Enderton. Finish ordering on natural numbers exercises.
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@ -6950,7 +6950,7 @@ Show that $<_L$ is a linear ordering on $A \times B$.
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\begin{proof}
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\lean{Init/Prelude}{Nat.succ}
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\lean{Std/Data/Nat/Lemmas}{Nat.succ\_eq\_one\_add}
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Let
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\begin{equation}
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@ -7605,7 +7605,7 @@ Formulate an analogue to Exercise 9 for a function
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\end{proof}
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\subsection{\sorry{Exercise 4.13}}%
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\subsection{\verified{Exercise 4.13}}%
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\hyperlabel{sub:exercise-4.13}
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Let $m$ and $n$ be natural numbers such that $m \cdot n = 0$.
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@ -7613,11 +7613,29 @@ Show that either $m = 0$ or $n = 0$.
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\begin{proof}
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TODO
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\lean{Bookshelf/Enderton/Set/Chapter\_4}
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{Enderton.Set.Chapter\_4.exercise\_4\_13}
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Suppose $m \cdot n = 0$.
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For the sake of contradiction, assume $m \neq 0$ and $n \neq 0$.
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By \nameref{sub:theorem-4c}, there exists some $p, q \in \omega$ such that
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$p^+ = m$ and $q^+ = n$.
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Thus
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\begin{align*}
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m \cdot n
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& = m \cdot q^+ \\
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& = m \cdot q + m & \textref{sub:theorem-4j} \\
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& = m \cdot q + p^+ \\
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& = (m \cdot q + p)^+. & \textref{sub:theorem-4i}
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\end{align*}
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By definition of a \nameref{ref:successor},
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$m \cdot n = (m \cdot q + p)^+ \neq \emptyset = 0$, a contradiction.
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Therefore our original assumption was wrong.
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Hence $m = 0$ or $n = 0$.
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\end{proof}
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\subsection{\sorry{Exercise 4.14}}%
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\subsection{\verified{Exercise 4.14}}%
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\hyperlabel{sub:exercise-4.14}
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Call a natural number \textit{even} if it has the form $2 \cdot m$ for some $m$.
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@ -7626,7 +7644,98 @@ Show that each natural number is either even or odd, but never both.
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\begin{proof}
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TODO
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\lean{Bookshelf/Enderton/Set/Chapter\_4}
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{Enderton.Set.Chapter\_4.exercise\_4\_14}
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Let $$S = \{n \in \omega \mid n \text{ is even or odd but not both}\}.$$
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We show that (i) $0 \in S$ and (ii) if $n \in S$ then $n^+ \in S$ as well.
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Afterward we prove (iii) that the theorem statement holds.
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\paragraph{(i)}%
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\hyperlabel{par:exercise-4.14a-i}
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$0$ is even since $2 \cdot 0 = 0$ by \nameref{sub:theorem-4j}.
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Furthermore, $0$ is not odd since that would imply there exists some
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$p$ such that $(2 \cdot p)^+ = 0$.
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By definition of \nameref{ref:successor}, this is not possible.
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Thus $0 \in S$.
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\paragraph{(ii)}%
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\hyperlabel{par:exercise-4.14a-ii}
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Suppose $n \in S$.
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Then $n$ is even or odd but not both.
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\subparagraph{Case 1}%
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Suppose $n$ is even and not odd.
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Then there exists some $m \in \omega$ such that $2 \cdot m = n$.
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Since the successor operation is one-to-one, $(2 \cdot m)^+ = n^+$.
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Thus $n^+$ is odd.
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For the sake of contradiction, suppose $n^+$ is even.
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Then there exists some $p$ such that $2 \cdot p = n^+$.
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We consider two additional cases:
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\vspace{8pt}\quad
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\textbf{Case 1a}: Suppose $p = 0$.
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Then, by \nameref{sub:theorem-4j}, $2 \cdot p = 0 = n^+$.
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By definition of \nameref{ref:successor}, this is not possible.
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\vspace{8pt}\quad
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\textbf{Case 1b}: Suppose $p \neq 0$.
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Then \nameref{sub:theorem-4c} implies there exists some $q$ such that
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$q^+ = p$.
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Thus
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\begin{align*}
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n^+
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& = 2 \cdot p \\
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& = 2 \cdot q^+ \\
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& = q^+ + q^+ \\
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& = (q^+ + q)^+. & \textref{sub:theorem-4i}
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\end{align*}
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Since the successor operation is one-to-one, $n = q^+ + q$.
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But then
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\begin{align*}
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n
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& = q^+ + q \\
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& = q + q^+ & \textref{sub:theorem-4k-2} \\
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& = (q + q)^+ & \textref{sub:theorem-4i} \\
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& = (2 \cdot q)^+,
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\end{align*}
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indicating $n$ is odd.
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This is a contradiction.
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\vspace{8pt}\quad
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\textbf{Conclusion}: Since the above two cases are exhaustive, it follows
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our original assumption is wrong.
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That is, $n^+$ is odd but not even.
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\subparagraph{Case 2}%
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Suppose $n$ is odd and not even.
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Then there exists some $p \in \omega$ such that $(2 \cdot p)^+ = n$.
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Since the successor operation is one-to-one,
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$(2 \cdot p)^{++} = (2 \cdot p^+) = n^+$.
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Thus $n^+$ is even.
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For the sake of contradiction, suppose $n^+$ is odd.
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Then there exists some $q$ such that $(2 \cdot q)^+ = n^+$.
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Since the successor operation is one-to-one, it follows $2 \cdot q = n$.
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But this implies $n$ is even, a contradiction.
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Thus our original assumption is wrong.
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That is, $n^+$ is even but not odd.
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\subparagraph{Conclusion}%
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Since the foregoing cases are exhaustive, it follows $n^+ \in S$.
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\paragraph{(iii)}%
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By \nameref{par:exercise-4.14a-i} and \nameref{par:exercise-4.14a-ii},
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$S$ is an \nameref{ref:inductive-set}.
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By \nameref{sub:theorem-4b}, $S = \omega$.
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Thus every natural number is either even or odd, but not both.
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\end{proof}
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@ -1,3 +1,4 @@
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import Common.Logic.Basic
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import Mathlib.Data.Set.Basic
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/-! # Enderton.Set.Chapter_4
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@ -24,4 +25,84 @@ Show that `1 ≠ 3` i.e., that `∅⁺ ≠ ∅⁺⁺⁺`.
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theorem exercise_4_1 : 1 ≠ 3 := by
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simp
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/-- #### Exercise 4.13
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Let `m` and `n` be natural numbers such that `m ⬝ n = 0`. Show that either
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`m = 0` or `n = 0`.
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-/
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theorem exercise_4_13 (m n : ℕ) (h : m * n = 0)
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: m = 0 ∨ n = 0 := by
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by_contra nh
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rw [not_or_de_morgan] at nh
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have ⟨p, hp⟩ : ∃ p, m = p.succ := Nat.exists_eq_succ_of_ne_zero nh.left
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have ⟨q, hq⟩ : ∃ q, n = q.succ := Nat.exists_eq_succ_of_ne_zero nh.right
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have : m * n = (m * q + p).succ := calc m * n
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_ = m * q.succ := by rw [hq]
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_ = m * q + m := rfl
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_ = m * q + p.succ := by rw [hp]
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_ = (m * q + p).succ := rfl
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rw [this] at h
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simp only [Nat.succ_ne_zero] at h
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/--
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Call a natural number *even* if it has the form `2 ⬝ m` for some `m`.
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-/
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def even (n : ℕ) : Prop := ∃ m, 2 * m = n
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/--
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Call a natural number *odd* if it has the form `(2 ⬝ p) + 1` for some `p`.
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-/
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def odd (n : ℕ) : Prop := ∃ p, (2 * p) + 1 = n
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/-- #### Exercise 4.14
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Show that each natural number is either even or odd, but never both.
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-/
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theorem exercise_4_14 (n : ℕ)
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: (even n ∧ ¬ odd n) ∨ (¬ even n ∧ odd n) := by
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induction n with
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| zero =>
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left
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refine ⟨⟨0, by simp⟩, ?_⟩
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intro ⟨p, hp⟩
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simp only [Nat.zero_eq, Nat.succ_ne_zero] at hp
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| succ n ih =>
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apply Or.elim ih
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· -- Assumes `n` is even meaning `n⁺` is odd.
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intro ⟨⟨m, hm⟩, h⟩
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right
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refine ⟨?_, ⟨m, by rw [← hm]⟩⟩
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by_contra nh
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have ⟨p, hp⟩ := nh
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by_cases hp' : p = 0
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· rw [hp'] at hp
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simp at hp
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· have ⟨q, hq⟩ := Nat.exists_eq_succ_of_ne_zero hp'
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rw [hq] at hp
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have hq₁ : (q.succ + q).succ = n.succ := calc (q.succ + q).succ
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_ = q.succ + q.succ := rfl
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_ = 2 * q.succ := by rw [Nat.two_mul]
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_ = n.succ := hp
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injection hq₁ with hq₂
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have : odd n := by
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refine ⟨q, ?_⟩
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calc 2 * q + 1
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_ = q + q + 1 := by rw [Nat.two_mul]
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_ = q + q.succ := rfl
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_ = q.succ + q := by rw [Nat.add_comm]
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_ = n := hq₂
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exact absurd this h
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· -- Assumes `n` is odd meaning `n⁺` is even.
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intro ⟨h, ⟨p, hp⟩⟩
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have hp' : 2 * p.succ = n.succ := congrArg Nat.succ hp
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left
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refine ⟨⟨p.succ, by rw [← hp']⟩, ?_⟩
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by_contra nh
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unfold odd at nh
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have ⟨q, hq⟩ := nh
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injection hq with hq'
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simp only [Nat.add_eq, Nat.add_zero] at hq'
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have : even n := ⟨q, hq'⟩
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exact absurd this h
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end Enderton.Set.Chapter_4
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