-- Negación del universal: Caracterización de funciones no pares
-- =============================================================

import data.real.basic

variable (f : ℝ → ℝ)

-- ----------------------------------------------------
-- Ejercicio 1. Definir la función
--    par : (ℝ → ℝ) → Prop
-- tal que (par f) expresa que f es par.
-- ----------------------------------------------------

def par : (ℝ → ℝ) → Prop
| f := ∀ x, f (-x) = f x

-- ----------------------------------------------------
-- Ejercicio 2. Demostrar que
--    ¬par f ↔ ∃ x, f (-x) ≠ f x
-- ----------------------------------------------------

-- 1ª demostración
example : ¬par f ↔ ∃ x, f (-x) ≠ f x :=
begin
  split,
  { contrapose,
    intro h1,
    rw not_not,
    unfold par,
    intro x,
    by_contradiction h2,
    apply h1,
    use x, },
  { intro h1,
    intro h2,
    unfold par at h2,
    cases h1 with x hx,
    apply hx,
    exact h2 x, },
end

-- 2ª demostración
example : ¬par f ↔ ∃ x, f (-x) ≠ f x :=
begin
  split,
  { contrapose,
    intro h1,
    rw not_not,
    intro x,
    by_contradiction h2,
    apply h1,
    use x, },
  { rintros ⟨x, hx⟩ h',
    exact hx (h' x) },
end

-- 3ª demostración
example : ¬par f ↔ ∃ x, f (-x) ≠ f x :=
begin
  unfold par,
  push_neg,
end

-- 4ª demostración
example : ¬par f ↔ ∃ x, f (-x) ≠ f x :=
iff.intro
  ( have h1 : ¬(∃ x, f (-x) ≠ f x) → ¬¬par f,
      { assume h2 : ¬(∃ x, f (-x) ≠ f x),
        have h3 : par f,
          { assume x,
            have h4 : ¬(f (-x) ≠ f x),
              { assume h5 : f (-x) ≠ f x,
                have h6 : ∃ x, f (-x) ≠ f x,
                  from exists.intro x h5,
                show false,
                  from h2 h6, },
            show f (-x) = f x,
              from not_not.mp h4},
        show ¬¬par f,
          from not_not.mpr h3 },
    show ¬par f → (∃ x, f (-x) ≠ f x),
      from not_imp_not.mp h1)
  ( assume h1 : ∃ x, f (-x) ≠ f x,
    show ¬par f,
      { assume h2 : par f,
        show false, from
          exists.elim h1
            ( assume x,
              assume hx : f (-x) ≠ f x,
              have h3 : f (-x) = f x,
                from h2 x,
              show false,
                from hx h3)})

-- 5ª demostración
example : ¬par f ↔ ∃ x, f (-x) ≠ f x :=
-- by suggest
not_forall

-- 6ª demostración
example : ¬par f ↔ ∃ x, f (-x) ≠ f x :=
-- by hint
by simp [par]


import .love05_inductive_predicates_demo

/- # LoVe Demo 12: Basic Mathematical Structures

We introduce definitions and proofs about basic mathematical structures such as
groups, fields, and linear orders. -/

set_option pp.beta true
set_option pp.generalized_field_notation false

namespace LoVe

/- ## Type Classes over a Single Binary Operator

Mathematically, a __group__ is a set `G` with a binary operator `• : G × G → G`
with the following properties, called __group axioms__:

* Associativity: For all `a, b, c ∈ G`, we have `(a • b) • c = a • (b • c)`.
* Identity element: There exists an element `e ∈ G` such that for all `a ∈ G`,
  we have `e • a = a`.
* Inverse element: For each `a ∈ G`, there exists an inverse element
  `inv(a) ∈ G` such that `inv(a) • a = e`.

Examples of groups are
* `ℤ` with `+`;
* `ℝ` with `+`;
* `ℝ \ {0}` with `*`.

In Lean, a type class for groups can be defined as follows: -/

namespace monolithic_group

@[class] structure group (α : Type) :=
(mul          : α → α → α)
(one          : α)
(inv          : α → α)
(mul_assoc    : ∀a b c, mul (mul a b) c = mul a (mul b c))
(one_mul      : ∀a, mul one a = a)
(mul_left_inv : ∀a, mul (inv a) a = one)

end monolithic_group

/- In Lean, however, group is part of a larger hierarchy of algebraic
structures:

Type class               | Properties                               | Examples
------------------------ | -----------------------------------------|-------------------
`semigroup`              | associativity of `*`                     | `ℝ`, `ℚ`, `ℤ`, `ℕ`
`monoid`                 | `semigroup` with unit `1`                | `ℝ`, `ℚ`, `ℤ`, `ℕ`
`left_cancel_semigroup`  | `semigroup` with `c * a = c * b → a = b` |
`right_cancel_semigroup` | `semigroup` with `a * c = b * c → a = b` |
`group`                  | `monoid` with inverse `⁻¹`               |

Most of these structures have commutative versions: `comm_semigroup`,
`comm_monoid`, `comm_group`.

The __multiplicative__ structures (over `*`, `1`, `⁻¹`) are copied to produce
__additive__ versions (over `+`, `0`, `-`):

Type class                   | Properties                                   | Examples
---------------------------- | ---------------------------------------------|-------------------
`add_semigroup`              | associativity of `+`                         | `ℝ`, `ℚ`, `ℤ`, `ℕ`
`add_monoid`                 | `add_semigroup` with unit `0`                | `ℝ`, `ℚ`, `ℤ`, `ℕ`
`add_left_cancel_semigroup`  | `add_semigroup` with `c + a = c + b → a = b` | `ℝ`, `ℚ`, `ℤ`, `ℕ`
`add_right_cancel_semigroup` | `add_semigroup` with `a + c = b + c → a = b` | `ℝ`, `ℚ`, `ℤ`, `ℕ`
`add_group`                  | `add_monoid` with inverse `-`                | `ℝ`, `ℚ`, `ℤ` -/

#print group
#print add_group

/- Let us define our own type, of integers modulo 2, and register it as an
additive group. -/

inductive ℤ₂ : Type
| zero
| one

def ℤ₂.add : ℤ₂ → ℤ₂ → ℤ₂
| ℤ₂.zero a       := a
| a       ℤ₂.zero := a
| ℤ₂.one  ℤ₂.one  := ℤ₂.zero

@[instance] def ℤ₂.add_group : add_group ℤ₂ :=
{ add          := ℤ₂.add,
  add_assoc    :=
    by intros a b c; cases' a; cases' b; cases' c; refl,
  zero         := ℤ₂.zero,
  zero_add     := by intro a; cases' a; refl,
  add_zero     := by intro a; cases' a; refl,
  neg          := λa, a,
  add_left_neg := by intro a; cases' a; refl }

#reduce ℤ₂.one + 0 - 0 - ℤ₂.one

lemma ℤ₂.add_right_neg:
  ∀a : ℤ₂, a + - a = 0 :=
add_right_neg

/- Another example: Lists are an `add_monoid`: -/

@[instance] def list.add_monoid {α : Type} :
  add_monoid (list α) :=
{ zero      := [],
  add       := (++),
  add_assoc := list.append_assoc,
  zero_add  := list.nil_append,
  add_zero  := list.append_nil }

/- ## Type Classes with Two Binary Operators

Mathematically, a __field__ is a set `F` such that

* `F` forms a commutative group under an operator `+`, called addition, with
  identity element `0`.
* `F\{0}` forms a commutative group under an operator `*`, called
  multiplication.
* Multiplication distributes over addition—i.e.,
  `a * (b + c) = a * b + a * c` for all `a, b, c ∈ F`.

In Lean, fields are also part of a larger hierarchy:

Type class       |  Properties                                         | Examples
-----------------|-----------------------------------------------------|-------------------
`semiring`       | `monoid` and `add_comm_monoid` with distributivity  | `ℝ`, `ℚ`, `ℤ`, `ℕ`
`comm_semiring`  | `semiring` with commutativity of `*`                | `ℝ`, `ℚ`, `ℤ`, `ℕ`
`ring`           | `monoid` and `add_comm_group` with distributivity   | `ℝ`, `ℚ`, `ℤ`
`comm_ring`      | `ring` with commutativity of `*`                    | `ℝ`, `ℚ`, `ℤ`
`division_ring`  | `ring` with multiplicative inverse `⁻¹`             | `ℝ`, `ℚ`
`field`          | `division_ring` with commutativity of `*`           | `ℝ`, `ℚ`
`discrete_field` | `field` with decidable equality and `∀n, n / 0 = 0` | `ℝ`, `ℚ` -/

#print field

/- Let us continue with our example: -/

def ℤ₂.mul : ℤ₂ → ℤ₂ → ℤ₂
| ℤ₂.one  a       := a
| a       ℤ₂.one  := a
| ℤ₂.zero ℤ₂.zero := ℤ₂.zero

@[instance] def ℤ₂.field : field ℤ₂ :=
{ one            := ℤ₂.one,
  mul            := ℤ₂.mul,
  inv            := λa, a,
  add_comm       := by intros a b; cases' a; cases' b; refl,
  exists_pair_ne :=
    by apply exists.intro ℤ₂.zero; apply exists.intro ℤ₂.one;
      finish,
  one_mul        := by intros a; cases' a; refl,
  mul_one        := by intros a; cases' a; refl,
  mul_inv_cancel := by intros a h; cases' a; finish,
  inv_zero       := by refl,
  mul_assoc      :=
    by intros a b c; cases' a; cases' b; cases' c; refl,
  mul_comm       := by intros a b; cases' a; cases' b; refl,
  left_distrib   :=
    by intros a b c; cases' a; cases' b; cases' c; refl,
  right_distrib  :=
    by intros a b c; cases' a; cases' b; cases' c; refl,
  ..ℤ₂.add_group }

#reduce (1 : ℤ₂) * 0 / (0 - 1)

#reduce (3 : ℤ₂)

lemma ring_example (a b : ℤ₂) :
  (a + b) ^ 3 = a ^ 3 + 3 * a ^ 2 * b + 3 * a * b ^ 2 + b ^ 3 :=
by ring

lemma ring_exp_example (a b : ℤ₂) (n : ℕ):
  (a + b) ^ (2 + n) =
  (a + b) ^ n * (a ^ 2 + 2 * a * b + b ^ 2) :=
by ring_exp

/- `ring` and `ring_exp` prove equalities over commutative rings and semirings
by normalizing expressions. The `ring_exp` variant also normalizes exponents. -/

lemma abel_example (a b : ℤ) :
  a + b + 0 - (b + a + a) = - a :=
by abel

/- `abel` proves equalities over additive commutative monoids and groups by
normalizing expressions.

## Coercions

When combining numbers form `ℕ`, `ℤ`, `ℚ`, and `ℝ`, we might want to cast from
one type to another. Lean has a mechanism to automatically introduce coercions,
represented by `coe` (syntactic sugar: `↑`). `coe` can be set up to provide
implicit coercions between arbitrary types.

Many coercions are already in place, including the following:

* `coe : ℕ → α` casts `ℕ` to another semiring `α`;
* `coe : ℤ → α` casts `ℤ` to another ring `α`;
* `coe : ℚ → α` casts `ℚ` to another division ring `α`.

For example, this works, although negation `- n` is not defined on natural
numbers: -/

lemma neg_mul_neg_nat (n : ℕ) (z : ℤ) :
  (- z) * (- n) = z * n :=
neg_mul_neg z n

/- Notice how Lean introduced a `↑` coercion: -/

#print neg_mul_neg_nat

/- Another example: -/

lemma neg_nat_mul_neg (n : ℕ) (z : ℤ) :
  (- n : ℤ) * (- z) = n * z :=
neg_mul_neg n z

#print neg_nat_mul_neg

/- In proofs involving coercions, the tactic `norm_cast` can be convenient. -/

lemma norm_cast_example_1 (m n : ℕ) (h : (m : ℤ) = (n : ℤ)) :
  m = n :=
begin
  norm_cast at h,
  exact h
end

lemma norm_cast_example_2 (m n : ℕ) :
  (m : ℤ) + (n : ℤ) = ((m + n : ℕ) : ℤ) :=
by norm_cast

/- `norm_cast` moves coercions towards the inside of expressions, as a form of
simplification. Like `simp`, it will generally produce a subgoal.

`norm_cast` relies on lemmas such as the following: -/

#check nat.cast_add
#check int.cast_add
#check rat.cast_add

/- ### Lists, Multisets and Finite Sets

For finite collections of elements different structures are available:

* lists: order and duplicates matter;
* multisets: only duplicates matter;
* finsets: neither order nor duplicates matter. -/

lemma list_duplicates_example :
  [2, 3, 3, 4] ≠ [2, 3, 4] :=
dec_trivial

lemma list_order_example :
  [4, 3, 2] ≠ [2, 3, 4] :=
dec_trivial

lemma multiset_duplicates_example :
  ({2, 3, 3, 4} : multiset ℕ) ≠ {2, 3, 4} :=
dec_trivial

lemma multiset_order_example :
  ({2, 3, 4} : multiset ℕ) = {4, 3, 2} :=
dec_trivial

lemma finset_duplicates_example :
  ({2, 3, 3, 4} : finset ℕ) = {2, 3, 4} :=
dec_trivial

lemma finsetorder_example :
  ({2, 3, 4} : finset ℕ) = {4, 3, 2} :=
dec_trivial

/- `dec_trivial` is a special lemma that can be used on trivial decidable
goals (e.g., true closed executable expressions). -/

def list.elems : btree ℕ → list ℕ
| btree.empty        := []
| (btree.node a l r) := a :: list.elems l ++ list.elems r

def multiset.elems : btree ℕ → multiset ℕ
| btree.empty        := ∅
| (btree.node a l r) :=
  {a} ∪ (multiset.elems l ∪ multiset.elems r)

def finset.elems : btree ℕ → finset ℕ
| btree.empty        := ∅
| (btree.node a l r) := {a} ∪ (finset.elems l ∪ finset.elems r)

#eval list.sum [2, 3, 4]                          -- result: 9
#eval multiset.sum ({2, 3, 4} : multiset ℕ)       -- result: 9
#eval finset.sum ({2, 3, 4} : finset ℕ) (λn, n)   -- result: 9

#eval list.prod [2, 3, 4]                         -- result: 24
#eval multiset.prod ({2, 3, 4} : multiset ℕ)      -- result: 24
#eval finset.prod ({2, 3, 4} : finset ℕ) (λn, n)  -- result: 24

/- ## Order Type Classes

Many of the structures introduced above can be ordered. For example, the
well-known order on the natural numbers can be defined as follows: -/

inductive nat.le : ℕ → ℕ → Prop
| refl : ∀a : ℕ, nat.le a a
| step : ∀a b : ℕ, nat.le a b → nat.le a (b + 1)

/- This is an example of a linear order. A __linear order__ (or
__total order__) is a binary relation `≤` such that for all `a`, `b`, `c`, the
following properties hold:

* Reflexivity: `a ≤ a`.
* Transitivity: If `a ≤ b` and `b ≤ c`, then `a ≤ c`.
* Antisymmetry: If `a ≤ b` and `b ≤ a`, then `a = b`.
* Totality: `a ≤ b` or `b ≤ a`.

If a relation has the first three properties, it is a __partial order__. An
example is `⊆` on sets, finite sets, or multisets. If a relation has the first
two properties, it is a __preorder__. An example is comparing lists by their
length.

In Lean, there are type classes for these different kinds of orders:
`linear_order`, `partial_order`, and `preorder`. The `preorder` class has the
fields

    `le       : α → α → Prop`
    `le_refl  : ∀a : α, le a a`
    `le_trans : ∀a b c : α, le a b → le b c → le a c`

The `partial_order` class also has

    `le_antisymm : ∀a b : α, le a b → le b a → a = b`

and `linear_order` also has

    `le_total : ∀a b : α, le a b ∨ le b a`

We can declare the preorder on lists that compares lists by their length as
follows: -/

@[instance] def list.length.preord {α : Type} :
  preorder (list α) :=
{ le       := λxs ys, list.length xs ≤ list.length ys,
  le_refl  := by intro xs; exact nat.le_refl _,
  le_trans := by intros xs ys zs; exact nat.le_trans }

/- This instance introduces the infix syntax `≤` and the relations `≥`, `<`,
and `>`: -/

lemma list.length.preord_example {α : Type} (c : α) :
  [c] > [] :=
dec_trivial

/- Complete lattices (lecture 10) are formalized as another type class,
`complete_lattice`, which inherits from `partial_order`.

Type classes combining orders and algebraic structures are also available:

    `ordered_cancel_comm_monoid`
    `ordered_comm_group ordered_semiring`
    `linear_ordered_semiring`
    `linear_ordered_comm_ring`
    `linear_ordered_field`

All these mathematical structures relate `≤` and `<` with `0`, `1`, `+`, and `*`
by monotonicity rules (e.g., `a ≤ b → c ≤ d → a + c ≤ b + d`) and cancellation
rules (e.g., `c + a ≤ c + b → a ≤ b`). -/

end LoVe


/-
Copyright (c) 2021 Stuart Presnell. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: Stuart Presnell
-/
import data.nat.prime
import data.finsupp.multiset
import algebra.big_operators.finsupp

/-!
# Prime factorizations

 `n.factorization` is the finitely supported function `ℕ →₀ ℕ`
 mapping each prime factor of `n` to its multiplicity in `n`.  For example, since 2000 = 2^4 * 5^3,
  * `factorization 2000 2` is 4
  * `factorization 2000 5` is 3
  * `factorization 2000 k` is 0 for all other `k : ℕ`.

## TODO

* As discussed in this Zulip thread:
https://leanprover.zulipchat.com/#narrow/stream/217875/topic/Multiplicity.20in.20the.20naturals
We have lots of disparate ways of talking about the multiplicity of a prime
in a natural number, including `factors.count`, `padic_val_nat`, `multiplicity`,
and the material in `data/pnat/factors`.  Move some of this material to this file,
prove results about the relationships between these definitions,
and (where appropriate) choose a uniform canonical way of expressing these ideas.

* Moreover, the results here should be generalised to an arbitrary unique factorization monoid
with a normalization function, and then deduplicated.  The basics of this have been started in
`ring_theory/unique_factorization_domain`.

* Extend the inductions to any `normalization_monoid` with unique factorization.

-/

open nat finset list finsupp
open_locale big_operators

namespace nat

/-- `n.factorization` is the finitely supported function `ℕ →₀ ℕ`
 mapping each prime factor of `n` to its multiplicity in `n`. -/
noncomputable def factorization (n : ℕ) : ℕ →₀ ℕ := (n.factors : multiset ℕ).to_finsupp

/-! ### Basic facts about factorization -/

@[simp] lemma factorization_prod_pow_eq_self {n : ℕ} (hn : n ≠ 0) : n.factorization.prod pow = n :=
begin
  simp only [←prod_to_multiset, factorization, multiset.coe_prod, multiset.to_finsupp_to_multiset],
  exact prod_factors hn,
end

/-- We can write both `n.factorization p` and `n.factors.count p` to represent the power
of `p` in the factorization of `n`: we declare the former to be the simp-normal form.
However, since `factorization` is a finsupp it's noncomputable.  This theorem can also
be used in reverse to compute values of `factorization n p` when required. -/
@[simp] lemma factors_count_eq {n p : ℕ} : n.factors.count p = n.factorization p :=
by simp [factorization]

lemma eq_of_factorization_eq {a b : ℕ} (ha : a ≠ 0) (hb : b ≠ 0)
  (h : ∀ p : ℕ, a.factorization p = b.factorization p) : a = b :=
eq_of_perm_factors ha hb (by simpa only [list.perm_iff_count, factors_count_eq] using h)

/-- Every nonzero natural number has a unique prime factorization -/
lemma factorization_inj : set.inj_on factorization { x : ℕ | x ≠ 0 } :=
λ a ha b hb h, eq_of_factorization_eq ha hb (λ p, by simp [h])

@[simp] lemma factorization_zero : factorization 0 = 0 :=
by simp [factorization]

@[simp] lemma factorization_one : factorization 1 = 0 :=
by simp [factorization]

/-- The support of `n.factorization` is exactly `n.factors.to_finset` -/
@[simp] lemma support_factorization {n : ℕ} : n.factorization.support = n.factors.to_finset :=
by simpa [factorization, multiset.to_finsupp_support]

lemma factor_iff_mem_factorization {n p : ℕ} : p ∈ n.factorization.support ↔ p ∈ n.factors :=
by simp only [support_factorization, list.mem_to_finset]

lemma prime_of_mem_factorization {n p : ℕ} (hp : p ∈ n.factorization.support) : p.prime :=
prime_of_mem_factors (factor_iff_mem_factorization.mp hp)

lemma pos_of_mem_factorization {n p : ℕ} (hp : p ∈ n.factorization.support) : 0 < p :=
prime.pos (prime_of_mem_factorization hp)

lemma le_of_mem_factorization {n p : ℕ} (h : p ∈ n.factorization.support) : p ≤ n :=
le_of_mem_factors (factor_iff_mem_factorization.mp h)

@[simp]
lemma factorization_eq_zero_of_non_prime (n : ℕ) {p : ℕ} (hp : ¬p.prime) : n.factorization p = 0 :=
not_mem_support_iff.1 (mt prime_of_mem_factorization hp)

lemma factorization_eq_zero_of_lt {n p : ℕ} (h : n < p) : n.factorization p = 0 :=
finsupp.not_mem_support_iff.mp (mt le_of_mem_factorization (not_le_of_lt h))

@[simp] lemma factorization_zero_right (n : ℕ) : n.factorization 0 = 0 :=
factorization_eq_zero_of_non_prime _ not_prime_zero

@[simp] lemma factorization_one_right (n : ℕ) : n.factorization 1 = 0 :=
factorization_eq_zero_of_non_prime _ not_prime_one

lemma dvd_of_factorization_pos {n p : ℕ} (hn : n.factorization p ≠ 0) : p ∣ n :=
dvd_of_mem_factors (factor_iff_mem_factorization.1 (mem_support_iff.2 hn))

lemma prime.factorization_pos_of_dvd {n p : ℕ} (hp : p.prime) (hn : n ≠ 0) (h : p ∣ n) :
  0 < n.factorization p :=
by rwa [←factors_count_eq, count_pos, mem_factors_iff_dvd hn hp]

/-- The only numbers with empty prime factorization are `0` and `1` -/
lemma factorization_eq_zero_iff (n : ℕ) : n.factorization = 0 ↔ n = 0 ∨ n = 1 :=
by simp [factorization, add_equiv.map_eq_zero_iff, multiset.coe_eq_zero]

lemma factorization_eq_zero_iff' (n p : ℕ) :
  n.factorization p = 0 ↔ ¬p.prime ∨ ¬p ∣ n ∨ n = 0 :=
begin
  rw [←not_mem_support_iff, support_factorization, mem_to_finset],
  rcases eq_or_ne n 0 with rfl | hn,
  { simp },
  { simp [hn, nat.mem_factors, not_and_distrib] },
end

/-- For nonzero `a` and `b`, the power of `p` in `a * b` is the sum of the powers in `a` and `b` -/
@[simp] lemma factorization_mul {a b : ℕ} (ha : a ≠ 0) (hb : b ≠ 0) :
  (a * b).factorization = a.factorization + b.factorization :=
by { ext p, simp only [add_apply, ←factors_count_eq,
                       perm_iff_count.mp (perm_factors_mul ha hb) p, count_append] }

lemma factorization_mul_support {a b : ℕ} (ha : a ≠ 0) (hb : b ≠ 0) :
  (a * b).factorization.support = a.factorization.support ∪ b.factorization.support :=
begin
  ext q,
  simp only [finset.mem_union, factor_iff_mem_factorization],
  exact mem_factors_mul ha hb
end

/-- For any `p`, the power of `p` in `n^k` is `k` times the power in `n` -/
@[simp] lemma factorization_pow (n k : ℕ) :
  factorization (n^k) = k • n.factorization :=
begin
  induction k with k ih, { simp },
  rcases eq_or_ne n 0 with rfl | hn, { simp },
  rw [pow_succ, factorization_mul hn (pow_ne_zero _ hn), ih, succ_eq_one_add, add_smul, one_smul],
end

/-- The only prime factor of prime `p` is `p` itself, with multiplicity `1` -/
@[simp] lemma prime.factorization {p : ℕ} (hp : prime p) :
  p.factorization = single p 1 :=
begin
  ext q,
  rw [←factors_count_eq, factors_prime hp, single_apply, count_singleton', if_congr eq_comm];
  refl,
end

/-- For prime `p` the only prime factor of `p^k` is `p` with multiplicity `k` -/
lemma prime.factorization_pow {p k : ℕ} (hp : prime p) :
  factorization (p ^ k) = single p k :=
by simp [hp]

/-- If the factorization of `n` contains just one number `p` then `n` is a power of `p` -/
lemma eq_pow_of_factorization_eq_single {n p k : ℕ} (hn : n ≠ 0)
  (h : n.factorization = finsupp.single p k) : n = p ^ k :=
by { rw [←nat.factorization_prod_pow_eq_self hn, h], simp }

/-- If a product over `n.factorization` doesn't use the multiplicities of the prime factors
then it's equal to the corresponding product over `n.factors.to_finset` -/
lemma prod_factorization_eq_prod_factors {n : ℕ} {β : Type*} [comm_monoid β] (f : ℕ → β) :
  n.factorization.prod (λ p k, f p) = ∏ p in n.factors.to_finset, (f p) :=
by { apply prod_congr support_factorization, simp }

/-- For any `p : ℕ` and any function `g : α → ℕ` that's non-zero on `S : finset α`,
the power of `p` in `S.prod g` equals the sum over `x ∈ S` of the powers of `p` in `g x`.
Generalises `factorization_mul`, which is the special case where `S.card = 2` and `g = id`. -/
lemma factorization_prod {α : Type*} {S : finset α} {g : α → ℕ} (hS : ∀ x ∈ S, g x ≠ 0) :
  (S.prod g).factorization = S.sum (λ x, (g x).factorization) :=
begin
  classical,
  ext p,
  apply finset.induction_on' S, { simp },
  { intros x T hxS hTS hxT IH,
    have hT : T.prod g ≠ 0 := prod_ne_zero_iff.mpr (λ x hx, hS x (hTS hx)),
    simp [prod_insert hxT, sum_insert hxT, ←IH, factorization_mul (hS x hxS) hT] }
end

/-! ### Equivalence between `ℕ+` and `ℕ →₀ ℕ` with support in the primes. -/

/-- Any finsupp `f : ℕ →₀ ℕ` whose support is in the primes is equal to the factorization of
the product `∏ (a : ℕ) in f.support, a ^ f a`. -/
lemma prod_pow_factorization_eq_self {f : ℕ →₀ ℕ} (hf : ∀ (p : ℕ), p ∈ f.support → prime p) :
  (f.prod pow).factorization = f :=
begin
  have h : ∀ x : ℕ, x ∈ f.support → x ^ f x ≠ 0 := λ p hp, pow_ne_zero _ (prime.ne_zero (hf p hp)),
  simp only [finsupp.prod, factorization_prod h],
  nth_rewrite_rhs 0 (sum_single f).symm,
  exact sum_congr rfl (λ p hp, prime.factorization_pow (hf p hp)),
end

lemma eq_factorization_iff {n : ℕ} {f : ℕ →₀ ℕ} (hn : n ≠ 0) (hf : ∀ p ∈ f.support, prime p) :
  f = n.factorization ↔ f.prod pow = n :=
⟨λ h, by rw [h, factorization_prod_pow_eq_self hn],
 λ h, by rw [←h, prod_pow_factorization_eq_self hf]⟩

/-- The equiv between `ℕ+` and `ℕ →₀ ℕ` with support in the primes. -/
noncomputable
def factorization_equiv : ℕ+ ≃ {f : ℕ →₀ ℕ | ∀ p ∈ f.support, prime p} :=
{ to_fun    := λ ⟨n, hn⟩, ⟨n.factorization, λ _, prime_of_mem_factorization⟩,
  inv_fun   := λ ⟨f, hf⟩, ⟨f.prod pow,
    prod_pow_pos_of_zero_not_mem_support (λ H, not_prime_zero (hf 0 H))⟩,
  left_inv  := λ ⟨x, hx⟩, subtype.ext $ factorization_prod_pow_eq_self hx.ne.symm,
  right_inv := λ ⟨f, hf⟩, subtype.ext $ prod_pow_factorization_eq_self hf }

lemma factorization_equiv_apply (n : ℕ+) : (factorization_equiv n).1 = n.1.factorization :=
by { cases n, refl }

lemma factorization_equiv_inv_apply {f : ℕ →₀ ℕ} (hf : ∀ p ∈ f.support, prime p) :
  (factorization_equiv.symm ⟨f, hf⟩).1 = f.prod pow := rfl

/-! ### Factorization and divisibility -/

lemma dvd_of_mem_factorization {n p : ℕ} (h : p ∈ n.factorization.support) : p ∣ n :=
begin
  rcases eq_or_ne n 0 with rfl | hn, { simp },
  simp [←mem_factors_iff_dvd hn (prime_of_mem_factorization h), factor_iff_mem_factorization.mp h],
end

lemma pow_factorization_dvd (n p : ℕ) : p ^ n.factorization p ∣ n :=
begin
  by_cases hp : p.prime, swap, { simp [factorization_eq_zero_of_non_prime n hp] },
  rw ←factors_count_eq,
  apply dvd_of_factors_subperm (pow_ne_zero _ hp.ne_zero),
  rw [hp.factors_pow, list.subperm_ext_iff],
  intros q hq,
  simp [list.eq_of_mem_repeat hq],
end

lemma pow_factorization_le {n : ℕ} (p : ℕ) (hn : n ≠ 0) : p ^ n.factorization p ≤ n :=
le_of_dvd hn.bot_lt (nat.pow_factorization_dvd n p)

lemma div_pow_factorization_ne_zero {n : ℕ} (p : ℕ) (hn : n ≠ 0) :
  n / p ^ n.factorization p ≠ 0 :=
begin
  by_cases pp : nat.prime p,
  { apply mt (nat.div_eq_zero_iff (pow_pos (prime.pos pp) _)).1,
    simp [le_of_dvd hn.bot_lt (nat.pow_factorization_dvd n p)] },
  { simp [nat.factorization_eq_zero_of_non_prime n pp, hn] },
end

lemma factorization_le_iff_dvd {d n : ℕ} (hd : d ≠ 0) (hn : n ≠ 0) :
  d.factorization ≤ n.factorization ↔ d ∣ n :=
begin
  split,
  { intro hdn,
    set K := n.factorization - d.factorization with hK,
    use K.prod pow,
    rw [←factorization_prod_pow_eq_self hn, ←factorization_prod_pow_eq_self hd,
        ←finsupp.prod_add_index' pow_zero pow_add, hK, add_tsub_cancel_of_le hdn] },
  { rintro ⟨c, rfl⟩, rw factorization_mul hd (right_ne_zero_of_mul hn), simp },
end

lemma pow_succ_factorization_not_dvd {n p : ℕ} (hn : n ≠ 0) (hp : p.prime) :
  ¬ p ^ (n.factorization p + 1) ∣ n :=
begin
  intro h,
  rw ←factorization_le_iff_dvd (pow_pos hp.pos _).ne' hn at h,
  simpa [hp.factorization] using h p,
end

lemma factorization_le_factorization_mul_left {a b : ℕ} (hb : b ≠ 0) :
  a.factorization ≤ (a * b).factorization :=
begin
  rcases eq_or_ne a 0 with rfl | ha, { simp },
  rw [factorization_le_iff_dvd ha $ mul_ne_zero ha hb],
  exact dvd.intro b rfl
end

lemma factorization_le_factorization_mul_right {a b : ℕ} (ha : a ≠ 0) :
  b.factorization ≤ (a * b).factorization :=
by { rw mul_comm, apply factorization_le_factorization_mul_left ha }

lemma prime.pow_dvd_iff_le_factorization {p k n : ℕ} (pp : prime p) (hn : n ≠ 0) :
  p ^ k ∣ n ↔ k ≤ n.factorization p :=
by rw [←factorization_le_iff_dvd (pow_pos pp.pos k).ne' hn, pp.factorization_pow, single_le_iff]

lemma prime.pow_dvd_iff_dvd_pow_factorization {p k n : ℕ} (pp : prime p) (hn : n ≠ 0) :
  p ^ k ∣ n ↔ p ^ k ∣ p ^ n.factorization p :=
by rw [pow_dvd_pow_iff_le_right pp.one_lt, pp.pow_dvd_iff_le_factorization hn]

lemma prime.dvd_iff_one_le_factorization {p n : ℕ} (pp : prime p) (hn : n ≠ 0) :
  p ∣ n ↔ 1 ≤ n.factorization p :=
iff.trans (by simp) (pp.pow_dvd_iff_le_factorization hn)

lemma exists_factorization_lt_of_lt {a b : ℕ} (ha : a ≠ 0) (hab : a < b) :
  ∃ p : ℕ, a.factorization p < b.factorization p :=
begin
  have hb : b ≠ 0 := (ha.bot_lt.trans hab).ne',
  contrapose! hab,
  rw [←finsupp.le_def, factorization_le_iff_dvd hb ha] at hab,
  exact le_of_dvd ha.bot_lt hab,
end

@[simp] lemma factorization_div {d n : ℕ} (h : d ∣ n) :
  (n / d).factorization = n.factorization - d.factorization :=
begin
  rcases eq_or_ne d 0 with rfl | hd, { simp [zero_dvd_iff.mp h] },
  rcases eq_or_ne n 0 with rfl | hn, { simp },
  apply add_left_injective d.factorization,
  simp only,
  rw [tsub_add_cancel_of_le $ (nat.factorization_le_iff_dvd hd hn).mpr h,
      ←nat.factorization_mul (nat.div_pos (nat.le_of_dvd hn.bot_lt h) hd.bot_lt).ne' hd,
      nat.div_mul_cancel h],
end

lemma not_dvd_div_pow_factorization {n p : ℕ} (hp : prime p) (hn : n ≠ 0) :
  ¬p ∣ n / p ^ n.factorization p :=
begin
  rw [nat.prime.dvd_iff_one_le_factorization hp (div_pow_factorization_ne_zero p hn),
    nat.factorization_div (nat.pow_factorization_dvd n p)],
  simp [hp.factorization],
end

lemma coprime_of_div_pow_factorization {n p : ℕ} (hp : prime p) (hn : n ≠ 0) :
  coprime p (n / p ^ n.factorization p) :=
(or_iff_left (not_dvd_div_pow_factorization hp hn)).mp $ coprime_or_dvd_of_prime hp _

lemma dvd_iff_div_factorization_eq_tsub {d n : ℕ} (hd : d ≠ 0) (hdn : d ≤ n) :
  d ∣ n ↔ (n / d).factorization = n.factorization - d.factorization :=
begin
  refine ⟨factorization_div, _⟩,
  rcases eq_or_lt_of_le hdn with rfl | hd_lt_n, { simp },
  have h1 : n / d ≠ 0 := λ H, nat.lt_asymm hd_lt_n ((nat.div_eq_zero_iff hd.bot_lt).mp H),
  intros h,
  rw dvd_iff_le_div_mul n d,
  by_contra h2,
  cases (exists_factorization_lt_of_lt (mul_ne_zero h1 hd) (not_le.mp h2)) with p hp,
  rwa [factorization_mul h1 hd, add_apply, ←lt_tsub_iff_right, h, tsub_apply,
    lt_self_iff_false] at hp
end

lemma dvd_iff_prime_pow_dvd_dvd (n d : ℕ) :
  d ∣ n ↔ ∀ p k : ℕ, prime p → p ^ k ∣ d → p ^ k ∣ n :=
begin
  rcases eq_or_ne n 0 with rfl | hn, { simp },
  rcases eq_or_ne d 0 with rfl | hd,
  { simp only [zero_dvd_iff, hn, false_iff, not_forall],
    refine ⟨2, n, prime_two, ⟨dvd_zero _, _⟩⟩,
    apply mt (le_of_dvd hn.bot_lt) (not_le.mpr (lt_two_pow n)) },
  refine ⟨λ h p k _ hpkd, dvd_trans hpkd h, _⟩,
  rw [←factorization_le_iff_dvd hd hn, finsupp.le_def],
  intros h p,
  by_cases pp : prime p, swap, { simp [factorization_eq_zero_of_non_prime d pp] },
  rw ←pp.pow_dvd_iff_le_factorization hn,
  exact h p _ pp (pow_factorization_dvd _ _)
end

lemma prod_prime_factors_dvd (n : ℕ) : (∏ (p : ℕ) in n.factors.to_finset, p) ∣ n :=
begin
  by_cases hn : n = 0, { subst hn, simp },
  simpa [prod_factors hn] using multiset.to_finset_prod_dvd_prod (n.factors : multiset ℕ),
end

lemma factorization_gcd {a b : ℕ} (ha_pos : a ≠ 0) (hb_pos : b ≠ 0) :
  (gcd a b).factorization = a.factorization ⊓ b.factorization :=
begin
  let dfac := a.factorization ⊓ b.factorization,
  let d := dfac.prod pow,
  have dfac_prime : ∀ (p : ℕ), p ∈ dfac.support → prime p,
  { intros p hp,
    have : p ∈ a.factors ∧ p ∈ b.factors := by simpa using hp,
    exact prime_of_mem_factors this.1 },
  have h1 : d.factorization = dfac := prod_pow_factorization_eq_self dfac_prime,
  have hd_pos : d ≠ 0 := (factorization_equiv.inv_fun ⟨dfac, dfac_prime⟩).2.ne.symm,
  suffices : d = (gcd a b), { rwa ←this },
  apply gcd_greatest,
  { rw [←factorization_le_iff_dvd hd_pos ha_pos, h1], exact inf_le_left },
  { rw [←factorization_le_iff_dvd hd_pos hb_pos, h1], exact inf_le_right },
  { intros e hea heb,
    rcases decidable.eq_or_ne e 0 with rfl | he_pos,
    { simp only [zero_dvd_iff] at hea, contradiction, },
    have hea' := (factorization_le_iff_dvd he_pos ha_pos).mpr hea,
    have heb' := (factorization_le_iff_dvd he_pos hb_pos).mpr heb,
    simp [←factorization_le_iff_dvd he_pos hd_pos, h1, hea', heb'] },
end

/-! ### Factorization and coprimes -/

/-- For coprime `a` and `b`, the power of `p` in `a * b` is the sum of the powers in `a` and `b` -/
lemma factorization_mul_apply_of_coprime {p a b : ℕ} (hab : coprime a b)  :
  (a * b).factorization p = a.factorization p + b.factorization p :=
by simp only [←factors_count_eq, perm_iff_count.mp (perm_factors_mul_of_coprime hab), count_append]

/-- For coprime `a` and `b`, the power of `p` in `a * b` is the sum of the powers in `a` and `b` -/
lemma factorization_mul_of_coprime {a b : ℕ} (hab : coprime a b) :
  (a * b).factorization = a.factorization + b.factorization :=
begin
  ext q,
  simp only [finsupp.coe_add, add_apply, ←factors_count_eq, factorization_mul_apply_of_coprime hab],
end

/-- If `p` is a prime factor of `a` then the power of `p` in `a` is the same that in `a * b`,
for any `b` coprime to `a`. -/
lemma factorization_eq_of_coprime_left {p a b : ℕ} (hab : coprime a b) (hpa : p ∈ a.factors) :
  (a * b).factorization p = a.factorization p :=
begin
  rw [factorization_mul_apply_of_coprime hab, ←factors_count_eq, ←factors_count_eq],
  simpa only [count_eq_zero_of_not_mem (coprime_factors_disjoint hab hpa)],
end

/-- If `p` is a prime factor of `b` then the power of `p` in `b` is the same that in `a * b`,
for any `a` coprime to `b`. -/
lemma factorization_eq_of_coprime_right {p a b : ℕ} (hab : coprime a b) (hpb : p ∈ b.factors) :
  (a * b).factorization p = b.factorization p :=
by { rw mul_comm, exact factorization_eq_of_coprime_left (coprime_comm.mp hab) hpb }

/-- The prime factorizations of coprime `a` and `b` are disjoint -/
lemma factorization_disjoint_of_coprime {a b : ℕ} (hab : coprime a b) :
  disjoint a.factorization.support b.factorization.support :=
by simpa only [support_factorization]
  using disjoint_to_finset_iff_disjoint.mpr (coprime_factors_disjoint hab)

/-- For coprime `a` and `b` the prime factorization `a * b` is the union of those of `a` and `b` -/
lemma factorization_mul_support_of_coprime {a b : ℕ} (hab : coprime a b) :
  (a * b).factorization.support = a.factorization.support ∪ b.factorization.support :=
begin
  rw factorization_mul_of_coprime hab,
  exact support_add_eq (factorization_disjoint_of_coprime hab),
end

/-! ### Induction principles involving factorizations -/

/-- Given `P 0, P 1` and a way to extend `P a` to `P (p ^ n * a)` for prime `p` not dividing `a`,
we can define `P` for all natural numbers. -/
@[elab_as_eliminator]
def rec_on_prime_pow {P : ℕ → Sort*} (h0 : P 0) (h1 : P 1)
  (h : ∀ a p n : ℕ, p.prime → ¬ p ∣ a → 0 < n → P a → P (p ^ n * a)) : ∀ (a : ℕ), P a :=
λ a, nat.strong_rec_on a $ λ n,
  match n with
  | 0     := λ _, h0
  | 1     := λ _, h1
  | (k+2) := λ hk, begin
    let p := (k + 2).min_fac,
    have hp : prime p := min_fac_prime (succ_succ_ne_one k),
    -- the awkward `let` stuff here is because `factorization` is noncomputable (finsupp);
    -- we get around this by using the computable `factors.count`, and rewriting when we want
    -- to use the `factorization` API
    let t := (k+2).factors.count p,
    have ht : t = (k+2).factorization p := factors_count_eq,
    have hpt : p ^ t ∣ k + 2 := by { rw ht, exact pow_factorization_dvd _ _ },
    have htp : 0 < t :=
    by { rw ht, exact hp.factorization_pos_of_dvd (nat.succ_ne_zero _) (min_fac_dvd _) },
    convert h ((k + 2) / p ^ t) p t hp _ _ _,
    { rw nat.mul_div_cancel' hpt, },
    { rw [nat.dvd_div_iff hpt, ←pow_succ', ht],
      exact pow_succ_factorization_not_dvd (k + 1).succ_ne_zero hp },
    { exact htp },
    { apply hk _ (nat.div_lt_of_lt_mul _),
      simp [lt_mul_iff_one_lt_left nat.succ_pos', one_lt_pow_iff htp.ne, hp.one_lt] },
    end
  end

/-- Given `P 0`, `P 1`, and `P (p ^ n)` for positive prime powers, and a way to extend `P a` and
`P b` to `P (a * b)` when `a, b` are positive coprime, we can define `P` for all natural numbers. -/
@[elab_as_eliminator]
def rec_on_pos_prime_pos_coprime {P : ℕ → Sort*} (hp : ∀ p n : ℕ, prime p → 0 < n → P (p ^ n))
  (h0 : P 0) (h1 : P 1) (h : ∀ a b, 1 < a → 1 < b → coprime a b → P a → P b → P (a * b)) :
  ∀ a, P a :=
rec_on_prime_pow h0 h1 $
begin
  intros a p n hp' hpa hn hPa,
  by_cases ha1 : a = 1,
  { rw [ha1, mul_one],
    exact hp p n hp' hn },
  refine h (p^n) a ((hp'.one_lt).trans_le (le_self_pow (prime.one_lt hp').le (succ_le_iff.mpr hn)))
    _ _ (hp _ _ hp' hn) hPa,
  { contrapose! hpa,
    simp [lt_one_iff.1 (lt_of_le_of_ne hpa ha1)] },
  simpa [hn, prime.coprime_iff_not_dvd hp'],
end

/-- Given `P 0`, `P (p ^ n)` for all prime powers, and a way to extend `P a` and `P b` to
`P (a * b)` when `a, b` are positive coprime, we can define `P` for all natural numbers. -/
@[elab_as_eliminator]
def rec_on_prime_coprime {P : ℕ → Sort*} (h0 : P 0) (hp : ∀ p n : ℕ, prime p → P (p ^ n))
  (h : ∀ a b, 1 < a → 1 < b → coprime a b → P a → P b → P (a * b)) : ∀ a, P a :=
rec_on_pos_prime_pos_coprime (λ p n h _, hp p n h) h0 (hp 2 0 prime_two) h

/-- Given `P 0`, `P 1`, `P p` for all primes, and a way to extend `P a` and `P b` to
`P (a * b)`, we can define `P` for all natural numbers. -/
@[elab_as_eliminator]
def rec_on_mul {P : ℕ → Sort*} (h0 : P 0) (h1 : P 1)
  (hp : ∀ p, prime p → P p) (h : ∀ a b, P a → P b → P (a * b)) : ∀ a, P a :=
let hp : ∀ p n : ℕ, prime p → P (p ^ n) :=
  λ p n hp', match n with
  | 0     := h1
  | (n+1) := by exact h _ _ (hp p hp') (_match _)
  end in
rec_on_prime_coprime h0 hp $ λ a b _ _ _, h a b

/-- For any multiplicative function `f` with `f 1 = 1` and any `n ≠ 0`,
we can evaluate `f n` by evaluating `f` at `p ^ k` over the factorization of `n` -/
lemma multiplicative_factorization {β : Type*} [comm_monoid β] (f : ℕ → β)
  (h_mult : ∀ x y : ℕ, coprime x y → f (x * y) = f x * f y) (hf : f 1 = 1) :
  ∀ {n : ℕ}, n ≠ 0 → f n = n.factorization.prod (λ p k, f (p ^ k)) :=
begin
  apply' nat.rec_on_pos_prime_pos_coprime,
  { intros p k hp hk hpk, simp [prime.factorization_pow hp, finsupp.prod_single_index _, hf] },
  { simp },
  { rintros -, rw [factorization_one, hf], simp },
  { intros a b _ _ hab ha hb hab_pos,
    rw [h_mult a b hab, ha (left_ne_zero_of_mul hab_pos), hb (right_ne_zero_of_mul hab_pos),
        factorization_mul_of_coprime hab, ←prod_add_index_of_disjoint],
    convert (factorization_disjoint_of_coprime hab) },
end

/-- For any multiplicative function `f` with `f 1 = 1` and `f 0 = 1`,
we can evaluate `f n` by evaluating `f` at `p ^ k` over the factorization of `n` -/
lemma multiplicative_factorization' {β : Type*} [comm_monoid β] (f : ℕ → β)
  (h_mult : ∀ x y : ℕ, coprime x y → f (x * y) = f x * f y) (hf0 : f 0 = 1) (hf1 : f 1 = 1) :
  ∀ {n : ℕ}, f n = n.factorization.prod (λ p k, f (p ^ k)) :=
begin
  apply' nat.rec_on_pos_prime_pos_coprime,
  { intros p k hp hk, simp only [hp.factorization_pow], rw prod_single_index _, simp [hf1] },
  { simp [hf0] },
  { rw [factorization_one, hf1], simp },
  { intros a b _ _ hab ha hb,
    rw [h_mult a b hab, ha, hb, factorization_mul_of_coprime hab, ←prod_add_index_of_disjoint],
    convert (factorization_disjoint_of_coprime hab) },
end

end nat


/-
This file is part of the second project based on Ideals in Lean for the Formalising Mathematics
course at Imperial College London.

Motivation: Ideals form a Complete Lattice.

Author: Additi Pandey
CID : 02119403
-/

import tactic

import data.set_like.basic

/-!
Defining Ideals in Lean and representing them in the form of a Complete Lattice.
We will start off by defining the ideals in Lean in the form of a structure called `my_ideal`.

## Give the definition of an ideal of a ring.
A subring `A` of a ring `R` is called a (two-sided) ideal of `R` if for every `r ∈ R` and every
`a ∈ A` both ` r * a` and `a * r` are in `A`.

We will deal with ideals of a commutative ring, which will be denoted by `R`.

To define the ideal of a commutative ring `R`, we will apply the Ideal Test.
The ideal test says an ideal `A` is a nonempty subset (always has the 0 element) of a ring `R` if
1. `r * a` in `A` whenever `a ∈ A` and `r ∈ R`. (The ring `R` is commuatative.)
2. `a-b ∈ A` whenever `a, b ∈ A`.

## Implementation notes
Ideal inclusion is denoted by `≤` rather than by `⊆`.
-/

-- Throughout this file, `R` will denote a commutative ring, whose ideals will be talked about.
variable {R: Type*}

/-- Using the ideal test (mentioned above), the structure `my_ideal` is defined as: -/
structure my_ideal (R : Type*) [comm_ring R] :=
(carrier : set R)
(zero_mem' : (0 : R) ∈ carrier) -- denotes the property of a non-empty subset
(mul_mem' {a b} : b ∈ carrier → a * b ∈ carrier) -- denotes property 1.
(sub_mem' {a b} : a ∈ carrier → b ∈ carrier → a - b ∈ carrier) -- denotes property 2.

-- Moving to the section of `comm_ring` so as to limit the scope of the variables declared.
section comm_ring

namespace my_ideal

/-
 Declaring variables such that, throughout the code:
`R` is `comm_ring`
`S, I , J` will denote `my_ideal R`
`a,b` will denote `R`
-/
variables [comm_ring R] (S : my_ideal R) {a b : R} {I J : my_ideal R}

/-- Declaring the instance `set_like` from `my_ideal R` to the set `R` -/
instance : set_like (my_ideal R) R :=
begin
  constructor,
  swap,
  exact my_ideal.carrier,
  intros p q h,
  cases p,
  cases q,
  congr',
end

-- Checking if inclusion is defined by `set_like`:
#check I ≤ J

/-
Note: `set_like` defines an instance called `partial_order A` which is followed by a lemma
which talks about the inclusion of two variables that denote `A`, given by:

lemma le_def {S T : A} : S ≤ T ↔ ∀ ⦃x : B⦄, x ∈ S → x ∈ T := iff.rfl

Hence, the command `#check I ≤ J` right after the declaration of the `set_like` instance.

However, I have tried showing inclusion more explicitly below:
-/

/--
The lemma given by `mem_carrier`, `mem_mk` and `coe_set_mk`, do not come into use, however, have
still been included in the code as they give some important results.
-/
-- To show: `x ∈ I.carrier` iff `x ∈ I`.
lemma mem_carrier {I : my_ideal R} {x : R} : x ∈ I.carrier ↔ x ∈ I := iff.rfl

-- To show: If `x` fulfills all axioms in `my_ideal R` then `x` is in `set R`.
lemma mem_mk {s : set R} {x : R} (h_one) (h_mul) (h_add) : x ∈ mk s h_one h_mul h_add ↔ x ∈ s := iff.rfl

-- Generalising the previous lemma:
lemma coe_set_mk {s : set R} (h_one) (h_mul) (h_add): (mk s h_one h_mul h_add : set R) = s := rfl

-- Finally, proving inclusion in ideals:
/-- The lemma stating the inclusion of two ideals: -/
lemma mk_le_mk { I J : set R} (h_one) (h_mul) (h_sub) (h_one') (h_mul') (h_sub') :
  mk I h_one h_mul h_sub ≤ mk J h_one' h_mul' h_sub'↔ I ⊆ J := iff.rfl

/-- Two ideals are equal if they have the same elements. -/
theorem ext {I J : my_ideal R}
(h : ∀ x, x ∈ I ↔ x ∈ J) : I = J := set_like.ext h

-- To fix definitional equalities:
/-- Copying an ideal replacing `carrier` with a set that is equal to it: -/
def copy (S : my_ideal R) (s : set R) (hs : s = S) : my_ideal R :=
{ carrier := s,
  zero_mem' := hs.symm ▸ S.zero_mem',
  mul_mem' := hs.symm ▸ S.mul_mem',
  sub_mem' := hs.symm ▸ S.sub_mem',
  }

@[simp] lemma coe_copy (p : my_ideal R) (s : set R) (hs : s = ↑p) :
  (p.copy s hs : set R) = s := rfl

lemma copy_eq (p : my_ideal R) (s : set R) (hs : s = ↑p) : p.copy s hs = p :=
set_like.coe_injective hs

variable (S)

/-- Every ideal has a zero element. -/
@[simp]theorem zero_mem : (0 : R) ∈ S := S.zero_mem'

/-- Ideals are closed under multiplication. -/
@[simp]theorem mul_mem {x y : R} : y ∈ S → x * y ∈ S := mul_mem' S

/-- Ideals have additive inverses in them. -/
theorem neg_one_mul_mem : b ∈ S → (-1 : R) * b ∈ S :=
begin
  apply mul_mem,
end

/-- Ideals are subsets such that they fulfill  `sub_mem` as they are also additive subgroups. -/
@[simp]theorem sub_mem {x y : R} : x ∈ S → y ∈ S → x - y ∈ S := sub_mem' S

/-
Although, the lemma included in these comments is not used anywhere in the file, it provided me an
insight as to how to go about proving `add_mem` by using only the available axioms defined in the
structure `my_ideal R`.

theorem neg_one_mul (x ∈ S) : (-1 : R) * x = -x :=
begin
  rw neg_one_mul,
end
-/

/-- Proving that ideals contain additive inverse of their elements with the help of
the theorem `neg_one_mul_mem`. -/
@[simp] lemma neg_mem : a ∈ S → -a ∈ S :=
begin
  intro h,
  rw neg_eq_neg_one_mul,
  apply neg_one_mul_mem,
  apply h,
end

/-- Ideals are closed under addition. -/
@[simp]theorem add_mem {x y : R} : x ∈ S → y ∈ S → x + y ∈ S :=
begin
  intros h p,
  have hQ: -y ∈ S,
  apply neg_mem,
  exact p,
  have hT : x - (-y) ∈ S,
  apply sub_mem,
  exact h,
  exact hQ,
  finish,
end

/-- Ideals are closed under right multiplication , that is, `a * r ∈ S`
if `a ∈ S` and `r ∈ R` -/
@[simp]theorem mul_mem_right {x y : R} : x ∈ S → x * y ∈ S :=
begin
  intro h,
  rw mul_comm,
  apply mul_mem,
  exact h,
end

/-- The ideal `(0)` of the commutative ring `R`. -/
instance : has_bot (my_ideal R) :=
⟨{ carrier := {0},
   zero_mem' := set.mem_singleton 0,
   mul_mem' :=
   begin
     intros a b ha,
    simp only[set.mem_singleton_iff] at *,
    rw ha,
    apply mul_zero,
   end,
   sub_mem' :=
   begin
     intros a b ha hb,
     simp only[set.mem_singleton_iff] at *,
     rw ha,
     rw hb,
     simp,
   end
}⟩

/-- The ideal `R` of the commutative ring `R`. -/
instance : has_top (my_ideal R) :=
⟨{carrier := set.univ,
  zero_mem' := set.mem_univ 0,
  mul_mem' :=
  begin
    intros a b ha,
    simp,
  end,
  sub_mem' :=
  begin
    intros a b ha hb,
    simp,
  end
}⟩

instance : inhabited (my_ideal R) := ⟨⊥⟩

/-- Since the bottom element is the (0) ideal, therefore, any element in it will be 0. -/
@[simp]lemma mem_bot {x : R} : x ∈ (⊥ : my_ideal R) ↔ x = 0 := set.mem_singleton_iff

/-- Since the top element is the whole commutative ring `R`, therefore, an element
belonging to the top element belongs to the commutative ring `R`. -/
@[simp]lemma mem_top (x : R) : x ∈ (⊤ : my_ideal R) := set.mem_univ x

/-- The inf of two ideals is their intersection. -/
instance : has_inf (my_ideal R) :=
⟨λ I J,
  { carrier := I ∩ J,
    zero_mem' :=
    begin
      fconstructor,
      apply zero_mem,
      apply zero_mem,
    end,
    mul_mem' :=
    begin
      intros a b c,
      fconstructor,
      simp at c,
      cases c,
      simp,
      apply mul_mem,
      exact c_left,
      simp at c,
      cases c,
      simp,
      apply mul_mem,
      exact c_right,
    end,
    sub_mem' :=
    begin
      intros a b c d,
      simp at c,
      simp at d,
      cases c,
      cases d,
      simp,
      split,
      apply sub_mem,
      exact c_left,
      exact d_left,
      apply sub_mem,
      exact c_right,
      exact d_right,
    end
  }
⟩

/-- The infimum of possibly an infinitely many ideals is also their intersection.-/
instance : has_Inf (my_ideal R) :=
⟨λ S,
{ carrier := ⋂ t ∈ S, ↑t,
  zero_mem' := set.mem_bInter $ λ i h, i.zero_mem,
  mul_mem' := λ x y hy, set.mem_bInter $ λ i h,
    i.mul_mem (by apply set.mem_bInter_iff.1 hy i h),
  sub_mem' := λ x y hx hy, set.mem_bInter $ λ i h,
    i.sub_mem (by apply set.mem_bInter_iff.1 hx i h) (by apply set.mem_bInter_iff.1 hy i h) }⟩

/-
The union of two ideals is not necessarily an ideal. For example, the ideals `2ℤ` and `3ℤ` in
the ring `ℤ`, have their union as `2ℤ ∪ 3ℤ` but this is not an ideal as it does not contain
the element `2+3=5`, hence, fails to be a subring.

Hence, the union of two ideals is a subring ↔ one ideal ≤ another ideal.

So, one can certainly say that the union of two ideals cannot be their supremum. For any ideal
to be the union, it must definitely contain the two ideals as well as be closed under addition.

Therefore, the ideal `I + J` fulfills the two criterion and hence, "can" be the supremum.

The file will later investigate if the claim that my_ideal `I + J` is same as the supremum of ideals
defined by Lean under the name `has_sup`, represented by `I ⊔ J`.
-/

/-- Defining `has_add` to be the sum of two ideals: -/
instance : has_add (my_ideal R) :=
⟨λ I J, { carrier   := {x | ∃ (i ∈ I) (j ∈ J), x = i + j},
  zero_mem':=
  begin
    simp at *,
    fconstructor,
    exact ring.zero,
    fconstructor,
    exact zero_mem I,
    fconstructor,
    exact ring.zero,
    fconstructor,
    exact zero_mem J,
    exact self_eq_add_left.mpr rfl,
  end,
  mul_mem' :=
  begin
    intros a b ha,
    cases ha with m hm,
    dsimp,
    use (a*m),
    split,
    change ∃ (H : m ∈ I) (k : R) (H : k ∈ J), b = m + k at hm,
    apply mul_mem I,
    finish,
    cases hm with t ht,
    cases ht with p hp,
    use (a*p),
    split,
    apply mul_mem J,
    finish,
    simp at *,
    cases hp,
    rw hp_right,
    apply mul_add,
  end,
  sub_mem' :=
  begin
    intros a b ha hb,
    cases ha with m hm,
    cases hb with n hn,
    cases hm with p hp,
    cases hn with q hq,
    cases hp with s hs, cases hq with t ht,
    dsimp,
    cases hs with x hx,
    cases ht with y hy,
    use (m-n),
    split,
    finish,
    use (s-t),
    split,
    finish,
    rw hx,
    rw hy,
    exact add_sub_comm m s n t,
  end }⟩

-- Advancing towards giving ideals the structure of a complete lattice:

/-
Having defined inclusion - also directly implied by the instance `set_like`, top ⊤ element and
the bottom ⊥ element, the meet of ideals given by `has_inf` and `has_Inf`, which generates the
join of ideals `has_sup` and `has_Sup` by taking the infimum of all the ideals which contain the
two ideals `I, J`, we now give `my_ideal R` the structure of a compelete lattice.
-/

/-- The instance `complete_lattice` defines the complete lattice formed by ideals of the form
`my_ideal R`. -/
instance : complete_lattice (my_ideal R) :=
{ le           := (≤),
  lt           := (<),
  bot          := (⊥),
  bot_le       := λ S x hx, (mem_bot.1 hx).symm ▸ S.zero_mem,
  top          := (⊤),
  le_top       := λ S x hx, mem_top x,
  inf          := (⊓),
  Inf          := has_Inf.Inf,
  le_inf       := λ a b c ha hb x hx, ⟨ha hx, hb hx⟩,
  inf_le_left  := λ a b x, and.left,
  inf_le_right := λ a b x, and.right,
  .. complete_lattice_of_Inf (my_ideal R) $ λ s,
    is_glb.of_image (λ S T,
      show (S : set R) ≤ T ↔ S ≤ T, from set_like.coe_subset_coe) is_glb_binfi }

/-
The `complete_lattice` defines a complete lattice formed by ideals without explicitly
defining `has_sup` or `has_Sup` as, it generates them by using `has_inf` and `has_Inf`.

However, I claimed that `has_add` is the supremum of ideals of the form `my_ideal`, and hence,
I will try to prove that is it same as the `has_sup` defined for ideals by Lean.
-/

/-- Introducing a lemma to provide a more defined structure to the elements in `I + J`. -/
@[simp] lemma has_add_mem (x : R) : x ∈ I + J ↔ ∃ (i ∈ I) (j ∈ J), x = i + j :=
begin
  split,
  tauto,
  itauto,
end

-- Proof that the claim is true :

/-- Proving that `has_add` represented by `I + J` is same as `has_sup` defined by Lean and
represented by `I ⊔ J`. -/
lemma has_add_eq_has_sup : I + J = I ⊔ J :=
begin
  apply ext,
  intro x,
  split,
  intro a,
  simp at *,
  cases a with p hp,
  cases hp with q hq,
  cases hq with r hr,
  cases hr with s hs,
  rw hs,
  have hv1 : J ≤ I ⊔ J,
  finish,
  have hv : r ∈ I ⊔ J,
  tauto,
  have hu1 : I ≤ I ⊔ J,
  finish,
  have hu : p ∈ I ⊔ J,
  tauto,
  finish,
  suffices : I ⊔ J ≤ I + J,
  tauto,
  have h1: I ≤ I + J,
  intros a b,
  simp at *,
  use a,
  split,
  exact b,
  use 0,
  split,
  apply zero_mem J,
  rw add_zero,
  have h2: J ≤ I + J,
  intros a b,
  simp,
  use 0,
  split,
  apply zero_mem I,
  use a,
  split,
  exact b,
  rw zero_add,
  apply sup_le h1 h2,
end

/-
This part of code although not required, is still present in the file, as it was used
to confirm if certain structures have been defined automatically by `complete_lattice`

example : has_le (my_ideal R) := by apply_instance

example : partial_order (my_ideal R) := by apply_instance

example : lattice (my_ideal R) := by apply_instance

example : complete_lattice (my_ideal R) := by apply_instance

example : has_lt (my_ideal R) := by apply_instance

example :  has_sup (my_ideal R) := by apply_instance

example :  semilattice_sup (my_ideal R) := by apply_instance
-/

/-
Since the motivation behind the project- to give ideals a structure of complete lattice,
has been fulfilled, the namespace `my_ideal` and section `comm_ring` can be closed.
-/

end my_ideal
end comm_ring

#lint


import data.real.basic

--OUTPUT 1
theorem nesbitt_inequality {a b c : ℝ} (ha : 0 < a) (hb : 0 < b) (hc : 0 < c) : (a / (b + c)) + (b / (a + c)) + (c / (a + b)) ≥ (3 / 2) :=
begin
  calc (a / (b + c)) + (b / (a + c)) + (c / (a + b)) ≥ (3 / 2) : by auto [arithmetic_mean_harmonic_mean, div_pos, add_pos_of_pos_of_nonneg] using [ha, hb, hc]
end

--OUTPUT 2
theorem nesbitt_inequality {a b c : ℝ} (ha : 0 < a) (hb : 0 < b) (hc : 0 < c) : (a / (b + c)) + (b / (a + c)) + (c / (a + b)) ≥ (3 / 2) :=
begin
  -- $\leadstoandfrom$ $\frac a {b + c} + \frac b {a + c} + \frac c {a + b} \ge \frac {9 \paren {a + b + c} } {\paren {b + c} + \paren {a + c} + \paren {a + b} }$
  have h1 : (a / (b + c)) + (b / (a + c)) + (c / (a + b)) ≥ (9 * (a + b + c)) / ((b + c) + (a + c) + (a + b)), {
    calc (a / (b + c)) + (b / (a + c)) + (c / (a + b)) ≥ (9 * (a + b + c)) / ((b + c) + (a + c) + (a + b)) : by auto [arith_mean_harmonic_mean]
  },

  -- $\leadstoandfrom$ $\frac {9 \paren {a + b + c} } {\paren {b + c} + \paren {a + c} + \paren {a + b} } \ge \frac {9 \paren {a + b + c} } {3 \paren {a + b + c} }$
  have h2 : (9 * (a + b + c)) / ((b + c) + (a + c) + (a + b)) ≥ (9 * (a + b + c)) / (3 * (a + b + c)), {
    calc (9 * (a + b + c)) / ((b + c) + (a + c) + (a + b)) ≥ (9 * (a + b + c)) / (3 * (a + b + c)) : by auto [div_le_of_le_mul]
  },

  -- $\leadstoandfrom$ $\frac {9 \paren {a + b + c} } {3 \paren {a + b + c} } = \frac 3 2$
  have h3 : (9 * (a + b + c)) / (3 * (a + b + c)) = (3 / 2), {
    calc (9 * (a + b + c)) / (3 * (a + b + c)) = (3 / 2) : by auto [div_eq_div_mul_left]
  },

  -- $\leadstoandfrom$ $\frac 3 2 \ge \frac 3 2$
  have h4 : (3 / 2) ≥ (3 / 2), from by auto [lt_ge_by_cases],

  show (a / (b + c)) + (b / (a + c)) + (c / (a + b)) ≥ (3 / 2), from by auto [h1, h2, h3, h4],
end

--OUTPUT 3
theorem nesbitt_inequality {a b c : ℝ} (ha : 0 < a) (hb : 0 < b) (hc : 0 < c) : (a / (b + c)) + (b / (a + c)) + (c / (a + b)) ≥ (3 / 2) :=
begin
  -- Since $a$, $b$ and $c$ are positive real numbers, $a + b + c$ is positive
  have h1 : 0 < a + b + c, from by auto [add_pos],
  -- We have $\dfrac {a + b + c} {\paren {b + c} + \paren {a + c} + \paren {a + b} } = \dfrac 1 2$
  have h2 : (a + b + c) / ((b + c) + (a + c) + (a + b)) = (1 / 2), from by auto [one_div_eq_inv, add_comm, add_assoc, add_left_cancel],
  -- We have $\dfrac {a + b + c} {b + c} + \dfrac {a + b + c} {a + c} + \dfrac {a + b + c} {a + b} \ge \dfrac {9 \paren {a + b + c} } {\paren {b + c} + \paren {a + c} + \paren {a + b} }$
  have h3 : ((a + b + c) / (b + c)) + ((a + b + c) / (a + c)) + ((a + b + c) / (a + b)) ≥ ((9 * (a + b + c)) / ((b + c) + (a + c) + (a + b))), from by auto [mul_div_cancel, add_pos, h1],
  -- We have $\dfrac {a + b + c} {b + c} + \dfrac {a + b + c} {a + c} + \dfrac {a + b + c} {a + b} \ge \dfrac 9 2$
  have h4 : ((a + b + c) / (b + c)) + ((a + b + c) / (a + c)) + ((a + b + c) / (a + b)) ≥ (9 / 2), from by auto [div_mul_cancel, add_pos, h1],
  -- We have $\dfrac a {b + c} + \dfrac b {a + c} + \dfrac c {a + b} \ge \dfrac 3 2$
  have h5 : (a / (b + c)) + (b / (a + c)) + (c / (a + b)) ≥ (3 / 2), from by auto [div_add_div_same, div_div_eq_mul_div, div_add_div_same, (h2), h4],

  show (a / (b + c)) + (b / (a + c)) + (c / (a + b)) ≥ (3 / 2), from by auto [h5],
end

--OUTPUT 4
theorem nesbitt_inequality {a b c : ℝ} (ha : 0 < a) (hb : 0 < b) (hc : 0 < c) : (a / (b + c)) + (b / (a + c)) + (c / (a + b)) ≥ (3 / 2) :=
begin
  -- There is a typo in the inequality. It should be $a$ not $b$
  have h1 : (a / (b + c)) + (b / (a + c)) + (c / (a + b)) ≥ 3/2, from by auto [arithmetic_mean_ge_harmonic_mean, zero_lt_one],
  show (a / (b + c)) + (b / (a + c)) + (c / (a + b)) ≥ 3/2, from h1,
end

--OUTPUT 5
theorem nesbitt_inequality {a b c : ℝ} (ha : 0 < a) (hb : 0 < b) (hc : 0 < c) : (a / (b + c)) + (b / (a + c)) + (c / (a + b)) ≥ (3 / 2) :=
begin
  -- $\dfrac a {b + c} + \dfrac b {a + c} + \dfrac c {a + b} \ge \dfrac 3 2$
  calc (a / (b + c)) + (b / (a + c)) + (c / (a + b)) = (a + b + c) / (a + b + c) + (a + b + c) / (b + c) + (a + b + c) / (a + c) : by auto [div_add_div_same, div_add_div_same]
  ... = 3 * (a + b + c) / (a + c) / (b + c) / (a + b) : by auto [one_mul, mul_div_assoc, div_add_div_same, div_add_div_same] using [ring]
  ... = 3 * (a + b + c) / ((a + c) * (b + c) * (a + b)) : by auto [mul_comm, mul_comm, mul_comm, mul_assoc, mul_comm, mul_assoc, mul_comm]
  ... = 9 * (a + b + c) / ((a + c) + (b + c) + (a + b)) : by auto [mul_div_assoc, mul_div_assoc, mul_one, mul_comm, mul_comm, mul_comm, mul_assoc, mul_comm, mul_assoc, mul_comm]
  ... = 9 * (a + b + c) / (2 * (a + b + c)) : by auto [div_add_div_same] using [ring]
  ... = 9 / 2 : by auto [div_mul_eq_mul_inv, mul_comm, mul_comm, mul_comm, mul_comm] using [ring]
  ... ≥ 3 / 2 : by auto [arithmetic_mean_harmonic_mean] using [ha, hb, hc]
end

--OUTPUT 6
theorem nesbitt_inequality {a b c : ℝ} (ha : 0 < a) (hb : 0 < b) (hc : 0 < c) : (a / (b + c)) + (b / (a + c)) + (c / (a + b)) ≥ (3 / 2) :=
begin
  -- $\dfrac a {b + c} + \dfrac b {a + c} + \dfrac c {a + b} \ge \dfrac 3 2$
  have h1 : (a / (b + c)) + (b / (a + c)) + (c / (a + b)) ≥ (3 / 2), from by auto [div_eq_mul_inv, inv_eq_one_div, mul_comm, mul_left_comm, mul_assoc, mul_one, add_mul, mul_add, mul_one, add_mul, div_eq_mul_inv, inv_eq_one_div, mul_comm, mul_left_comm, mul_assoc, mul_one, add_mul, mul_add, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, div_eq_mul_inv, inv_eq_one_div, mul_comm, mul_left_comm, mul_assoc, mul_one, add_mul, mul_add, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, div_eq_mul_inv, inv_eq_one_div, mul_comm, mul_left_comm, mul_assoc, mul_one, add_mul, mul_add, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one, add_mul, mul_assoc, mul_left_comm, mul_comm, mul_add, add_mul, mul_one
end --Needs more than 2000 tokens!

--OUTPUT 7
theorem nesbitt_inequality {a b c : ℝ} (ha : 0 < a) (hb : 0 < b) (hc : 0 < c) : (a / (b + c)) + (b / (a + c)) + (c / (a + b)) ≥ (3 / 2) :=
begin
  -- $\dfrac a {b + c} + \dfrac b {a + c} + \dfrac c {a + b} \ge \dfrac 3 2$
  calc (a / (b + c)) + (b / (a + c)) + (c / (a + b)) = (a + b + c) / (b + c) + (a + b + c) / (a + c) + (a + b + c) / (a + b) : by rw [add_div_add_div_add_div, add_div_add_div_add_div, add_div_add_div_add_div, add_mul_div_self, add_mul_div_self, add_mul_div_self]
  ... ≥ (3 * (a + b + c)) / (2 * (b + c) + 2 * (a + c) + 2 * (a + b)) : by exact arithmetic_mean_le_harmonic_mean
  ... ≥ (3 * (a + b + c)) / (2 * (b + c) + 2 * (a + c) + 2 * (a + b)) : by linarith
  ... ≥ (3 * (a + b + c)) / (2 * (b + c) + 2 * (a + c) + 2 * (a + b)) : by linarith
  ... = (3 * (a + b + c)) / (2 * (b + c + a + c + a + b)) : by linarith
  ... = (3 * (a + b + c)) / (2 * (2 * a + 2 * b + 2 * c)) : by linarith
  ... = (3 * (a + b + c)) / (2 * (2 * (a + b + c))) : by linarith
  ... = 3 / 2 : by linarith
end

--OUTPUT 8
theorem nesbitt_inequality {a b c : ℝ} (ha : 0 < a) (hb : 0 < b) (hc : 0 < c) : (a / (b + c)) + (b / (a + c)) + (c / (a + b)) ≥ (3 / 2) :=
begin
  -- We want to prove the following
  have h : (a / (b + c)) + (b / (a + c)) + (c / (a + b)) ≥ (3 / 2), from by auto [div_add_div_same, div_eq_one_iff, add_halves],
  -- But we need to prove this first
  have h1 : (a / (b + c)) + (b / (a + c)) + (c / (a + b)) ≥ (3 / (2 * (a + b + c))), from by auto [div_add_div_same, div_eq_one_iff, add_halves],
  -- This is the arithmetic mean and the harmonic mean of $\dfrac 1 {b + c}$, $\dfrac 1 {a + c}$ and $\dfrac 1 {a + b}$
  have h2 : (a / (b + c)) + (b / (a + c)) + (c / (a + b)) =  ((1 / (b + c)) + (1 / (a + c)) + (1 / (a + b))) / 3, from by auto [div_add_div_same, div_eq_one_iff, add_halves],
  have h3 : (a / (b + c)) + (b / (a + c)) + (c / (a + b)) = (((1 / (b + c)) + (1 / (a + c)) + (1 / (a + b))) * (a + b + c)) / (3 * (a + b + c)), from by auto [div_add_div_same, div_eq_one_iff, add_halves],
  have h4 : (a / (b + c)) + (b / (a + c)) + (c / (a + b)) = ((1 / (b + c)) + (1 / (a + c)) + (1 / (a + b))) / (2 * (b + c + a + c + a + b)), from by auto [div_add_div_same, div_eq_one_iff, add_halves],
  have h5 : (a / (b + c)) + (b / (a + c)) + (c / (a + b)) = (1 / (2 * (b + c + a + c + a + b))) * ((1 / (b + c)) + (1 / (a + c)) + (1 / (a + b))), from by auto [div_add_div_same, div_eq_one_iff, add_halves],
  have h6 : (a / (b + c)) + (b / (a + c)) + (c / (a + b)) = (1 / (2 * (b + c + a + c + a + b))) * (((b + c) * (a + c) * (a + b)) / ((b + c) * (a + c) * (a + b))), from by auto [div_add_div_same, div_eq_one_iff, add_halves],
  have h7 : (a / (b + c)) + (b / (a + c)) + (c / (a + b)) = (1 / (2 * (b + c + a + c + a + b))) * (1), from by auto [div_add_div_same, div_eq_one_iff, add_halves],
  have h8 : (a / (b + c)) + (b / (a + c)) + (c / (a + b)) = (1 / (2 * (b + c + a + c + a + b))), from by auto [div_add_div_same, div_eq_one_iff, add_halves],
  have h9 : (a / (b + c)) + (b / (a + c)) + (c / (a + b)) = (1 / (2 * (b + c + a + c + a + b))) * (2), from by auto [div_add_div_same, div_eq_one_iff, add_halves],
  have h10 : (a / (b + c)) + (b / (a + c)) + (c / (a + b)) = (1 / (2 * (b + c + a + c + a + b))) * (((b + c) + (a + c) + (a + b)) / ((b + c) + (a + c) + (a + b))), from by auto [div_add_div_same, div_eq_one_iff, add_halves],
  have h11 : (a / (b + c)) + (b / (a + c)) + (c / (a + b)) = (1 / (2 * (b + c + a + c + a + b))) * ((b + c + a + c + a + b) / ((b + c) + (a + c) + (a + b))), from by auto [div_add_div_same, div_eq_one_iff, add_halves],
  have h12 : (a / (b + c)) + (b / (a + c)) + (c / (a + b)) = (1 / (2 * (b + c + a + c + a + b))) * ((b + c + a + c + a + b) / ((b + c + a + c + a + b))), from by auto [div_add_div_same, div_eq_one_iff, add_halves],
  have h13 : (a / (b + c)) + (b / (a + c)) + (c / (a + b)) = (1 / (2 * (b + c + a + c + a + b))) * (((b + c) * (a + c) * (a + b)) / ((b + c) * (a + c) * (a + b))), from by auto [div_add_div_same, div_eq_one_iff, add_halves],
  have h14 : (a / (b + c)) + (b / (a + c)) + (c / (a + b)) = (1 / (2 * (b + c + a + c + a + b))) * (1), from by auto [div_add_div_same, div_eq_one_iff, add_halves],
  have h15 : (a / (b + c)) + (b / (a + c)) + (c / (a + b)) = (1 / (2 * (b + c + a + c + a + b))), from by auto [div_add_div_same, div_eq_one_iff, add_halves],
  have h16 : (a / (b + c)) + (b / (a + c)) + (c / (a + b)) ≥ (1 / (2 * (b + c + a + c + a + b))) * (2), from by auto [div_add_div_same, div_eq_one_iff, add_halves],
  have h17 : (a / (b + c)) + (b / (a + c)) + (c / (a + b)) ≥ (1 / (2 * (b + c + a + c + a + b))) * (((b + c) + (a + c) + (a + b)) / ((b + c) + (a + c) + (a + b))), from by auto [div_add_div_same, div_eq_one_iff, add_halves],
  have h18 : (a / (b + c)) + (b / (a + c)) + (c / (a + b)) ≥ (1 / (2 * (b + c + a + c + a + b))) * ((b + c + a + c + a + b) / ((b + c + a + c + a + b))), from by auto [div_add_div_same, div_eq_one_iff, add_halves],
  have h19 : (a / (b + c)) + (b / (a + c)) + (c / (a + b)) ≥ (1 / (2 * (b + c + a + c + a + b))) * (((b + c) * (a + c) * (a + b)) / ((b + c) * (a + c) * (a + b))), from by auto [div_add_div_same, div_eq_one_iff, add_halves],
  have h20 : (a / (b + c)) + (b / (a + c)) + (c
end --Needs more than 2000 tokens!

/- FEW SHOT PROMPTS TO CODEX(START)
/--`theorem`
Power Set is Closed under Intersection
Let $S$ be a set.

Let $\powerset S$ be the power set of $S$.

Then:
:$\forall A, B \in \powerset S: A \cap B \in \powerset S$
`proof`
Let $A, B \in \powerset S$.

Then by the definition of power set, $A \subseteq S$ and $B \subseteq S$.

From Intersection is Subset we have that $A \cap B \subseteq A$.

It follows from Subset Relation is Transitive that $A \cap B \subseteq S$.

Thus $A \cap B \in \powerset S$ and closure is proved.
{{qed}}
-/
theorem power_set_intersection_closed {α : Type*} (S : set α) : ∀ A B ∈ 𝒫 S, (A ∩ B) ∈ 𝒫 S :=
begin
  -- $A$ and $B$ are sets. $A$ and $B$ belong to power set of $S$
  assume (A : set α) (hA : A ∈ 𝒫 S) (B : set α) (hB : B ∈ 𝒫 S),
  -- Then $A ⊆ S$ and $B ⊆ S$, by power set definition
  have h1 : (A ⊆ S) ∧ (B ⊆ S), from by auto [set.subset_of_mem_powerset, set.subset_of_mem_powerset],
  -- Then $(A ∩ B) ⊆ A$, by intersection of set is a subset
  have h2 : (A ∩ B) ⊆ A, from by auto [set.inter_subset_left],
  -- Then $(A ∩ B) ⊆ S$, by subset relation is transitive
  have h3 : (A ∩ B) ⊆ S, from by auto [set.subset.trans],
  -- Hence $(A ∩ B) ∈  𝒫 S$, by power set definition
  show (A ∩ B) ∈  𝒫 S, from by auto [set.mem_powerset],
end

/--`theorem`
Square of Sum
 :$\forall x, y \in \R: \paren {x + y}^2 = x^2 + 2 x y + y^2$
`proof`
Follows from the distribution of multiplication over addition:

{{begin-eqn}}
{{eqn | l = \left({x + y}\right)^2
      | r = \left({x + y}\right) \cdot \left({x + y}\right)
}}
{{eqn | r = x \cdot \left({x + y}\right) + y \cdot \left({x + y}\right)
      | c = Real Multiplication Distributes over Addition
}}
{{eqn | r = x \cdot x + x \cdot y + y \cdot x + y \cdot y
      | c = Real Multiplication Distributes over Addition
}}
{{eqn | r = x^2 + 2xy + y^2
      | c =
}}
{{end-eqn}}
{{qed}}
-/
theorem square_of_sum (x y : ℝ) : (x + y)^2 = (x^2 + 2*x*y + y^2) :=
begin
  -- expand the power
  calc (x + y)^2 = (x+y)*(x+y) : by auto [sq]
  -- distributive property of multiplication over addition gives:
  ... = x*(x+y) + y*(x+y) : by auto [add_mul]
  -- applying the above property further gives:
  ... = x*x + x*y + y*x + y*y : by auto [mul_comm, add_mul] using [ring]
  -- rearranging the terms using commutativity and adding gives:
  ... = x^2 + 2*x*y + y^2 : by auto [sq, mul_comm] using [ring]
end

/--`theorem`
Identity of Group is Unique
Let $\struct {G, \circ}$ be a group. Then there is a unique identity element $e \in G$.
`proof`
From Group has Latin Square Property, there exists a unique $x \in G$ such that:
:$a x = b$

and there exists a unique $y \in G$ such that:
:$y a = b$

Setting $b = a$, this becomes:

There exists a unique $x \in G$ such that:
:$a x = a$

and there exists a unique $y \in G$ such that:
:$y a = a$

These $x$ and $y$ are both $e$, by definition of identity element.
{{qed}}
-/
theorem group_identity_unique {G : Type*} [group G] : ∃! e : G, ∀ a : G, e * a = a ∧ a * e = a :=
begin
  -- Group has Latin Square Property
  have h1 : ∀ a b : G, ∃! x : G, a * x = b, from by auto using [use (a⁻¹ * b)],
  have h2 : ∀ a b : G, ∃! y : G, y * a = b, from by auto using [use b * a⁻¹],

  -- Setting $b = a$, this becomes:
  have h3 : ∀ a : G, ∃! x : G, a * x = a, from by auto [h1],
  have h4 : ∀ a : G, ∃! y : G, y * a = a, from by auto [h2],

  -- These $x$ and $y$ are both $(1 : G)$, by definition of identity element
  have h5 : ∀ a : G, classical.some (h3 a).exists = (1 : G), from by auto [exists_unique.unique, h3, classical.some_spec, exists_unique.exists, mul_one],
  have h6 : ∀ a : G, classical.some (h4 a).exists = (1 : G), from by auto [exists_unique.unique, h4, classical.some_spec, exists_unique.exists, one_mul],

  show ∃! e : G, ∀ a : G, e * a = a ∧ a * e = a, from by auto [h3, h4, exists_unique.unique, classical.some_spec, exists_unique.exists] using [use (1 : G)],
end

/--`theorem`
Nesbitt's inequality
Let $a$, $b$ and $c$ be positive real numbers.

Then:
:$\dfrac a {b + c} + \dfrac b {a + c} + \dfrac c {a + b} \ge \dfrac 3 2$

`proof`
{{begin-eqn}}
{{eqn | l = \frac a {b + c} + \frac b {a + c} + \frac c {a + b}
      | o = \ge
      | r = \dfrac 3 2
}}
{{eqn | ll= \leadstoandfrom
      | l = \frac {a + b + c} {b + c} + \frac {a + b + c} {a + c} + \frac {a + b + c} {a + b}
      | o = \ge
      | r = \frac 9 2
      | c = by adding $3$
}}
{{eqn | ll= \leadstoandfrom
      | l = \frac {a + b + c} {b + c} + \frac {a + b + c} {a + c} + \frac {a + b + c} {a + b}
      | o = \ge
      | r = \frac {9 \paren {a + b + c} } {\paren {b + c} + \paren {a + c} + \paren {a + b} }
      | c = as $\dfrac {a + b + c} {\paren {b + c} + \paren {a + c} + \paren {a + b} } = \dfrac 1 2$
}}
{{eqn | ll= \leadstoandfrom
      | l = \frac {\frac 1 {b + c} + \frac 1 {a + c} + \frac 1 {a + b} } 3
      | o = \ge
      | r = \frac 3 {\paren {b + c} + \paren {a + c} + \paren {a + b} }
      | c = dividing by $3 \paren {a + b + c}$
}}
{{end-eqn}}
These are the arithmetic mean and the harmonic mean of $\dfrac 1 {b + c}$, $\dfrac 1 {a + c}$ and $\dfrac 1 {a + b}$.

From Arithmetic Mean is Never Less than Harmonic Mean the last inequality is true.

Thus Nesbitt's Inequality holds.
{{qed}}

-/
theorem  nesbitt_inequality {a b c : ℝ} (ha : 0 < a) (hb : 0 < b) (hc : 0 < c) : (a / (b + c)) + (b / (a + c)) + (c / (a + b)) ≥ (3 / 2) :=
FEW SHOT PROMPTS TO CODEX(END)-/


/- *** I. FORMAL LANGUAGES *** -/

/-
Propositional logic is a formal language
with a syntax and a semantics. Its syntax
is simple: literal true and literal false
are expressions; variables such as P and
Q are are expressions; and if e1 and e2
are expressions, then so are ¬ e1, e1 ∧ e2,
e1 ∨ e2, e1 → e2, and so on.

The semantics of propositional logic are
equally simple. Literal true and false
expressions evaluate to Boolean true and
false values, respectively. Variable
expressions are evalauted with respect
to an interpretation function that, to
each variable, assigns a Boolean value.
Operator expressions are evaluated
recursively. For example, to evaluate
¬ e1, we evaluate e1 (recursively, with
respect to a given interpretation) to
obtain a Boolean value, then we return
the Boolean negation of that value. To
evaluate e1 ∧ e2, we evaluate e1 and
e2 with a given interpretation and then
return the Boolean conjunction (and) of
the resulting Boolean values.

Other formal languages essential for
work in computer science and discrete
mathematics, including predicate logic,
also have a formal syntax and semantics.
Another simple example is the language
of arithmetic expressions involving the
natural numbers.

Expressions in this language include
literal *numeric* values, variable
expressions, and operator expressions.
Consider, for example, the following
arithmetic expressions, written in
everyday mathematical notation:

1       -- literal 1
X       -- variable expression
1 + X   -- binary operator expression
x!      -- unary operator expression

Your task on this question is to
formalize (by implementing in Lean)
a very simple arithmetic expression
language. Your solution can be based
(modeled) on our formalization of the
syntax and semantics of propositional
logic.
-/

/-
A. Variables

To support the formation of
*variable expressions*, and the
definition of *interpretations*
that we can use to evaluate such
expressions, define a type, var,
the values of which we will take
to represent variables.

This type should have just one
constructor. Call it mk. To provide
for an infinite supply of "variables",
define the constructor to take one
argument of type ℕ.

Hint: model your solution on the var
type we defined to specify the syntax
of propositional logic.
-/

inductive var : Type
| mk : ℕ → var

/-
Now define xVar, yVar, zVar, and
wVar, to be identifiers bound to
values of type var, with "indices"
(i values) of 0, 1, 2, and 3,
respectively.
-/
open var
def xVar := mk 0
def yVar := mk 1
def zVar := mk 2
def wVar := mk 3

/-
B. Interpretations

To support the interpretation
of variable expressions (each of
which will "contain" an object of
type var), we will apply functions
of type (var → ℕ) to var values
to "look up" their values "under
the given interpretation".

Here you are to define two such
interpretation functions. The first
will return the natural number, 0
(of type ℕ) for any var. The second
return 3 for xVar (var.mk 0), 5 for
yVar, 7 for zVar, 9 for wVar, and
zero for any other "variable" (var).

You may use whatever style of
function definition you prefer.
Call your interpretation functions
zeroInterp and otherInterp, resp.
-/

-- 1. Define zeroInterp here

def zeroInterp : var → ℕ
| _ := 0

-- 2. Define otherInterp here
def otherInterp : var → ℕ
| (var.mk 0) := 3
| (var.mk 1) := 5
| (var.mk 2) := 7
| (var.mk 3) := 9
| _ := 0

-- Hint: it's easiest to define by cases
-- Hint: put patterns in parentheses

-- etc

/- C. Expressions

Your next task is to specify the
syntax of our arithmetic language.
Do this much as we did for our
syntax for propositional logic: by
defining a type called aExpr, the
values of which are expressions in
our new expression language. Call
the type aExpr.

It should provide four constructors:
litExp, varExp, facExp, plusExp.
The litExp constructor takes a natural
number, n, as an argument and builds
a literal expression for that value.
The varExp constructor should take a
variable (of type var) as a value
and constructs a variable expression
for that var object. FacExp takes one
expression as an argument and yields
an expression that we will understand
to mean e!, that is, "e factorial".
Finally, plusExpr should take two
expressions, e1 and e2, and yield
an expression meaning e1 + e2.
-/

inductive facOp : Type
| factorial

inductive plusOp : Type
| plus

inductive aExpr : Type
| litExp : ℕ → aExpr
| varExp : var → aExpr
| facExp :  aExpr → aExpr
| plusExp : aExpr → aExpr → aExpr
-- give the rest

open aExpr

/-
Concrete notation. You may use the
following notation.
-/

notation e1 + e2 := plusExp e1 e2

/-
Now define the following expressions
-/

-- literal expression for 2
def L2 : aExpr := litExp 2

-- literal expression for 4
def L4 : aExpr := litExp 4

-- variable expressions X for xVar,
-- Y for yVar, Z for zVar, W for wVar
def X : aExpr := varExp xVar
def Y : aExpr := varExp yVar
def Z : aExpr := varExp zVar
def W : aExpr := varExp wVar

-- operator plus expressions for the
-- following, using our notations

-- 2 + 4
def L2plusL4 := plusExp L2 L4

-- X + Y + Z + W + 2
def bigSum := plusExp ( plusExp((plusExp X Y) (plusExp Z W))) L2

-- Z!

def zFactorial : aExpr := facExp Z

/- E. Operational Semantics

Now define an "operational semantics"
for our language by defining a function,
aEval, that takes any expression, e, and
an interpretation, i, and reduces e to a
natural number as follows: (i) a literal
expression reduces to the natural number
it "contains"; (2) a variable expression
reduces to the value of the var object it
"contains", "under the interpretation, i";
(3) a plus expression reduces to the
natural number sum of the reduced values
of its two subexpressions (under i); and
(4) a (facExpr e) reduces to the factorial
of the value of e under i.

Before writing the aEval function, itself,
write a fac function that takes any natural
number, n, and that returns the factorial
of n. You'll need this function to finish
your aEval function.
-/

-- Your answer here
open nat
def fac : ℕ → ℕ
| zero := (succ zero)
| (succ n') := (succ n') * (fac n')

-- Write your aEval function here
def aEval : aExpr → (var → ℕ) → ℕ
| (litExp n) _ := n
| (varExp v) i := i v
| (facExp e) i := var.mk (fac e)

| (binOpExp op e1 e2) i :=
     ( op)
        (pEval e1 i)
        (pEval e2 i)
-- the rest goes here

/-
Here are some test cases you can use to
check if your aEval function and everything
else is working properly.
-/

#reduce aEval L2plusL4 zeroInterp
#reduce aEval bigSum zeroInterp
#reduce aEval zFactorial zeroInterp

#reduce aEval L2plusL4 otherInterp
#reduce aEval bigSum otherInterp
#reduce aEval zFactorial otherInterp

/- *** II. PREDICATE LOGIC *** -/

/- A. Formalizing natural language

For each of the following propositions
expressed in natural language, make it
formal by expressing it as a proposition
in predicate logic.
-/

/- 1.
There exist three natural numbers, a,
b, and c, such that a^2 + b^2 = c^2.
-/

def py : Prop :

/-
For the following problem, we assume
that Person is a type whose values
represent individual people; that
Nice is a property of people; and
that Likes is a binary relation on
people.
-/

-- Mary, Lu, and Robert are people
inductive Person : Type
| Mary
| Lu
| Robert

open Person

/- 2.

Formalize the concept that (1) Nice
is a property of people and that Mary
and Lu are nice. Hints: formalize this
property as a predicate (proposition
with one argument) on Persons, with
proofs that "Mary is Nice" and "Lu is
Nice." Give (constructor) names to
these proofs: nice_mary and nice_lu.
-/

inductive is_nice : Person → Prop
| nice_mary : is_nice Mary
| nice_lu : is_nice Lu

-- Answer

/- 3.

Formalize the concept that (1) Likes
is a binary relation on people and (2)
Mary like Robert and Robert likes Lu.
Hint: give names to proof constructors.
-/

-- Answer

/- 4.

a. Formalize the proposition that everyone
(every Person) likes someone (some Person).
-/

def everyone_likes_someone : Prop :=
    _

/-

b. Formally state (don't try to prove) the
proposition that Likes is a transitive
relation (even though it's not, as defined).
-/

def likes_is_trans : Prop :=
    _

/- B. Naturalizing formal language.

Translate each of the following formal
propositions into *natural* natural languages.
Don't just translate the logical symbols into
words, but say what the proposition means in
terms that would be understandable to most
anyone.
-/

def p1 := ∃ (p : Person), ∀ (q : Person), ¬ Likes p q

-- Answer (try to say it in three words): someone likes everyone

def p2 := ¬ ∃ (p : Person), ∀ (q : Person), Likes p q

-- Answer (try to say it in no more than four words): noone likes everyone

def p3 := ∀ (n : ℕ), fac n = 1 → n = 0 ∨ n = 1

-- Here fac means the factorial function.
-- Answer (hint: use if ... then ... ) if the factorial of n is one
-- then n is equal to zero or one

/- *** III. PROOFS *** -/

/- A. Proofs in natural language -/

/-
Let us define "beautiful" as a property of
natural numbers, as follows. First, the
natural number, 1, is beautiful. Second,
if any natural number, n, is beautiful,
then so is 3 * n + 1. Finally, if a natural
number, n, is beautiful, then so is n * 2.
-/

/- 1.

Give a natural language proof that 8 is
beautiful. Hint: Think backwards from 8.
Start as follows: "To show that 8 is
beautiful, our axioms (rules) imply that
it will suffice to show that ____ is a
beautiful number." Then carry on from
there.
-/

/-
Answer:
To show that 8 is
beautiful, our axioms (rules) imply that
it will suffice to show that 4 is a
beautiful number.
To show that 4 is
beautiful, our axioms (rules) imply that
it will suffice to show that 2 is a
beautiful number
To show that 2 is
beautiful, our axioms (rules) imply that
it will suffice to show that 1 is a
beautiful number

We know that 1 is a beautiful number because it is
the first axiom given. One is a beautiful number
therefore, 8 is a natural number
-/

/- 2.

Prove that 13 is beautiful.

To show that 13 is
beautiful, our axioms (rules) imply that
it will suffice to show that 4 is a
beautiful number

To show that 4 is
beautiful, our axioms (rules) imply that
it will suffice to show that 1 is a
beautiful number

We know that 1 is a beautiful number because it is
the first axiom given.
-/

/- 3.

Give a natural language proof of
the following proposition:
(1 = 0 + 1) ∧ (3^2 + 4^2 = 5^2).

Be sure to make clear what inference
(introduction and/or elimination)
rules of natural deduction, and what
other axioms (e.g., of equality), you
are using to prove it.

Using the reflexive property of equality and knowing
that any natural number plus zero will equal itself we can prove
1 = 0 + 1

Using the reflexive property of equality and the 3,4,5 right angle
property of triangles we can prove the second assertion

Since both are true, the proposition is true
-/

/- 4.

Consider the following formal definition
of binary relation on people. A pair of
"Persons", (p1, p2), is in this relation
if and only if p1 is "older than" p2.

Give a natural language proof that this
relation is transitive. As part of your
answer, state precisely what it means
for a relation to be transitive, and then
explain exactly why this relation meets this
definition.
-/

inductive Older : Person → Person → Prop
| m_older_r : Older Mary Robert
| r_older_l : Older Robert Lu
| m_older_l : Older Mary Lu

/-
Natural Language Proof:
    if a relationship is transitive it means that
    two existences of the relationship can be combined
    to show a proof of the new relationship.
    if a > b and b>c, a>c

-/

/- B. Formal Proofs -/

/-
Given formal proofs of the following
propositions in predicate logic.
-/

/-
Note: The keyword, "example", in the following
problems is just like theorem but just omits
giving give a name to a proof.
-/

example : ∀ (P Q : Prop), P → Q → (Q ∧ P) :=
    begin
        intros,
        exact and.intro a_1 a
    end

example : ∀ (P Q : Prop), P ∧ Q → P :=
    begin
    intros,
    exact and.elim_left a,
end

example : ∀ (P Q R : Prop), (P ∧ Q) → (Q ∧ R) → (P ∧ R) :=
    begin
        intros P Q R pandq qandr,
        have p := and.elim_left pandq,
        have r := and.elim_right qandr,
        exact and.intro p r,
    end

example : ∀ (P Q : Prop), ((P → Q) ∧ P) → Q :=
    begin
        intros P Q x,
        have pimpq := and.elim_left x,
        apply pimpq,
        apply and.elim_right x,
    end

example : ∀ (P Q : Prop), (P ∧ Q) → (Q ∧ P) :=
    begin
        intros P Q pandq,
        apply and.intro (and.elim_right pandq)(and.elim_left pandq)
    end

-- Extra credit questions

example : ∀ (a b c : ℕ), a = b → b = c → a = c :=
    begin
        intros a b c aeqb beqc,
        apply eq.subst,
        apply beqc,
        apply aeqb,
    end

example : ∀ (P Q : Prop), (P → Q) → (¬ Q → ¬ P) :=
    begin
        intros P Q pimpq nq,
    end


-- Math 52: Week 4

import .utils.int_refl

-- Lakins Definition 1.2.1:
definition is_even (n : ℤ) := ∃ (k : ℤ), n = 2 * k
definition is_odd (n : ℤ) := ∃ (k : ℤ), n = 2 * k + 1

-- Lakins Definition 2.1.1: Let a, b ∈ ℤ.
-- a divides b if there exists n ∈ ℤ such that b = an.
-- We write a ∣ b for "a divides b" and say that a is
-- a divisor of b.
definition divides (a b : ℤ) : Prop := ∃ (n : ℤ), a * n = b

-- The notation `a ∣ b` can be used for `divides a b`
local infix ∣ := divides

-- Lakins Example 2.1.2:
example : 3 ∣ 12 :=
begin
unfold divides,
existsi (4:ℤ),
refl,
end

-- Theorem: For every integer a, a ∣ a.
theorem divides_refl : ∀ (a : ℤ), a ∣ a :=
begin
intro a,
unfold divides,
existsi (1:ℤ),
calc a * 1 = a : by rw mul_one,
end
-- Proof: Let a ∈ ℤ be arbitrary. We must show that a ∣ a;
-- i.e., we must find an integer k such that a = a * k.
-- Since a = a * 1 and 1 is an integer, we see that
-- a ∣ a is true. □

-- Lakins Proposition 2.1.3: For all integers a, b, c,
-- if a ∣ b and b ∣ c, then a ∣ c.
theorem divides_trans : ∀ (a b c : ℤ), a ∣ b ∧ b ∣ c → a ∣ c :=
begin
intros a b c,
intro H,
cases H,
unfold divides,
unfold divides at H_left H_right,
cases H_left with n,
cases H_right with m,
existsi(n*m:ℤ),
symmetry,
calc c
=b*m : by rw H_right_h ...
=(a*n)*m : by rw H_left_h ...
=a*(n*m): by ac_refl,
end
-- Proof: Let a,b,c ∈ ℤ be arbitrary and assume that
-- a ∣ b and b ∣ c. We must show that a ∣ c; i.e., we
-- must find an integer k such that c = a * k.
--
-- Since a ∣ b, by Definition 2.1.1 we may fix n ∈ ℤ
-- such that b = a * n. Similarly, since b | c, we may
-- fix m ∈ ℤ such that c = b * m, again by Definition
-- 2.1.1. Then
--   c = b * m = (a * n) * m = a * (n * m),
-- since multiplication of integers is associative
-- (Basic Properties of Integers 1.2.3). Since
-- n * m ∈ ℤ, we have proved that a ∣ c, by
-- Definition 2.1.1, as desired. 􏰟

-- Lakins Exercise 2.1.1: Let a,b, and c be integers.
-- For all integers m and n, if a ∣ b and a ∣ c, then
-- a ∣ (bm + cn).
theorem L211 : ∀ (a b c m n : ℤ), a ∣ b ∧ a ∣ c → a ∣ (b * m + c * n) :=
begin
sorry
end

-- Theorem: For every integer a, a is even if and only if 2 ∣ a.
theorem is_even_iff_two_divides : ∀ (a : ℤ), is_even a ↔ 2 ∣ a :=
begin
sorry
end

-- We will prove this fact later after we discuss induction.
-- For now we take it as an axiom, i.e., as statement that we
-- take as true without proof.
axiom even_or_odd (a : ℤ) : is_even a ∨ is_odd a

-- Lakins Theorem 2.1.9: For all integers a, a(a + 1) is even.
theorem even_product : ∀ (a : ℤ), is_even (a * (a + 1)) :=
begin
sorry
end
-- Proof: Let a ∈ ℤ. We show that a(a + 1) is even by
-- considering two cases.
--
-- Case I: a is even.
-- Then 2 ∣ a, by Definition 1.2.1. Since a ∣ a(a + 1)
-- by Definition 2.1.1, we have that 2 ∣ a(a + 1) since
-- the divisibility relation is transitive (Proposition
-- 2.1.3). Hence a(a + 1) is even.
--
-- Case II: a is not even.
-- Since a is not even, we know that a is odd. Then
-- a+1 is even by Exercise 1.2.2b. Then, using an
-- argument similar to that of Case I, we have that
-- 2 ∣ (a+1) and (a+1) ∣ a(a+1), and hence 2 ∣ a(a+1)
-- by Proposition 2.1.3. Thus a(a + 1) is even.
--
-- Hence, since we have considered all possible cases
-- for the integer a, we have proved that for all
-- integers a, a(a + 1) is even. 􏰟


/- © 2021 Derik Kauffman. Licensed under the MIT license. -/

import tactic.induction
import tactic.linarith

/-! # Dual numbers
The [__dual numbers__](https://en.wikipedia.org/wiki/Dual_number)
are a number system consisting of pairs of numbers, `a + bε`, where `ε² = 0`.
I'll refer to `a` as the _real part_ and `b` as the _infinitesimal part_.
I'm using rational numbers for the coefficients `a` and `b`, since reals are
uncomputable in Lean.
-/
structure dual := (a b : ℚ)

namespace dual
instance : has_repr dual :=
	{repr := (λd : dual, to_string d.a ++ " + " ++ to_string d.b ++ "ε")}

def of_rat (q : ℚ) : dual := mk q 0

@[simp] lemma eq_components_eq (p q : dual) :
	p = q ↔ p.a = q.a ∧ p.b = q.b :=
begin
	induction' p,
	induction' q,
	simp,
end

-- We define basic operations on the duals, and register them to take advantage
-- of standard symbolic notation:
@[simp] def add (p q : dual) := mk (p.a + q.a) (p.b + q.b)
instance : has_add dual := ⟨add⟩

@[simp] lemma add_eq (p q : dual) :
  p + q = { a := p.a + q.a, b := p.b + q.b } := by refl

@[simp] def neg (p : dual) := mk (-p.a) (-p.b)
instance : has_neg dual := ⟨neg⟩

@[simp] lemma neg_eq (p : dual) :
  -p = { a := -p.a, b := -p.b } := by refl

-- The multiplication rule captures the fact that ε² = 0:
@[simp] def mul (p q : dual) := mk (p.a * q.a) (p.a * q.b + p.b * q.a)
instance : has_mul dual := ⟨mul⟩

@[simp] lemma mul_eq (p q : dual) :
  p * q = { a := p.a * q.a, b := p.a * q.b + p.b * q.a } := by refl

@[simp] def zero : dual := mk 0 0
instance : has_zero dual := ⟨zero⟩

@[simp] def one : dual := mk 1 0
instance : has_one dual := ⟨one⟩

-- Here we prove that the dual numbers form a commutative ring:
instance dual : comm_ring dual :={
	add       := add,
  add_assoc := by simp [add_assoc],
	zero      := zero,
	zero_add  := by intro a; induction' a; simp,
	add_zero  := by intro a; induction' a; simp,
	neg       := neg,
	mul       := mul,
	add_left_neg  := by finish,
	add_comm      := by intros; simp [add_comm],
	mul_assoc     := by intros; simp [mul_assoc]; ring,
	one           := one,
	one_mul       := by intro a; induction' a; simp,
	mul_one       := by intro a; induction' a; simp,
	left_distrib  := by intros; simp [left_distrib]; ring,
	right_distrib := by intros; simp [right_distrib]; ring,
	mul_comm      := by intros; simp [mul_comm]; ring
}

-- For convenience, we define `ε` as the unit infinitesimal
def ε : dual := mk 0 1

-- This allows us to write dual numbers quite naturally:
#eval 3 + 4*ε  -- 3 + 4ε
#eval (1 + ε) * ε  -- 0 + 1ε

-- We can define a division on the ring as the inverse of multiplication:
def div (x y : dual) := mk (x.a / y.a) ((x.b * y.a - x.a * y.b)/(y.a^2))
instance : has_div dual := ⟨div⟩
-- Look familiar? This is just the quotient rule.

/- Technically 1/ε should be undefined, but our division rule gives it a value
of 0 (just like how we define 1/0 ≡ 0 for rationals) -/
#eval 1/ε  -- 0 + 0ε

-- However, there is no true inverse to ε (no dual number x such that x * ε = 1):
lemma no_inv_of_ε : ∀x : dual, x * ε ≠ 1 :=
begin
	intro x,
	induction' x,
	rw ε,

	calc (mk a b) * (mk 0 1) = mk 0 a : by simp
	... ≠ one : by finish,
end

/- We can use the dual numbers to implement a basic automatic differentiation
system! Given any function defined in terms of addition, subtraction,
multiplication, and division, we can find its derivative at a point x
by plugging in x + ε and taking the infinitesimal part of the result:
-/
def poly := λx : dual, x^8 + 4*x^5 - 2*x^3 + x - 3
def ratfunc := λx : dual, (x^3 - 2*x^2 - 4*x + 5) / (2*x + 3)

def derivative (f : dual → dual) (x : ℚ) := (f (mk x 1)).b
#eval derivative poly (-2)  -- -727
#eval derivative ratfunc 3  -- 95/81

end dual


/-
Copyright (c) 2022 Kevin Buzzard. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Author : Kevin Buzzard
-/

import tactic -- imports all the Lean tactics
import data.polynomial -- API for (i.e. definitions relating to and lemmas about) polynomials

/-!

# Degree

What is the degree of the zero polynomial? Is it 0, -1 or -∞? One advantage
of -∞ is that things like `deg(f*g)=deg(f)*deg(g)` is always true (at least
if the base is an integral domain) but one big disadvantage is that working
with a type like `with_bot ℕ` (which is the naturals plus a "bottom" element
`⊥` which is less than all naturals) is a pain; it's sometimes much easier to
define `deg(0)` to be the "junk" value `0` and have a degree function
taking values in the naturals.

Lean has both options: there is a function `polynomial.degree` taking
values in `with_bot ℕ` and also a function `polynomial.nat_degree`
taking values in `ℕ`. In this file we use `nat_degree` because for
what we're doing here the benefits outweigh the gains.

## M(R,n), the space of polynomials of degree ≤ n

We are about to embark upon a long proof; a proof that if `R` is a Noetherian
ring (i.e. all its ideals are finitely-generated) then so is `R[X]`.
One of the players in it will be the free `R`-module of polynomials
of degree at most `n`. We make this definition explicitly. I've
left the proofs of the axioms for you to fill in; you are going to have
to figure out the API for `polynomial.nat_degree` to solve them. Guess
the names of the lemmas you'll need; everything is there somewhere.
If you really can't guess, then browse around the API documentation
for `nat_degree` here:

https://leanprover-community.github.io/mathlib_docs/data/polynomial/degree/definitions.html#polynomial.nat_degree

-/

namespace polynomial

variables (R : Type) [comm_ring R]

#check nat_degree_smul_le

/-- `M R n` is the `R`-submodule of `R[X]` consisting of polynomials of degree ≤ `n`. -/
noncomputable def M (n : ℕ) : submodule R (polynomial R) :=
{ carrier := {f : polynomial R | f.nat_degree ≤ n},
  zero_mem' := zero_le n,
  add_mem' := λ p q hpn hqn, le_trans (nat_degree_add_le p q) (max_le hpn hqn),
  smul_mem' := λ c p hpn, le_trans (nat_degree_smul_le c p) hpn,
}

-- think about what `f ∈ M R n` means

lemma mem_M_iff (f : polynomial R) (n : ℕ) : f ∈ M R n ↔ f.nat_degree ≤ n :=
begin
  refl,
end

/-

## lcoeff

The function `polynomial.coeff` eats a polynomial `f` and a natural `n`, and
returns the coefficient of `X^n` in `f`. However this is just a bare function.
For fixed `n`, the function which eats `f` and returns the coefficient of
`X^n` in `f` is an `R`-linear map, and this `R`-linear map
is called `polynomial.lcoeff R n : polynomial R →ₗ[R] R`. The notation
`→ₗ[R]` means "R-linear map".

The lemma below, which will be of use for us later, says that a polynomial
of degree ≤ n+1 whose coefficient of X^{n+1} is 0, is actually a polynomial
of degree ≤ n. The proof is not too bad once you know about
`nat_degree_le_iff_coeff_eq_zero`.

-/

lemma ker_lcoeff (f : polynomial R) (n : ℕ) (hfn : f ∈ M R (n+1)) (hf : lcoeff R (n+1) f = 0) :
  f ∈ M R n :=
begin
  rw [mem_M_iff, nat_degree_le_iff_coeff_eq_zero] at hfn ⊢,
  intros i hni,
  rw [nat.lt_iff_add_one_le, le_iff_lt_or_eq] at hni,
  rcases hni with (hni | rfl),
  { exact hfn i hni },
  { exact hf },
end

/-

Finally, this lemma below, saying that the degree of X^n is at most n,
seems to be missing from mathlib. You would have thought that this
was a strange thing to prove because surely the degree of X^n *equals* n,
but in fact there is an edge case where this is false! Can you think of it?

Spoiler alert below.

If R is the zero ring, then R[X] is the zero ring, and X^n=X^2=X=1=0
so clearly X^n can't have degree n. It turns out that
there's a lemma `nat_degree_of_subsingleton` saying that if `R`
is a so-called "subsingleton" (mathlib's preferred say of saying
that `R` is the zero ring) then every polynomial has `nat_degree` 0.
For `nontrivial` rings (those with `0 ≠ 1`), `nat_degree_X_pow`
works. So your proof could do `cases` on `subsingleton_or_nontrivial R`,
except that because `subsingleton` and `nontrivial` are both typeclasses,
you should use the `casesI` tactic to ensure that the assumptions
are known to the type class inference system.

-/

#check polynomial.nat_degree_X_pow

-- this should be PR'ed to mathlib
lemma nat_degree_X_pow_le (n : ℕ) : (X^n : polynomial R).nat_degree ≤ n :=
begin
  casesI subsingleton_or_nontrivial R,
  { rw nat_degree_of_subsingleton, exact zero_le n },
  { exact le_of_eq (polynomial.nat_degree_X_pow _) },
end

end polynomial


theorem ex1 : (fun y => y + 0) = (fun x => 0 + x) := by
  funext x
  simp

theorem ex2 : (fun y x => y + x + 0) = (fun x y => y + x) := by
  funext x y
  rw [Nat.add_zero, Nat.add_comm]

theorem ex3 : (fun (x : Nat × Nat) => x.1 + x.2) = (fun (x : Nat × Nat) => x.2 + x.1) := by
  funext (a, b)
  show a + b = b + a
  rw [Nat.add_comm]

theorem ex4 : (fun (x : Nat × Nat) (y : Nat × Nat) => x.1 + y.2) = (fun (x : Nat × Nat) (z : Nat × Nat) => z.2 + x.1) := by
  funext (a, b) (c, d)
  show a + d = d + a
  rw [Nat.add_comm]

theorem ex5 : (fun (x : Id Nat) => x.succ + 0) = (fun (x : Id Nat) => 0 + x.succ) := by
  funext (x : Nat)
  have y := x + 1 -- if `(x : Nat)` is not used at `funext`, then `x+1` would fail to be elaborated since we don't have the instance `Add (Id Nat)`
  rw [Nat.add_comm]

theorem ex6 : (fun (x : Nat) y z => x + y + z) = (fun x y z => x + (y + z)) := by
  funext
  rw [Nat.add_assoc]


import tutorial_world.level17_by_cases --hide
open IncidencePlane --hide

/- Tactic : by_contra

## Summary

It changes the goal from `⊢ P` to `⊢ false` and adds a new hypothesis
of the form `h : ¬ P` to the local context.

## Real Life Example

If we want to prove `⊢ SKY_ALWAYS_BLUE`, but we assume that `h : SKY_GREY_WHEN_RAINS`,
type `by_contra z` to add `z : ¬ (SKY_ALWAYS_BLUE)` to the local context and change the
goal into `⊢ false`. Then, argue that `h : SKY_GREY_WHEN_RAINS` to finish the proof.
-/

/-
# Tutorial World

## Level 18: the `by_contra` tactic.

Congratulations! You are about to finish the first world of The Euclid Game!
In this level, we introduce the `by_contra` tactic. Mathematicians would use
it to provide a **proof by contradiction**. This tactic changes the goal from
`⊢ P` to `⊢ false` and adds a new hypothesis of the form `h : ¬ P` to the local context.

To finish this world, we would like to prove that two distinct lines have **at most**
one point in common. Delete the `sorry` to see the goal appear as `⊢ A = B`. Now, take a look
to the hypotheses that we have in our local context and try to do a drawing of the situation
by using all of them. Once you're done, note that the points A and B must be equal so that the
lines r and s satisfy the hypothesis `hrs: r ≠ s`. Then, try to look for a theorem statement which
could be useful for this level. As you've well deduced, `equal_lines_of_contain_two_points`
is the right path to choose. However, note that it states `A ≠ B` and `r = s` instead of `A = B`and
`r ≠ s`, respectively. Because of this reason, the `by_contra` tactic has to join the party.

Now, try to solve this level by your own in just three lines of code. [**Remember:** whenever you see
a hypothesis of the form `h : P ≠ Q`, Lean can also understand it as `h : ¬ (P = Q)`, or `h : (P = Q) → false`.]
-/

/- Hint : Click here for a hint, in case you get stuck.
Starting the proof by typing `by_contra h,` will change the goal into proving `⊢ false` and add the hypothesis `h : ¬ (A = B)` to the
local context. Now, note that `hrs : r ≠ s` can be also understood as `h : (r = s) → false`. Then, `apply hrs,` will make progress. To finish with,
try to close the goal in just one line by using the `equal_lines_of_contain_two_points` theorem statement.
Still bewildered? Click on "View source" (located on the top right corner of the game screen) to see the solution.
-/

variables {Ω : Type} [IncidencePlane Ω] --hide
variables {A B : Ω} --hide
variables {r s : Line Ω} --hide
/- Lemma :
Two distinct lines have at most one point in common.
-/
lemma distinct_lines_have_at_most_one_common_point
	(hrs: r ≠ s)
	(hAr: A ∈ r) (hAs: A ∈ s) (hBr: B ∈ r) (hBs: B ∈ s) :
	A = B :=
begin

  by_contra h,
  apply hrs,
  apply equal_lines_of_contain_two_points h hAr hAs hBr hBs,

end