import Common.Logic.Basic import Common.Set.Basic import Common.Set.Peano import Mathlib.Data.Set.Basic import Mathlib.SetTheory.Ordinal.Basic /-! # Enderton.Set.Chapter_4 Natural Numbers -/ namespace Enderton.Set.Chapter_4 /-- ### Theorem 4C Every natural number except `0` is the successor of some natural number. -/ theorem theorem_4c (n : ℕ) : n = 0 ∨ (∃ m : ℕ, n = m.succ) := by match n with | 0 => simp | m + 1 => simp #check Nat.exists_eq_succ_of_ne_zero /-- ### Theorem 4I For natural numbers `m` and `n`, ``` m + 0 = m, m + n⁺ = (m + n)⁺ ``` -/ theorem theorem_4i (m n : ℕ) : m + 0 = m ∧ m + n.succ = (m + n).succ := ⟨rfl, rfl⟩ /-- ### Theorem 4J For natural numbers `m` and `n`, ``` m ⬝ 0 = 0, m ⬝ n⁺ = m ⬝ n + m . ``` -/ theorem theorem_4j (m n : ℕ) : m * 0 = 0 ∧ m * n.succ = m * n + m := ⟨rfl, rfl⟩ /-- ### Left Additive Identity For all `n ∈ ω`, `A₀(n) = n`. In other words, `0 + n = n`. -/ lemma left_additive_identity (n : ℕ) : 0 + n = n := by induction n with | zero => simp | succ n ih => calc 0 + n.succ _ = (0 + n).succ := rfl _ = n.succ := by rw [ih] #check Nat.zero_add /-- ### Lemma 2 For all `m, n ∈ ω`, `Aₘ₊(n) = Aₘ(n⁺)`. In other words, `m⁺ + n = m + n⁺`. -/ lemma lemma_2 (m n : ℕ) : m.succ + n = m + n.succ := by induction n with | zero => rfl | succ n ih => calc m.succ + n.succ _ = (m.succ + n).succ := rfl _ = (m + n.succ).succ := by rw [ih] _ = m + n.succ.succ := rfl #check Nat.succ_add_eq_succ_add /-- ### Theorem 4K-1 Associatve law for addition. For `m, n, p ∈ ω`, ``` m + (n + p) = (m + n) + p. ``` -/ theorem theorem_4k_1 {m n p : ℕ} : m + (n + p) = (m + n) + p := by induction m with | zero => simp | succ m ih => calc m.succ + (n + p) _ = m + (n + p).succ := by rw [lemma_2] _ = (m + (n + p)).succ := rfl _ = ((m + n) + p).succ := by rw [ih] _ = (m + n) + p.succ := rfl _ = (m + n).succ + p := by rw [lemma_2] _ = (m + n.succ) + p := rfl _ = (m.succ + n) + p := by rw [lemma_2] #check Nat.add_assoc /-- ### Theorem 4K-2 Commutative law for addition. For `m, n ∈ ω`, ``` m + n = n + m. ``` -/ theorem theorem_4k_2 {m n : ℕ} : m + n = n + m := by induction m with | zero => simp | succ m ih => calc m.succ + n _ = m + n.succ := by rw [lemma_2] _ = (m + n).succ := rfl _ = (n + m).succ := by rw [ih] _ = n + m.succ := by rfl #check Nat.add_comm /-- ### Zero Multiplicand For all `n ∈ ω`, `M₀(n) = 0`. In other words, `0 ⬝ n = 0`. -/ theorem zero_multiplicand (n : ℕ) : 0 * n = 0 := by induction n with | zero => simp | succ n ih => calc 0 * n.succ _ = 0 * n + 0 := rfl _ = 0 * n := rfl _ = 0 := by rw [ih] #check Nat.zero_mul /-- ### Successor Distribution For all `m, n ∈ ω`, `Mₘ₊(n) = Mₘ(n) + n`. In other words, ``` m⁺ ⬝ n = m ⬝ n + n. ``` -/ theorem succ_distrib (m n : ℕ) : m.succ * n = m * n + n := by induction n with | zero => simp | succ n ih => calc m.succ * n.succ _ = m.succ * n + m.succ := rfl _ = (m * n + n) + m.succ := by rw [ih] _ = m * n + (n + m.succ) := by rw [theorem_4k_1] _ = m * n + (n.succ + m) := by rw [lemma_2] _ = m * n + (m + n.succ) := by conv => left; arg 2; rw [theorem_4k_2] _ = (m * n + m) + n.succ := by rw [theorem_4k_1] _ = m * n.succ + n.succ := rfl #check Nat.succ_mul /-- ### Theorem 4K-3 Distributive law. For `m, n, p ∈ ω`, ``` m ⬝ (n + p) = m ⬝ n + m ⬝ p. ``` -/ theorem theorem_4k_3 (m n p : ℕ) : m * (n + p) = m * n + m * p := by induction m with | zero => simp | succ m ih => calc m.succ * (n + p) _ = m * (n + p) + (n + p) := by rw [succ_distrib] _ = m * (n + p) + n + p := by rw [← theorem_4k_1] _ = m * n + m * p + n + p := by rw [ih] _ = m * n + (m * p + n) + p := by rw [theorem_4k_1] _ = m * n + (n + m * p) + p := by conv => left; arg 1; arg 2; rw [theorem_4k_2] _ = (m * n + n) + (m * p + p) := by rw [theorem_4k_1, theorem_4k_1] _ = m.succ * n + m.succ * p := by rw [succ_distrib, succ_distrib] /-- ### Successor Identity For all `m ∈ ω`, `Aₘ(1) = m⁺`. In other words, `m + 1 = m⁺`. -/ theorem succ_identity (m : ℕ) : m + 1 = m.succ := by induction m with | zero => simp | succ m ih => calc m.succ + 1 _ = m + (Nat.succ Nat.zero).succ := by rw [lemma_2] _ = (m + 1).succ := rfl _ = m.succ.succ := by rw [ih] #check Nat.succ_eq_one_add /-- ### Right Multiplicative Identity For all `m ∈ ω`, `Mₘ(1) = m`. In other words, `m ⬝ 1 = m`. -/ theorem right_mul_id (m : ℕ) : m * 1 = m := by induction m with | zero => simp | succ m ih => calc m.succ * 1 _ = m * 1 + 1 := by rw [succ_distrib] _ = m + 1 := by rw [ih] _ = m.succ := by rw [succ_identity] #check Nat.mul_one /-- ### Theorem 4K-5 Commutative law for multiplication. For `m, n ∈ ω`, `m ⬝ n = n ⬝ m`. -/ theorem theorem_4k_5 (m n : ℕ) : m * n = n * m := by induction m with | zero => simp | succ m ih => calc m.succ * n _ = m * n + n := by rw [succ_distrib] _ = n * m + n := by rw [ih] _ = n * m + n * 1 := by conv => left; arg 2; rw [← right_mul_id n] _ = n * (m + 1) := by rw [← theorem_4k_3] _ = n * m.succ := by rw [succ_identity] #check Nat.mul_comm /-- ### Theorem 4K-4 Associative law for multiplication. For `m, n, p ∈ ω`, ``` m ⬝ (n ⬝ p) = (m ⬝ n) ⬝ p. ``` -/ theorem theorem_4k_4 (m n p : ℕ) : m * (n * p) = (m * n) * p := by induction p with | zero => simp | succ p ih => calc m * (n * p.succ) _ = m * (n * p + n) := rfl _ = m * (n * p) + m * n := by rw [theorem_4k_3] _ = (m * n) * p + m * n := by rw [ih] _ = p * (m * n) + m * n := by rw [theorem_4k_5] _ = p.succ * (m * n) := by rw [succ_distrib] _ = (m * n) * p.succ := by rw [theorem_4k_5] #check Nat.mul_assoc /-- ### Lemma 4L(b) No natural number is a member of itself. -/ lemma lemma_4l_b (n : ℕ) : ¬ n < n := by induction n with | zero => simp | succ n ih => by_contra nh rw [Nat.succ_lt_succ_iff] at nh exact absurd nh ih #check Nat.lt_irrefl /-- ### Lemma 10 For every natural number `n ≠ 0`, `0 ∈ n`. -/ theorem zero_least_nat (n : ℕ) : 0 = n ∨ 0 < n := by by_cases h : n = 0 · left rw [h] · right have ⟨m, hm⟩ := Nat.exists_eq_succ_of_ne_zero h rw [hm] exact Nat.succ_pos m #check Nat.pos_of_ne_zero /-! ### Theorem 4N For any natural numbers `n`, `m`, and `p`, ``` m ∈ n ↔ m ⬝ p ∈ n ⬝ p. ``` If, in addition, `p ≠ 0`, then ``` m ∈ n ↔ m ⬝ p ∈ n ⬝ p. ``` -/ theorem theorem_4n_i (m n p : ℕ) : m < n ↔ m + p < n + p := by /- > Let `m` and `n` be natural numbers. > > ##### (⇒) > Suppose `m ∈ n`. Let > > `S = {p ∈ ω | m + p ∈ n + p}` -/ have hf : ∀ m n : ℕ, m < n → m + p < n + p := by induction p with /- > It trivially follows that `0 ∈ S`. -/ | zero => simp /- > Next, suppose `p ∈ S`. That is, suppose `m + p ∈ n + p`. By *Lemma 4L(a)*, > this holds if and only if `(m + p)⁺ ∈ (n + p)⁺`. *Theorem 4I* then implies > that `m + p⁺ ∈ n + p⁺` meaning `p⁺ ∈ S`. -/ | succ p ih => intro m n hp have := ih m n hp rw [← Nat.succ_lt_succ_iff] at this have h₁ : (m + p).succ = m + p.succ := rfl have h₂ : (n + p).succ = n + p.succ := rfl rwa [← h₁, ← h₂] apply Iff.intro /- > Thus `S` is an inductive set. Hence *Theorem 4B* implies `S = ω`. Therefore, > for all `p ∈ ω`, `m ∈n` implies `m + p ∈ n + p`. -/ · exact hf m n /- > ##### (⇐) > Let `p` be a natural number and suppose `m + p ∈ n + p`. By the > *Trichotomy Law for `ω`*, there are two cases to consider regarding how `m` > and `n` relate to one another: -/ · intro h match @trichotomous ℕ LT.lt _ m n with | Or.inr (Or.inl h₁) => /- > ###### Case 1 > Suppose `m = n`. Then `m + p ∈ n + p = m + p`. *Lemma 4L(b)* shows this is > impossible. -/ rw [← h₁] at h exact absurd h (lemma_4l_b (m + p)) | Or.inr (Or.inr h₁) => /- > ###### Case 2 > Suppose `n ∈ m`. Then *(⇒)* indicates `n + p ∈ m + p`. But this contradicts > the *Trichotomy Law for `ω`* since, by hypothesis, `m + p ∈ n + p`. -/ have := hf n m h₁ exact absurd this (Nat.lt_asymm h) | Or.inl h₁ => /- > ###### Conclusion > By trichotomy, it follows `m ∈ n`. -/ exact h₁ #check Nat.add_lt_add_iff_right theorem theorem_4n_ii (m n p : ℕ) : m < n ↔ m * p.succ < n * p.succ := by /- > Let `m` and `n` be natural numbers. > > ##### (⇒) > Suppose `m ∈ n`. Let > > `S = {p ∈ ω | m ⬝ p⁺ ∈ n ⬝ p⁺}`. -/ have hf : ∀ m n : ℕ, m < n → m * p.succ < n * p.succ := by intro m n hp₁ induction p with | zero => /- > `0 ∈ S` by *Right Multiplicative Identity*. -/ simp only [Nat.mul_one] exact hp₁ | succ p ih => /- > Next, suppose `p ∈ S`. That is, `m ⬝ p⁺ ∈ n ⬝ p⁺`. Then > > `m ⬝ p⁺⁺ = m ⬝ p⁺ + m` *Theorem 4J* > ` ∈ n ⬝ p⁺ + m` *(i)* > ` = m + n ⬝ p⁺` *Theorem 4K-2* > ` ∈ n + n ⬝ p⁺` *(i)* > ` = n ⬝p⁺ + n` *Theorem 4K-2* > ` n ⬝ p⁺⁺` *Theorem 4J* > > Therefore `p⁺ ∈ S`. -/ have hp₂ : m * p.succ < n * p.succ := by by_cases hp₃ : p = 0 · rw [hp₃] at * simp only [Nat.mul_one] at * exact hp₁ · exact ih calc m * p.succ.succ _ = m * p.succ + m := rfl _ < n * p.succ + m := (theorem_4n_i (m * p.succ) (n * p.succ) m).mp hp₂ _ = m + n * p.succ := by rw [theorem_4k_2] _ < n + n * p.succ := (theorem_4n_i m n (n * p.succ)).mp hp₁ _ = n * p.succ + n := by rw [theorem_4k_2] _ = n * p.succ.succ := rfl apply Iff.intro /- > Thus `S` is an inductive set. Hence *Theorem 4B* implies `S = ω`. By > *Theorem 4C*, every natural number except `0` is the successor of some natural > number. Therefore, for all `p ∈ ω` such that `p ≠ 0`, `m ∈ n` implies `m ⬝ p ∈ n ⬝ p`. -/ · exact hf m n /- > ##### (⇐) > Let `p ≠ 0` be a natural number and suppose `m ⬝ p ∈ n ⬝ p`. By the > *Trichotomy Law for `ω`*, there are two cases to consider regarding how `m` > and `n` relate to one another. -/ intro hp match @trichotomous ℕ LT.lt _ m n with | Or.inr (Or.inl h₁) => /- > ###### Case 1 > Suppose `m = n`. Then `m ⬝ p ∈ n ⬝ p = m ⬝ p`. *Lemma 4L(b)* shows this is > impossible. -/ rw [← h₁] at hp exact absurd hp (lemma_4l_b (m * p.succ)) | Or.inr (Or.inr h₁) => /- > ###### Case 2 > Suppose `n ∈ m`. Then *(⇒)* indicates `n ⬝ p ∈ m ⬝ p`. But this contradicts > *Trichotomy Law for `ω`* since, by hypothesis, `m ⬝ p ∈ n ⬝ p`. -/ have := hf n m h₁ exact absurd this (Nat.lt_asymm hp) | Or.inl h₁ => /- > ###### Conclusion > By trichotomy, it follows `m ∈ n`. -/ exact h₁ #check Nat.mul_lt_mul_of_pos_right /-! ### Corollary 4P The following cancellation laws hold for `m`, `n`, and `p` in `ω`: ``` m + p = n + p ⇒ m = n, m ⬝ p = n ⬝ p ∧ p ≠ 0 ⇒ m = n. ``` -/ theorem corollary_4p_i (m n p : ℕ) (h : m + p = n + p) : m = n := by /- > Suppose `m + p = n + p`. By the *Trichotomy Law for `ω`*, there are two cases > to consider regarding how `m` and `n` relate to one another. -/ match @trichotomous ℕ LT.lt _ m n with | Or.inl h₁ => /- > 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