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| − | == RA_Relacion_1 ==
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| − | theory Relacion_2
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| − | imports Main
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| − | begin
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| − |
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| − | (* Demostrar o refutar los siguientes lemas usando sólo las reglas básicas
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| − | de deducción natural de la lógica proposicional y de la lógica de
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| − | predicados (allI, allE, exI y exE). *)
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| − |
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| − |
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| − | lemma "(∃x. ∀y. P x y) ⟶ (∀y. ∃x. P x y)"
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| − | proof (rule impI)
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| − | assume 1: "(∃x. ∀y. P x y)"
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| − | show "(∀y. ∃x. P x y)"
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| − | proof (rule allI)
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| − | from 1 obtain x where 2: "∀y. P x y" by (rule exE)
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| − | fix y
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| − | from 2 have 3: "P x y" by (rule allE)
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| − | from 3 show 4: "∃x. P x y" by (rule exI)
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| − | qed
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| − | qed
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| − |
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| − | lemma "(∀x. P x ⟶ Q) = ((∃x. P x) ⟶ Q)"
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| − | proof
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| − | { assume 1: "∀x. P x ⟶ Q"
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| − | show "(∃x. P x) ⟶ Q"
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| − | proof
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| − | assume "∃x. P x"
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| − | then obtain a where "P a" by (rule exE)
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| − | have "P a ⟶ Q" using 1 by (rule allE)
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| − | thus "Q" using `P a` by (rule mp)
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| − | qed }
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| − | next
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| − | { assume 2: "(∃x. P x) ⟶ Q"
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| − | show "∀x. P x ⟶ Q"
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| − | proof
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| − | fix a
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| − | show "P a ⟶ Q"
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| − | proof
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| − | assume "P a"
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| − | hence 3: "∃x. P x" by (rule exI)
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| − | from 2 and 3 show "Q" by (rule mp)
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| − | qed
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| − | qed }
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| − | qed
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| − |
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| − | lemma "((∀ x. P x) ∧ (∀ x. Q x)) = (∀ x. (P x ∧ Q x))"
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| − | proof
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| − | {
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| − | assume 1: "(∀ x. P x) ∧ (∀ x. Q x)"
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| − | show "(∀ x. (P x ∧ Q x))"
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| − | proof (rule allI)
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| − | from 1 have 2: "(∀ x. P x)" by (rule conjE)
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| − | from 1 have 3: "(∀ x. Q x)" by (rule conjE)
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| − | fix a
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| − | from 2 have 4: "P a" by (rule allE)
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| − | from 3 have 5: "Q a" by (rule allE)
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| − | from 4 and 5 show 6: "(P a) ∧ (Q a)" by (rule conjI)
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| − | qed
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| − | }
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| − | next
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| − | {
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| − | assume 7: "∀ x. (P x ∧ Q x)"
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| − | show "(∀ x. P x) ∧ (∀ x. Q x)"
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| − | proof
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| − | fix a
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| − | from 7 have 8: "P a ∧ Q a" by (rule allE)
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| − | from 8 have 9: "P a" by (rule conjE)
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| − | (* seguir... *)
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| − | from 8 have 11: "Q a" by (rule allE)
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| − | (* seguir... *)
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| − | qed
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| − | }
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| − | qed
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| − |
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| − | oops
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| − |
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| − | lemma "((∀ x. P x) ∨ (∀ x. Q x)) = (∀ x. (P x ∨ Q x))"
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| − | oops
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| − |
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| − | lemma "((∃ x. P x) ∨ (∃ x. Q x)) = (∃ x. (P x ∨ Q x))"
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| − | oops
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| − |
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| − | lemma "(∀x. ∃y. P x y) ⟶ (∃y. ∀x. P x y)"
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| − | oops
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| − | (*
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| − | proof (rule impI)
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| − | assume 1: "∀x. ∃y. P x y"
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| − | show "∃y. ∀x. P x y"
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| − | proof (rule exI)
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| − | show " ∀x. P x y"
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| − | proof (rule allI)
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| − | fix x
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| − | from 1 have 2: "∃y. P x y" by (rule allE)
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| − | from 2 obtain y where 3: "P x y" by (rule exE)
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| − | thus ?thesis ..
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| − | qed
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| − | *)
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| − |
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| − | lemma "(¬ (∀ x. P x)) = (∃ x. ¬ P x)"
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| − | proof
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| − | {
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| − | assume 1: "¬ (∀ x. P x)"
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| − | show "∃ x. ¬ P x"
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| − | proof
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| − |
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| − |
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| − | qed
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| − | }
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| − | next
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| − | {
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| − | assume 2: "∃ x. ¬ P x"
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| − | show "¬ (∀ x. P x)"
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| − | proof
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| − |
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| − | qed
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| − | }
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| − | qed
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| − |
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| − | end
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