bookshelf/Bookshelf/Enderton/Set/Chapter_4.lean

393 lines
9.5 KiB
Plaintext
Raw Normal View History

import Common.Logic.Basic
import Mathlib.Data.Set.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 = 0 (∃ m : , 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 : )
: 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 : )
: 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 : )
: 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 : )
: 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 : )
: 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 : }
: 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 : )
: 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 : )
: 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 : )
: 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 : )
: 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 : )
: 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 : )
: 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 : )
: 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 := 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 : )
: 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
/-- #### Trichotomy Law for ω
For any natural numbers `m` and `n`, exactly one of the three conditions
```
m ∈ n, m = n, n ∈ m
```
holds.
-/
theorem trichotomy_law_for_nat
: IsAsymm LT.lt ∧ IsTrichotomous LT.lt :=
⟨instIsAsymmLtToLT, instIsTrichotomousLtToLTToPreorderToPartialOrder⟩
/-- #### Linear Ordering on ω
Relation
```
∈_ω = {⟨m, n⟩ ∈ ω × ω | m ∈ n}
```
is a linear ordering on `ω`.
-/
theorem linear_ordering_on_nat
: IsStrictTotalOrder LT.lt := isStrictTotalOrder_of_linearOrder
/-- #### 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 : ) (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 : ) : Prop := ∃ m, 2 * m = n
/--
Call a natural number *odd* if it has the form `(2 ⬝ p) + 1` for some `p`.
-/
def odd (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 : )
: (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
end Enderton.Set.Chapter_4