ForMatUS: Pruebas en Lean de la propiedad distributiva de la intersección sobre la unión
He añadido a la lista Lógica con Lean el vídeo en el que se comentan pruebas en Lean de la propiedad distributiva de la intersección sobre la unión:
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A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) |
usando los estilos declarativos, aplicativos, funcional y automático.
A continuación, se muestra el vídeo
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import data.set open set variable {U : Type} variables A B C : set U -- ---------------------------------------------------- -- Ej. 1. Demostrar -- A ∩ (B ∪ C) ⊆ (A ∩ B) ∪ (A ∩ C) -- ---------------------------------------------------- -- 1ª demostración example : A ∩ (B ∪ C) ⊆ (A ∩ B) ∪ (A ∩ C) := begin intros x h, cases h with ha hbc, cases hbc with hb hc, { left, split, { exact ha, }, { exact hb, }}, { right, split, { exact ha, }, { exact hc, }}, end -- 2ª demostración example : A ∩ (B ∪ C) ⊆ (A ∩ B) ∪ (A ∩ C) := begin intros x h, cases h with ha hbc, cases hbc with hb hc, { left, split, { assumption, }, { assumption, }}, { right, split, { assumption, }, { assumption, }}, end -- 3ª demostración example : A ∩ (B ∪ C) ⊆ (A ∩ B) ∪ (A ∩ C) := begin intros x h, cases h with ha hbc, cases hbc with hb hc, { left, split, assumption', }, { right, split, assumption', }, end -- 4ª demostración example : A ∩ (B ∪ C) ⊆ (A ∩ B) ∪ (A ∩ C) := begin rintros x ⟨ha, (hb | hc)⟩, { left, split, assumption', }, { right, split, assumption', }, end -- 5ª demostración example : A ∩ (B ∪ C) ⊆ (A ∩ B) ∪ (A ∩ C) := assume x, assume h : x ∈ A ∩ (B ∪ C), have x ∈ A, from and.left h, have x ∈ B ∪ C, from and.right h, or.elim (‹x ∈ B ∪ C›) ( assume : x ∈ B, have x ∈ A ∩ B, from and.intro ‹x ∈ A› ‹x ∈ B›, show x ∈ (A ∩ B) ∪ (A ∩ C), from or.inl this) ( assume : x ∈ C, have x ∈ A ∩ C, from and.intro ‹x ∈ A› ‹x ∈ C›, show x ∈ (A ∩ B) ∪ (A ∩ C), from or.inr this) -- 6ª demostración lemma inter_union_l1 : A ∩ (B ∪ C) ⊆ (A ∩ B) ∪ (A ∩ C) := assume x, assume h : x ∈ A ∩ (B ∪ C), have ha : x ∈ A, from and.left h, have hbc : x ∈ B ∪ C, from and.right h, or.elim hbc ( assume hb : x ∈ B, have hab: x ∈ A ∩ B, from and.intro ha hb, show x ∈ (A ∩ B) ∪ (A ∩ C), from or.inl hab) ( assume hc : x ∈ C, have hac : x ∈ A ∩ C, from and.intro ha hc, show x ∈ (A ∩ B) ∪ (A ∩ C), from or.inr hac) -- ---------------------------------------------------- -- Ej. 2. Demostrar -- (A ∩ B) ∪ (A ∩ C) ⊆ A ∩ (B ∪ C) -- ---------------------------------------------------- -- 1ª demostración example : (A ∩ B) ∪ (A ∩ C) ⊆ A ∩ (B ∪ C) := begin intros x h, cases h with hab hac, { split, { exact hab.left, }, { left, exact hab.right, }}, { split, { exact hac.left, }, { right, exact hac.right, }}, end -- 2ª demostración example : (A ∩ B) ∪ (A ∩ C) ⊆ A ∩ (B ∪ C) := begin rintros x (⟨ha, hb⟩ | ⟨ha, hc⟩), { split, { exact ha, }, { left, exact hb, }}, { split, { exact ha, }, { right, exact hc, }}, end -- 3ª demostración lemma inter_union_l2 : (A ∩ B) ∪ (A ∩ C) ⊆ A ∩ (B ∪ C) := assume x, assume : x ∈ (A ∩ B) ∪ (A ∩ C), or.elim this ( assume h : x ∈ A ∩ B, have x ∈ A, from and.left h, have x ∈ B, from and.right h, have x ∈ B ∪ C, from or.inl this, show x ∈ A ∩ (B ∪ C), from and.intro ‹x ∈ A› this) ( assume h : x ∈ A ∩ C, have x ∈ A, from and.left h, have x ∈ C, from and.right h, have x ∈ B ∪ C, from or.inr this, show x ∈ A ∩ (B ∪ C), from and.intro ‹x ∈ A› this) -- ---------------------------------------------------- -- Ej. 3. Demostrar -- (A ∩ B) ∪ (A ∩ C) = A ∩ (B ∪ C) -- ---------------------------------------------------- -- 1ª demostración example : A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) := -- by library_search inter_distrib_left A B C -- 2ª demostración theorem inter_union : A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) := eq_of_subset_of_subset (inter_union_l1 A B C) (inter_union_l2 A B C) -- 3ª demostración example : A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) := begin ext, simp, exact and_or_distrib_left, end -- 4ª demostración example : A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) := begin ext, exact and_or_distrib_left, end -- 5ª demostración example : A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) := ext (λ x, and_or_distrib_left) |