145 lines
3.8 KiB
Plaintext
145 lines
3.8 KiB
Plaintext
import Common.Logic.Basic
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import Mathlib.Data.Set.Basic
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/-! # Enderton.Set.Chapter_4
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Natural Numbers
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-/
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namespace Enderton.Set.Chapter_4
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/-- #### Theorem 4C
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Every natural number except `0` is the successor of some natural number.
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-/
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theorem theorem_4c (n : ℕ)
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: n = 0 ∨ (∃ m : ℕ, n = m.succ) := by
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match n with
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| 0 => simp
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| m + 1 => simp
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/-- #### Exercise 4.1
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Show that `1 ≠ 3` i.e., that `∅⁺ ≠ ∅⁺⁺⁺`.
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-/
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theorem exercise_4_1 : 1 ≠ 3 := by
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simp
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/-- #### Exercise 4.13
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Let `m` and `n` be natural numbers such that `m ⬝ n = 0`. Show that either
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`m = 0` or `n = 0`.
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-/
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theorem exercise_4_13 (m n : ℕ) (h : m * n = 0)
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: m = 0 ∨ n = 0 := by
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by_contra nh
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rw [not_or_de_morgan] at nh
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have ⟨p, hp⟩ : ∃ p, m = p.succ := Nat.exists_eq_succ_of_ne_zero nh.left
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have ⟨q, hq⟩ : ∃ q, n = q.succ := Nat.exists_eq_succ_of_ne_zero nh.right
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have : m * n = (m * q + p).succ := calc m * n
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_ = m * q.succ := by rw [hq]
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_ = m * q + m := rfl
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_ = m * q + p.succ := by rw [hp]
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_ = (m * q + p).succ := rfl
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rw [this] at h
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simp only [Nat.succ_ne_zero] at h
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/--
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Call a natural number *even* if it has the form `2 ⬝ m` for some `m`.
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-/
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def even (n : ℕ) : Prop := ∃ m, 2 * m = n
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/--
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Call a natural number *odd* if it has the form `(2 ⬝ p) + 1` for some `p`.
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-/
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def odd (n : ℕ) : Prop := ∃ p, (2 * p) + 1 = n
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/-- #### Exercise 4.14
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Show that each natural number is either even or odd, but never both.
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-/
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theorem exercise_4_14 (n : ℕ)
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: (even n ∧ ¬ odd n) ∨ (¬ even n ∧ odd n) := by
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induction n with
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| zero =>
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left
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refine ⟨⟨0, by simp⟩, ?_⟩
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intro ⟨p, hp⟩
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simp only [Nat.zero_eq, Nat.succ_ne_zero] at hp
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| succ n ih =>
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apply Or.elim ih
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· -- Assumes `n` is even meaning `n⁺` is odd.
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intro ⟨⟨m, hm⟩, h⟩
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right
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refine ⟨?_, ⟨m, by rw [← hm]⟩⟩
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by_contra nh
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have ⟨p, hp⟩ := nh
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by_cases hp' : p = 0
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· rw [hp'] at hp
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simp at hp
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· have ⟨q, hq⟩ := Nat.exists_eq_succ_of_ne_zero hp'
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rw [hq] at hp
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have hq₁ : (q.succ + q).succ = n.succ := calc (q.succ + q).succ
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_ = q.succ + q.succ := rfl
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_ = 2 * q.succ := by rw [Nat.two_mul]
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_ = n.succ := hp
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injection hq₁ with hq₂
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have : odd n := by
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refine ⟨q, ?_⟩
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calc 2 * q + 1
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_ = q + q + 1 := by rw [Nat.two_mul]
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_ = q + q.succ := rfl
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_ = q.succ + q := by rw [Nat.add_comm]
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_ = n := hq₂
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exact absurd this h
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· -- Assumes `n` is odd meaning `n⁺` is even.
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intro ⟨h, ⟨p, hp⟩⟩
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have hp' : 2 * p.succ = n.succ := congrArg Nat.succ hp
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left
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refine ⟨⟨p.succ, by rw [← hp']⟩, ?_⟩
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by_contra nh
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unfold odd at nh
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have ⟨q, hq⟩ := nh
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injection hq with hq'
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simp only [Nat.add_eq, Nat.add_zero] at hq'
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have : even n := ⟨q, hq'⟩
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exact absurd this h
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/-- #### Lemma 10
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For every natural number `n ≠ 0`, `0 ∈ n`.
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-/
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theorem zero_least_nat (n : ℕ)
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: 0 = n ∨ 0 < n := by
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by_cases h : n = 0
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· left
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rw [h]
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· right
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have ⟨m, hm⟩ := Nat.exists_eq_succ_of_ne_zero h
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rw [hm]
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exact Nat.succ_pos m
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/-- #### Trichotomy Law for ω
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For any natural numbers `m` and `n`, exactly one of the three conditions
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```
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m ∈ n, m = n, n ∈ m
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```
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holds.
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-/
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theorem trichotomy_law_for_nat
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: IsAsymm ℕ LT.lt ∧ IsTrichotomous ℕ LT.lt :=
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⟨instIsAsymmLtToLT, instIsTrichotomousLtToLTToPreorderToPartialOrder⟩
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/-- #### Linear Ordering on ω
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Relation
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```
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∈_ω = {⟨m, n⟩ ∈ ω × ω | m ∈ n}
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```
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is a linear ordering on `ω`.
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-/
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theorem linear_ordering_on_nat
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: IsStrictTotalOrder ℕ LT.lt := isStrictTotalOrder_of_linearOrder
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end Enderton.Set.Chapter_4 |