832 lines
22 KiB
Plaintext
832 lines
22 KiB
Plaintext
import Common.Logic.Basic
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import Common.Set.Basic
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import Common.Set.Peano
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import Mathlib.Data.Set.Basic
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import Mathlib.SetTheory.Ordinal.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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#check Nat.exists_eq_succ_of_ne_zero
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/-- ### Theorem 4I
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For natural numbers `m` and `n`,
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```
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m + 0 = m,
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m + n⁺ = (m + n)⁺
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```
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-/
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theorem theorem_4i (m n : ℕ)
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: m + 0 = m ∧ m + n.succ = (m + n).succ := ⟨rfl, rfl⟩
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/-- ### Theorem 4J
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For natural numbers `m` and `n`,
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```
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m ⬝ 0 = 0,
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m ⬝ n⁺ = m ⬝ n + m .
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```
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-/
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theorem theorem_4j (m n : ℕ)
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: m * 0 = 0 ∧ m * n.succ = m * n + m := ⟨rfl, rfl⟩
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/-- ### Left Additive Identity
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For all `n ∈ ω`, `A₀(n) = n`. In other words, `0 + n = n`.
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-/
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lemma left_additive_identity (n : ℕ)
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: 0 + n = n := by
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induction n with
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| zero => simp
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| succ n ih =>
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calc 0 + n.succ
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_ = (0 + n).succ := rfl
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_ = n.succ := by rw [ih]
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#check Nat.zero_add
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/-- ### Lemma 2
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For all `m, n ∈ ω`, `Aₘ₊(n) = Aₘ(n⁺)`. In other words, `m⁺ + n = m + n⁺`.
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-/
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lemma lemma_2 (m n : ℕ)
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: m.succ + n = m + n.succ := by
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induction n with
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| zero => rfl
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| succ n ih =>
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calc m.succ + n.succ
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_ = (m.succ + n).succ := rfl
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_ = (m + n.succ).succ := by rw [ih]
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_ = m + n.succ.succ := rfl
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#check Nat.succ_add_eq_succ_add
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/-- ### Theorem 4K-1
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Associatve law for addition. For `m, n, p ∈ ω`,
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```
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m + (n + p) = (m + n) + p.
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```
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-/
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theorem theorem_4k_1 {m n p : ℕ}
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: m + (n + p) = (m + n) + p := by
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induction m with
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| zero => simp
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| succ m ih =>
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calc m.succ + (n + p)
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_ = m + (n + p).succ := by rw [lemma_2]
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_ = (m + (n + p)).succ := rfl
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_ = ((m + n) + p).succ := by rw [ih]
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_ = (m + n) + p.succ := rfl
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_ = (m + n).succ + p := by rw [lemma_2]
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_ = (m + n.succ) + p := rfl
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_ = (m.succ + n) + p := by rw [lemma_2]
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#check Nat.add_assoc
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/-- ### Theorem 4K-2
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Commutative law for addition. For `m, n ∈ ω`,
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```
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m + n = n + m.
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```
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-/
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theorem theorem_4k_2 {m n : ℕ}
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: m + n = n + m := by
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induction m with
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| zero => simp
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| succ m ih =>
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calc m.succ + n
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_ = m + n.succ := by rw [lemma_2]
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_ = (m + n).succ := rfl
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_ = (n + m).succ := by rw [ih]
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_ = n + m.succ := by rfl
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#check Nat.add_comm
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/-- ### Zero Multiplicand
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For all `n ∈ ω`, `M₀(n) = 0`. In other words, `0 ⬝ n = 0`.
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-/
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theorem zero_multiplicand (n : ℕ)
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: 0 * n = 0 := by
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induction n with
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| zero => simp
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| succ n ih =>
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calc 0 * n.succ
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_ = 0 * n + 0 := rfl
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_ = 0 * n := rfl
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_ = 0 := by rw [ih]
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#check Nat.zero_mul
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/-- ### Successor Distribution
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For all `m, n ∈ ω`, `Mₘ₊(n) = Mₘ(n) + n`. In other words,
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```
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m⁺ ⬝ n = m ⬝ n + n.
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```
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-/
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theorem succ_distrib (m n : ℕ)
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: m.succ * n = m * n + n := by
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induction n with
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| zero => simp
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| succ n ih =>
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calc m.succ * n.succ
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_ = m.succ * n + m.succ := rfl
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_ = (m * n + n) + m.succ := by rw [ih]
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_ = m * n + (n + m.succ) := by rw [theorem_4k_1]
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_ = m * n + (n.succ + m) := by rw [lemma_2]
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_ = m * n + (m + n.succ) := by
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conv => left; arg 2; rw [theorem_4k_2]
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_ = (m * n + m) + n.succ := by rw [theorem_4k_1]
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_ = m * n.succ + n.succ := rfl
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#check Nat.succ_mul
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/-- ### Theorem 4K-3
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Distributive law. For `m, n, p ∈ ω`,
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```
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m ⬝ (n + p) = m ⬝ n + m ⬝ p.
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```
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-/
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theorem theorem_4k_3 (m n p : ℕ)
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: m * (n + p) = m * n + m * p := by
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induction m with
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| zero => simp
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| succ m ih =>
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calc m.succ * (n + p)
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_ = m * (n + p) + (n + p) := by rw [succ_distrib]
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_ = m * (n + p) + n + p := by rw [← theorem_4k_1]
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_ = m * n + m * p + n + p := by rw [ih]
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_ = m * n + (m * p + n) + p := by rw [theorem_4k_1]
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_ = m * n + (n + m * p) + p := by
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conv => left; arg 1; arg 2; rw [theorem_4k_2]
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_ = (m * n + n) + (m * p + p) := by rw [theorem_4k_1, theorem_4k_1]
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_ = m.succ * n + m.succ * p := by rw [succ_distrib, succ_distrib]
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/-- ### Successor Identity
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For all `m ∈ ω`, `Aₘ(1) = m⁺`. In other words, `m + 1 = m⁺`.
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-/
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theorem succ_identity (m : ℕ)
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: m + 1 = m.succ := by
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induction m with
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| zero => simp
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| succ m ih =>
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calc m.succ + 1
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_ = m + (Nat.succ Nat.zero).succ := by rw [lemma_2]
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_ = (m + 1).succ := rfl
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_ = m.succ.succ := by rw [ih]
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#check Nat.succ_eq_one_add
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/-- ### Right Multiplicative Identity
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For all `m ∈ ω`, `Mₘ(1) = m`. In other words, `m ⬝ 1 = m`.
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-/
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theorem right_mul_id (m : ℕ)
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: m * 1 = m := by
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induction m with
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| zero => simp
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| succ m ih =>
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calc m.succ * 1
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_ = m * 1 + 1 := by rw [succ_distrib]
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_ = m + 1 := by rw [ih]
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_ = m.succ := by rw [succ_identity]
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#check Nat.mul_one
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/-- ### Theorem 4K-5
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Commutative law for multiplication. For `m, n ∈ ω`, `m ⬝ n = n ⬝ m`.
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-/
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theorem theorem_4k_5 (m n : ℕ)
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: m * n = n * m := by
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induction m with
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| zero => simp
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| succ m ih =>
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calc m.succ * n
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_ = m * n + n := by rw [succ_distrib]
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_ = n * m + n := by rw [ih]
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_ = n * m + n * 1 := by
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conv => left; arg 2; rw [← right_mul_id n]
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_ = n * (m + 1) := by rw [← theorem_4k_3]
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_ = n * m.succ := by rw [succ_identity]
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#check Nat.mul_comm
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/-- ### Theorem 4K-4
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Associative law for multiplication. For `m, n, p ∈ ω`,
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```
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m ⬝ (n ⬝ p) = (m ⬝ n) ⬝ p.
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```
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-/
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theorem theorem_4k_4 (m n p : ℕ)
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: m * (n * p) = (m * n) * p := by
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induction p with
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| zero => simp
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| succ p ih =>
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calc m * (n * p.succ)
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_ = m * (n * p + n) := rfl
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_ = m * (n * p) + m * n := by rw [theorem_4k_3]
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_ = (m * n) * p + m * n := by rw [ih]
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_ = p * (m * n) + m * n := by rw [theorem_4k_5]
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_ = p.succ * (m * n) := by rw [succ_distrib]
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_ = (m * n) * p.succ := by rw [theorem_4k_5]
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#check Nat.mul_assoc
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/-- ### Lemma 4L(b)
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No natural number is a member of itself.
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-/
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lemma lemma_4l_b (n : ℕ)
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: ¬ n < n := by
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induction n with
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| zero => simp
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| succ n ih =>
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by_contra nh
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rw [Nat.succ_lt_succ_iff] at nh
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exact absurd nh ih
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#check Nat.lt_irrefl
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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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#check Nat.pos_of_ne_zero
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/-! ### Theorem 4N
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For any natural numbers `n`, `m`, and `p`,
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```
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m ∈ n ↔ m ⬝ p ∈ n ⬝ p.
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```
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If, in addition, `p ≠ 0`, then
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```
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m ∈ n ↔ m ⬝ p ∈ n ⬝ p.
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```
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-/
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theorem theorem_4n_i (m n p : ℕ)
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: m < n ↔ m + p < n + p := by
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/-
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> Let `m` and `n` be natural numbers.
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>
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> ##### (⇒)
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> Suppose `m ∈ n`. Let
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>
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> `S = {p ∈ ω | m + p ∈ n + p}`
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-/
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have hf : ∀ m n : ℕ, m < n → m + p < n + p := by
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induction p with
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/-
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> It trivially follows that `0 ∈ S`.
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-/
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| zero => simp
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/-
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> Next, suppose `p ∈ S`. That is, suppose `m + p ∈ n + p`. By *Lemma 4L(a)*,
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> this holds if and only if `(m + p)⁺ ∈ (n + p)⁺`. *Theorem 4I* then implies
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> that `m + p⁺ ∈ n + p⁺` meaning `p⁺ ∈ S`.
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-/
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| succ p ih =>
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intro m n hp
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have := ih m n hp
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rw [← Nat.succ_lt_succ_iff] at this
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have h₁ : (m + p).succ = m + p.succ := rfl
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have h₂ : (n + p).succ = n + p.succ := rfl
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rwa [← h₁, ← h₂]
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apply Iff.intro
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/-
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> Thus `S` is an inductive set. Hence *Theorem 4B* implies `S = ω`. Therefore,
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> for all `p ∈ ω`, `m ∈n` implies `m + p ∈ n + p`.
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-/
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· exact hf m n
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/-
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> ##### (⇐)
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> Let `p` be a natural number and suppose `m + p ∈ n + p`. By the
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> *Trichotomy Law for `ω`*, there are two cases to consider regarding how `m`
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> and `n` relate to one another:
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-/
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· intro h
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match @trichotomous ℕ LT.lt _ m n with
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| Or.inr (Or.inl h₁) =>
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/-
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> ###### Case 1
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> Suppose `m = n`. Then `m + p ∈ n + p = m + p`. *Lemma 4L(b)* shows this is
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> impossible.
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-/
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rw [← h₁] at h
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exact absurd h (lemma_4l_b (m + p))
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| Or.inr (Or.inr h₁) =>
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/-
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> ###### Case 2
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> Suppose `n ∈ m`. Then *(⇒)* indicates `n + p ∈ m + p`. But this contradicts
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> the *Trichotomy Law for `ω`* since, by hypothesis, `m + p ∈ n + p`.
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-/
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have := hf n m h₁
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exact absurd this (Nat.lt_asymm h)
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| Or.inl h₁ =>
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/-
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> ###### Conclusion
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> By trichotomy, it follows `m ∈ n`.
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-/
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exact h₁
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#check Nat.add_lt_add_iff_right
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theorem theorem_4n_ii (m n p : ℕ)
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: m < n ↔ m * p.succ < n * p.succ := by
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/-
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> Let `m` and `n` be natural numbers.
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>
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> ##### (⇒)
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> Suppose `m ∈ n`. Let
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>
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> `S = {p ∈ ω | m ⬝ p⁺ ∈ n ⬝ p⁺}`.
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-/
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have hf : ∀ m n : ℕ, m < n → m * p.succ < n * p.succ := by
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intro m n hp₁
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induction p with
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| zero =>
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/-
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> `0 ∈ S` by *Right Multiplicative Identity*.
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-/
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simp only [Nat.mul_one]
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exact hp₁
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| succ p ih =>
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/-
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> Next, suppose `p ∈ S`. That is, `m ⬝ p⁺ ∈ n ⬝ p⁺`. Then
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>
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> `m ⬝ p⁺⁺ = m ⬝ p⁺ + m` *Theorem 4J*
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> ` ∈ n ⬝ p⁺ + m` *(i)*
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> ` = m + n ⬝ p⁺` *Theorem 4K-2*
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> ` ∈ n + n ⬝ p⁺` *(i)*
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> ` = n ⬝p⁺ + n` *Theorem 4K-2*
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> ` n ⬝ p⁺⁺` *Theorem 4J*
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>
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> Therefore `p⁺ ∈ S`.
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-/
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have hp₂ : m * p.succ < n * p.succ := by
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by_cases hp₃ : p = 0
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· rw [hp₃] at *
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simp only [Nat.mul_one] at *
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exact hp₁
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· exact ih
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calc m * p.succ.succ
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_ = m * p.succ + m := rfl
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_ < n * p.succ + m := (theorem_4n_i (m * p.succ) (n * p.succ) m).mp hp₂
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_ = m + n * p.succ := by rw [theorem_4k_2]
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_ < n + n * p.succ := (theorem_4n_i m n (n * p.succ)).mp hp₁
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_ = n * p.succ + n := by rw [theorem_4k_2]
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_ = n * p.succ.succ := rfl
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apply Iff.intro
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/-
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> Thus `S` is an inductive set. Hence *Theorem 4B* implies `S = ω`. By
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> *Theorem 4C*, every natural number except `0` is the successor of some natural
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> number. Therefore, for all `p ∈ ω` such that `p ≠ 0`, `m ∈ n` implies
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`m ⬝ p ∈ n ⬝ p`.
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-/
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· exact hf m n
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/-
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> ##### (⇐)
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> Let `p ≠ 0` be a natural number and suppose `m ⬝ p ∈ n ⬝ p`. By the
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> *Trichotomy Law for `ω`*, there are two cases to consider regarding how `m`
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> and `n` relate to one another.
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-/
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intro hp
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match @trichotomous ℕ LT.lt _ m n with
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| Or.inr (Or.inl h₁) =>
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/-
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> ###### Case 1
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> Suppose `m = n`. Then `m ⬝ p ∈ n ⬝ p = m ⬝ p`. *Lemma 4L(b)* shows this is
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> impossible.
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-/
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rw [← h₁] at hp
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exact absurd hp (lemma_4l_b (m * p.succ))
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| Or.inr (Or.inr h₁) =>
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/-
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> ###### Case 2
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> Suppose `n ∈ m`. Then *(⇒)* indicates `n ⬝ p ∈ m ⬝ p`. But this contradicts
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> *Trichotomy Law for `ω`* since, by hypothesis, `m ⬝ p ∈ n ⬝ p`.
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-/
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have := hf n m h₁
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exact absurd this (Nat.lt_asymm hp)
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| Or.inl h₁ =>
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/-
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> ###### Conclusion
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> By trichotomy, it follows `m ∈ n`.
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-/
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exact h₁
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#check Nat.mul_lt_mul_of_pos_right
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/-! ### Corollary 4P
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The following cancellation laws hold for `m`, `n`, and `p` in `ω`:
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```
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m + p = n + p ⇒ m = n,
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m ⬝ p = n ⬝ p ∧ p ≠ 0 ⇒ m = n.
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```
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-/
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theorem corollary_4p_i (m n p : ℕ) (h : m + p = n + p)
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: m = n := by
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/-
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> Suppose `m + p = n + p`. By the *Trichotomy Law for `ω`*, there are two cases
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> to consider regarding how `m` and `n` relate to one another.
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-/
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match @trichotomous ℕ LT.lt _ m n with
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| Or.inl h₁ =>
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/-
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> If `m ∈n`, then *Theorem 4N* implies `m + p ∈ n + p`.
|
||
-/
|
||
rw [theorem_4n_i m n p, h] at h₁
|
||
exact absurd h₁ (lemma_4l_b (n + p))
|
||
| Or.inr (Or.inr h₁) =>
|
||
/-
|
||
> If `n ∈ m`, then *Theorem 4N* implies `n + p ∈ m + p`.
|
||
-/
|
||
rw [theorem_4n_i n m p, h] at h₁
|
||
exact absurd h₁ (lemma_4l_b (n + p))
|
||
/-
|
||
> Both of these contradict the *Trichotomy Law for `ω`* of `m + p` and `n + p`.
|
||
> Thus `m = n` is the only remaining possibility.
|
||
-/
|
||
| Or.inr (Or.inl h₁) =>
|
||
exact h₁
|
||
|
||
|
||
#check Nat.add_right_cancel
|
||
|
||
/-- ### Well Ordering of ω
|
||
|
||
Let `A` be a nonempty subset of `ω`. Then there is some `m ∈ A` such that
|
||
`m ≤ n` for all `n ∈ A`.
|
||
-/
|
||
theorem well_ordering_nat {A : Set ℕ} (hA : Set.Nonempty A)
|
||
: ∃ m ∈ A, ∀ n, n ∈ A → m ≤ n := by
|
||
/-
|
||
> Let `A` be a nonempty subset of `ω`. For the sake of contradiciton, suppose
|
||
> `A` does not have a least element.
|
||
-/
|
||
by_contra nh
|
||
simp only [not_exists, not_and, not_forall, not_le, exists_prop] at nh
|
||
/-
|
||
> It then suffices to prove that the complement of `A` equals `ω`. If we do so,
|
||
> then `A = ∅`, a contradiction.
|
||
-/
|
||
suffices A.compl = Set.univ by
|
||
have h := Set.univ_diff_compl_eq_self A
|
||
rw [this] at h
|
||
simp only [sdiff_self, Set.bot_eq_empty] at h
|
||
exact absurd h.symm (Set.Nonempty.ne_empty hA)
|
||
/-
|
||
> Define
|
||
>
|
||
> `S = {n ∈ ω | (∀ m ∈ n)m ∉ A}`.
|
||
>
|
||
> We prove `S` is an inductive set by showing that (i) `0 ∈ S` and (ii) if
|
||
> `n ∈ S`, then `n⁺ ∈ S`. Afterward we show that `ω - A = ω`, completing the
|
||
> proof.
|
||
-/
|
||
have : ∀ n : ℕ, (∀ m, m < n → m ∈ A.compl) := by
|
||
intro n
|
||
induction n with
|
||
| zero =>
|
||
/-
|
||
> #### (i)
|
||
> It vacuously holds that `0 ∈ S`.
|
||
-/
|
||
intro m hm
|
||
exact False.elim (Nat.not_lt_zero m hm)
|
||
| succ n ih =>
|
||
/-
|
||
> #### (ii)
|
||
> Suppose `n ∈ S`. We want to prove that
|
||
>
|
||
> `∀ m, m ∈ n⁺ ⇒ m ∉ A`.
|
||
>
|
||
> To this end, let `m ∈ ω` such that `m ∈ n⁺`. By definition of the successor,
|
||
> `m ∈ n` or `m = n`. If the former, `n ∈ S` implies `m ∉ A`. If the latter, it
|
||
> isn't possible for `n ∈ A` since the *Trichotomy Law for `ω`* would otherwise
|
||
> imply `n` is the least element of `A`, which is assumed to not exist. Hence
|
||
> `n⁺ ∈ S`.
|
||
-/
|
||
intro m hm
|
||
have hm' : m < n ∨ m = n := by
|
||
rw [Nat.lt_succ] at hm
|
||
exact Nat.lt_or_eq_of_le hm
|
||
apply Or.elim hm'
|
||
· intro h
|
||
exact ih m h
|
||
· intro h
|
||
have : ∀ x : ℕ, x ∈ A → n ≤ x := by
|
||
intro x hx
|
||
exact match @trichotomous ℕ LT.lt _ n x with
|
||
| Or.inl h₁ => Nat.le_of_lt h₁
|
||
| Or.inr (Or.inl h₁) => Nat.le_of_eq h₁
|
||
| Or.inr (Or.inr h₁) => False.elim (ih x h₁ hx)
|
||
by_cases hn : n ∈ A
|
||
· have ⟨p, hp⟩ := nh n hn
|
||
exact absurd hp.left (ih p hp.right)
|
||
· rw [h]
|
||
exact hn
|
||
/-
|
||
> #### Conclusion
|
||
> By *(i)* and *(ii)*, `S` is an inductive set. Since `S ⊆ ω`, *Theorem 4B*
|
||
> implies `S = ω`. Bu this immediately implies `ω = ω - A` meaning `A` is the
|
||
> empty set.
|
||
-/
|
||
ext x
|
||
simp only [Set.mem_univ, iff_true]
|
||
by_contra nh'
|
||
have ⟨y, hy₁, hy₂⟩ := nh x (show x ∈ A from Set.not_not_mem.mp nh')
|
||
exact absurd hy₁ (this x y hy₂)
|
||
|
||
#check WellOrder
|
||
|
||
/-- ### Strong Induction Principle for ω
|
||
|
||
Let `A` be a subset of `ω`, and assume that for every `n ∈ ω`, if every number
|
||
less than `n` is in `A`, then `n ∈ A`. Then `A = ω`.
|
||
-/
|
||
theorem strong_induction_principle_nat (A : Set ℕ)
|
||
(h : ∀ n : ℕ, (∀ x : ℕ, x < n → x ∈ A) → n ∈ A)
|
||
: A = Set.univ := by
|
||
suffices A.compl = ∅ by
|
||
have h' := Set.univ_diff_compl_eq_self A
|
||
rw [this] at h'
|
||
simp only [Set.diff_empty] at h'
|
||
exact h'.symm
|
||
/-
|
||
> For the sake of contradiction, suppose `ω - A` is a nonempty set. By
|
||
> *Well Ordering of `ω`*, there exists a least element `m ∈ ω - A`.
|
||
-/
|
||
by_contra nh
|
||
have ⟨m, hm⟩ := well_ordering_nat (Set.nmem_singleton_empty.mp nh)
|
||
refine absurd (h m ?_) hm.left
|
||
/-
|
||
> Then every number less than `m` is in `A`. But then *(4.23)* implies `m ∈ A`,
|
||
> a contradiction. Thus `ω - A` is an empty set meaning `A = ω`.
|
||
-/
|
||
-- Show that every number less than `m` is in `A`.
|
||
intro x hx
|
||
by_contra nx
|
||
have : x < x := Nat.lt_of_lt_of_le hx (hm.right x nx)
|
||
simp at this
|
||
|
||
/-- ### Exercise 4.1
|
||
|
||
Show that `1 ≠ 3` i.e., that `∅⁺ ≠ ∅⁺⁺⁺`.
|
||
-/
|
||
theorem exercise_4_1 : 1 ≠ 3 := by
|
||
simp
|
||
|
||
/-- ### Exercise 4.13
|
||
|
||
Let `m` and `n` be natural numbers such that `m ⬝ n = 0`. Show that either
|
||
`m = 0` or `n = 0`.
|
||
-/
|
||
theorem exercise_4_13 (m n : ℕ) (h : m * n = 0)
|
||
: m = 0 ∨ n = 0 := by
|
||
by_contra nh
|
||
rw [not_or_de_morgan] at nh
|
||
have ⟨p, hp⟩ : ∃ p, m = p.succ := Nat.exists_eq_succ_of_ne_zero nh.left
|
||
have ⟨q, hq⟩ : ∃ q, n = q.succ := Nat.exists_eq_succ_of_ne_zero nh.right
|
||
have : m * n = (m * q + p).succ := calc m * n
|
||
_ = m * q.succ := by rw [hq]
|
||
_ = m * q + m := rfl
|
||
_ = m * q + p.succ := by rw [hp]
|
||
_ = (m * q + p).succ := rfl
|
||
rw [this] at h
|
||
simp only [Nat.succ_ne_zero] at h
|
||
|
||
/--
|
||
Call a natural number *even* if it has the form `2 ⬝ m` for some `m`.
|
||
-/
|
||
def even (n : ℕ) : Prop := ∃ m, 2 * m = n
|
||
|
||
/--
|
||
Call a natural number *odd* if it has the form `(2 ⬝ p) + 1` for some `p`.
|
||
-/
|
||
def odd (n : ℕ) : Prop := ∃ p, (2 * p) + 1 = n
|
||
|
||
/-- ### Exercise 4.14
|
||
|
||
Show that each natural number is either even or odd, but never both.
|
||
-/
|
||
theorem exercise_4_14 (n : ℕ)
|
||
: (even n ∧ ¬ odd n) ∨ (¬ even n ∧ odd n) := by
|
||
induction n with
|
||
| zero =>
|
||
left
|
||
refine ⟨⟨0, by simp⟩, ?_⟩
|
||
intro ⟨p, hp⟩
|
||
simp only [Nat.zero_eq, Nat.succ_ne_zero] at hp
|
||
| succ n ih =>
|
||
apply Or.elim ih
|
||
· -- Assumes `n` is even meaning `n⁺` is odd.
|
||
intro ⟨⟨m, hm⟩, h⟩
|
||
right
|
||
refine ⟨?_, ⟨m, by rw [← hm]⟩⟩
|
||
by_contra nh
|
||
have ⟨p, hp⟩ := nh
|
||
by_cases hp' : p = 0
|
||
· rw [hp'] at hp
|
||
simp at hp
|
||
· have ⟨q, hq⟩ := Nat.exists_eq_succ_of_ne_zero hp'
|
||
rw [hq] at hp
|
||
have hq₁ : (q.succ + q).succ = n.succ := calc (q.succ + q).succ
|
||
_ = q.succ + q.succ := rfl
|
||
_ = 2 * q.succ := by rw [Nat.two_mul]
|
||
_ = n.succ := hp
|
||
injection hq₁ with hq₂
|
||
have : odd n := by
|
||
refine ⟨q, ?_⟩
|
||
calc 2 * q + 1
|
||
_ = q + q + 1 := by rw [Nat.two_mul]
|
||
_ = q + q.succ := rfl
|
||
_ = q.succ + q := by rw [Nat.add_comm]
|
||
_ = n := hq₂
|
||
exact absurd this h
|
||
· -- Assumes `n` is odd meaning `n⁺` is even.
|
||
intro ⟨h, ⟨p, hp⟩⟩
|
||
have hp' : 2 * p.succ = n.succ := congrArg Nat.succ hp
|
||
left
|
||
refine ⟨⟨p.succ, by rw [← hp']⟩, ?_⟩
|
||
by_contra nh
|
||
unfold odd at nh
|
||
have ⟨q, hq⟩ := nh
|
||
injection hq with hq'
|
||
simp only [Nat.add_eq, Nat.add_zero] at hq'
|
||
have : even n := ⟨q, hq'⟩
|
||
exact absurd this h
|
||
|
||
/-- ### Exercise 4.17
|
||
|
||
Prove that `mⁿ⁺ᵖ = mⁿ ⬝ mᵖ.`
|
||
-/
|
||
theorem exercise_4_17 (m n p : ℕ)
|
||
: m ^ (n + p) = m ^ n * m ^ p := by
|
||
induction p with
|
||
| zero => calc m ^ (n + 0)
|
||
_ = m ^ n := rfl
|
||
_ = m ^ n * 1 := by rw [right_mul_id]
|
||
_ = m ^ n * m ^ 0 := rfl
|
||
| succ p ih => calc m ^ (n + p.succ)
|
||
_ = m ^ (n + p).succ := rfl
|
||
_ = m ^ (n + p) * m := rfl
|
||
_ = m ^ n * m ^ p * m := by rw [ih]
|
||
_ = m ^ n * (m ^ p * m) := by rw [theorem_4k_4]
|
||
_ = m ^ n * m ^ p.succ := rfl
|
||
|
||
/-- ### Exercise 4.19
|
||
|
||
Prove that if `m` is a natural number and `d` is a nonzero number, then there
|
||
exist numbers `q` and `r` such that `m = (d ⬝ q) + r` and `r` is less than `d`.
|
||
-/
|
||
theorem exercise_4_19 (m d : ℕ) (hd : d ≠ 0)
|
||
: ∃ q r : ℕ, m = (d * q) + r ∧ r < d := by
|
||
induction m with
|
||
| zero =>
|
||
refine ⟨0, 0, ?_⟩
|
||
simp only [Nat.zero_eq, mul_zero, add_zero, true_and]
|
||
exact Nat.pos_of_ne_zero hd
|
||
| succ m ih =>
|
||
have ⟨q, r, hm, hr⟩ := ih
|
||
have hm' := calc m.succ
|
||
_ = ((d * q) + r).succ := by rw [hm]
|
||
_ = (d * q) + r.succ := rfl
|
||
match @trichotomous ℕ LT.lt _ r.succ d with
|
||
| Or.inl h₁ =>
|
||
exact ⟨q, r.succ, hm', h₁⟩
|
||
| Or.inr (Or.inl h₁) =>
|
||
refine ⟨q.succ, 0, ?_, Nat.pos_of_ne_zero hd⟩
|
||
calc Nat.succ m
|
||
_ = d * q + Nat.succ r := hm'
|
||
_ = d * q + d := by rw [h₁]
|
||
_ = d * q.succ := rfl
|
||
_ = d * q.succ + 0 := rfl
|
||
| Or.inr (Or.inr h₁) =>
|
||
have : d < d := calc d
|
||
_ ≤ r := Nat.lt_succ.mp h₁
|
||
_ < d := hr
|
||
simp at this
|
||
|
||
/-- ### Exercise 4.22
|
||
|
||
Show that for any natural numbers `m` and `p` we have `m ∈ m + p⁺`.
|
||
-/
|
||
theorem exercise_4_22 (m p : ℕ)
|
||
: m < m + p.succ := by
|
||
induction p with
|
||
| zero => simp
|
||
| succ p ih => calc m
|
||
_ < m + p.succ := ih
|
||
_ < m + p.succ.succ := Nat.lt.base (m + p.succ)
|
||
|
||
/-- ### Exercise 4.23
|
||
|
||
Assume that `m` and `n` are natural numbers with `m` less than `n`. Show that
|
||
there is some `p` in `ω` for which `m + p⁺ = n`. (It follows from this and the
|
||
preceding exercise that `m` is less than `n` iff (`∃p ∈ ω) m + p⁺ = n`.)
|
||
-/
|
||
theorem exercise_4_23 {m n : ℕ} (hm : m < n)
|
||
: ∃ p : ℕ, m + p.succ = n := by
|
||
induction n with
|
||
| zero => simp at hm
|
||
| succ n ih =>
|
||
have : m < n ∨ m = n := by
|
||
rw [Nat.lt_succ] at hm
|
||
exact Nat.lt_or_eq_of_le hm
|
||
apply Or.elim this
|
||
· intro hm₁
|
||
have ⟨p, hp⟩ := ih hm₁
|
||
refine ⟨p.succ, ?_⟩
|
||
exact Eq.symm $ calc n.succ
|
||
_ = (m + p.succ).succ := by rw [← hp]
|
||
_ = m + p.succ.succ := rfl
|
||
· intro hm₁
|
||
refine ⟨0, ?_⟩
|
||
rw [hm₁]
|
||
|
||
/-- ### Exercise 4.24
|
||
|
||
Assume that `m + n = p + q`. Show that
|
||
```
|
||
m ∈ p ↔ q ∈ n.
|
||
```
|
||
-/
|
||
theorem exercise_4_24 (m n p q : ℕ) (h : m + n = p + q)
|
||
: m < p ↔ q < n := by
|
||
apply Iff.intro
|
||
· intro hm
|
||
have hr : m + n < p + n := (theorem_4n_i m p n).mp hm
|
||
rw [h] at hr
|
||
conv at hr => left; rw [add_comm]
|
||
conv at hr => right; rw [add_comm]
|
||
exact (theorem_4n_i q n p).mpr hr
|
||
· intro hq
|
||
have hr : q + p < n + p := (theorem_4n_i q n p).mp hq
|
||
conv at hr => left; rw [add_comm]
|
||
conv at hr => right; rw [add_comm]
|
||
rw [← h] at hr
|
||
exact (theorem_4n_i m p n).mpr hr
|
||
|
||
/-- ### Exercise 4.25
|
||
|
||
Assume that `n ∈ m` and `q ∈ p`. Show that
|
||
```
|
||
(m ⬝ q) + (n ⬝ p) ∈ (m ⬝ p) + (n ⬝ q).
|
||
```
|
||
-/
|
||
theorem exercise_4_25 (m n p q : ℕ) (h₁ : n < m) (h₂ : q < p)
|
||
: (m * q) + (n * p) < (m * p) + (n * q) := by
|
||
have ⟨r, hr⟩ : ∃ r : ℕ, q + r.succ = p := exercise_4_23 h₂
|
||
rw [
|
||
theorem_4n_ii n m r,
|
||
theorem_4n_i (n * r.succ) (m * r.succ) ((m * q) + (n * q))
|
||
] at h₁
|
||
conv at h₁ => left; rw [theorem_4k_2, ← theorem_4k_1]
|
||
conv at h₁ => right; rw [theorem_4k_2]; arg 1; rw [theorem_4k_2]
|
||
conv at h₁ => right; rw [← theorem_4k_1]
|
||
rw [
|
||
← theorem_4k_3 n q r.succ,
|
||
← theorem_4k_3 m q r.succ,
|
||
hr
|
||
] at h₁
|
||
conv at h₁ => right; rw [add_comm]
|
||
exact h₁
|
||
|
||
end Enderton.Set.Chapter_4
|