(s \ t) ∪ (t \ s) = (s ∪ t) \ (s ∩ t)
Demostrar con Lean4 que
\[ (s \setminus t) ∪ (t \setminus s) = (s ∪ t) \setminus (s ∩ t) \]
Para ello, completar la siguiente teoría de Lean4:
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import Mathlib.Data.Set.Basic open Set variable {α : Type} variable (s t : Set α) example : (s \\ t) ∪ (t \\ s) = (s ∪ t) \\ (s ∩ t) := by sorry |
1. Demostración en lenguaje natural
Tenemos que demostrar que, para todo \(x\),
\[ x ∈ (s \setminus t) ∪ (t \setminus s) ↔ x ∈ (s ∪ t) \setminus (s ∩ t) \]
Se demuestra mediante la siguiente cadena de equivalencias:
\begin{align}
&x ∈ (s \setminus t) ∪ (t \setminus s) \\
↔ &x ∈ (s \setminus t) ∨ x ∈ (t \setminus s) \\
↔ &(x ∈ s ∧ x ∉ t) ∨ x ∈ (t \setminus s) \\
↔ &(x ∈ s ∨ x ∈ (t \ s)) ∧ (x ∉ t ∨ x ∈ (t \setminus s)) \\
↔ &(x ∈ s ∨ (x ∈ t ∧ x ∉ s)) ∧ (x ∉ t ∨ (x ∈ t ∧ x ∉ s)) \\
↔ &((x ∈ s ∨ x ∈ t) ∧ (x ∈ s ∨ x ∉ s)) ∧ ((x ∉ t ∨ x ∈ t) ∧ (x ∉ t ∨ x ∉ s)) \\
↔ &(x ∈ s ∨ x ∈ t) ∧ (x ∉ t ∨ x ∉ s) \\
↔ &(x ∈ s ∪ t) ∧ (x ∉ t ∨ x ∉ s) \\
↔ &(x ∈ s ∪ t) ∧ (x ∉ s ∨ x ∉ t) \\
↔ &(x ∈ s ∪ t) ∧ ¬(x ∈ s ∧ x ∈ t) \\
↔ &(x ∈ s ∪ t) ∧ ¬(x ∈ s ∩ t) \\
↔ &x ∈ (s ∪ t) \setminus (s ∩ t)
\end{align}
2. Demostraciones con Lean4
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import Mathlib.Data.Set.Basic open Set variable {α : Type} variable (s t : Set α) -- 1ª demostración -- =============== example : (s \ t) ∪ (t \ s) = (s ∪ t) \ (s ∩ t) := by ext x -- x : α -- ⊢ x ∈ (s \ t) ∪ (t \ s) ↔ x ∈ (s ∪ t) \ (s ∩ t) calc x ∈ (s \ t) ∪ (t \ s) ↔ x ∈ (s \ t) ∨ x ∈ (t \ s) := by exact mem_union x (s \ t) (t \ s) _ ↔ (x ∈ s ∧ x ∉ t) ∨ x ∈ (t \ s) := by simp only [mem_diff] _ ↔ (x ∈ s ∨ x ∈ (t \ s)) ∧ (x ∉ t ∨ x ∈ (t \ s)) := by exact and_or_right _ ↔ (x ∈ s ∨ (x ∈ t ∧ x ∉ s)) ∧ (x ∉ t ∨ (x ∈ t ∧ x ∉ s)) := by simp only [mem_diff] _ ↔ ((x ∈ s ∨ x ∈ t) ∧ (x ∈ s ∨ x ∉ s)) ∧ ((x ∉ t ∨ x ∈ t) ∧ (x ∉ t ∨ x ∉ s)) := by simp_all only [or_and_left] _ ↔ ((x ∈ s ∨ x ∈ t) ∧ True) ∧ (True ∧ (x ∉ t ∨ x ∉ s)) := by simp only [em (x ∈ s), em' (x ∈ t)] _ ↔ (x ∈ s ∨ x ∈ t) ∧ (x ∉ t ∨ x ∉ s) := by simp only [and_true_iff (x ∈ s ∨ x ∈ t), true_and_iff (¬x ∈ t ∨ ¬x ∈ s)] _ ↔ (x ∈ s ∪ t) ∧ (x ∉ t ∨ x ∉ s) := by simp only [mem_union] _ ↔ (x ∈ s ∪ t) ∧ (x ∉ s ∨ x ∉ t) := by simp only [or_comm] _ ↔ (x ∈ s ∪ t) ∧ ¬(x ∈ s ∧ x ∈ t) := by simp only [not_and_or] _ ↔ (x ∈ s ∪ t) ∧ ¬(x ∈ s ∩ t) := by simp only [mem_inter_iff] _ ↔ x ∈ (s ∪ t) \ (s ∩ t) := by simp only [mem_diff] -- 2ª demostración -- =============== example : (s \ t) ∪ (t \ s) = (s ∪ t) \ (s ∩ t) := by ext x -- x : α -- ⊢ x ∈ (s \ t) ∪ (t \ s) ↔ x ∈ (s ∪ t) \ (s ∩ t) constructor . -- ⊢ x ∈ (s \ t) ∪ (t \ s) → x ∈ (s ∪ t) \ (s ∩ t) rintro (⟨xs, xnt⟩ | ⟨xt, xns⟩) . -- xs : x ∈ s -- xnt : ¬x ∈ t -- ⊢ x ∈ (s ∪ t) \ (s ∩ t) constructor . -- ⊢ x ∈ s ∪ t left -- ⊢ x ∈ s exact xs . -- ⊢ ¬x ∈ s ∩ t rintro ⟨-, xt⟩ -- xt : x ∈ t -- ⊢ False exact xnt xt . -- xt : x ∈ t -- xns : ¬x ∈ s -- ⊢ x ∈ (s ∪ t) \ (s ∩ t) constructor . -- ⊢ x ∈ s ∪ t right -- ⊢ x ∈ t exact xt . -- ⊢ ¬x ∈ s ∩ t rintro ⟨xs, -⟩ -- xs : x ∈ s -- ⊢ False exact xns xs . -- ⊢ x ∈ (s ∪ t) \ (s ∩ t) → x ∈ (s \ t) ∪ (t \ s) rintro ⟨xs | xt, nxst⟩ . -- xs : x ∈ s -- ⊢ x ∈ (s \ t) ∪ (t \ s) left -- ⊢ x ∈ s \ t use xs -- ⊢ ¬x ∈ t intro xt -- xt : x ∈ t -- ⊢ False apply nxst -- ⊢ x ∈ s ∩ t constructor . -- ⊢ x ∈ s exact xs . -- ⊢ x ∈ t exact xt . -- nxst : ¬x ∈ s ∩ t -- xt : x ∈ t -- ⊢ x ∈ (s \ t) ∪ (t \ s) right -- ⊢ x ∈ t \ s use xt -- ⊢ ¬x ∈ s intro xs -- xs : x ∈ s -- ⊢ False apply nxst -- ⊢ x ∈ s ∩ t constructor . -- ⊢ x ∈ s exact xs . -- ⊢ x ∈ t exact xt -- 3ª demostración -- =============== example : (s \ t) ∪ (t \ s) = (s ∪ t) \ (s ∩ t) := by ext x -- x : α -- ⊢ x ∈ (s \ t) ∪ (t \ s) ↔ x ∈ (s ∪ t) \ (s ∩ t) constructor . -- ⊢ x ∈ (s \ t) ∪ (t \ s) → x ∈ (s ∪ t) \ (s ∩ t) rintro (⟨xs, xnt⟩ | ⟨xt, xns⟩) . -- xt : x ∈ t -- xns : ¬x ∈ s -- ⊢ x ∈ (s ∪ t) \ (s ∩ t) aesop . -- xt : x ∈ t -- xns : ¬x ∈ s -- ⊢ x ∈ (s ∪ t) \ (s ∩ t) aesop . rintro ⟨xs | xt, nxst⟩ . -- xs : x ∈ s -- ⊢ x ∈ (s \ t) ∪ (t \ s) aesop . -- nxst : ¬x ∈ s ∩ t -- xt : x ∈ t -- ⊢ x ∈ (s \ t) ∪ (t \ s) aesop -- 4ª demostración -- =============== example : (s \ t) ∪ (t \ s) = (s ∪ t) \ (s ∩ t) := by ext x -- x : α -- ⊢ x ∈ (s \ t) ∪ (t \ s) ↔ x ∈ (s ∪ t) \ (s ∩ t) constructor . -- ⊢ x ∈ (s \ t) ∪ (t \ s) → x ∈ (s ∪ t) \ (s ∩ t) rintro (⟨xs, xnt⟩ | ⟨xt, xns⟩) <;> aesop . -- ⊢ x ∈ (s ∪ t) \ (s ∩ t) → x ∈ (s \ t) ∪ (t \ s) rintro ⟨xs | xt, nxst⟩ <;> aesop -- 5ª demostración -- =============== example : (s \ t) ∪ (t \ s) = (s ∪ t) \ (s ∩ t) := by ext constructor . aesop . aesop -- 6ª demostración -- =============== example : (s \ t) ∪ (t \ s) = (s ∪ t) \ (s ∩ t) := by ext constructor <;> aesop -- 7ª demostración -- =============== example : (s \ t) ∪ (t \ s) = (s ∪ t) \ (s ∩ t) := by rw [ext_iff] -- ⊢ ∀ (x : α), x ∈ (s \ t) ∪ (t \ s) ↔ x ∈ (s ∪ t) \ (s ∩ t) intro -- x : α -- ⊢ x ∈ (s \ t) ∪ (t \ s) ↔ x ∈ (s ∪ t) \ (s ∩ t) rw [iff_def] -- ⊢ (x ∈ (s \ t) ∪ (t \ s) → x ∈ (s ∪ t) \ (s ∩ t)) ∧ -- (x ∈ (s ∪ t) \ (s ∩ t) → x ∈ (s \ t) ∪ (t \ s)) aesop -- Lemas usados -- ============ -- variable (x : α) -- variable (a b c : Prop) -- #check (mem_union x s t : x ∈ s ∪ t ↔ x ∈ s ∨ x ∈ t) -- #check (mem_diff x : x ∈ s \ t ↔ x ∈ s ∧ ¬x ∈ t) -- #check (and_or_right : (a ∧ b) ∨ c ↔ (a ∨ c) ∧ (b ∨ c)) -- #check (or_and_left : a ∨ (b ∧ c) ↔ (a ∨ b) ∧ (a ∨ c)) -- #check (em a : a ∨ ¬ a) -- #check (em' a : ¬ a ∨ a) -- #check (and_true_iff a : a ∧ True ↔ a) -- #check (true_and_iff a : True ∧ a ↔ a) -- #check (or_comm : a ∨ b ↔ b ∨ a) -- #check (not_and_or : ¬(a ∧ b) ↔ ¬a ∨ ¬b) -- #check (mem_inter_iff x s t : x ∈ s ∩ t ↔ x ∈ s ∧ x ∈ t) -- #check (ext_iff : s = t ↔ ∀ (x : α), x ∈ s ↔ x ∈ t) -- #check (iff_def : (a ↔ b) ↔ (a → b) ∧ (b → a)) |
Se puede interactuar con las demostraciones anteriores en Lean 4 Web.
3. Demostraciones con Isabelle/HOL
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theory Diferencia_de_union_e_interseccion imports Main begin (* 1 demostración *) lemma "(s - t) ∪ (t - s) = (s ∪ t) - (s ∩ t)" proof (rule equalityI) show "(s - t) ∪ (t - s) ⊆ (s ∪ t) - (s ∩ t)" proof (rule subsetI) fix x assume "x ∈ (s - t) ∪ (t - s)" then show "x ∈ (s ∪ t) - (s ∩ t)" proof (rule UnE) assume "x ∈ s - t" then show "x ∈ (s ∪ t) - (s ∩ t)" proof (rule DiffE) assume "x ∈ s" assume "x ∉ t" have "x ∈ s ∪ t" using ‹x ∈ s› by (simp only: UnI1) moreover have "x ∉ s ∩ t" proof (rule notI) assume "x ∈ s ∩ t" then have "x ∈ t" by (simp only: IntD2) with ‹x ∉ t› show False by (rule notE) qed ultimately show "x ∈ (s ∪ t) - (s ∩ t)" by (rule DiffI) qed next assume "x ∈ t - s" then show "x ∈ (s ∪ t) - (s ∩ t)" proof (rule DiffE) assume "x ∈ t" assume "x ∉ s" have "x ∈ s ∪ t" using ‹x ∈ t› by (simp only: UnI2) moreover have "x ∉ s ∩ t" proof (rule notI) assume "x ∈ s ∩ t" then have "x ∈ s" by (simp only: IntD1) with ‹x ∉ s› show False by (rule notE) qed ultimately show "x ∈ (s ∪ t) - (s ∩ t)" by (rule DiffI) qed qed qed next show "(s ∪ t) - (s ∩ t) ⊆ (s - t) ∪ (t - s)" proof (rule subsetI) fix x assume "x ∈ (s ∪ t) - (s ∩ t)" then show "x ∈ (s - t) ∪ (t - s)" proof (rule DiffE) assume "x ∈ s ∪ t" assume "x ∉ s ∩ t" note ‹x ∈ s ∪ t› then show "x ∈ (s - t) ∪ (t - s)" proof (rule UnE) assume "x ∈ s" have "x ∉ t" proof (rule notI) assume "x ∈ t" with ‹x ∈ s› have "x ∈ s ∩ t" by (rule IntI) with ‹x ∉ s ∩ t› show False by (rule notE) qed with ‹x ∈ s› have "x ∈ s - t" by (rule DiffI) then show "x ∈ (s - t) ∪ (t - s)" by (simp only: UnI1) next assume "x ∈ t" have "x ∉ s" proof (rule notI) assume "x ∈ s" then have "x ∈ s ∩ t" using ‹x ∈ t› by (rule IntI) with ‹x ∉ s ∩ t› show False by (rule notE) qed with ‹x ∈ t› have "x ∈ t - s" by (rule DiffI) then show "x ∈ (s - t) ∪ (t - s)" by (rule UnI2) qed qed qed qed (* 2 demostración *) lemma "(s - t) ∪ (t - s) = (s ∪ t) - (s ∩ t)" proof show "(s - t) ∪ (t - s) ⊆ (s ∪ t) - (s ∩ t)" proof fix x assume "x ∈ (s - t) ∪ (t - s)" then show "x ∈ (s ∪ t) - (s ∩ t)" proof assume "x ∈ s - t" then show "x ∈ (s ∪ t) - (s ∩ t)" proof assume "x ∈ s" assume "x ∉ t" have "x ∈ s ∪ t" using ‹x ∈ s› by simp moreover have "x ∉ s ∩ t" proof assume "x ∈ s ∩ t" then have "x ∈ t" by simp with ‹x ∉ t› show False by simp qed ultimately show "x ∈ (s ∪ t) - (s ∩ t)" by simp qed next assume "x ∈ t - s" then show "x ∈ (s ∪ t) - (s ∩ t)" proof assume "x ∈ t" assume "x ∉ s" have "x ∈ s ∪ t" using ‹x ∈ t› by simp moreover have "x ∉ s ∩ t" proof assume "x ∈ s ∩ t" then have "x ∈ s" by simp with ‹x ∉ s› show False by simp qed ultimately show "x ∈ (s ∪ t) - (s ∩ t)" by simp qed qed qed next show "(s ∪ t) - (s ∩ t) ⊆ (s - t) ∪ (t - s)" proof fix x assume "x ∈ (s ∪ t) - (s ∩ t)" then show "x ∈ (s - t) ∪ (t - s)" proof assume "x ∈ s ∪ t" assume "x ∉ s ∩ t" note ‹x ∈ s ∪ t› then show "x ∈ (s - t) ∪ (t - s)" proof assume "x ∈ s" have "x ∉ t" proof assume "x ∈ t" with ‹x ∈ s› have "x ∈ s ∩ t" by simp with ‹x ∉ s ∩ t› show False by simp qed with ‹x ∈ s› have "x ∈ s - t" by simp then show "x ∈ (s - t) ∪ (t - s)" by simp next assume "x ∈ t" have "x ∉ s" proof assume "x ∈ s" then have "x ∈ s ∩ t" using ‹x ∈ t› by simp with ‹x ∉ s ∩ t› show False by simp qed with ‹x ∈ t› have "x ∈ t - s" by simp then show "x ∈ (s - t) ∪ (t - s)" by simp qed qed qed qed (* 3ª demostración *) lemma "(s - t) ∪ (t - s) = (s ∪ t) - (s ∩ t)" proof show "(s - t) ∪ (t - s) ⊆ (s ∪ t) - (s ∩ t)" proof fix x assume "x ∈ (s - t) ∪ (t - s)" then show "x ∈ (s ∪ t) - (s ∩ t)" proof assume "x ∈ s - t" then show "x ∈ (s ∪ t) - (s ∩ t)" by simp next assume "x ∈ t - s" then show "x ∈ (s ∪ t) - (s ∩ t)" by simp qed qed next show "(s ∪ t) - (s ∩ t) ⊆ (s - t) ∪ (t - s)" proof fix x assume "x ∈ (s ∪ t) - (s ∩ t)" then show "x ∈ (s - t) ∪ (t - s)" proof assume "x ∈ s ∪ t" assume "x ∉ s ∩ t" note ‹x ∈ s ∪ t› then show "x ∈ (s - t) ∪ (t - s)" proof assume "x ∈ s" then show "x ∈ (s - t) ∪ (t - s)" using ‹x ∉ s ∩ t› by simp next assume "x ∈ t" then show "x ∈ (s - t) ∪ (t - s)" using ‹x ∉ s ∩ t› by simp qed qed qed qed (* 4ª demostración *) lemma "(s - t) ∪ (t - s) = (s ∪ t) - (s ∩ t)" proof show "(s - t) ∪ (t - s) ⊆ (s ∪ t) - (s ∩ t)" proof fix x assume "x ∈ (s - t) ∪ (t - s)" then show "x ∈ (s ∪ t) - (s ∩ t)" by auto qed next show "(s ∪ t) - (s ∩ t) ⊆ (s - t) ∪ (t - s)" proof fix x assume "x ∈ (s ∪ t) - (s ∩ t)" then show "x ∈ (s - t) ∪ (t - s)" by auto qed qed (* 5ª demostración *) lemma "(s - t) ∪ (t - s) = (s ∪ t) - (s ∩ t)" proof show "(s - t) ∪ (t - s) ⊆ (s ∪ t) - (s ∩ t)" by auto next show "(s ∪ t) - (s ∩ t) ⊆ (s - t) ∪ (t - s)" by auto qed (* 6ª demostración *) lemma "(s - t) ∪ (t - s) = (s ∪ t) - (s ∩ t)" by auto end |