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Enderton. Prove ordering theorems and fix injective successor proof.

finite-set-exercises
Joshua Potter 2023-08-04 18:26:15 -06:00
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356
Bookshelf/Enderton/Set.tex
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@ -119,7 +119,8 @@ The \textbf{composition} of sets $F$ and $G$ is
\section{\defined{Connected}}%
\hyperlabel{ref:connected}
A binary relation $R$ on $A$ is \textbf{connected} if for distinct $x, y \in A$, either $xRy$ or $yRx$.
A binary relation $R$ on $A$ is \textbf{connected} if for distinct $x, y \in A$,
either $xRy$ or $yRx$.
\begin{definition}
@ -340,7 +341,8 @@ The \textbf{inverse} of a set $F$ is the set
\section{\defined{Irreflexive}}%
\hyperlabel{ref:irreflexive}
A binary relation $R$ on set $A$ is \textbf{irreflexive} if there is no $x \in A$ for which $xRx$.
A binary relation $R$ on set $A$ is \textbf{irreflexive} if there is no
$x \in A$ for which $xRx$.
\begin{definition}
@ -3096,7 +3098,7 @@ If not, then under what conditions does equality hold?
\hyperlabel{sub:lemma-1}
\hyperlabel{sub:one-to-one-inverse}
\begin{lemma}[1]
\begin{lemma}
For any one-to-one function $F$, $F^{-1}$ is also one-to-one.
@ -3814,7 +3816,7 @@ If not, then under what conditions does equality hold?
Let $R$ be a linear ordering on $A$.
\begin{enumerate}[(i)]
\item There is no $x$ for which $xRx$.
\item For distinct $x$ and $y$ in $A$, either $xRy$ or $yRx$.
\item For distinct $x$ and $y$ in $A$, either $xRy$ or $yRx$ (but not both).
\end{enumerate}
\end{theorem}
@ -3838,7 +3840,7 @@ If not, then under what conditions does equality hold?
Let $x, y \in A$ such that $x \neq y$.
By definition, $R$ is \nameref{ref:trichotomous}.
Thus only one of $$xRy, \quad x = y, \quad yRx$$ hold.
By hypothesis $x \neq y$ meaning either $xRy$ or $yRx$.
By hypothesis $x \neq y$ meaning either $xRy$ or $yRx$ (but not both).
\end{proof}
@ -6134,44 +6136,6 @@ Show that $<_L$ is a linear ordering on $A \times B$.
\section{Peano's Postulates}%
\hyperlabel{sec:peanos-postulates}
\subsection{\verified{Theorem 4D}}%
\hyperlabel{sub:theorem-4d}
\begin{theorem}[4D]
$\langle \omega, \sigma, 0 \rangle$ is a Peano system.
\end{theorem}
\begin{proof}
\lean{Common/Set/Peano}{Peano.instSystemNatUnivSuccOfNatInstOfNatNat}
Note $\sigma$ is defined as $\sigma = \{\pair{n, n^+} \mid n \in \omega\}$.
To prove $\langle \omega, \sigma, 0 \rangle$ is a \nameref{ref:peano-system},
we must show that (i) $0 \not\in \ran{S}$, (ii) $\sigma$ is one-to-one, and
(iii) every subset $A$ of $\omega$ containing $0$ and closed under $\sigma$
is $\omega$ itself.
\paragraph{(i)}%
This follows immediately from \nameref{sub:theorem-4c}.
\paragraph{(ii)}%
Let $n^+ \in \ran{\sigma}$.
By construction, there exists some $m_1 \in \omega$ such that $m_1 = n^+$.
Suppose there exists some $m_2 \in \omega$ such that $m_2 = n^+$.
By definition of the \nameref{ref:successor}, $m_1 = n \cup \{n\} = m_2$.
By the \nameref{ref:extensionality-axiom}, $m_1 = m_2$.
Thus $\sigma$ is one-to-one.
\paragraph{(iii)}%
This follows immediately from \nameref{sub:theorem-4b}.
\end{proof}
\subsection{\unverified{Theorem 4E}}%
\hyperlabel{sub:theorem-4e}
@ -6260,6 +6224,50 @@ Show that $<_L$ is a linear ordering on $A \times B$.
\end{proof}
\subsection{\verified{Theorem 4D}}%
\hyperlabel{sub:theorem-4d}
\begin{theorem}[4D]
$\langle \omega, \sigma, 0 \rangle$ is a Peano system.
\end{theorem}
\begin{note}
Notice this theorem comes after \nameref{sub:theorem-4e}. Enderton introduces
this theorem before \nameref{sub:theorem-4e} but its proof relies on
\nameref{sub:theorem-4e}.
\end{note}
\begin{proof}
\lean{Common/Set/Peano}{Peano.instSystemNatUnivSuccOfNatInstOfNatNat}
Note $\sigma$ is defined as $\sigma = \{\pair{n, n^+} \mid n \in \omega\}$.
To prove $\langle \omega, \sigma, 0 \rangle$ is a \nameref{ref:peano-system},
we must show that (i) $0 \not\in \ran{S}$, (ii) $\sigma$ is one-to-one, and
(iii) every subset $A$ of $\omega$ containing $0$ and closed under $\sigma$
is $\omega$ itself.
\paragraph{(i)}%
This follows immediately from \nameref{sub:theorem-4c}.
\paragraph{(ii)}%
Let $m, n \in \omega$ and suppose $m^+ = n^+$.
Then $\bigcup \left(m^+\right) = \bigcup \left(n^+\right)$.
By \nameref{sub:theorem-4f}, every natural number is a
\nameref{ref:transitive-set}.
Therefore, by \nameref{sub:theorem-4e},
$$\bigcup \left(m^+\right) = m = \bigcup \left(n^+\right) = n.$$
\paragraph{(iii)}%
This follows immediately from \nameref{sub:theorem-4b}.
\end{proof}
\subsection{\unverified{Theorem 4G}}%
\hyperlabel{sub:theorem-4g}
@ -6580,7 +6588,7 @@ Show that $<_L$ is a linear ordering on $A \times B$.
\subsection{\verified{Left Additive Identity}}%
\hyperlabel{sub:left-additive-identity}
\begin{lemma}[2]
\begin{lemma}
For all $n \in \omega$, $A_0(n) = n$.
In other words, $$0 + n = n.$$
@ -6623,7 +6631,7 @@ Show that $<_L$ is a linear ordering on $A \times B$.
\hyperlabel{sub:lemma-3}
\hyperlabel{sub:succ-add-eq-add-succ}
\begin{lemma}[3]
\begin{lemma}
For all $m, n \in \omega$, $A_{m^+}(n) = A_m(n^+)$.
In other words, $$m^+ + n = m + n^+.$$
@ -6774,7 +6782,7 @@ Show that $<_L$ is a linear ordering on $A \times B$.
\subsection{\verified{Zero Multiplicand}}%
\hyperlabel{sub:zero-multiplicand}
\begin{lemma}[4]
\begin{lemma}
For all $n \in \omega$, $M_0(n) = 0$.
In other words, $$0 \cdot n = 0.$$
@ -6824,7 +6832,7 @@ Show that $<_L$ is a linear ordering on $A \times B$.
\subsection{\verified{Successor Distribution}}%
\hyperlabel{sub:successor-distribution}
\begin{lemma}[5]
\begin{lemma}
For all $m, n \in \omega$, $M_{m^+}(n) = M_m(n) + n$.
In other words, $$m^+ \cdot n = m \cdot n + n.$$
@ -6941,7 +6949,7 @@ Show that $<_L$ is a linear ordering on $A \times B$.
\subsection{\verified{Successor Identity}}%
\hyperlabel{sub:successor-identity}
\begin{lemma}[6]
\begin{lemma}
For all $m \in \omega$, $A_m(1) = m^+$.
In other words, $$m + 1 = m^+.$$
@ -6994,7 +7002,7 @@ Show that $<_L$ is a linear ordering on $A \times B$.
\subsection{\verified{Right Multiplicative Identity}}%
\hyperlabel{sub:right-multiplicative-identity}
\begin{lemma}[7]
\begin{lemma}
For all $m \in \omega$, $M_m(1) = m$.
In other words, $$m \cdot 1 = m.$$
@ -7160,10 +7168,10 @@ Show that $<_L$ is a linear ordering on $A \times B$.
\section{Ordering on \texorpdfstring{$\omega$}{Natural Numbers}}%
\hyperlabel{sec:ordering-natural-numbers}
\subsection{\unverified{Ordering on Successor}}%
\subsection{\verified{Ordering on Successor}}%
\hyperlabel{sub:ordering-successor}
\begin{lemma}[8]
\begin{lemma}
Let $m, n \in \omega$.
Then $m < n^+ \iff m \leq n$.
@ -7172,6 +7180,8 @@ Show that $<_L$ is a linear ordering on $A \times B$.
\begin{proof}
\lean{Std/Data/Nat/Lemmas}{Nat.lt\_succ}
Let $m, n \in \omega$.
By \nameref{ref:ordering-natural-numbers},
\begin{align*}
@ -7185,27 +7195,144 @@ Show that $<_L$ is a linear ordering on $A \times B$.
\end{proof}
\subsection{\sorry{Lemma 4L}}%
\hyperlabel{sub:lemma-4l}
\subsection{\unverified{Members of Natural Numbers}}%
\hyperlabel{sub:members-natural-numbers}
\begin{lemma}[4L]
\begin{lemma}
\begin{enumerate}[(a)]
\item For any natural numbers $m$ and $n$, $$m \in n \iff m^+ \in n^+.$$
\item No natural number is a member of itself.
\end{enumerate}
Every natural number is the set of all smaller natural numbers.
\end{lemma}
\begin{proof}
Let $n \in \omega$.
Consider $m \in n$.
By \nameref{sub:theorem-4b}, $\omega$ is a \nameref{ref:transitive-set}.
Thus $m \in n$ implies $m \in \omega$.
Thus $m \in n \iff m \in \omega \land m \in n$.
\end{proof}
\subsection{\pending{Lemma 4L(a)}}%
\hyperlabel{sub:lemma-4l-a}
\begin{lemma}[4L(a)]
For any natural numbers $m$ and $n$, $$m \in n \iff m^+ \in n^+.$$
\end{lemma}
\begin{proof}
\lean{Std/Data/Nat/Lemmas}{Nat.succ\_lt\_succ\_iff}
Let $m$ and $n$ be \nameref{ref:natural-number}s.
\paragraph{($\Rightarrow$)}%
Suppose $m \in n$.
TODO
\paragraph{($\Leftarrow$)}%
Suppose $m^+ \in n^+$.
TODO
\end{proof}
\subsection{\sorry{%
\subsection{\verified{Lemma 4L(b)}}%
\hyperlabel{sub:lemma-4l-b}
\begin{lemma}[4L(b)]
No natural number is a member of itself.
\end{lemma}
\begin{proof}
\lean{Init/Prelude}{Nat.lt\_irrefl}
Define
\begin{equation}
\hyperlabel{sub:lemma-4l-b-eq1}
S = \{n \in \omega \mid n \not\in n\}.
\end{equation}
We prove that (i) $0 \in S$ and (ii) if $n \in S$ then $n^+ \in S$.
Afterwards we show that (iii) our theorem holds.
\paragraph{(i)}%
\hyperlabel{par:lemma-4l-b-i}
By definition, $0 = \emptyset$.
It obviously holds that $\emptyset \not\in \emptyset$ since $\emptyset$,
by definition, has no members.
Thus $0 \in S$.
\paragraph{(ii)}%
\hyperlabel{par:lemma-4l-b-ii}
Suppose $n \in S$.
By \eqref{sub:lemma-4l-b-eq1}, $n \not\in n$.
By \nameref{sub:lemma-4l-a}, it follows $n^+ \not\in n^+$.
Thus $n^+ \in S$.
\paragraph{(iii)}%
By \nameref{par:lemma-4l-b-i} and \nameref{par:lemma-4l-b-ii}, $S$ is an
\nameref{ref:inductive-set}.
Hence \nameref{sub:theorem-4b} implies $S = \omega$.
Thus for all $n \in \omega$, $n \not\in n$.
\end{proof}
\subsection{\verified{\texorpdfstring{$0$}{Zero} is the Least Natural Number}}%
\hyperlabel{sub:zero-least-natural-number}
\begin{lemma}
For every natural number $n \neq 0$, $0 \in n$.
\end{lemma}
\begin{proof}
\lean{Bookshelf/Enderton/Set/Chapter\_4}
{Enderton.Set.Chapter\_4.zero\_least\_nat}
Let $$S = \{n \in \omega \mid n = 0 \lor 0 \in n\}.$$
We prove that (i) $0 \in S$ and (ii) if $n \in S$ then $n^+ \in S$.
Afterwards we show that (iii) our theorem holds.
\paragraph{(i)}%
\hyperlabel{par:zero-least-natural-number-i}
This trivially holds by definition of $S$.
\paragraph{(ii)}%
\hyperlabel{par:zero-least-natural-number-ii}
Suppose $n \in S$.
By definition of the \nameref{ref:successor} function, $n^+ = n \cup \{n\}$.
Thus $n \in n^+$.
By \nameref{sub:theorem-4f}, $n^+$ is a \nameref{ref:transitive-set}.
Since $0 \in n$ and $n \in n^+$, it follows $0 \in n^+$.
Thus $n^+ \in S$.
\paragraph{(iii)}%
By \nameref{par:zero-least-natural-number-i} and
\nameref{par:zero-least-natural-number-ii}, $S$ is an
\nameref{ref:inductive-set}.
Hence \nameref{sub:theorem-4b} implies $S = \omega$.
Thus for all $n \in \omega$, either $n = 0$ or $0 \in n$.
\end{proof}
\subsection{\verified{%
Trichotomy Law for \texorpdfstring{$\omega$}{Natural Numbers}}}%
\hyperlabel{sub:trichotomy-law-natrual-numbers}
\hyperlabel{sub:trichotomy-law-natural-numbers}
\begin{theorem}
@ -7216,25 +7343,111 @@ Show that $<_L$ is a linear ordering on $A \times B$.
\begin{proof}
TODO
\lean{Bookshelf/Enderton/Set/Chapter\_4}
{Enderton.Set.Chapter\_4.trichotomy\_law\_for\_nat}
Let $n \in \omega$ and define
\begin{equation}
\hyperlabel{sub:trichotomy-law-natural-numbers-eq1}
S = \{m \in \omega \mid m \in n \lor m = n \lor n \in m\}.
\end{equation}
We prove that (i) $0 \in S$ and (ii) if $m \in S$ then $m^+ \in S$.
Afterwards we show that (iii) our theorem holds.
\paragraph{(i)}%
\hyperlabel{par:trichotomy-law-natural-numbers-i}
If $n = 0$, then it trivially follows $0 \in S$.
Otherwise \nameref{sub:zero-least-natural-number} implies $0 \in n$.
Thus $0 \in S$.
\paragraph{(ii)}%
\hyperlabel{par:trichotomy-law-natural-numbers-ii}
Suppose $m \in S$.
By \eqref{sub:trichotomy-law-natural-numbers-eq1}, there are three cases to
consider:
\subparagraph{Case 1}%
Suppose $m \in n$.
By \nameref{sub:lemma-4l-a}, $m^+ \in n^+ = n \cup \{n\}$.
By definition of the \nameref{ref:successor}, $m^+ \in n$ or $m^+ = n$.
Either way, $m^+ \in S$.
\subparagraph{Case 2}%
Suppose $m = n$.
Since $m \in m^+$, it follows $n \in m^+$.
Thus $m^+ \in S$.
\subparagraph{Case 3}%
Suppose $n \in m$.
Then $n \in m \cup \{m\} = m^+$.
Thus $m^+ \in S$.
\subparagraph{Conclusion}%
Since the above three cases are exhaustive, it follows $m^+ \in S$.
\paragraph{(iii)}%
By \nameref{par:trichotomy-law-natural-numbers-i} and
\nameref{par:trichotomy-law-natural-numbers-ii}, $S$ is an
\nameref{ref:inductive-set}.
Hence \nameref{sub:theorem-4b} implies $S = \omega$.
Thus for all $m, n \in \omega$, $$m \in n \lor m = n \lor n \in m.$$
We now prove that
$$\in_\omega = \{\pair{m, n} \in \omega \times \omega \mid m \in n\}$$
is \nameref{ref:irreflexive} and \nameref{ref:connected}.
Irreflexivity immediately follows from \nameref{sub:lemma-4l-b}.
Connectivity follows immediately from the fact $S = \omega$.
Furthermore, it is not possible both $m \in n$ and $n \in m$ since, by
\nameref{sub:theorem-4f}, $m$ and $n$ are \nameref{ref:transitive-set}s.
This would otherwise imply $m \in m$, an immediate contradiction to
irreflexivity.
Thus $\in_\omega$ is a \nameref{ref:trichotomous} relation.
\end{proof}
\subsection{\sorry{%
\subsection{\verified{%
Linear Ordering on \texorpdfstring{$\omega$}{Natural Numbers}}}%
\hyperlabel{sub:linear-ordering-natural-numbers}
\begin{theorem}
Relation
$$\in_\omega = \{\pair{m, n} \in \omega \times \omega \mid m \in n\}$$
\begin{equation}
\hyperlabel{sub:linear-ordering-natural-numbers-eq1}
\in_\omega = \{\pair{m, n} \in \omega \times \omega \mid m \in n\}
\end{equation}
is a linear ordering on $\omega$.
\end{theorem}
\begin{proof}
TODO
\lean{Bookshelf/Enderton/Set/Chapter\_4}
{Enderton.Set.Chapter\_4.linear\_ordering\_on\_nat}
By definition, \eqref{sub:linear-ordering-natural-numbers-eq1} is a
\nameref{ref:linear-ordering} on $\omega$ if it is (i) transitive and
(ii) trichotomous.
\paragraph{(i)}%
Suppose $p, q, r \in \omega$ such that $p \in q$ and $q \in r$.
By \nameref{sub:theorem-4f}, $p$, $q$, and $r$ are
\nameref{ref:transitive-set}s.
By definition of a transitive set, it follows $p \in r$.
Hence \eqref{sub:linear-ordering-natural-numbers-eq1} is
\nameref{ref:transitive}.
\paragraph{(ii)}%
By \nameref{sub:trichotomy-law-natural-numbers},
\eqref{sub:linear-ordering-natural-numbers-eq1} is trichotomous.
\end{proof}
@ -7670,8 +7883,8 @@ Show that each natural number is either even or odd, but never both.
Suppose $n$ is even and not odd.
Then there exists some $m \in \omega$ such that $2 \cdot m = n$.
Since the successor operation is one-to-one, $(2 \cdot m)^+ = n^+$.
Thus $n^+$ is odd.
Therefore $(2 \cdot m)^+ = n^+$.
Hence $n^+$ is odd.
For the sake of contradiction, suppose $n^+$ is even.
Then there exists some $p$ such that $2 \cdot p = n^+$.
@ -7694,7 +7907,8 @@ Show that each natural number is either even or odd, but never both.
& = q^+ + q^+ \\
& = (q^+ + q)^+. & \textref{sub:theorem-4i}
\end{align*}
Since the successor operation is one-to-one, $n = q^+ + q$.
By \nameref{sub:theorem-4d}, the \nameref{ref:successor} operation is
one-to-one meaning $n = q^+ + q$.
But then
\begin{align*}
n
@ -7715,13 +7929,13 @@ Show that each natural number is either even or odd, but never both.
Suppose $n$ is odd and not even.
Then there exists some $p \in \omega$ such that $(2 \cdot p)^+ = n$.
Since the successor operation is one-to-one,
$(2 \cdot p)^{++} = (2 \cdot p^+) = n^+$.
Thus $n^+$ is even.
Therefore $(2 \cdot p)^{++} = (2 \cdot p^+) = n^+$.
Hence $n^+$ is even.
For the sake of contradiction, suppose $n^+$ is odd.
Then there exists some $q$ such that $(2 \cdot q)^+ = n^+$.
Since the successor operation is one-to-one, it follows $2 \cdot q = n$.
By \nameref{sub:theorem-4d}, the \nameref{ref:successor} operation is
one-to-one meaning $2 \cdot q = n$.
But this implies $n$ is even, a contradiction.
Thus our original assumption is wrong.
That is, $n^+$ is even but not odd.

37
Bookshelf/Enderton/Set/Chapter_4.lean
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@ -105,4 +105,41 @@ theorem exercise_4_14 (n : ℕ)
have : even n := ⟨q, hq'⟩
exact absurd this h
/-- #### Lemma 10
For every natural number `n ≠ 0`, `0 ∈ n`.
-/
theorem zero_least_nat (n : ℕ)
: 0 = n ∨ 0 < n := by
by_cases h : n = 0
· left
rw [h]
· right
have ⟨m, hm⟩ := Nat.exists_eq_succ_of_ne_zero h
rw [hm]
exact Nat.succ_pos m
/-- #### Trichotomy Law for ω
For any natural numbers `m` and `n`, exactly one of the three conditions
```
m ∈ n, m = n, n ∈ m
```
holds.
-/
theorem trichotomy_law_for_nat
: IsAsymm ℕ LT.lt ∧ IsTrichotomous ℕ LT.lt :=
⟨instIsAsymmLtToLT, instIsTrichotomousLtToLTToPreorderToPartialOrder⟩
/-- #### Linear Ordering on ω
Relation
```
∈_ω = {⟨m, n⟩ ∈ ω × ω | m ∈ n}
```
is a linear ordering on `ω`.
-/
theorem linear_ordering_on_nat
: IsStrictTotalOrder ℕ LT.lt := isStrictTotalOrder_of_linearOrder
end Enderton.Set.Chapter_4