ForMatUS: Pruebas en Lean de ∀x (P(x) ∧ Q(x)) ↔ ∀x P(x) ∧ ∀x Q(x)
He añadido a la lista Lógica con Lean el vídeo en el que se comentan pruebas en Lean de la propiedad
1 |
∀x (P(x) ∧ Q(x)) ↔ ∀x P(x) ∧ ∀x Q(x) |
usando los estilos declarativos, aplicativos, funcional y automático.
A continuación, se muestra el vídeo
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import tactic section variable {U : Type} variables {P Q : U -> Prop} -- ------------------------------------------------------ -- Ej. 1. Demostrar -- ∀x (P(x) ∧ Q(x)) ⊢ ∀x P(x) ∧ ∀x Q(x) -- ------------------------------------------------------ -- 1ª demostración example (h1 : ∀x, P x ∧ Q x) : (∀x, P x) ∧ (∀x, Q x) := have h5 : ∀x, P x, from assume x₀, have h3 : P x₀ ∧ Q x₀, from h1 x₀, show P x₀, from and.elim_left h3, have h9 : ∀x, Q x, from assume x₁, have h7 : P x₁ ∧ Q x₁, from h1 x₁, show Q x₁, from and.elim_right h7, show (∀x, P x) ∧ (∀x, Q x), from and.intro h5 h9 -- 2ª demostración example (h1 : ∀x, P x ∧ Q x) : (∀x, P x) ∧ (∀x, Q x) := have h5 : ∀x, P x, from assume x₀, have h3 : P x₀ ∧ Q x₀, from h1 x₀, show P x₀, from h3.left, have h9 : ∀x, Q x, from assume x₁, have h7 : P x₁ ∧ Q x₁, from h1 x₁, show Q x₁, from h7.right, show (∀x, P x) ∧ (∀x, Q x), from ⟨h5, h9⟩ -- 3ª demostración example (h1 : ∀x, P x ∧ Q x) : (∀x, P x) ∧ (∀x, Q x) := have h5 : ∀x, P x, from assume x₀, have h3 : P x₀ ∧ Q x₀, from h1 x₀, h3.left, have h9 : ∀x, Q x, from assume x₁, have h7 : P x₁ ∧ Q x₁, from h1 x₁, h7.right, show (∀x, P x) ∧ (∀x, Q x), from ⟨h5, h9⟩ -- 4ª demostración example (h1 : ∀x, P x ∧ Q x) : (∀x, P x) ∧ (∀x, Q x) := have h5 : ∀x, P x, from assume x₀, (h1 x₀).left, have h9 : ∀x, Q x, from assume x₁, (h1 x₁).right, show (∀x, P x) ∧ (∀x, Q x), from ⟨h5, h9⟩ -- 5ª demostración example (h1 : ∀x, P x ∧ Q x) : (∀x, P x) ∧ (∀x, Q x) := have h5 : ∀x, P x, from λ x₀, (h1 x₀).left, have h9 : ∀x, Q x, from λ x₁, (h1 x₁).right, show (∀x, P x) ∧ (∀x, Q x), from ⟨h5, h9⟩ -- 6ª demostración example (h1 : ∀x, P x ∧ Q x) : (∀x, P x) ∧ (∀x, Q x) := have h5 : ∀x, P x, from λ x₀, (h1 x₀).left, have h9 : ∀x, Q x, from λ x₁, (h1 x₁).right, ⟨h5, h9⟩ -- 7ª demostración example (h1 : ∀x, P x ∧ Q x) : (∀x, P x) ∧ (∀x, Q x) := ⟨λ x₀, (h1 x₀).left, λ x₁, (h1 x₁).right⟩ -- 8ª demostración example (h1 : ∀x, P x ∧ Q x) : (∀x, P x) ∧ (∀x, Q x) := -- by library_search forall_and_distrib.mp h1 -- 9ª demostración example (h1 : ∀x, P x ∧ Q x) : (∀x, P x) ∧ (∀x, Q x) := begin split, { intro x₀, specialize h1 x₀, exact h1.left, }, { intro x₁, specialize h1 x₁, exact h1.right, }, end -- 9ª demostración lemma aux1 (h1 : ∀x, P x ∧ Q x) : (∀x, P x) ∧ (∀x, Q x) := -- by hint by finish -- ------------------------------------------------------ -- Ej. 2. Demostrar -- ∀x P(x) ∧ ∀x Q(x) ⊢ ∀x (P(x) ∧ Q(x)) -- ------------------------------------------------------ -- 1ª demostración example (h1 : (∀x, P x) ∧ (∀x, Q x)) : ∀x, P x ∧ Q x := assume x₀, have h3 : ∀x, P x, from and.elim_left h1, have h4 : P x₀, from h3 x₀, have h5 : ∀x, Q x, from and.elim_right h1, have h6 : Q x₀, from h5 x₀, show P x₀ ∧ Q x₀, from and.intro h4 h6 -- 2ª demostración example (h1 : (∀x, P x) ∧ (∀x, Q x)) : ∀x, P x ∧ Q x := assume x₀, have h3 : ∀x, P x, from h1.left, have h4 : P x₀, from h3 x₀, have h5 : ∀x, Q x, from h1.right, have h6 : Q x₀, from h5 x₀, show P x₀ ∧ Q x₀, from ⟨h4, h6⟩ -- 3ª demostración example (h1 : (∀x, P x) ∧ (∀x, Q x)) : ∀x, P x ∧ Q x := assume x₀, have h3 : ∀x, P x, from h1.left, have h4 : P x₀, from h3 x₀, have h5 : ∀x, Q x, from h1.right, have h6 : Q x₀, from h5 x₀, ⟨h4, h6⟩ -- 4ª demostración example (h1 : (∀x, P x) ∧ (∀x, Q x)) : ∀x, P x ∧ Q x := assume x₀, have h3 : ∀x, P x, from h1.left, have h4 : P x₀, from h3 x₀, have h5 : ∀x, Q x, from h1.right, ⟨h4, h5 x₀⟩ -- 5ª demostración example (h1 : (∀x, P x) ∧ (∀x, Q x)) : ∀x, P x ∧ Q x := assume x₀, have h3 : ∀x, P x, from h1.left, have h4 : P x₀, from h3 x₀, ⟨h4, h1.right x₀⟩ -- 6ª demostración example (h1 : (∀x, P x) ∧ (∀x, Q x)) : ∀x, P x ∧ Q x := assume x₀, have h3 : ∀x, P x, from h1.left, ⟨h3 x₀, h1.right x₀⟩ -- 7ª demostración example (h1 : (∀x, P x) ∧ (∀x, Q x)) : ∀x, P x ∧ Q x := assume x₀, ⟨h1.left x₀, h1.right x₀⟩ -- 8ª demostración example (h1 : (∀x, P x) ∧ (∀x, Q x)) : ∀x, P x ∧ Q x := λ x₀, ⟨h1.left x₀, h1.right x₀⟩ -- 9ª demostración example (h1 : (∀x, P x) ∧ (∀x, Q x)) : ∀x, P x ∧ Q x := -- by library_search forall_and_distrib.mpr h1 -- 10ª demostración example (h1 : (∀x, P x) ∧ (∀x, Q x)) : ∀x, P x ∧ Q x := begin cases h1 with h2 h3, intro x₀, split, { apply h2, }, { apply h3, }, end -- 11ª demostración example (h1 : (∀x, P x) ∧ (∀x, Q x)) : ∀x, P x ∧ Q x := -- by hint by tauto -- 12ª demostración lemma aux2 (h1 : (∀x, P x) ∧ (∀x, Q x)) : ∀x, P x ∧ Q x := by finish -- ------------------------------------------------------ -- Ej. 3. Demostrar -- ∀x (P(x) ∧ Q(x)) ↔ ∀x P(x) ∧ ∀x Q(x) -- ------------------------------------------------------ -- 1ª demostración example : (∀x, P x ∧ Q x) ↔ (∀x, P x) ∧ (∀x, Q x) := iff.intro aux1 aux2 -- 2ª demostración example : (∀x, P x ∧ Q x) ↔ (∀x, P x) ∧ (∀x, Q x) := -- by library_search forall_and_distrib -- 3ª demostración example : (∀x, P x ∧ Q x) ↔ (∀x, P x) ∧ (∀x, Q x) := -- by hint by finish end |