LMF2013: Ejercicios de deducción natural en lógica de primer orden con Isabelle/HOL (1)
En la clase de hoy del curso Lógica matemática y fundamentos se han resuelto los ejercicios 10, 11, 12, 13 y 16 de la relación 7 sobre deducción natural en lógica de primer orden con Isabelle/HOL.
Las soluciones de los ejercicios resueltos se muestran a continuación:
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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 11. Demostrar (∃x. P x) ⟶ Q a ⊢ ∀x. P x ⟶ Q a ------------------------------------------------------------------ *} -- "La demostración automática es" lemma ejercicio_11a: "(∃x. P x) ⟶ Q a ⟹ ∀x. P x ⟶ Q a" by auto -- "La demostración estructurada es" lemma ejercicio_11b: assumes "(∃x. P x) ⟶ Q a" shows "∀x. P x ⟶ Q a" proof fix b show "P b ⟶ Q a" proof assume "P b" hence "∃x. P x" .. with assms show "Q a" .. qed qed -- "La demostración detallada es" lemma ejercicio_11c: assumes "(∃x. P x) ⟶ Q a" shows "∀x. P x ⟶ Q a" proof (rule allI) fix b show "P b ⟶ Q a" proof (rule impI) assume "P b" hence "∃x. P x" by (rule exI) with assms show "Q a" by (rule mp) qed qed text {* --------------------------------------------------------------- Ejercicio 12. Demostrar ∀x. P x ⟶ Q a ⊢ ∃ x. P x ⟶ Q a ------------------------------------------------------------------ *} -- "La demostración automática es" lemma ejercicio_12a: "∀x. P x ⟶ Q a ⟹ ∃x. P x ⟶ Q a" by auto -- "La demostración estructurada es" lemma ejercicio_12b: assumes "∀x. P x ⟶ Q a" shows "∃x. P x ⟶ Q a" proof - have "P b ⟶ Q a" using assms .. thus "∃x. P x ⟶ Q a" .. qed -- "La demostración detallada es" lemma ejercicio_12c: assumes "∀x. P x ⟶ Q a" shows "∃x. P x ⟶ Q a" proof - have "P b ⟶ Q a" using assms by (rule allE) thus "∃x. P x ⟶ Q a" by (rule exI) qed 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 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 |