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 distributiva del existencial sobre la disyunción
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∃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) := exists.elim h1 ( assume x₀ (h2 : P x₀ ∨ Q x₀), or.elim h2 ( assume h3 : P x₀, have h4 : ∃x, P x, from exists.intro x₀ h3, show (∃x, P x) ∨ (∃x, Q x), from or.inl h4 ) ( assume h6 : Q x₀, have h7 : ∃x, Q x, from exists.intro x₀ h6, show (∃x, P x) ∨ (∃x, Q x), from or.inr h7 )) -- 2ª demostración example (h1 : ∃x, P x ∨ Q x) : (∃x, P x) ∨ (∃x, Q x) := exists.elim h1 ( assume x₀ (h2 : P x₀ ∨ Q x₀), or.elim h2 ( assume h3 : P x₀, have h4 : ∃x, P x, from ⟨x₀, h3⟩, show (∃x, P x) ∨ (∃x, Q x), from or.inl h4 ) ( assume h6 : Q x₀, have h7 : ∃x, Q x, from ⟨x₀, h6⟩, show (∃x, P x) ∨ (∃x, Q x), from or.inr h7 )) -- 3ª demostración example (h1 : ∃x, P x ∨ Q x) : (∃x, P x) ∨ (∃x, Q x) := exists.elim h1 ( assume x₀ (h2 : P x₀ ∨ Q x₀), or.elim h2 ( assume h3 : P x₀, have h4 : ∃x, P x, from ⟨x₀, h3⟩, or.inl h4 ) ( assume h6 : Q x₀, have h7 : ∃x, Q x, from ⟨x₀, h6⟩, or.inr h7 )) -- 4ª demostración example (h1 : ∃x, P x ∨ Q x) : (∃x, P x) ∨ (∃x, Q x) := exists.elim h1 ( assume x₀ (h2 : P x₀ ∨ Q x₀), or.elim h2 ( assume h3 : P x₀, or.inl ⟨x₀, h3⟩ ) ( assume h6 : Q x₀, or.inr ⟨x₀, h6⟩ )) -- 5ª demostración example (h1 : ∃x, P x ∨ Q x) : (∃x, P x) ∨ (∃x, Q x) := exists.elim h1 ( assume x₀ (h2 : P x₀ ∨ Q x₀), or.elim h2 ( λ h3, or.inl ⟨x₀, h3⟩ ) ( λ h6, or.inr ⟨x₀, h6⟩ )) -- 6ª demostración example (h1 : ∃x, P x ∨ Q x) : (∃x, P x) ∨ (∃x, Q x) := exists.elim h1 (λ x₀ h2, h2.elim (λ h3, or.inl ⟨x₀, h3⟩) (λ h6, or.inr ⟨x₀, h6⟩)) -- 7ª demostración example (h1 : ∃x, P x ∨ Q x) : (∃x, P x) ∨ (∃x, Q x) := -- by library_search exists_or_distrib.mp h1 -- 8ª demostración example (h1 : ∃x, P x ∨ Q x) : (∃x, P x) ∨ (∃x, Q x) := match h1 with ⟨x₀, (h2 : P x₀ ∨ Q x₀)⟩ := ( or.elim h2 ( assume h3 : P x₀, have h4 : ∃x, P x, from exists.intro x₀ h3, show (∃x, P x) ∨ (∃x, Q x), from or.inl h4 ) ( assume h6 : Q x₀, have h7 : ∃x, Q x, from exists.intro x₀ h6, show (∃x, P x) ∨ (∃x, Q x), from or.inr h7 )) end -- 9ª demostración example (h1 : ∃x, P x ∨ Q x) : (∃x, P x) ∨ (∃x, Q x) := begin cases h1 with x₀ h3, cases h3 with hp hq, { left, use x₀, exact hp, }, { right, use x₀, exact hq, }, end -- 10ª demostración example (h1 : ∃x, P x ∨ Q x) : (∃x, P x) ∨ (∃x, Q x) := begin rcases h1 with ⟨x₀, hp | hq⟩, { left, use x₀, exact hp, }, { right, use x₀, exact hq, }, end -- 11ª 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 := or.elim h1 ( assume h2 : ∃x, P x, exists.elim h2 ( assume x₀ (h3 : P x₀), have h4 : P x₀ ∨ Q x₀, from or.inl h3, show ∃x, P x ∨ Q x, from exists.intro x₀ h4 )) ( assume h2 : ∃x, Q x, exists.elim h2 ( assume x₀ (h3 : Q x₀), have h4 : P x₀ ∨ Q x₀, from or.inr h3, show ∃x, P x ∨ Q x, from exists.intro x₀ h4 )) -- 2ª demostración example (h1 : (∃x, P x) ∨ (∃x, Q x)) : ∃x, P x ∨ Q x := h1.elim ( assume ⟨x₀, (h3 : P x₀)⟩, have h4 : P x₀ ∨ Q x₀, from or.inl h3, show ∃x, P x ∨ Q x, from ⟨x₀, h4⟩ ) ( assume ⟨x₀, (h3 : Q x₀)⟩, have h4 : P x₀ ∨ Q x₀, from or.inr h3, show ∃x, P x ∨ Q x, from ⟨x₀, h4⟩ ) -- 3ª demostración example (h1 : (∃x, P x) ∨ (∃x, Q x)) : ∃x, P x ∨ Q x := h1.elim ( assume ⟨x₀, (h3 : P x₀)⟩, have h4 : P x₀ ∨ Q x₀, from or.inl h3, ⟨x₀, h4⟩ ) ( assume ⟨x₀, (h3 : Q x₀)⟩, have h4 : P x₀ ∨ Q x₀, from or.inr h3, ⟨x₀, h4⟩ ) -- 4ª demostración example (h1 : (∃x, P x) ∨ (∃x, Q x)) : ∃x, P x ∨ Q x := h1.elim ( assume ⟨x₀, (h3 : P x₀)⟩, ⟨x₀, or.inl h3⟩ ) ( assume ⟨x₀, (h3 : Q x₀)⟩, ⟨x₀, or.inr h3⟩ ) -- 5ª demostración example (h1 : (∃x, P x) ∨ (∃x, Q x)) : ∃x, P x ∨ Q x := h1.elim (λ ⟨x₀, h3⟩, ⟨x₀, or.inl h3⟩) (λ ⟨x₀, h3⟩, ⟨x₀, or.inr h3⟩) -- 6ª demostración example (h1 : (∃x, P x) ∨ (∃x, Q x)) : ∃x, P x ∨ Q x := -- by library_search exists_or_distrib.mpr h1 -- 7ª demostración example (h1 : (∃x, P x) ∨ (∃x, Q x)) : ∃x, P x ∨ Q x := begin cases h1 with hp hq, { cases hp with x₀ hx₀, use x₀, left, exact hx₀, }, { cases hq with x₁ hx₁, use x₁, right, exact hx₁, }, end -- 8ª demostración example (h1 : (∃x, P x) ∨ (∃x, Q x)) : ∃x, P x ∨ Q x := begin rcases h1 with ⟨x₀, hx₀⟩ | ⟨x₁, hx₁⟩, { use x₀, left, exact hx₀, }, { use x₁, right, exact hx₁, }, end -- 9ª demostración lemma aux2 (h1 : (∃x, P x) ∨ (∃x, Q x)) : ∃x, P x ∨ Q x := -- by hint 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) := ⟨aux1, aux2⟩ -- 3ª demostración example : (∃x, P x ∨ Q x) ↔ (∃x, P x) ∨ (∃x, Q x) := -- by library_search exists_or_distrib end |