Praeclarum theorema
Demostrar el Praeclarum theorema de Leibniz:
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theory Praeclarum_theorema imports Main begin lemma "(p ⟶ q) ∧ (r ⟶ s) ⟶ ((p ∧ r) ⟶ (q ∧ s))" oops end |
Soluciones con Isabelle/HOL
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theory Praeclarum_theorema imports Main begin (* 1ª demostración: automática *) lemma "(p ⟶ q) ∧ (r ⟶ s) ⟶ ((p ∧ r) ⟶ (q ∧ s))" by simp (* 2ª demostración: aplicativa *) lemma "(p ⟶ q) ∧ (r ⟶ s) ⟶ ((p ∧ r) ⟶ (q ∧ s))" apply (rule impI) apply (rule impI) apply (erule conjE)+ apply (rule conjI) apply (erule mp) apply assumption apply (erule mp) apply assumption done (* 3ª demostración: estructurada *) lemma "(p ⟶ q) ∧ (r ⟶ s) ⟶ ((p ∧ r) ⟶ (q ∧ s))" proof assume "(p ⟶ q) ∧ (r ⟶ s)" show "(p ∧ r) ⟶ (q ∧ s)" proof assume "p ∧ r" show "q ∧ s" proof have "p ⟶ q" using ‹(p ⟶ q) ∧ (r ⟶ s)› .. moreover have "p" using ‹p ∧ r› .. ultimately show "q" .. next have "r ⟶ s" using ‹(p ⟶ q) ∧ (r ⟶ s)› .. moreover have "r" using ‹p ∧ r› .. ultimately show "s" .. qed qed qed (* 4ª demostración: detallada *) lemma "(p ⟶ q) ∧ (r ⟶ s) ⟶ ((p ∧ r) ⟶ (q ∧ s))" proof (rule impI) assume "(p ⟶ q) ∧ (r ⟶ s)" show "(p ∧ r) ⟶ (q ∧ s)" proof (rule impI) assume "p ∧ r" show "q ∧ s" proof (rule conjI) have "p ⟶ q" using ‹(p ⟶ q) ∧ (r ⟶ s)› by (rule conjunct1) moreover have "p" using ‹p ∧ r› by (rule conjunct1) ultimately show "q" by (rule mp) next have "r ⟶ s" using ‹(p ⟶ q) ∧ (r ⟶ s)› by (rule conjunct2) moreover have "r" using ‹p ∧ r› by (rule conjunct2) ultimately show "s" by (rule mp) qed qed qed end |
Otras soluciones
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