LMF2014: Ejercicios de deducción en lógica de primer orden con Isabelle/HOL (2)
En la clase de hoy del curso Lógica matemática y fundamentos se ha explicado cómo demostrar mediante deducción natural teoremas de primer orden con Isabelle/HOL. En concreto, se han visto los ejercicios 13, 15, 16, 19, 21, 32 y 35 de la relación 6.
Los ejercicios y sus soluciones se muestran a continuación:
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theory R6 imports Main begin text {* --------------------------------------------------------------- Ejercicio 13. Demostrar (∀x. P x) ∨ (∀x. Q x) ⊢ ∀x. P x ∨ Q x ------------------------------------------------------------------ *} -- "La demostración automática es" lemma ejercicio_13a: "(∀x. P x) ∨ (∀x. Q x) ⟹ ∀x. P x ∨ Q x" by auto -- "La demostración estructurada es" lemma ejercicio_13b: assumes "(∀x. P x) ∨ (∀x. Q x)" shows "∀x. P x ∨ Q x" proof fix a note assms thus "P a ∨ Q a" proof assume "∀x. P x" hence "P a" .. thus "P a ∨ Q a" .. next assume "∀x. Q x" hence "Q a" .. thus "P a ∨ Q a" .. qed qed -- "La demostración detallada es" lemma ejercicio_13c: assumes "(∀x. P x) ∨ (∀x. Q x)" shows "∀x. P x ∨ Q x" proof (rule allI) fix a note assms thus "P a ∨ Q a" proof (rule disjE) assume "∀x. P x" hence "P a" by (rule allE) thus "P a ∨ Q a" by (rule disjI1) next assume "∀x. Q x" hence "Q a" by (rule allE) thus "P a ∨ Q a" by (rule disjI2) qed qed text {* --------------------------------------------------------------- Ejercicio 15. Demostrar ∀x y. P y ⟶ Q x ⊢ (∃y. P y) ⟶ (∀x. Q x) ------------------------------------------------------------------ *} -- "La demostración automática es" lemma ejercicio_15a: "∀x y. P y ⟶ Q x ⟹ (∃y. P y) ⟶ (∀x. Q x)" by auto -- "La demostración estructurada es" lemma ejercicio_15b: assumes "∀x y. P y ⟶ Q x" shows "(∃y. P y) ⟶ (∀x. Q x)" proof assume "∃y. P y" then obtain b where "P b" .. show "∀x. Q x" proof fix a have "∀y. P y ⟶ Q a" using assms .. hence "P b ⟶ Q a" .. thus "Q a" using `P b` .. qed qed -- "La demostración detallada es" lemma ejercicio_15c: assumes "∀x y. P y ⟶ Q x" shows "(∃y. P y) ⟶ (∀x. Q x)" proof (rule impI) assume "∃y. P y" then obtain b where "P b" by (rule exE) show "∀x. Q x" proof (rule allI) fix a have "∀y. P y ⟶ Q a" using assms by (rule allE) hence "P b ⟶ Q a" by (rule allE) thus "Q a" using `P b` by (rule mp) qed qed text {* --------------------------------------------------------------- Ejercicio 16. Demostrar ¬(∀x. ¬(P x)) ⊢ ∃x. P x ------------------------------------------------------------------ *} -- "La demostración automática es" lemma ejercicio_16a: "¬(∀x. ¬(P x)) ⟹ ∃x. P x" by auto -- "La demostración estructurada es" lemma ejercicio_16b: assumes "¬(∀x. ¬(P x))" shows "∃x. P x" proof (rule ccontr) assume "¬(∃x. P x)" have "∀x. ¬(P x)" proof fix a show "¬(P a)" proof assume "P a" hence "∃x. P x" .. with `¬(∃x. P x)` show False .. qed qed with assms show False .. qed -- "La demostración detallada es" lemma ejercicio_16c: assumes "¬(∀x. ¬(P x))" shows "∃x. P x" proof (rule ccontr) assume "¬(∃x. P x)" have "∀x. ¬(P x)" proof (rule allI) fix a show "¬(P a)" proof assume "P a" hence "∃x. P x" by (rule exI) with `¬(∃x. P x)` show False by (rule notE) qed qed with assms show False by (rule notE) qed text {* --------------------------------------------------------------- Ejercicio 19. Demostrar P a ⟶ (∀x. Q x) ⊢ ∀x. P a ⟶ Q x ------------------------------------------------------------------ *} -- "La demostración automática es" lemma ejercicio_19a: "P a ⟶ (∀x. Q x) ⟹ ∀x. P a ⟶ Q x" by auto -- "La demostración estructurada es" lemma ejercicio_19b: assumes "P a ⟶ (∀x. Q x)" shows "∀x. P a ⟶ Q x" proof fix b show "P a ⟶ Q b" proof assume "P a" with assms have "∀x. Q x" .. thus "Q b" .. qed qed -- "La demostración detallada es" lemma ejercicio_19c: assumes "P a ⟶ (∀x. Q x)" shows "∀x. P a ⟶ Q x" proof (rule allI) fix b show "P a ⟶ Q b" proof (rule impI) assume "P a" with assms have "∀x. Q x" by (rule mp) thus "Q b" by (rule allE) qed qed text {* --------------------------------------------------------------- Ejercicio 21. Demostrar {∀x. P x ∨ Q x, ∃x. ¬(Q x), ∀x. R x ⟶ ¬(P x)} ⊢ ∃x. ¬(R x) ------------------------------------------------------------------ *} -- "La demostración automática es" lemma ejercicio_21a: "⟦∀x. P x ∨ Q x; ∃x. ¬(Q x); ∀x. R x ⟶ ¬(P x)⟧ ⟹ ∃x. ¬(R x)" by auto -- "La demostración estructurada es" lemma ejercicio_21b: assumes "∀x. P x ∨ Q x" "∃x. ¬(Q x)" "∀x. R x ⟶ ¬(P x)" shows "∃x. ¬(R x)" proof - obtain a where "¬(Q a)" using assms(2) .. have "P a ∨ Q a" using assms(1) .. hence "P a" proof assume "P a" thus "P a" . next assume "Q a" with `¬(Q a)` show "P a" .. qed hence "¬¬(P a)" by (rule notnotI) have "R a ⟶ ¬(P a)" using assms(3) .. hence "¬(R a)" using `¬¬(P a)` by (rule mt) thus "∃x. ¬(R x)" .. qed -- "La demostración detallada es" lemma ejercicio_21c: assumes "∀x. P x ∨ Q x" "∃x. ¬(Q x)" "∀x. R x ⟶ ¬(P x)" shows "∃x. ¬(R x)" proof - obtain a where "¬(Q a)" using assms(2) by (rule exE) have "P a ∨ Q a" using assms(1) by (rule allE) hence "P a" proof (rule disjE) assume "P a" thus "P a" by this next assume "Q a" with `¬(Q a)` show "P a" by (rule notE) qed hence "¬¬(P a)" by (rule notnotI) have "R a ⟶ ¬(P a)" using assms(3) by (rule allE) hence "¬(R a)" using `¬¬(P a)` by (rule mt) thus "∃x. ¬(R x)" by (rule exI) qed text {* --------------------------------------------------------------- Ejercicio 32. Demostrar o refutar ∃x y. R x y ∨ R y x; ¬(∃x. R x x)⟧ ⟹ ∃x y. x ≠ y ------------------------------------------------------------------ *} -- "La demostración automática es" lemma ejercicio_32a: fixes R :: "'c ⇒ 'c ⇒ bool" assumes "∃x y. R x y ∨ R y x" "¬(∃x. R x x)" shows "∃(x::'c) y. x ≠ y" using assms by metis -- "La demostración estructurada es" lemma ejercicio_32b: fixes R :: "'c ⇒ 'c ⇒ bool" assumes "∃x y. R x y ∨ R y x" "¬(∃x. R x x)" shows "∃(x::'c) y. x ≠ y" proof - obtain a where "∃y. R a y ∨ R y a" using assms(1) .. then obtain b where "R a b ∨ R b a" .. hence "a ≠ b" proof assume "R a b" show "a ≠ b" proof assume "a = b" hence "R b b" using `R a b` by (rule subst) hence "∃x. R x x" .. with assms(2) show False .. qed next assume "R b a" show "a ≠ b" proof assume "a = b" hence "R a a" using `R b a` by (rule ssubst) hence "∃x. R x x" .. with assms(2) show False .. qed qed hence "∃y. a ≠ y" .. thus "∃(x::'c) y. x ≠ y" .. qed -- "La demostración detallada es" lemma ejercicio_32c: fixes R :: "'c ⇒ 'c ⇒ bool" assumes "∃x y. R x y ∨ R y x" "¬(∃x. R x x)" shows "∃(x::'c) y. x ≠ y" proof - obtain a where "∃y. R a y ∨ R y a" using assms(1) by (rule exE) then obtain b where "R a b ∨ R b a" by (rule exE) hence "a ≠ b" proof (rule disjE) assume "R a b" show "a ≠ b" proof (rule notI) assume "a = b" hence "R b b" using `R a b` by (rule subst) hence "∃x. R x x" by (rule exI) with assms(2) show False by (rule notE) qed next assume "R b a" show "a ≠ b" proof (rule notI) assume "a = b" hence "R a a" using `R b a` by (rule ssubst) hence "∃x. R x x" by (rule exI) with assms(2) show False by (rule notE) qed qed hence "∃y. a ≠ y" by (rule exI) thus "∃(x::'c) y. x ≠ y" by (rule exI) qed text {* --------------------------------------------------------------- Ejercicio 35. Demostrar o refutar {∀y. Q a y, ∀x y. Q x y ⟶ Q (s x) (s y)} ⊢ ∃z. Qa z ∧ Q z (s (s a)) ------------------------------------------------------------------ *} -- "La demostración automática es" lemma ejercicio_35a: "⟦∀y. Q a y; ∀x y. Q x y ⟶ Q (s x) (s y)⟧ ⟹ ∃z. Q a z ∧ Q z (s (s a))" by auto -- "La demostración estructura es" lemma ejercicio_35b: assumes "∀y. Q a y" "∀x y. Q x y ⟶ Q (s x) (s y)" shows "∃z. Q a z ∧ Q z (s (s a))" proof - have "Q a (s a)" using assms(1) .. have "∀y. Q a y ⟶ Q (s a) (s y)" using assms(2) .. hence "Q a (s a) ⟶ Q (s a) (s (s a))" .. hence "Q (s a) (s (s a))" using `Q a (s a)` .. with `Q a (s a)` have "Q a (s a) ∧ Q (s a) (s (s a))" .. thus "∃z. Q a z ∧ Q z (s (s a))" .. qed -- "La demostración detallada es" lemma ejercicio_35c: assumes "∀y. Q a y" "∀x y. Q x y ⟶ Q (s x) (s y)" shows "∃z. Q a z ∧ Q z (s (s a))" proof - have "Q a (s a)" using assms(1) by (rule allE) have "∀y. Q a y ⟶ Q (s a) (s y)" using assms(2) by (rule allE) hence "Q a (s a) ⟶ Q (s a) (s (s a))" by (rule allE) hence "Q (s a) (s (s a))" using `Q a (s a)` by (rule mp) with `Q a (s a)` have "Q a (s a) ∧ Q (s a) (s (s a))" by (rule conjI) thus "∃z. Q a z ∧ Q z (s (s a))" by (rule exI) qed end |