercicio 42. Demostrar p ∨ q, ¬p ⊢ q ------------------------------------------------------------------ *} -- "La demostración estructurada es" lemma ejercicio_43_1: assumes "p ∨ q" "¬p" shows "q" using assms(1) proof assume "p" with assms(2) show "q" .. next assume "q" thus "q" . qed -- "La demostración detallada es" lemma ejercicio_43_2: assumes "p ∨ q" "¬p" shows "q" using assms(1) proof (rule disjE) assume "p" show "q" using assms(2) `p` by (rule notE) next assume "q" thus "q" by this qed -- "La demostración automática es" lemma ejercicio_43_3: assumes "p ∨ q" "¬p" shows "q" using assms by auto text {* --------------------------------------------------------------- Ejercicio 44. Demostrar p ∨ q ⊢ ¬(¬p ∧ ¬q) ------------------------------------------------------------------ *} -- "La demostración estructurada es" lemma ejercicio_44_1: assumes "p ∨ q" shows "¬(¬p ∧ ¬q)" proof assume "¬p ∧ ¬q" note `p ∨ q` thus "False" proof assume "p" have "¬p" using `¬p ∧ ¬q` .. thus "False" using `p` .. next assume "q" have "¬q" using `¬p ∧ ¬q` .. thus "False" using `q` .. qed qed -- "La demostración detallada es" lemma ejercicio_44_2: assumes "p ∨ q" shows "¬(¬p ∧ ¬q)" proof (rule notI) assume "¬p ∧ ¬q" note `p ∨ q` thus "False" proof assume "p" have "¬p" using `¬p ∧ ¬q` by (rule conjunct1) thus "False" using `p` by (rule notE) next assume "q" have "¬q" using `¬p ∧ ¬q` by (rule conjunct2) thus "False" using `q` by (rule notE) qed qed -- "La demostración automática es" lemma ejercicio_44_3: assumes "p ∨ q" shows "¬(¬p ∧ ¬q)" using assms by auto text {* --------------------------------------------------------------- Ejercicio 45. Demostrar p ∧ q ⊢ ¬(¬p ∨ ¬q) ------------------------------------------------------------------ *} -- "La demostración estructurada es" lemma ejercicio_45_1: assumes "p ∧ q" shows "¬(¬p ∨ ¬q)" proof assume "¬p ∨ ¬q" thus "False" proof assume "¬p" have "p" using assms .. with `¬p` show "False" .. next assume "¬q" have "q" using assms .. with `¬q` show "False" .. qed qed -- "La demostración detallada es" lemma ejercicio_45_2: assumes "p ∧ q" shows "¬(¬p ∨ ¬q)" proof (rule notI) assume "¬p ∨ ¬q" thus "False" proof assume "¬p" have "p" using assms by (rule conjunct1) show "False" using `¬p` `p` by (rule notE) next assume "¬q" have "q" using assms by (rule conjunct2) show "False" using `¬q` `q` by (rule notE) qed qed -- "La demostración automática es" lemma ejercicio_45_3: assumes "p ∧ q" shows "¬(¬p ∨ ¬q)" using assms by auto text {* --------------------------------------------------------------- Ejercicio 46. Demostrar ¬(p ∨ q) ⊢ ¬p ∧ ¬q ------------------------------------------------------------------ *} -- "La demostración estructurada es" lemma ejercicio_46_1: assumes "¬(p ∨ q)" shows "¬p ∧ ¬q" proof show "¬p" proof assume "p" hence "p ∨ q" .. with assms show "False" .. qed next show "¬q" proof assume "q" hence "p ∨ q" .. with assms show "False" .. qed qed -- "La demostración detallada es" lemma ejercicio_46_2: assumes "¬(p ∨ q)" shows "¬p ∧ ¬q" proof (rule conjI) show "¬p" proof (rule notI) assume "p" hence "p ∨ q" by (rule disjI1) show "False" using assms `p ∨ q` by (rule notE) qed next show "¬q" proof (rule notI) assume "q" hence "p ∨ q" by (rule disjI2) show "False" using assms `p ∨ q` by (rule notE) qed qed -- "La demostración automática es" lemma ejercicio_46_3: assumes "¬(p ∨ q)" shows "¬p ∧ ¬q" using assms by auto text {* --------------------------------------------------------------- Ejercicio 47. Demostrar ¬p ∧ ¬q ⊢ ¬(p ∨ q) ------------------------------------------------------------------ *} -- "La demostración estructurada es" lemma ejercicio_47_1: assumes "¬p ∧ ¬q" shows "¬(p ∨ q)" proof assume "p ∨ q" thus False proof assume "p" have "¬p" using assms .. thus False using `p` .. next assume "q" have "¬q" using assms .. thus False using `q` .. qed qed -- "La demostración detallada es" lemma ejercicio_47_2: assumes "¬p ∧ ¬q" shows "¬(p ∨ q)" proof (rule notI) assume "p ∨ q" thus False proof (rule disjE) assume "p" have "¬p" using assms by (rule conjunct1) thus False using `p` by (rule notE) next assume "q" have "¬q" using assms by (rule conjunct2) thus False using `q` by (rule notE) qed qed -- "La demostración automática es" lemma ejercicio_47_3: assumes "¬p ∧ ¬q" shows "¬(p ∨ q)" using assms by auto text {* --------------------------------------------------------------- Ejercicio 48. Demostrar ¬p ∨ ¬q ⊢ ¬(p ∧ q) ------------------------------------------------------------------ *} -- "La demostración estructurada es" lemma ejercicio_48_1: assumes "¬p ∨ ¬q" shows "¬(p ∧ q)" proof assume "p ∧ q" note `¬p ∨ ¬ q` thus False proof assume "¬p" have "p" using `p ∧ q` .. with `¬p` show False .. next assume "¬q" have "q" using `p ∧ q` .. with `¬q` show False .. qed qed -- "La demostración detallada es" lemma ejercicio_48_2: assumes "¬p ∨ ¬q" shows "¬(p ∧ q)" proof (rule notI) assume "p ∧ q" note `¬p ∨ ¬ q` thus False proof (rule disjE) assume "¬p" have "p" using `p ∧ q` by (rule conjunct1) show False using `¬p` `p` by (rule notE) next assume "¬q" have "q" using `p ∧ q` by (rule conjunct2) show False using `¬q` `q` by (rule notE) qed qed -- "La demostración automática es" lemma ejercicio_48_3: assumes "¬p ∨ ¬q" shows "¬(p ∧ q)" using assms by auto text {* --------------------------------------------------------------- Ejercicio 49. Demostrar ⊢ ¬(p ∧ ¬p) ------------------------------------------------------------------ *} -- "La demostración estructurada es" lemma ejercicio_49_1: "¬(p ∧ ¬p)" proof assume "p ∧ ¬p" hence "p" .. have "¬p" using `p ∧ ¬p` .. thus False using `p` .. qed -- "La demostración detallada es" lemma ejercicio_49_2: "¬(p ∧ ¬p)" proof (rule notI) assume "p ∧ ¬p" hence "p" by (rule conjunct1) have "¬p" using `p ∧ ¬p` by (rule conjunct2) thus False using `p` by (rule notE) qed -- "La demostración automática es" lemma ejercicio_49_3: "¬(p ∧ ¬p)" by auto text {* --------------------------------------------------------------- Ejercicio 50. Demostrar p ∧ ¬p ⊢ q ------------------------------------------------------------------ *} -- "La demostración estructurada es" lemma ejercicio_50_1: assumes "p ∧ ¬p" shows "q" proof - have "p" using assms .. have "¬p" using assms .. thus "q" using `p` .. qed -- "La demostración detallada es" lemma ejercicio_50_2: assumes "p ∧ ¬p" shows "q" proof - have "p" using assms by (rule conjunct1) have "¬p" using assms by (rule conjunct2) thus "q" using `p` by (rule notE) qed -- "La demostración automática es" lemma ejercicio_50_3: assumes "p ∧ ¬p" shows "q" using assms by auto text {* --------------------------------------------------------------- Ejercicio 51. Demostrar ¬¬p ⊢ p ------------------------------------------------------------------ *} -- "La demostración detallada es" lemma ejercicio_51_1: assumes "¬¬p" shows "p" using assms by (rule notnotD) -- "La demostración automática es" lemma ejercicio_51_2: assumes "¬¬p" shows "p" using assms by auto text {* --------------------------------------------------------------- Ejercicio 52. Demostrar ⊢ p ∨ ¬p ------------------------------------------------------------------ *} -- "La demostración estructurada es" lemma ejercicio_52_1: "p ∨ ¬p" proof - have "¬¬p ∨ ¬p" .. thus "p ∨ ¬p" by simp qed -- "La demostración detallada es" lemma ejercicio_52_2: "p ∨ ¬p" proof - have "¬¬p ∨ ¬p" by (rule excluded_middle) thus "p ∨ ¬p" by simp qed -- "La demostración automática es" lemma ejercicio_52_3: "p ∨ ¬p" by auto text {* --------------------------------------------------------------- Ejercicio 53. Demostrar ⊢ ((p ⟶ q) ⟶ p) ⟶ p ------------------------------------------------------------------ *} -- "La demostración estructurada es" lemma ejercicio_53_1: "((p ⟶ q) ⟶ p) ⟶ p" proof assume "(p ⟶ q) ⟶ p" show "p" proof (rule ccontr) assume "¬p" have "¬(p ⟶ q)" using `(p ⟶ q) ⟶ p` `¬p` by (rule mt) have "p ⟶ q" proof assume "p" with `¬p` show "q" .. qed show False using `¬(p ⟶ q)` `p ⟶ q` .. qed qed -- "La demostración detallada es" lemma ejercicio_53_2: "((p ⟶ q) ⟶ p) ⟶ p" proof (rule impI) assume "(p ⟶ q) ⟶ p" show "p" proof (rule ccontr) assume "¬p" have "¬(p ⟶ q)" using `(p ⟶ q) ⟶ p` `¬p` by (rule mt) have "p ⟶ q" proof (rule impI) assume "p" show "q" using `¬p` `p` by (rule notE) qed show False using `¬(p ⟶ q)` `p ⟶ q` by (rule notE) qed qed -- "La demostración automática es" lemma ejercicio_53_3: "((p ⟶ q) ⟶ p) ⟶ p" by auto text {* --------------------------------------------------------------- Ejercicio 54. Demostrar ¬q ⟶ ¬p ⊢ p ⟶ q ------------------------------------------------------------------ *} -- "La demostración estructurada es" lemma ejercicio_54_1: assumes "¬q ⟶ ¬p" shows "p ⟶ q" proof assume "p" show "q" proof (rule ccontr) assume "¬q" with assms have "¬p" .. thus False using `p` .. qed qed -- "La demostración detallada es" lemma ejercicio_54_2: assumes "¬q ⟶ ¬p" shows "p ⟶ q" proof (rule impI) assume "p" show "q" proof (rule ccontr) assume "¬q" have "¬p" using assms `¬q` by (rule mp) thus False using `p` by (rule notE) qed qed -- "La demostración automática es" lemma ejercicio_54_3: assumes "¬q ⟶ ¬p" shows "p ⟶ q" using assms by auto text {* --------------------------------------------------------------- Ejercicio 55. Demostrar ¬(¬p ∧ ¬q) ⊢ p ∨ q ------------------------------------------------------------------ *} -- "La demostración estructurada es" lemma ejercicio_55_1: assumes "¬(¬p ∧ ¬q)" shows "p ∨ q" proof - have "¬p ∨ p" .. thus "p ∨ q" proof assume "¬p" have "¬q ∨ q" .. thus "p ∨ q" proof assume "¬q" with `¬p` have "¬p ∧ ¬q" .. with assms show "p ∨ q" .. next assume "q" thus "p ∨ q" .. qed next assume "p" thus "p ∨ q" .. qed qed -- "La demostración detallada es" lemma ejercicio_55_2: assumes "¬(¬p ∧ ¬q)" shows "p ∨ q" proof - have "¬p ∨ p" by (rule excluded_middle) thus "p ∨ q" proof assume "¬p" have "¬q ∨ q" by (rule excluded_middle) thus "p ∨ q" proof assume "¬q" have "¬p ∧ ¬q" using `¬p` `¬q` by (rule conjI) show "p ∨ q" using assms `¬p ∧ ¬q` by (rule notE) next assume "q" thus "p ∨ q" by (rule disjI2) qed next assume "p" thus "p ∨ q" by (rule disjI1) qed qed -- "La demostración automática es" lemma ejercicio_55_3: assumes "¬(¬p ∧ ¬q)" shows "p ∨ q" using assms by auto text {* --------------------------------------------------------------- Ejercicio 56. Demostrar ¬(¬p ∨ ¬q) ⊢ p ∧ q ------------------------------------------------------------------ *} -- "La demostración estructurada es" lemma ejercicio_56_1: assumes "¬(¬p ∨ ¬q)" shows "p ∧ q" proof show "p" proof (rule ccontr) assume "¬p" hence "¬p ∨ ¬q" .. with assms show False .. qed next show "q" proof (rule ccontr) assume "¬q" hence "¬p ∨ ¬q" .. with assms show False .. qed qed -- "La demostración detallada es" lemma ejercicio_56_2: assumes "¬(¬p ∨ ¬q)" shows "p ∧ q" proof (rule conjI) show "p" proof (rule ccontr) assume "¬p" hence "¬p ∨ ¬q" by (rule disjI1) show False using assms `¬p ∨ ¬q` by (rule notE) qed next show "q" proof (rule ccontr) assume "¬q" hence "¬p ∨ ¬q" by (rule disjI2) show False using assms `¬p ∨ ¬q` by (rule notE) qed qed -- "La demostración automática es" lemma ejercicio_56_3: assumes "¬(¬p ∨ ¬q)" shows "p ∧ q" using assms by auto text {* --------------------------------------------------------------- Ejercicio 57. Demostrar ¬(p ∧ q) ⊢ ¬p ∨ ¬q ------------------------------------------------------------------ *} -- "La demostración estructurada es" lemma ejercicio_57_1: assumes "¬(p ∧ q)" shows "¬p ∨ ¬q" proof - have "¬p ∨ p" .. thus "¬p ∨ ¬q" proof assume "¬p" thus "¬p ∨ ¬q" .. next assume "p" have "¬q ∨ q" .. thus "¬p ∨ ¬q" proof assume "¬q" thus "¬p ∨ ¬q" .. next assume "q" with `p` have "p ∧ q" .. with assms show "¬p ∨ ¬q" .. qed qed qed -- "La demostración detallada es" lemma ejercicio_57_2: assumes "¬(p ∧ q)" shows "¬p ∨ ¬q" proof - have "¬p ∨ p" by (rule excluded_middle) thus "¬p ∨ ¬q" proof (rule disjE) assume "¬p" thus "¬p ∨ ¬q" by (rule disjI1) next assume "p" have "¬q ∨ q" by (rule excluded_middle) thus "¬p ∨ ¬q" proof assume "¬q" thus "¬p ∨ ¬q" by (rule disjI2) next assume "q" have "p ∧ q" using `p` `q` by (rule conjI) show "¬p ∨ ¬q" using assms `p ∧ q` by (rule notE) qed qed qed -- "La demostración automática es" lemma ejercicio_57_3: assumes "¬(p ∧ q)" shows "¬p ∨ ¬q" using assms by auto text {* --------------------------------------------------------------- Ejercicio 58. Demostrar ⊢ (p ⟶ q) ∨ (q ⟶ p) ------------------------------------------------------------------ *} -- "La demostración estructurada es" lemma ejercicio_58_1: "(p ⟶ q) ∨ (q ⟶ p)" proof - have "¬p ∨ p" .. thus "(p ⟶ q) ∨ (q ⟶ p)" proof assume "¬p" have "p ⟶ q" proof assume "p" with `¬p` show "q" .. qed thus "(p ⟶ q) ∨ (q ⟶ p)" .. next assume "p" have "q ⟶ p" proof assume "q" show "p" using `p` . qed thus "(p ⟶ q) ∨ (q ⟶ p)" .. qed qed -- "La demostración detallada es" lemma ejercicio_58_2: "(p ⟶ q) ∨ (q ⟶ p)" proof - have "¬p ∨ p" by (rule excluded_middle) thus "(p ⟶ q) ∨ (q ⟶ p)" proof assume "¬p" have "p ⟶ q" proof (rule impI) assume "p" show "q" using `¬p` `p` by (rule notE) qed thus "(p ⟶ q) ∨ (q ⟶ p)" by (rule disjI1) next assume "p" have "q ⟶ p" proof assume "q" show "p" using `p` by this qed thus "(p ⟶ q) ∨ (q ⟶ p)" by (rule disjI2) qed qed -- "La demostración automática es" lemma ejercicio_58_3: "(p ⟶ q) ∨ (q ⟶ p)" by auto end |