LMF2014: Ejercicios de deducción en lógica de primer orden con Isabelle/HOL
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 1, 5, 9, 10, 20 y 27 de la relación 6.
Los ejercicios y sus soluciones se muestran a continuación:
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theory R6 imports Main begin text {* --------------------------------------------------------------- Ejercicio 1. Demostrar ∀x. P x ⟶ Q x ⊢ (∀x. P x) ⟶ (∀x. Q x) ------------------------------------------------------------------ *} -- "La demostración automática es" lemma ejercicio_1a: "∀x. P x ⟶ Q x ⟹ (∀x. P x) ⟶ (∀x. Q x)" by auto -- "La demostración estructurada es" lemma ejercicio_1b: assumes "∀x. P x ⟶ Q x" shows "(∀x. P x) ⟶ (∀x. Q x)" proof assume "∀x. P x" show "∀x. Q x" proof fix a have "P a" using `∀x. P x` .. have "P a ⟶ Q a" using assms(1) .. thus "Q a" using `P a` .. qed qed -- "La demostración detallada es" lemma ejercicio_1c: assumes "∀x. P x ⟶ Q x" shows "(∀x. P x) ⟶ (∀x. Q x)" proof (rule impI) assume "∀x. P x" show "∀x. Q x" proof (rule allI) fix a have "P a" using `∀x. P x` by (rule allE) have "P a ⟶ Q a" using assms(1) by (rule allE) thus "Q a" using `P a` by (rule mp) qed qed text {* --------------------------------------------------------------- Ejercicio 5. Demostrar ∀x. P x ⟶ ¬(Q x) ⊢ ¬(∃x. P x ∧ Q x) ------------------------------------------------------------------ *} -- "La demostración automática es" lemma ejercicio_5a: "∀x. P x ⟶ ¬(Q x) ⟹ ¬(∃x. P x ∧ Q x)" by auto -- "La demostración estructurada es" lemma ejercicio_5b: assumes "∀x. P x ⟶ ¬(Q x)" shows "¬(∃x. P x ∧ Q x)" proof assume "∃x. P x ∧ Q x" then obtain a where "P a ∧ Q a" .. hence "P a" .. have "P a ⟶ ¬(Q a)" using assms .. hence "¬(Q a)" using `P a` .. have "Q a" using `P a ∧ Q a` .. with `¬(Q a)` show False .. qed -- "La demostración estructurada es" lemma ejercicio_5c: assumes "∀x. P x ⟶ ¬(Q x)" shows "¬(∃x. P x ∧ Q x)" proof (rule notI) assume "∃x. P x ∧ Q x" then obtain a where "P a ∧ Q a" by (rule exE) hence "P a" by (rule conjunct1) have "P a ⟶ ¬(Q a)" using assms by (rule allE) hence "¬(Q a)" using `P a` by (rule mp) have "Q a" using `P a ∧ Q a` by (rule conjunct2) with `¬(Q a)` show False by (rule notE) qed text {* --------------------------------------------------------------- Ejercicio 9. Demostrar ∃x. P a ⟶ Q x ⊢ P a ⟶ (∃x. Q x) ------------------------------------------------------------------ *} -- "La demostración automática es" lemma ejercicio_9a: "∃x. P a ⟶ Q x ⟹ P a ⟶ (∃x. Q x)" by auto -- "La demostración estructurada es" lemma ejercicio_9b: assumes "∃x. P a ⟶ Q x" shows "P a ⟶ (∃x. Q x)" proof assume "P a" obtain b where "P a ⟶ Q b" using assms .. hence "Q b" using `P a` .. thus "∃x. Q x" .. qed -- "La demostración detallada es" lemma ejercicio_9c: assumes "∃x. P a ⟶ Q x" shows "P a ⟶ (∃x. Q x)" proof (rule impI) assume "P a" obtain b where "P a ⟶ Q b" using assms by (rule exE) hence "Q b" using `P a` by (rule mp) thus "∃x. Q x" by (rule exI) qed text {* --------------------------------------------------------------- Ejercicio 10. Demostrar P a ⟶ (∃x. Q x) ⊢ ∃x. P a ⟶ Q x ------------------------------------------------------------------ *} -- "La demostración automática es" lemma ejercicio_10a: "P a ⟶ (∃x. Q x) ⟹ ∃x. P a ⟶ Q x" by auto -- "La demostración estructurada es" lemma ejercicio_10b: fixes P Q :: "'b ⇒ bool" assumes "P a ⟶ (∃x. Q x)" shows "∃x. P a ⟶ Q x" proof - have "¬(P a) ∨ P a" .. thus "∃x. P a ⟶ Q x" proof assume "¬(P a)" have "P a ⟶ Q a" proof assume "P a" with `¬(P a)` show "Q a" .. qed thus "∃x. P a ⟶ Q x" .. next assume "P a" with assms have "∃x. Q x" by (rule mp) then obtain b where "Q b" .. have "P a ⟶ Q b" proof assume "P a" note `Q b` thus "Q b" . qed thus "∃x. P a ⟶ Q x" .. qed qed -- "La demostración detallada es" lemma ejercicio_10c: fixes P Q :: "'b ⇒ bool" assumes "P a ⟶ (∃x. Q x)" shows "∃x. P a ⟶ Q x" proof - have "¬(P a) ∨ P a" by (rule excluded_middle) thus "∃x. P a ⟶ Q x" proof (rule disjE) assume "¬(P a)" have "P a ⟶ Q a" proof (rule impI) assume "P a" with `¬(P a)` show "Q a" by (rule notE) qed thus "∃x. P a ⟶ Q x" by (rule exI) next assume "P a" with assms have "∃x. Q x" by (rule mp) then obtain b where "Q b" by (rule exE) have "P a ⟶ Q b" proof (rule impI) assume "P a" note `Q b` thus "Q b" by this qed thus "∃x. P a ⟶ Q x" by (rule exI) qed qed text {* --------------------------------------------------------------- Ejercicio 20. Demostrar {∀x y z. R x y ∧ R y z ⟶ R x z, ∀x. ¬(R x x)} ⊢ ∀x y. R x y ⟶ ¬(R y x) ------------------------------------------------------------------ *} -- "La demostración automática es" lemma ejercicio_20a: "⟦∀x y z. R x y ∧ R y z ⟶ R x z; ∀x. ¬(R x x)⟧ ⟹ ∀x y. R x y ⟶ ¬(R y x)" by metis -- "La demostración estructurada es" lemma ejercicio_20b: assumes "∀x y z. R x y ∧ R y z ⟶ R x z" "∀x. ¬(R x x)" shows "∀x y. R x y ⟶ ¬(R y x)" proof (rule allI)+ fix a b show "R a b ⟶ ¬(R b a)" proof assume "R a b" show "¬(R b a)" proof assume "R b a" show False proof - have "R a b ∧ R b a" using `R a b` `R b a` .. have "∀y z. R a y ∧ R y z ⟶ R a z" using assms(1) .. hence "∀z. R a b ∧ R b z ⟶ R a z" .. hence "R a b ∧ R b a ⟶ R a a" .. hence "R a a" using `R a b ∧ R b a` .. have "¬(R a a)" using assms(2) .. thus False using `R a a` .. qed qed qed qed -- "La demostración detallada es" lemma ejercicio_20c: assumes "∀x y z. R x y ∧ R y z ⟶ R x z" "∀x. ¬(R x x)" shows "∀x y. R x y ⟶ ¬(R y x)" proof (rule allI)+ fix a b show "R a b ⟶ ¬(R b a)" proof (rule impI) assume "R a b" show "¬(R b a)" proof (rule notI) assume "R b a" show False proof - have "R a b ∧ R b a" using `R a b` `R b a` by (rule conjI) have "∀y z. R a y ∧ R y z ⟶ R a z" using assms(1) by (rule allE) hence "∀z. R a b ∧ R b z ⟶ R a z" by (rule allE) hence "R a b ∧ R b a ⟶ R a a" by (rule allE) hence "R a a" using `R a b ∧ R b a` by (rule mp) have "¬(R a a)" using assms(2) by (rule allE) thus False using `R a a` by (rule notE) qed qed qed qed text {* --------------------------------------------------------------- Ejercicio 27. Demostrar o refutar ((\<forall>x. P x) \<or> (\<forall>x. Q x)) \<longleftrightarrow> (\<forall>x. P x \<or> Q x) ------------------------------------------------------------------ *} lemma ejercicio_27: "((\<forall>x. P x) \<or> (\<forall>x. Q x)) \<longleftrightarrow> (\<forall>x. P x \<or> Q x)" oops (* Auto Quickcheck found a counterexample: P = {a1} Q = {a2} *) end |