bookshelf/Bookshelf/Enderton/Set/Chapter_4.lean

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