Un número es par si y solo si lo es su cuadrado
Demostrar con Lean4 que un número es par si y solo si lo es su cuadrado.
Para ello, completar la siguiente teoría de Lean4:
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import Mathlib.Data.Int.Parity import Mathlib.Tactic open Int variable (n : ℤ) example : Even (n^2) ↔ Even n := by sorry |
1. Demostración en lenguaje natural
Sea \(n ∈ ℤ\). Tenemos que demostrar que \(n^2\) es par si y solo si n es par. Lo haremos probando las dos implicaciones.
(⟹) Lo demostraremos por contraposición. Para ello, supongamos que \(n\) no es par. Entonces, existe un \(k ∈ Z\) tal que
\[ n = 2k+1 \tag{1} \]
Luego,
\begin{align}
n^2 &= (2k+1)^2 &&\text{[por (1)]} \\
&= 4k^2+4k+1 \\
&= 2(2k(k+1))+1
\end{align}
Por tanto, \(n^2\) es impar.
(⟸) Supongamos que \(n\) es par. Entonces, existe un \(k ∈ ℤ\) tal que
\[ n = 2k \tag{2} \]
Luego,
\begin{align}
n^2 &= (2k)^2 &&\text{[por (2)]} \\
&= 2(2k^2)
\end{align}
Por tanto, \(n^2\) es par.
2. Demostraciones con Lean4
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import Mathlib.Data.Int.Parity import Mathlib.Tactic open Int variable (n : ℤ) -- 1ª demostración -- =============== example : Even (n^2) ↔ Even n := by constructor . -- ⊢ Even (n ^ 2) → Even n contrapose -- ⊢ ¬Even n → ¬Even (n ^ 2) intro h -- h : ¬Even n -- ⊢ ¬Even (n ^ 2) rw [←odd_iff_not_even] at * -- h : Odd n -- ⊢ Odd (n ^ 2) cases' h with k hk -- k : ℤ -- hk : n = 2 * k + 1 use 2*k*(k+1) -- ⊢ n ^ 2 = 2 * (2 * k * (k + 1)) + 1 calc n^2 = (2*k+1)^2 := by rw [hk] _ = 4*k^2+4*k+1 := by ring _ = 2*(2*k*(k+1))+1 := by ring . -- ⊢ Even n → Even (n ^ 2) intro h -- h : Even n -- ⊢ Even (n ^ 2) cases' h with k hk -- k : ℤ -- hk : n = k + k use 2*k^2 -- ⊢ n ^ 2 = 2 * k ^ 2 + 2 * k ^ 2 calc n^2 = (k + k)^2 := by rw [hk] _ = 2*k^2 + 2*k^2 := by ring -- 2ª demostración -- =============== example : Even (n^2) ↔ Even n := by constructor . -- ⊢ Even (n ^ 2) → Even n contrapose -- ⊢ ¬Even n → ¬Even (n ^ 2) rw [←odd_iff_not_even] -- ⊢ Odd n → ¬Even (n ^ 2) rw [←odd_iff_not_even] -- ⊢ Odd n → Odd (n ^ 2) unfold Odd -- ⊢ (∃ k, n = 2 * k + 1) → ∃ k, n ^ 2 = 2 * k + 1 intro h -- h : ∃ k, n = 2 * k + 1 -- ⊢ ∃ k, n ^ 2 = 2 * k + 1 cases' h with k hk -- k : ℤ -- hk : n = 2 * k + 1 use 2*k*(k+1) -- ⊢ n ^ 2 = 2 * (2 * k * (k + 1)) + 1 rw [hk] -- ⊢ (2 * k + 1) ^ 2 = 2 * (2 * k * (k + 1)) + 1 ring . -- ⊢ Even n → Even (n ^ 2) unfold Even -- ⊢ (∃ r, n = r + r) → ∃ r, n ^ 2 = r + r intro h -- h : ∃ r, n = r + r -- ⊢ ∃ r, n ^ 2 = r + r cases' h with k hk -- k : ℤ -- hk : n = k + k use 2*k^2 -- ⊢ n ^ 2 = 2 * k ^ 2 + 2 * k ^ 2 rw [hk] -- ⊢ (k + k) ^ 2 = 2 * k ^ 2 + 2 * k ^ 2 ring -- 3ª demostración -- =============== example : Even (n^2) ↔ Even n := by constructor . -- ⊢ Even (n ^ 2) → Even n contrapose -- ⊢ ¬Even n → ¬Even (n ^ 2) rw [←odd_iff_not_even] -- ⊢ Odd n → ¬Even (n ^ 2) rw [←odd_iff_not_even] -- ⊢ Odd n → Odd (n ^ 2) rintro ⟨k, rfl⟩ -- k : ℤ -- ⊢ Odd ((2 * k + 1) ^ 2) use 2*k*(k+1) -- ⊢ (2 * k + 1) ^ 2 = 2 * (2 * k * (k + 1)) + 1 ring . -- ⊢ Even n → Even (n ^ 2) rintro ⟨k, rfl⟩ -- k : ℤ -- ⊢ Even ((k + k) ^ 2) use 2*k^2 -- ⊢ (k + k) ^ 2 = 2 * k ^ 2 + 2 * k ^ 2 ring -- 4ª demostración -- =============== example : Even (n^2) ↔ Even n := calc Even (n^2) ↔ Even (n * n) := iff_of_eq (congrArg Even (sq n)) _ ↔ (Even n ∨ Even n) := even_mul _ ↔ Even n := or_self_iff (Even n) -- 5ª demostración -- =============== example : Even (n^2) ↔ Even n := calc Even (n^2) ↔ Even (n * n) := by ring_nf _ ↔ (Even n ∨ Even n) := even_mul _ ↔ Even n := by simp -- Lemas usados -- ============ -- variable (a b : Prop) -- variable (m : ℤ) -- #check (even_mul : Even (m * n) ↔ Even m ∨ Even n) -- #check (iff_of_eq : a = b → (a ↔ b)) -- #check (odd_iff_not_even : Odd n ↔ ¬Even n) -- #check (or_self_iff a : a ∨ a ↔ a) |
Se puede interactuar con las demostraciones anteriores en Lean 4 Web.
3. Demostraciones con Isabelle/HOL
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theory Un_numero_es_par_syss_lo_es_su_cuadrado imports Main begin (* 1ª demostración *) lemma fixes n :: int shows "even (n⇧2) ⟷ even n" proof (rule iffI) assume "even (n⇧2)" show "even n" proof (rule ccontr) assume "odd n" then obtain k where "n = 2*k+1" by (rule oddE) then have "n⇧2 = 2*(2*k*(k+1))+1" proof - have "n⇧2 = (2*k+1)⇧2" by (simp add: ‹n = 2 * k + 1›) also have "… = 4*k⇧2+4*k+1" by algebra also have "… = 2*(2*k*(k+1))+1" by algebra finally show "n⇧2 = 2*(2*k*(k+1))+1" . qed then have "∃k'. n⇧2 = 2*k'+1" by (rule exI) then have "odd (n⇧2)" by fastforce then show False using ‹even (n⇧2)› by blast qed next assume "even n" then obtain k where "n = 2*k" by (rule evenE) then have "n⇧2 = 2*(2*k⇧2)" by simp then show "even (n⇧2)" by simp qed (* 2ª demostración *) lemma fixes n :: int shows "even (n⇧2) ⟷ even n" proof assume "even (n⇧2)" show "even n" proof (rule ccontr) assume "odd n" then obtain k where "n = 2*k+1" by (rule oddE) then have "n⇧2 = 2*(2*k*(k+1))+1" by algebra then have "odd (n⇧2)" by simp then show False using ‹even (n⇧2)› by blast qed next assume "even n" then obtain k where "n = 2*k" by (rule evenE) then have "n⇧2 = 2*(2*k⇧2)" by simp then show "even (n⇧2)" by simp qed (* 3ª demostración *) lemma fixes n :: int shows "even (n⇧2) ⟷ even n" proof - have "even (n⇧2) = (even n ∧ (0::nat) < 2)" by (simp only: even_power) also have "… = (even n ∧ True)" by (simp only: less_numeral_simps) also have "… = even n" by (simp only: HOL.simp_thms(21)) finally show "even (n⇧2) ⟷ even n" by this qed (* 4ª demostración *) lemma fixes n :: int shows "even (n⇧2) ⟷ even n" proof - have "even (n⇧2) = (even n ∧ (0::nat) < 2)" by (simp only: even_power) also have "… = even n" by simp finally show "even (n⇧2) ⟷ even n" . qed (* 5ª demostración *) lemma fixes n :: int shows "even (n⇧2) ⟷ even n" by simp end |
