bookshelf/Bookshelf/Enderton/Logic/Chapter_1.lean

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import Common.Logic.Basic
import Common.Nat.Basic
import Mathlib.Algebra.Parity
import Mathlib.Data.Nat.Basic
import Mathlib.Data.Real.Basic
import Mathlib.Data.Setoid.Partition
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 →
| 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 →
| 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 →
| 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 →
| 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 →
| 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
induction φ with
| SS _ =>
unfold sententialSymbolCount binarySymbolCount
rfl
| Not _ _ =>
unfold hasNotSymbol at h
exfalso
exact h trivial
| And e₁ e₂ ih₁ ih₂
| Or e₁ e₂ ih₁ ih₂
| Cond e₁ e₂ ih₁ ih₂
| Iff e₁ e₂ ih₁ ih₂ =>
unfold hasNotSymbol at h
rw [not_or_de_morgan] at h
unfold sententialSymbolCount binarySymbolCount
rw [ih₁ h.left, ih₂ h.right]
/-- ### 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
/-! ### Exercise 1.1.2
Show that there are no wffs of length `2`, `3`, or `6`, but that any other
positive length is possible.
-/
section Exercise_1_1_2
/--
An enumeration of values `m` and `n` can take on in equation `m + n = 3`.
-/
private lemma eq_3_by_cases (m 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⟩
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
private def recNot : → Wff → Wff
| 0, φ => φ
| n + 1, φ => Wff.Not (recNot n φ)
theorem exercise_1_1_2_ii (n : ) (hn : n ≠ 2 ∧ n ≠ 3 ∧ n ≠ 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)
let S : Set (Set { n : // n ≠ 2 ∧ n ≠ 3 ∧ n ≠ 6 }) := {
{ m | ∃ n : , (recNot n φ₁).length = m.1 },
{ m | ∃ n : , (recNot n φ₂).length = m.1 },
{ m | ∃ n : , (recNot n φ₃).length = m.1 }
}
have hS : Setoid.IsPartition S := by
sorry
sorry
end Exercise_1_1_2
/-- ### 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_5a (α : Wff) (hα : ¬α.hasNotSymbol)
: Odd α.length := by
suffices ∃ k : , 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_5b (α : Wff) (hα : ¬α.hasNotSymbol)
: α.sentenceSymbolCount > (Nat.cast α.length : ) / 4 := by
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
simp only [
Nat.cast_add,
Nat.cast_one,
Nat.cast_mul,
Nat.cast_ofNat,
gt_iff_lt
]
ring_nf
simp only [
Int.ofNat_eq_coe,
Nat.cast_one,
Int.cast_one,
Nat.cast_ofNat,
one_div,
add_lt_add_iff_right
]
exact inv_lt_one (by norm_num)
/-! ### Exercise 1.2.1
Show that neither of the following two formulas tautologically implies the
other:
```
(A ↔ (B ↔ C))
((A ∧ (B ∧ C)) ((¬ A) ∧ ((¬ B) ∧ (¬ C)))).
```
*Suggestion*: Only two truth assignments are needed, not eight.
-/
section Exercise_1_2_1
private def f₁ (A B C : Prop) : Prop :=
A ↔ (B ↔ C)
private def f₂ (A B C : Prop) : Prop :=
((A ∧ (B ∧ C)) ((¬ A) ∧ ((¬ B) ∧ (¬ C))))
theorem exercise_1_2_1_i
: f₁ True False False ≠ f₂ True False False := by
unfold f₁ f₂
simp
theorem exercise_1_2_1_ii
: f₁ False False False ≠ f₂ False False False := by
unfold f₁ f₂
simp
end Exercise_1_2_1
section Exercise_1_2_2
/-- #### Exercise 1.2.2 (a)
Is `(((P → Q) → P) → P)` a tautology?
-/
theorem exercise_1_2_2a (P Q : Prop)
: (((P → Q) → P) → P) := by
tauto
/-! ### Exercise 1.2.2 (b)
Define `σₖ` recursively as follows: `σ₀ = (P → Q)` and `σₖ₊₁ = (σₖ → P)`. For
which values of `k` is `σₖ` a tautology? (Part (a) corresponds to `k = 2`.)
-/
private def σ (P Q : Prop) : → Prop
| 0 => P → Q
| n + 1 => σ P Q n → P
theorem exercise_1_2_2b_i (P Q : Prop) {k : } (h : k > 0)
: σ P Q (2 * k) := by
induction k with
| zero => simp at h
| succ k ih =>
by_cases hk : k = 0
· rw [hk]
simp only [Nat.mul_one]
unfold σ σ σ
exact exercise_1_2_2a P Q
· have := ih (Nat.pos_of_ne_zero hk)
unfold σ σ
have hk₁ := calc 2 * k.succ
_ = 2 * (k + 1) := rfl
_ = 2 * k + 2 * 1 := rfl
_ = 2 * k + 2 := by simp
rw [hk₁]
simp only [Nat.add_eq, add_zero]
tauto
theorem exercise_1_2_2b_ii
: ¬ σ True False 0 := by
unfold σ
simp
theorem exercise_1_2_2b_iii {k : } (h : Odd k)
: ¬ σ False Q k := by
by_cases hk : k = 1
· unfold σ σ
rw [hk]
simp
· have ⟨n, hn₁, hn₂⟩ : ∃ n : , k = (2 * n) + 1 ∧ n > 0 := by
have ⟨r, hr⟩ := h
refine ⟨r, hr, ?_⟩
by_contra nr
have : r = 0 := Nat.eq_zero_of_not_pos nr
rw [this] at hr
simp only [mul_zero, zero_add] at hr
exact absurd hr hk
unfold σ
rw [hn₁]
2023-12-14 19:49:31 +00:00
simp only [Nat.add_eq, add_zero, imp_false, not_not]
exact exercise_1_2_2b_i False Q hn₂
end Exercise_1_2_2
/-- ### Exercise 1.2.3 (a)
Determine whether or not `((P → Q)) (Q → P)` is a tautology.
-/
theorem exercise_1_2_3a (P Q : Prop)
: ((P → Q) (Q → P)) := by
tauto
/-- ### Exercise 1.2.3 (b)
Determine whether or not `((P ∧ Q) → R))` tautologically implies
`((P → R) (Q → R))`.
-/
theorem exercise_1_2_3b (P Q R : Prop)
: ((P ∧ Q) → R) ↔ ((P → R) (Q → R)) := by
tauto
/-! ### Exercise 1.2.5
Prove or refute each of the following assertions:
(a) If either `Σ ⊨ α` or `Σ ⊨ β`, then `Σ ⊨ (α β)`.
(b) If `Σ ⊨ (α β)`, then either `Σ ⊨ α` or `Σ ⊨ β`.
-/
theorem exercise_1_2_5a (P α β : Prop)
: ((P → α) (P → β)) → (P → (α β)) := by
tauto
theorem exercise_1_2_6b
: (False True) ∧ ¬ False := by
simp
/-! ### Exercise 1.2.15
Of the following three formulas, which tautologically implies which?
(a) `(A ↔ B)`
(b) `(¬((A → B) →(¬(B → A))))`
(c) `(((¬ A) B) ∧ (A (¬ B)))`
-/
theorem exercise_1_2_15_i (A B : Prop)
: (A ↔ B) ↔ (¬((A → B) → (¬(B → A)))) := by
tauto
theorem exercise_1_2_15_ii (A B : Prop)
: (A ↔ B) ↔ (((¬ A) B) ∧ (A (¬ B))) := by
tauto
end Enderton.Logic.Chapter_1