Enderton (logic). Continue Exercises 1.1.
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@ -602,7 +602,7 @@
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\end{proof}
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\subsection{\pending{Exercise 1.1.5}}%
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\subsection{\verified{Exercise 1.1.5}}%
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\hyperlabel{sub:exercise-1.1.5}
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Suppose that $\alpha$ is a wff not containing the negation symbol $\neg$.
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@ -675,7 +675,7 @@
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\end{proof}
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\subsubsection{\pending{Exercise 1.1.5b}}%
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\subsubsection{\verified{Exercise 1.1.5b}}%
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\hyperlabel{ssub:exercise-1.1.5-b}
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Show that more than a quarter of the symbols are sentence symbols.
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@ -2,6 +2,8 @@ import Common.Logic.Basic
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import Common.Nat.Basic
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import Mathlib.Algebra.Parity
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import Mathlib.Data.Nat.Basic
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import Mathlib.Data.Real.Basic
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import Mathlib.Data.Setoid.Partition
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import Mathlib.Tactic.NormNum
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import Mathlib.Tactic.Ring
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@ -112,7 +114,22 @@ negation symbol is the only non-binary sentential connective.
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-/
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lemma no_neg_sentential_count_eq_binary_count {φ : Wff} (h : ¬φ.hasNotSymbol)
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: φ.sententialSymbolCount = φ.binarySymbolCount := by
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sorry
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induction φ with
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| SS _ =>
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unfold sententialSymbolCount binarySymbolCount
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rfl
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| Not _ _ =>
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unfold hasNotSymbol at h
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exfalso
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exact h trivial
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| And e₁ e₂ ih₁ ih₂
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| Or e₁ e₂ ih₁ ih₂
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| Cond e₁ e₂ ih₁ ih₂
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| Iff e₁ e₂ ih₁ ih₂ =>
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unfold hasNotSymbol at h
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rw [not_or_de_morgan] at h
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unfold sententialSymbolCount binarySymbolCount
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rw [ih₁ h.left, ih₂ h.right]
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/-- #### Parentheses Count
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@ -163,6 +180,14 @@ theorem length_eq_sum_symbol_count (φ : Wff)
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end Wff
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/-! #### Exercise 1.1.2
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Show that there are no wffs of length `2`, `3`, or `6`, but that any other
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positive length is possible.
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-/
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section Exercise_1_1_2
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/--
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An enumeration of values `m` and `n` can take on in equation `m + n = 3`.
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-/
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@ -202,12 +227,6 @@ private lemma eq_3_by_cases (m n : ℕ) (h : m + n = 3)
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right; right; right
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exact ⟨m_eq_3, h⟩
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/-! #### Exercise 1.1.2
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Show that there are no wffs of length `2`, `3`, or `6`, but that any other
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positive length is possible.
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-/
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theorem exercise_1_1_2_i (φ : Wff)
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: φ.length ≠ 2 ∧ φ.length ≠ 3 ∧ φ.length ≠ 6 := by
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induction φ with
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@ -255,13 +274,28 @@ theorem exercise_1_1_2_i (φ : Wff)
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intro h₃
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exact absurd h₃.left ih₁.right.left
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theorem exercise_1_1_2_ii (n : ℕ) (h : n ∈ Set.univ \ {2, 3, 6})
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private def recNot : ℕ → Wff → Wff
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| 0, φ => φ
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| n + 1, φ => Wff.Not (recNot n φ)
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theorem exercise_1_1_2_ii (n : ℕ) (hn : n ≠ 2 ∧ n ≠ 3 ∧ n ≠ 6)
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: ∃ φ : Wff, φ.length = n := by
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let φ₁ := Wff.SS 1
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let φ₂ := Wff.And φ₁ (Wff.SS 2)
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let φ₃ := Wff.And φ₂ (Wff.SS 3)
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let S : Set (Set { n : ℕ // n ≠ 2 ∧ n ≠ 3 ∧ n ≠ 6 }) := {
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{ m | ∃ n : ℕ, (recNot n φ₁).length = m.1 },
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{ m | ∃ n : ℕ, (recNot n φ₂).length = m.1 },
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{ m | ∃ n : ℕ, (recNot n φ₃).length = m.1 }
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}
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have hS : Setoid.IsPartition S := by
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sorry
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sorry
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end Exercise_1_1_2
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/-- #### Exercise 1.1.3
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Let `α` be a wff; let `c` be the number of places at which binary connective
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@ -330,10 +364,7 @@ than a quarter of the symbols are sentence symbols.
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`k + 1`.
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-/
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theorem exercise_1_1_5_b (α : Wff) (hα : ¬α.hasNotSymbol)
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: α.sentenceSymbolCount > α.length / 4 := by
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have h₁ := α.sentenceSymbolCount
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have h₂ := α.sententialSymbolCount
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have h₃ := α.parenCount
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: α.sentenceSymbolCount > (Nat.cast α.length : ℝ) / 4 := by
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rw [
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α.length_eq_sum_symbol_count,
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Wff.paren_count_double_sentential_count α,
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@ -341,9 +372,22 @@ theorem exercise_1_1_5_b (α : Wff) (hα : ¬α.hasNotSymbol)
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exercise_1_1_3 α
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]
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generalize Wff.binarySymbolCount α = k
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calc k + 1
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_ = 4 * (k + 1) / 4 := Eq.symm $ Nat.mul_div_cancel_left (k + 1) (by simp)
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_ = (k + 4 + k + 2 * k) / 4 := by ring_nf
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_ > (k + 1 + k + 2 * k) / 4 := by sorry
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simp only [
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Nat.cast_add,
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Nat.cast_one,
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Nat.cast_mul,
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Nat.cast_ofNat,
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gt_iff_lt
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]
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ring_nf
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simp only [
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Int.ofNat_eq_coe,
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Nat.cast_one,
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Int.cast_one,
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Nat.cast_ofNat,
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one_div,
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add_lt_add_iff_right
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]
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exact inv_lt_one (by norm_num)
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end Enderton.Logic.Chapter_1
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