Diferencia de diferencia de conjuntos
Demostrar que (s \ t) \ u ⊆ s \ (t ∪ u)
Para ello, completar la siguiente teoría de Lean:
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import data.set.basic open set variable {α : Type} variables s t u : set α example : (s \ t) \ u ⊆ s \ (t ∪ u) := sorry |
Soluciones con Lean
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import data.set.basic open set variable {α : Type} variables s t u : set α -- 1ª demostración -- =============== example : (s \ t) \ u ⊆ s \ (t ∪ u) := begin intros x hx, cases hx with hxst hxnu, cases hxst with hxs hxnt, split, { exact hxs }, { dsimp, by_contradiction hxtu, cases hxtu with hxt hxu, { apply hxnt, exact hxt, }, { apply hxnu, exact hxu, }}, end -- 2ª demostración -- =============== example : (s \ t) \ u ⊆ s \ (t ∪ u) := begin rintros x ⟨⟨hxs, hxnt⟩, hxnu⟩, split, { exact hxs }, { by_contradiction hxtu, cases hxtu with hxt hxu, { exact hxnt hxt, }, { exact hxnu hxu, }}, end -- 3ª demostración -- =============== example : (s \ t) \ u ⊆ s \ (t ∪ u) := begin rintros x ⟨⟨xs, xnt⟩, xnu⟩, use xs, rintros (xt | xu), { contradiction, }, { contradiction, }, end -- 4ª demostración -- =============== example : (s \ t) \ u ⊆ s \ (t ∪ u) := begin rintros x ⟨⟨xs, xnt⟩, xnu⟩, use xs, rintros (xt | xu); contradiction, end -- 5ª demostración -- =============== example : (s \ t) \ u ⊆ s \ (t ∪ u) := begin intros x xstu, simp at *, finish, end -- 6ª demostración -- =============== example : (s \ t) \ u ⊆ s \ (t ∪ u) := begin intros x xstu, finish, end -- 7ª demostración -- =============== example : (s \ t) \ u ⊆ s \ (t ∪ u) := by rw diff_diff -- 8ª demostración -- =============== example : (s \ t) \ u ⊆ s \ (t ∪ u) := by tidy |
El código de las demostraciones se encuentra en GitHub y puede ejecutarse con el Lean Web editor.
La construcción de las demostraciones se muestra en el siguiente vídeo
Soluciones con Isabelle/HOL
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theory Diferencia_de_diferencia_de_conjuntos imports Main begin (* 1ª demostración *) lemma "(s - t) - u ⊆ s - (t ∪ u)" proof (rule subsetI) fix x assume hx : "x ∈ (s - t) - u" then show "x ∈ s - (t ∪ u)" proof (rule DiffE) assume xst : "x ∈ s - t" assume xnu : "x ∉ u" note xst then show "x ∈ s - (t ∪ u)" proof (rule DiffE) assume xs : "x ∈ s" assume xnt : "x ∉ t" have xntu : "x ∉ t ∪ u" proof (rule notI) assume xtu : "x ∈ t ∪ u" then show False proof (rule UnE) assume xt : "x ∈ t" with xnt show False by (rule notE) next assume xu : "x ∈ u" with xnu show False by (rule notE) qed qed show "x ∈ s - (t ∪ u)" using xs xntu by (rule DiffI) qed qed qed (* 2ª demostración *) lemma "(s - t) - u ⊆ s - (t ∪ u)" proof fix x assume hx : "x ∈ (s - t) - u" then have xst : "x ∈ (s - t)" by simp then have xs : "x ∈ s" by simp have xnt : "x ∉ t" using xst by simp have xnu : "x ∉ u" using hx by simp have xntu : "x ∉ t ∪ u" using xnt xnu by simp then show "x ∈ s - (t ∪ u)" using xs by simp qed (* 3ª demostración *) lemma "(s - t) - u ⊆ s - (t ∪ u)" proof fix x assume "x ∈ (s - t) - u" then show "x ∈ s - (t ∪ u)" by simp qed (* 4ª demostración *) lemma "(s - t) - u ⊆ s - (t ∪ u)" by auto |
