bookshelf/Bookshelf/Enderton/Logic/Chapter_1.lean

import Common.Logic.Basic
import Common.Nat.Basic
import Mathlib.Algebra.Parity
import Mathlib.Data.Nat.Basic
import Mathlib.Tactic.NormNum
import Mathlib.Tactic.Ring

/-! # Enderton.Logic.Chapter_1

Sentential Logic
-/

namespace Enderton.Logic.Chapter_1

/--
An abstract representation of a well-formed formula as defined by Enderton.
-/
inductive Wff where
  |   SS : Nat → Wff        -- e.g. **S**entence **S**ymbol `Aₙ`
  |  Not : Wff → Wff        -- e.g. `(¬ α)`
  |  And : Wff → Wff → Wff  -- e.g. `(α ∧ β)`
  |   Or : Wff → Wff → Wff  -- e.g. `(α ∨ β)`
  | Cond : Wff → Wff → Wff  -- e.g. `(α → β)`
  |  Iff : Wff → Wff → Wff  -- e.g. `(α ↔ β)`

namespace Wff

/--
Returns the length of the expression, i.e. a count of all symbols..
-/
def length : Wff → ℕ
  | Wff.SS   _     => 1
  | Wff.Not  e     => length e + 3
  | Wff.And  e₁ e₂
  | Wff.Or   e₁ e₂
  | Wff.Cond e₁ e₂
  | Wff.Iff  e₁ e₂ => length e₁ + length e₂ + 3

/--
Every `Wff` has a positive length.
-/
theorem length_gt_zero (φ : Wff)
  : length φ > 0 := by
  unfold length
  match φ with
  |   SS _
  |  Not _
  |  And _ _
  |   Or _ _
  | Cond _ _
  |  Iff _ _ => simp

/--
The number of sentence symbols found in the provided `Wff`.
-/
def sentenceSymbolCount : Wff → ℕ
  | Wff.SS   _     => 1
  | Wff.Not  e     => sentenceSymbolCount e
  | Wff.And  e₁ e₂
  | Wff.Or   e₁ e₂
  | Wff.Cond e₁ e₂
  | Wff.Iff  e₁ e₂ => sentenceSymbolCount e₁ + sentenceSymbolCount e₂

/--
The number of sentential connective symbols found in the provided `Wff`.
-/
def sententialSymbolCount : Wff → ℕ
  | Wff.SS   _     => 0
  | Wff.Not  e     => sententialSymbolCount e + 1
  | Wff.And  e₁ e₂
  | Wff.Or   e₁ e₂
  | Wff.Cond e₁ e₂
  | Wff.Iff  e₁ e₂ => sententialSymbolCount e₁ + sententialSymbolCount e₂ + 1

/--
The number of binary connective symbols found in the provided `Wff`.
-/
def binarySymbolCount : Wff → ℕ
  | Wff.SS   _     => 0
  | Wff.Not  e     => binarySymbolCount e
  | Wff.And  e₁ e₂
  | Wff.Or   e₁ e₂
  | Wff.Cond e₁ e₂
  | Wff.Iff  e₁ e₂ => binarySymbolCount e₁ + binarySymbolCount e₂ + 1

/--
The number of parentheses found in the provided `Wff`.
-/
def parenCount : Wff → ℕ
  | Wff.SS   _     => 0
  | Wff.Not  e     => 2 + parenCount e
  | Wff.And  e₁ e₂
  | Wff.Or   e₁ e₂
  | Wff.Cond e₁ e₂
  | Wff.Iff  e₁ e₂ => 2 + parenCount e₁ + parenCount e₂

/--
Whether or not the `Wff` contains a `¬`.
-/
def hasNotSymbol : Wff → Prop
  | Wff.SS   _     => False
  | Wff.Not  _     => True
  | Wff.And  e₁ e₂
  | Wff.Or   e₁ e₂
  | Wff.Cond e₁ e₂
  | Wff.Iff  e₁ e₂ => hasNotSymbol e₁ ∨ hasNotSymbol e₂

/--
If a `Wff` does not contain the `¬` symbol, it has the same number of sentential
connective symbols as it does binary connective symbols. In other words, the
negation symbol is the only non-binary sentential connective.
-/
lemma no_neg_sentential_count_eq_binary_count {φ : Wff} (h : ¬φ.hasNotSymbol)
  : φ.sententialSymbolCount = φ.binarySymbolCount := by
  sorry

/-- #### Parentheses Count

Let `φ` be a well-formed formula and `c` be the number of places at which a
sentential connective symbol exists. Then there is `2c` parentheses in `φ`.
-/
theorem paren_count_double_sentential_count (φ : Wff)
  : φ.parenCount = 2 * φ.sententialSymbolCount := by
  induction φ with
  | SS _ =>
    unfold parenCount sententialSymbolCount
    simp
  | Not e ih =>
    unfold parenCount sententialSymbolCount
    rw [ih]
    ring
  | And  e₁ e₂ ih₁ ih₂
  | Or   e₁ e₂ ih₁ ih₂
  | Cond e₁ e₂ ih₁ ih₂
  | Iff  e₁ e₂ ih₁ ih₂ =>
    unfold parenCount sententialSymbolCount
    rw [ih₁, ih₂]
    ring

/--
The length of a `Wff` corresponds to the summation of sentence symbols,
sentential connective symbols, and parentheses.
-/
theorem length_eq_sum_symbol_count (φ : Wff)
  : φ.length = φ.sentenceSymbolCount +
               φ.sententialSymbolCount +
               φ.parenCount := by
  induction φ with
  | SS _ =>
    unfold length sentenceSymbolCount sententialSymbolCount parenCount
    simp
  | Not e ih =>
    unfold length sentenceSymbolCount sententialSymbolCount parenCount
    rw [ih]
    ac_rfl
  | And  e₁ e₂ ih₁ ih₂
  | Or   e₁ e₂ ih₁ ih₂
  | Cond e₁ e₂ ih₁ ih₂
  | Iff  e₁ e₂ ih₁ ih₂ =>
    unfold length sentenceSymbolCount sententialSymbolCount parenCount
    rw [ih₁, ih₂]
    ac_rfl

end Wff

/--
An enumeration of values `m` and `n` can take on in equation `m + n = 3`.
-/
private lemma eq_3_by_cases (m n : ℕ) (h : m + n = 3)
  : m = 0 ∧ n = 3 ∨
    m = 1 ∧ n = 2 ∨
    m = 2 ∧ n = 1 ∨
    m = 3 ∧ n = 0 := by
  have m_le_3 : m ≤ 3 := by
    have : m = 3 - n := Eq.symm $ Nat.sub_eq_of_eq_add (Eq.symm h)
    rw [this]
    norm_num
  apply Or.elim (Nat.lt_or_eq_of_le m_le_3)
  · intro hm₁
    apply Or.elim (Nat.lt_or_eq_of_lt hm₁)
    · intro hm₂
      apply Or.elim (Nat.lt_or_eq_of_lt hm₂)
      · intro hm₃
        refine Or.elim (Nat.lt_or_eq_of_lt hm₃) (by simp) ?_
        intro m_eq_0
        rw [m_eq_0, zero_add] at h
        left
        exact ⟨m_eq_0, h⟩
      · intro m_eq_1
        rw [m_eq_1, add_comm] at h
        norm_num at h
        right; left
        exact ⟨m_eq_1, h⟩
    · intro m_eq_2
      rw [m_eq_2, add_comm] at h
      norm_num at h
      right; right; left
      exact ⟨m_eq_2, h⟩
  · intro m_eq_3
    rw [m_eq_3, add_comm] at h
    norm_num at h
    right; right; right
    exact ⟨m_eq_3, h⟩

/-! #### Exercise 1.1.2

Show that there are no wffs of length `2`, `3`, or `6`, but that any other
positive length is possible.
-/

theorem exercise_1_1_2_i (φ : Wff)
  : φ.length ≠ 2 ∧ φ.length ≠ 3 ∧ φ.length ≠ 6 := by
  induction φ with
  | SS _ =>
    unfold Wff.length
    simp
  | Not e ih =>
    unfold Wff.length
    refine ⟨by norm_num, ?_, ?_⟩
    · intro h
      norm_num at h
      have := e.length_gt_zero
      rw [h] at this
      simp at this
    · intro h
      norm_num at h
      rw [h] at ih
      simp at ih
  | And  e₁ e₂ ih₁ ih₂
  | Or   e₁ e₂ ih₁ ih₂
  | Cond e₁ e₂ ih₁ ih₂
  | Iff  e₁ e₂ ih₁ ih₂ =>
    unfold Wff.length
    refine ⟨by norm_num, ?_, ?_⟩
    · intro h
      norm_num at h
      have := e₁.length_gt_zero
      rw [h.left] at this
      simp at this
    · intro h
      norm_num at h
      apply Or.elim (eq_3_by_cases e₁.length e₂.length h)
      · intro h₁
        have := e₁.length_gt_zero
        rw [h₁.left] at this
        simp at this
      · intro h₁
        apply Or.elim h₁
        · intro h₂
          exact absurd h₂.right ih₂.left
        intro h₂
        apply Or.elim h₂
        · intro h₃
          exact absurd h₃.left ih₁.left
        intro h₃
        exact absurd h₃.left ih₁.right.left

theorem exercise_1_1_2_ii (n : ℕ) (h : n ∈ Set.univ \ {2, 3, 6})
  : ∃ φ : Wff, φ.length = n := by
  let φ₁ := Wff.SS 1
  let φ₂ := Wff.And φ₁ (Wff.SS 2)
  let φ₃ := Wff.And φ₂ (Wff.SS 3)
  sorry

/-- #### Exercise 1.1.3

Let `α` be a wff; let `c` be the number of places at which binary connective
symbols (`∧`, `∨`, `→`, `↔`) occur in `α`; let `s` be the number of places at
which sentence symbols occur in `α`. (For example, if `α` is `(A → (¬ A))` then
`c = 1` and `s = 2`.) Show by using the induction principle that `s = c + 1`.
-/
theorem exercise_1_1_3 (φ : Wff)
  : φ.sentenceSymbolCount = φ.binarySymbolCount + 1 := by
  induction φ with
  | SS _ =>
    unfold Wff.sentenceSymbolCount Wff.binarySymbolCount
    simp
  | Not e ih =>
    unfold Wff.sentenceSymbolCount Wff.binarySymbolCount
    exact ih
  | And  e₁ e₂ ih₁ ih₂
  | Or   e₁ e₂ ih₁ ih₂
  | Cond e₁ e₂ ih₁ ih₂
  | Iff  e₁ e₂ ih₁ ih₂ =>
    unfold Wff.sentenceSymbolCount Wff.binarySymbolCount
    rw [ih₁, ih₂]
    ring

/-- #### Exercise 1.1.5 (a)

Suppose that `α` is a wff not containing the negation symbol `¬`. Show that the
length of `α` (i.e., the number of symbols in the string) is odd.

*Suggestion*: Apply induction to show that the length is of the form `4k + 1`.
-/
theorem exercise_1_1_5_a (α : Wff) (hα : ¬α.hasNotSymbol)
  : Odd α.length := by
  suffices ∃ k : ℕ, α.length = 4 * k + 1 by
    have ⟨k, hk⟩ := this
    unfold Odd
    exact ⟨2 * k, by rw [hk, ← mul_assoc]; norm_num⟩
  induction α with
  | SS _ =>
    refine ⟨0, ?_⟩
    unfold Wff.length
    simp
  | Not e _ =>
    unfold Wff.hasNotSymbol at hα
    exfalso
    exact hα trivial
  | And  e₁ e₂ ih₁ ih₂
  | Or   e₁ e₂ ih₁ ih₂
  | Cond e₁ e₂ ih₁ ih₂
  | Iff  e₁ e₂ ih₁ ih₂  =>
    unfold Wff.hasNotSymbol at hα
    rw [not_or_de_morgan] at hα
    have ⟨k₁, hk₁⟩ := ih₁ hα.left
    have ⟨k₂, hk₂⟩ := ih₂ hα.right
    refine ⟨k₁ + k₂ + 1, ?_⟩
    unfold Wff.length
    rw [hk₁, hk₂]
    ring

/-- #### Exercise 1.1.5 (b)

Suppose that `α` is a wff not containing the negation symbol `¬`. Show that more
than a quarter of the symbols are sentence symbols.

*Suggestion*: Apply induction to show that the number of sentence symbols is
`k + 1`.
-/
theorem exercise_1_1_5_b (α : Wff) (hα : ¬α.hasNotSymbol)
  : α.sentenceSymbolCount > α.length / 4 := by
  have h₁ := α.sentenceSymbolCount
  have h₂ := α.sententialSymbolCount
  have h₃ := α.parenCount
  rw [
    α.length_eq_sum_symbol_count,
    Wff.paren_count_double_sentential_count α,
    Wff.no_neg_sentential_count_eq_binary_count hα,
    exercise_1_1_3 α
  ]
  generalize Wff.binarySymbolCount α = k
  calc k + 1
    _ = 4 * (k + 1) / 4 := Eq.symm $ Nat.mul_div_cancel_left (k + 1) (by simp)
    _ = (k + 4 + k + 2 * k) / 4 := by ring_nf
    _ > (k + 1 + k + 2 * k) / 4 := by sorry

end Enderton.Logic.Chapter_1