ForMatUS: Pruebas en Lean de ¬∀x P(x) ↔ ∃x ¬P(x)
He añadido a la lista Lógica con Lean el vídeo en el que se comentan pruebas en Lean de la propiedad
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¬∀x P(x) ↔ ∃x ¬P(x) |
usando los estilos declarativos, aplicativos, funcional y automático.
A continuación, se muestra el vídeo
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import tactic variable {U : Type} variable {P : U -> Prop} -- ------------------------------------------------------ -- Ej. 1. Demostrar que -- ¬∀x P(x) ⊢ ∃x ¬P(x) -- ------------------------------------------------------ -- 1ª demostración example (h1 : ¬∀x, P x) : ∃x, ¬P x := by_contra ( assume h2 : ¬∃x, ¬P x, have h8 : ∀x, P x, from ( assume x₀, show P x₀, from by_contra ( assume h4 : ¬P x₀, have h5 : ∃x, ¬P x, from exists.intro x₀ h4, show false, from h2 h5 )), show false, from h1 h8) -- 2ª demostración example (h1 : ¬∀x, P x) : ∃x, ¬P x := by_contra ( assume h2 : ¬∃x, ¬P x, have h8 : ∀x, P x, from ( assume x₀, show P x₀, from by_contra ( assume h4 : ¬P x₀, have h5 : ∃x, ¬P x, from exists.intro x₀ h4, show false, from h2 h5 )), h1 h8) -- 3ª demostración example (h1 : ¬∀x, P x) : ∃x, ¬P x := by_contra ( assume h2 : ¬∃x, ¬P x, have h8 : ∀x, P x, from ( assume x₀, show P x₀, from by_contra ( assume h4 : ¬P x₀, have h5 : ∃x, ¬P x, from exists.intro x₀ h4, h2 h5 )), h1 h8) -- 4ª demostración example (h1 : ¬∀x, P x) : ∃x, ¬P x := by_contra ( assume h2 : ¬∃x, ¬P x, have h8 : ∀x, P x, from ( assume x₀, show P x₀, from by_contra ( assume h4 : ¬P x₀, have h5 : ∃x, ¬P x, from ⟨x₀, h4⟩, h2 h5 )), h1 h8) -- 5ª demostración example (h1 : ¬∀x, P x) : ∃x, ¬P x := by_contra ( assume h2 : ¬∃x, ¬P x, have h8 : ∀x, P x, from ( assume x₀, show P x₀, from by_contra ( assume h4 : ¬P x₀, h2 ⟨x₀, h4⟩ )), h1 h8) -- 6ª demostración example (h1 : ¬∀x, P x) : ∃x, ¬P x := by_contra ( assume h2 : ¬∃x, ¬P x, have h8 : ∀x, P x, from ( assume x₀, show P x₀, from by_contra (λ h4, h2 ⟨x₀, h4⟩)), h1 h8) -- 7ª demostración example (h1 : ¬∀x, P x) : ∃x, ¬P x := by_contra ( assume h2 : ¬∃x, ¬P x, have h8 : ∀x, P x, from ( assume x₀, by_contra (λ h4, h2 ⟨x₀, h4⟩)), h1 h8) -- 8ª demostración example (h1 : ¬∀x, P x) : ∃x, ¬P x := by_contra ( assume h2 : ¬∃x, ¬P x, have h8 : ∀x, P x, from (λ x₀, by_contra (λ h4, h2 ⟨x₀, h4⟩)), h1 h8) -- 9ª demostración example (h1 : ¬∀x, P x) : ∃x, ¬P x := by_contra ( assume h2 : ¬∃x, ¬P x, h1 (λ x₀, by_contra (λ h4, h2 ⟨x₀, h4⟩))) -- 10ª demostración example (h1 : ¬∀x, P x) : ∃x, ¬P x := by_contra (λ h2, h1 (λ x₀, by_contra (λ h4, h2 ⟨x₀, h4⟩))) -- 11ª demostración example (h1 : ¬∀x, P x) : ∃x, ¬P x := -- by library_search not_forall.mp h1 -- 12ª demostración lemma aux1 (h1 : ¬∀x, P x) : ∃x, ¬P x := -- by hint by finish -- ------------------------------------------------------ -- Ej. 2. Demostrar que -- ∃x ¬P(x) ⊢ ¬∀x P(x) -- ------------------------------------------------------ -- 1ª demostración example (h1 : ∃x, ¬P x) : ¬∀x, P x := assume h2 : ∀x, P x, exists.elim h1 ( assume x₀ (h3 : ¬P x₀), have h4 : P x₀, from h2 x₀, show false, from h3 h4) -- 2ª demostración example (h1 : ∃x, ¬P x) : ¬∀x, P x := assume h2 : ∀x, P x, exists.elim h1 ( assume x₀ (h3 : ¬P x₀), have h4 : P x₀, from h2 x₀, h3 h4) -- 3ª demostración example (h1 : ∃x, ¬P x) : ¬∀x, P x := assume h2 : ∀x, P x, exists.elim h1 ( assume x₀ (h3 : ¬P x₀), h3 (h2 x₀)) -- 4ª demostración example (h1 : ∃x, ¬P x) : ¬∀x, P x := assume h2 : ∀x, P x, exists.elim h1 (λ x₀ h3, h3 (h2 x₀)) -- 5ª demostración example (h1 : ∃x, ¬P x) : ¬∀x, P x := λ h2, exists.elim h1 (λ x₀ h3, h3 (h2 x₀)) -- 6ª demostración example (h1 : ∃x, ¬P x) : ¬∀x, P x := -- by library_search not_forall.mpr h1 -- 7ª demostración example (h1 : ∃x, ¬P x) : ¬∀x, P x := assume h2 : ∀x, P x, match h1 with ⟨x₀, (h3 : ¬P x₀)⟩ := ( have h4 : P x₀, from h2 x₀, show false, from h3 h4) end -- 8ª demostración example (h1 : ∃x, ¬P x) : ¬∀x, P x := begin intro h2, cases h1 with x₀ h3, apply h3, apply h2, end example (h1 : ∃x, ¬P x) : ¬∀x, P x := begin intro h2, obtain ⟨x₀, h3⟩ := h1, apply h3, apply h2, end -- 9ª demostración example (h1 : ∃x, ¬P x) : ¬∀x, P x := -- by hint by tauto -- 10ª demostración lemma aux2 (h1 : ∃x, ¬P x) : ¬∀x, P x := by finish #print axioms aux2 -- ------------------------------------------------------ -- Ej. 3. Demostrar que -- ¬∀x P(x) ↔ ∃x ¬P(x) -- ------------------------------------------------------ -- 1ª demostración example : (¬∀x, P x) ↔ (∃x, ¬P x) := iff.intro ( assume h1 : ¬∀x, P x, show ∃x, ¬P x, from aux1 h1) ( assume h2 : ∃x, ¬P x, show ¬∀x, P x, from aux2 h2) -- 2ª demostración example : (¬∀x, P x) ↔ (∃x, ¬P x) := iff.intro aux1 aux2 -- 3ª demostración example : (¬∀x, P x) ↔ (∃x, ¬P x) := -- by library_search not_forall -- 4ª demostración example : (¬∀x, P x) ↔ (∃x, ¬P x) := begin split, { exact aux1, }, { exact aux2, }, end -- 5ª demostración example : (¬∀x, P x) ↔ (∃x, ¬P x) := -- by hint by finish |