DAO2011: Deducción natural en lógica de primer orden con Isabelle/HOL
En la clase de hoy del curso de Demostración asistida por ordenador se se ha estudiado cómo escribir demostraciones mediante deducción natural en lógica de primer orden usando Isabelle/HOL/Isar.
La teoría correspondiente a la clase es Tema_3.thy cuyo contenido se muestra a continuación
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header {* Deducción natural en la lógica de primer orden *} theory LogicaDePrimerOrden imports Main begin text {* En esta teoría se presentan los ejemplos del tema de deducción natural proposicional siguiendo la presentación de Huth y Ryan en su libro "Logic in Computer Science" http://www.cs.bham.ac.uk/research/projects/lics/ y, más concretamente, a la forma como se explica en la asignatura de "Lógica informática" y que puede verse en http://www.cs.us.es/~jalonso/cursos/li-10/temas/tema-7.pdf La página al lado de cada teorema indica la página de las anteriores transparencias donde se encuentra la demostración. *} section {* Reglas del cuantificador universal *} text {* La regla de eliminación del cuantificador universal es · allI: ⟦∀x. P x; P x ⟹ R⟧ ⟹ R La regla de introducción del cuantificador universal es · allE: (⋀x. P x) ⟹ ∀x. P x *} lemma -- "p. 10" assumes 1: "P(c)" and 2: "∀x. P(x) ⟶ ¬Q(x)" shows "¬Q(c)" proof (rule notI) assume 3: "Q(c)" have "P(c) ⟶ ¬Q(c)" using 2 by (rule allE) hence "¬Q(c)" using 1 by (rule mp) thus False using 3 by (rule notE) qed lemma -- "p. 11" assumes 1: "∀x. P(x) ⟶ Q(x)" and 2: "∀x. P(x)" shows "∀x. Q(x)" proof (rule allI) fix x have 3: "P(x) ⟶ Q(x)" using 1 by (rule allE) have 4: "P(x)" using 2 by (rule allE) show "Q(x)" using 3 4 by (rule mp) qed section {* Reglas del cuantificador existencial *} text {* La regla de eliminación del cuantificador existencial es · exI: P x ⟹ ∃x. P x La regla de introducción del cuantificador existencial es · exE: ⟦∃x. P x; ⋀x. P x ⟹ Q⟧ ⟹ Q *} lemma -- "p. 12" assumes 1: "∀x. P(x)" shows "∃x. P(x)" proof - have 2: "P(x)" using 1 by (rule allE) show "∃x. P(x)" using 2 by (rule exI) qed lemma -- "p. 12" assumes 1: "∀x. P(x)" shows "∃x. P(x)" proof (rule exI) show "P(x)" using 1 by (rule allE) qed lemma -- "p. 12" assumes 1: "∀x. P(x)" shows "∃x. P(x)" proof show "P(x)" using 1 .. qed lemma -- "p. 13" assumes 1: "∀x. P(x) ⟶ Q(x)" and 2: "∃x. P(x)" shows "∃x. Q(x)" proof - obtain "a" where 3: "P(a)" using 2 by (rule exE) have 4: "P(a) ⟶ Q(a)" using 1 by (rule allE) have 5: "Q(a)" using 4 3 by (rule mp) show "∃x. Q(x)" using 5 by (rule exI) qed lemma assumes 1: "∀x. Q(x) ⟶ R(x)" and 2: "∃x. P(x) ∧ Q(x)" shows "∃x. P(x) ∧ R(x)" proof - obtain x where 3: "P(x) ∧ Q(x)" using 2 by (rule exE) have 5: "P(x)" using 3 by (rule conjunct1) have 6: "Q(x) ⟶ R(x)" using 1 by (rule allE) have 7: "Q(x)" using 3 by (rule conjunct2) have 8: "R(x)" using 6 7 by (rule mp) have 9: "P(x) ∧ R(x)" using 5 8 by (rule conjI) show "∃x. P(x) ∧ R(x)" using 9 by (rule exI) qed lemma assumes 1: "∃x. P(x)" and 2: "∀x.∀y. P(x) ⟶ Q(y)" shows "∀y. Q(y)" proof (rule allI) fix y obtain x where 3: "P(x)" using 1 by (rule exE) have 4: "∀y. P(x) ⟶ Q(y)" using 2 by (rule allE) have 5: "P(x) ⟶ Q(y)" using 4 by (rule allE) show "Q(y)" using 5 3 by (rule mp) qed section {* Equivalencias *} -- "p. 15" lemma equivalencia_1a1: assumes 1: "¬(∀x. P(x))" shows "∃x. ¬P(x)" proof (rule ccontr) assume 2: "¬(∃x. ¬P(x))" note 1 thus False proof (rule notE) show "∀x. P(x)" proof (rule allI) fix x show "P(x)" proof (rule ccontr) assume 3: "¬P(x)" have 4: "∃x. ¬P(x)" using 3 by (rule exI) show False using 2 4 by (rule notE) qed qed qed qed -- "p. 16" lemma equivalencia_1a2: assumes 1: "∃x. ¬P(x)" shows "¬(∀x. P(x))" proof (rule ccontr) assume 2: "¬¬(∀x. P(x))" obtain x where 3: "¬P(x)" using 1 by (rule exE) have 4: "∀x. P(x)" using 2 by (rule notnotD) have 5: "P(x)" using 4 by (rule allE) show False using 3 5 by (rule notE) qed -- "p. 17" theorem equivalencia_1a: "(¬(∀x. P(x))) = (∃x. ¬P(x))" proof (rule iffI) { assume "¬(∀x. P x)" thus "∃x. ¬P x" by (rule equivalencia_1a1)} next { assume "∃x. ¬P x" thus "¬(∀x. P x)" by (rule equivalencia_1a2) } qed -- "p. 18" lemma equivalencia_3a1: assumes 1: "∀x. P(x) ∧ Q(x)" shows "(∀x. P(x)) ∧ (∀x. Q(x))" proof (rule conjI) show "∀x. P(x)" proof (rule allI) fix x have 2: "P(x) ∧ Q(x)" using 1 by (rule allE) show "P(x)" using 2 by (rule conjunct1) qed next show "∀x. Q(x)" proof (rule allI) fix x have 3: "P(x) ∧ Q(x)" using 1 by (rule allE) show "Q(x)" using 3 by (rule conjunct2) qed qed -- "p. 19" lemma equivalencia_3a2: assumes 1: "(∀x. P(x)) ∧ (∀x. Q(x))" shows "∀x. P(x) ∧ Q(x)" proof (rule allI) fix x have 2: "∀x. P(x)" using 1 by (rule conjunct1) have 3: "∀x. Q(x)" using 1 by (rule conjunct2) have 4: "P(x)" using 2 by (rule allE) have 5: "Q(x)" using 3 by (rule allE) show "P(x) ∧ Q(x)" using 4 5 by (rule conjI) qed -- "p. 20" lemma equivalencia_3a: "(∀x. P(x) ∧ Q(x)) = ((∀x. P(x)) ∧ (∀x. Q(x)))" proof (rule iffI) { assume "∀x. P(x) ∧ Q(x)" thus "(∀x. P(x)) ∧ (∀x. Q(x))" by (rule equivalencia_3a1) } next { assume "(∀x. P(x)) ∧ (∀x. Q(x))" thus "∀x. P(x) ∧ Q(x)" by (rule equivalencia_3a2) } qed -- "p. 21" lemma equivalencia_3b1: assumes 1: "(∃x. P(x)) ∨ (∃x. Q(x))" shows "∃x. P(x) ∨ Q(x)" using 1 proof (rule disjE) { assume "∃x. P(x)" then obtain x where "P(x)" by (rule exE) hence "P(x) ∨ Q(x)" by (rule disjI1) thus "∃x. P(x) ∨ Q(x)" by (rule exI) } next { assume "∃x. Q(x)" then obtain x where "Q(x)" by (rule exE) hence "P(x) ∨ Q(x)" by (rule disjI2) thus "∃x. P(x) ∨ Q(x)" by (rule exI) } qed -- "p. 22" lemma equivalencia_3b2: assumes 1: "∃x. P(x) ∨ Q(x)" shows "(∃x. P(x)) ∨ (∃x. Q(x))" proof - obtain x where 2: "P(x) ∨ Q(x)" using 1 by (rule exE) thus "(∃x. P(x)) ∨ (∃x. Q(x))" proof (rule disjE) { assume "P(x)" hence "∃x. P(x)" by (rule exI) thus "(∃x. P(x)) ∨ (∃x. Q(x))" by (rule disjI1) } next { assume "Q(x)" hence "∃x. Q(x)" by (rule exI) thus "(∃x. P(x)) ∨ (∃x. Q(x))" by (rule disjI2) } qed qed -- "p. 23" lemma equivalencia_3b: "((∃x. P(x)) ∨ (∃x. Q(x))) = (∃x. P(x) ∨ Q(x))" proof (rule iffI) { assume "(∃x. P(x)) ∨ (∃x. Q(x))" thus "∃x. P(x) ∨ Q(x)" by (rule equivalencia_3b1) } next { assume "∃x. P(x) ∨ Q(x)" thus "(∃x. P(x)) ∨ (∃x. Q(x))" by (rule equivalencia_3b2) } qed -- "p. 24" lemma equivalencia_4b1: assumes 1: "∃x.∃y. P x y" shows "∃y.∃x. P x y" proof - obtain x where "∃y. P x y" using 1 by (rule exE) then obtain y where "P x y" by (rule exE) hence "∃x. P x y" by (rule exI) thus "∃y.∃x. P x y" by (rule exI) qed -- "p. 25" theorem equivalencia_4b: "(∃x.∃y. P x y) = (∃y.∃x. P x y)" proof (rule iffI) { assume "∃x.∃y. P x y" thus "∃y.∃x. P x y" by (rule equivalencia_4b1) } next {assume "∃y.∃x. P x y" thus "∃x.∃y. P x y" by (rule equivalencia_4b1) } qed end |