DAO2011: Ejercicios de deducción natural en lógica de primer orden con Isabelle/HOL
En la clase de hoy del curso de Demostración asistida por ordenador se han comentado las soluciones de los ejercicios de deducción natural en lógica de primer orden con Isabelle/HOL/Isar.
A continuación se muestra la teoría correspondiente a las soluciones de los ejercicios
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header {* Deducción natural en la lógica de primer orden (Ejercicios) *} theory LogicaDePrimerOrdenEj imports Main begin text {* Los siguientes ejercicios son los del tema 7 del curso de LI demostrado usando las reglas de deducción natural. *} text {* Se usarán las reglas del modus tollens y de la introducción de la doble negación. *} lemma mt: "⟦F ⟶ G; ¬G⟧ ⟹ ¬F" by auto lemma notnotI: "P ⟹ ¬¬P" by auto lemma assumes "∀x. P x ⟶ Q x" shows "(∀x. P x) ⟶ (∀x. Q x)" proof (rule impI) assume 1: "∀x. P x" show "∀x. Q x" proof (rule allI) fix x have 2: "P x ⟶ Q x" using assms by (rule allE) have 3: "P x" using 1 by (rule allE) show "Q x" using 2 3 by (rule mp) qed qed lemma assumes "∃x. ¬P x" shows " ¬(∀x. P x)" proof (rule notI) assume 1: "∀x. P x" obtain a where 2: "¬P a" using assms(1) by (rule exE) have 3: "P a" using 1 by (rule allE) show False using 2 3 by (rule notE) qed lemma assumes "∀x. P x" shows "∀y. P y" proof (rule allI) fix a show "P a" using assms(1) by (rule allE) qed lemma assumes "∀x. P x ⟶ Q x" shows "(∀x. ¬Q x) ⟶ (∀x. ¬P x)" proof (rule impI) assume 1: "∀x. ¬Q x" show "∀x. ¬P x" proof (rule allI) fix x have 2: "P x ⟶ Q x" using assms(1) by (rule allE) have 3: "¬Q x" using 1 by (rule allE) show "¬P x" using 2 3 by (rule mt) qed qed lemma 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 1: "P a ∧ Q a" by (rule exE) have 2: "P a ⟶ ¬Q a" using assms(1) by (rule allE) have 3: "P a" using 1 by (rule conjunct1) have 4: "¬Q a" using 2 3 by (rule mp) have 5: "Q a" using 1 by (rule conjunct2) show False using 4 5 by (rule notE) qed lemma assumes "∀x. ∀y. P x y" shows "∀u. ∀v. P u v" proof (rule allI) fix u show "∀v. P u v" proof (rule allI) fix v have "∀y. P u y" using assms(1) by (rule allE) thus "P u v" by (rule allE) qed qed lemma assumes "∃x. ∃y. P x y" shows "∃u. ∃v. P u v" proof - obtain a where "∃y. P a y" using assms(1) by (rule exE) then obtain b where "P a b" by (rule exE) hence "∃v. P a v" by (rule exI) thus "∃u. ∃v. P u v" by (rule exI) qed lemma assumes "∃x. ∀y. P x y" shows "∀y. ∃x. P x y" proof (rule allI) fix y obtain a where "∀y. P a y" using assms(1) by (rule exE) hence "P a y" by (rule allE) thus "∃x. P x y" by (rule exI) qed lemma assumes "∃x. P a ⟶ Q x" shows "P a ⟶ (∃x. Q x)" proof (rule impI) assume 1: "P a" obtain b where "P a ⟶ Q b" using assms(1) by (rule exE) hence "Q b" using 1 by (rule mp) thus "∃x. Q x" by (rule exI) qed lemma 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" hence "P a ⟶ Q b" by simp thus "∃x. P a ⟶ Q x" by (rule exI) } next { assume "P a" hence "∃x. Q x" using assms(1) by simp then obtain c where "Q c" by (rule exE) hence "P a ⟶ Q c" by simp thus "∃x. P a ⟶ Q x" by (rule exI) } qed qed lemma assumes "(∃x. P x) ⟶ Q a" shows "∀x. P x ⟶ Q a" proof (rule allI) fix x show "P x ⟶ Q a" proof assume "P x" hence 1: "∃x. P x" by (rule exI) show "Q a" using assms(1) 1 by (rule mp) qed qed lemma assumes "∀x. P x ⟶ Q a" shows "∃x. P x ⟶ Q a" proof (rule exI) show "P b ⟶ Q a" using assms(1) .. qed lemma assumes "(∀x. P x) ∨ (∀x. Q x)" shows "∀x. P x ∨ Q x" using assms proof (rule disjE) { assume 1: "∀x. P x" show "∀x. P x ∨ Q x" proof (rule allI) fix x have "P x" using 1 by (rule allE) thus "P x ∨ Q x" by (rule disjI1) qed } next { assume 2: "∀x. Q x" show "∀x. P x ∨ Q x" proof (rule allI) fix x have "Q x" using 2 by (rule allE) thus "P x ∨ Q x" by (rule disjI2) qed } qed lemma assumes "∃x. P x ∧ Q x" shows "(∃x. P x) ∧ (∃x. Q x)" proof - obtain a where 1: "P a ∧ Q a" using assms(1) by (rule exE) hence "P a" by (rule conjunct1) hence 2: "∃x. P x" by (rule exI) have "Q a" using 1 by (rule conjunct2) hence 3: "∃x. Q x" by (rule exI) show "(∃x. P x) ∧ (∃x. Q x)" using 2 3 by (rule conjI) qed lemma assumes "∀x.∀y. P y ⟶ Q x" shows "(∃y. P y) ⟶ (∀x. Q x)" proof (rule impI) assume 1: "∃y. P y" show "∀x. Q x" proof (rule allI) fix x have 2: "∀y. P y ⟶ Q x" using assms by (rule allE) obtain b where 3: "P b" using 1 by (rule exE) have "P b ⟶ Q x" using 2 by (rule allE) thus "Q x" using 3 by (rule mp) qed qed lemma assumes "¬(∀x. ¬P x)" shows "∃x. P x" proof (rule ccontr) assume 1: "¬(∃x. P x)" have 2: "∀x. ¬P x" proof fix x show "¬P x" proof assume "P x" hence 3: "∃x. P x" by (rule exI) show False using 1 3 by (rule notE) qed qed show False using assms(1) 2 by (rule notE) qed lemma assumes "∀x. ¬P x" shows "¬(∃x. P x)" proof (rule notI) assume "∃x. P x" then obtain a where 1: "P a" by (rule exE) have "¬P a" using assms(1) by (rule allE) thus False using 1 by (rule notE) qed lemma assumes "∃x. P x" shows "¬(∀x. ¬P x)" proof (rule notI) assume 1: "∀x. ¬P x" obtain a where 2: "P a" using assms(1) by (rule exE) have "¬P a" using 1 by (rule allE) thus False using 2 by (rule notE) qed lemma assumes "P a ⟶ (∀x. Q x)" shows "∀x. P a ⟶ Q x" proof (rule allI) fix x show "P a ⟶ Q x" proof assume 1: "P a" have "∀x. Q x" using assms(1) 1 by (rule mp) thus "Q x" by (rule allE) qed qed lemma assumes "∀x.∀y.∀z. R x y ∧ R y z ⟶ R x z" and "∀x. ¬R x x" shows "∀x.∀y. R x y ⟶ ¬R y x" proof (rule allI) fix x show "∀y. R x y ⟶ ¬R y x" proof (rule allI) fix y show "R x y ⟶ ¬R y x" proof (rule impI) assume 1: "R x y" show "¬R y x" proof assume 2: "R y x" have "∀y.∀z. R x y ∧ R y z ⟶ R x z" using assms(1) by (rule allE) hence "∀z. R x y ∧ R y z ⟶ R x z" by (rule allE) hence 3: "R x y ∧ R y x ⟶ R x x" by (rule allE) have 4: "R x y ∧ R y x" using 1 2 by (rule conjI) have 5: "R x x" using 3 4 by (rule mp) have "¬R x x" using assms(2) by (rule allE) thus False using 5 by (rule notE) qed qed qed qed lemma assumes "∀x. P x ∨ Q x" and "∃x. ¬Q x" and "∀x. R x ⟶ ¬P x" shows "∃x. ¬R x" proof - obtain a where 1: "¬Q a" using assms(2) .. have "P a ∨ Q a" using assms(1) .. thus "∃x. ¬R x" proof (rule disjE) { assume "P a" hence 2: "¬¬P a" by (rule notnotI) have "R a ⟶ ¬P a" using assms(3) by (rule allE) hence "¬R a" using 2 by (rule mt) thus "∃x. ¬R x" by (rule exI) } next { assume 3: "Q a" show "∃x. ¬R x" using 1 3 by (rule notE) } qed qed lemma assumes "∀x. P x ⟶ Q x ∨ R x" and "¬(∃x. P x ∧ R x)" shows "∀x. P x ⟶ Q x" proof fix x show "P x ⟶ Q x" proof assume 1: "P x" have "P x ⟶ Q x ∨ R x" using assms(1) by (rule allE) hence "Q x ∨ R x" using 1 by (rule mp) thus "Q x" proof (rule disjE) { assume "Q x" thus "Q x" by this } next { assume 2: "R x" have "P x ∧ R x" using 1 2 by (rule conjI) hence 3: "∃x. P x ∧ R x" by (rule exI) show "Q x" using assms(2) 3 by (rule notE) } qed qed qed lemma assumes "∃x.∃y. R x y ∨ R y x" shows "∃x.∃y. R 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) thus "∃x.∃y. R x y" proof { assume "R a b" hence "∃y. R a y" by (rule exI) thus "∃x.∃y. R x y" by (rule exI) } next { assume "R b a" hence "∃y. R b y" by (rule exI) thus "∃x.∃y. R x y" by (rule exI) } qed qed end |