Enderton. Begin adding theorem statements around ordering.

finite-set-exercises
Joshua Potter 2023-08-04 08:54:11 -06:00
parent 03f51f3097
commit 11d25dd28c
2 changed files with 148 additions and 3 deletions

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@ -3,4 +3,4 @@ import Bookshelf.Avigad.Chapter_3
import Bookshelf.Avigad.Chapter_4 import Bookshelf.Avigad.Chapter_4
import Bookshelf.Avigad.Chapter_5 import Bookshelf.Avigad.Chapter_5
import Bookshelf.Avigad.Chapter_7 import Bookshelf.Avigad.Chapter_7
import Bookshelf.Avigad.Chapter_8 import Bookshelf.Avigad.Chapter_8

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@ -433,6 +433,27 @@ For any sets $u$ and $v$, the \textbf{ordered pair} $\pair{u, v}$ is
\end{definition} \end{definition}
\section{\defined{Ordering on \texorpdfstring{$\omega$}{Natural Numbers}}}%
\hyperlabel{ref:ordering-natural-numbers}
For \nameref{ref:natural-number}s $m$ and $n$, define $m$ to be
\textbf{less than} $n$ if and only if $m \in n$.
That is, $$m < n \iff m \in n.$$
Likewise, define $m$ to be \textbf{less than or equal to} $n$ if and only if
$m \in n \lor m = n$.
That is,
\begin{align*}
m \leq n
& \iff m \mathop{\underline{\in}} n \\
& \iff m < n \lor m = n.
\end{align*}
\begin{definition}
\lean*{Init/Prelude}{Nat.lt}
\end{definition}
\section{\defined{Pair Set}}% \section{\defined{Pair Set}}%
\hyperlabel{ref:pair-set} \hyperlabel{ref:pair-set}
@ -7136,6 +7157,87 @@ Show that $<_L$ is a linear ordering on $A \times B$.
\end{proof} \end{proof}
\section{Ordering on \texorpdfstring{$\omega$}{Natural Numbers}}%
\hyperlabel{sec:ordering-natural-numbers}
\subsection{\unverified{Ordering on Successor}}%
\hyperlabel{sub:ordering-successor}
\begin{lemma}[8]
Let $m, n \in \omega$.
Then $m < n^+ \iff m \leq n$.
\end{lemma}
\begin{proof}
Let $m, n \in \omega$.
By \nameref{ref:ordering-natural-numbers},
\begin{align*}
m < n^+
& \iff m \in n^+ \\
& \iff m \in n \cup \{n\} & \textref{ref:successor} \\
& \iff m \in n \lor m \in \{n\} \\
& \iff m \in n \lor m = n \\
& \iff m \leq n.
\end{align*}
\end{proof}
\subsection{\sorry{Lemma 4L}}%
\hyperlabel{sub:lemma-4l}
\begin{lemma}[4L]
\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}
\end{lemma}
\begin{proof}
TODO
\end{proof}
\subsection{\sorry{%
Trichotomy Law for \texorpdfstring{$\omega$}{Natural Numbers}}}%
\hyperlabel{sub:trichotomy-law-natrual-numbers}
\begin{theorem}
For any natural numbers $m$ and $n$, exactly one of the three conditions
$$m \in n, \quad m = n, \quad n \in m$$ holds.
\end{theorem}
\begin{proof}
TODO
\end{proof}
\subsection{\sorry{%
Linear Ordering on \texorpdfstring{$\omega$}{Natural Numbers}}}%
\hyperlabel{sub:linear-ordering-natural-numbers}
\begin{theorem}
Relation
$$\in_\omega = \{\pair{m, n} \in \omega \times \omega \mid m \in n\}$$
is a linear ordering on $\omega$.
\end{theorem}
\begin{proof}
TODO
\end{proof}
\section{Exercises 4}% \section{Exercises 4}%
\hyperlabel{sec:exercises-4} \hyperlabel{sec:exercises-4}
@ -7550,14 +7652,57 @@ Complete the proof of \nameref{sub:theorem-4k-5}.
\end{proof} \end{proof}
\subsection{\sorry{Exercise 4.17}}% \subsection{\verified{Exercise 4.17}}%
\hyperlabel{sub:exercise-4.17} \hyperlabel{sub:exercise-4.17}
Prove that $m^{n+p} = m^n \cdot m^p$. Prove that $m^{n+p} = m^n \cdot m^p$.
\begin{proof} \begin{proof}
TODO \lean{Data/Nat/Lemmas}{Nat.pow\_add}
Let $m$ and $n$ be \nameref{ref:natural-number}s and define
\begin{equation}
\hyperlabel{sub:exercise-4.17-eq1}
S = \{p \in \omega \mid m^{n+p} = m^n \cdot m^p\}.
\end{equation}
We prove that (i) $0 \in S$ and (ii) if $p \in S$ then $p^+ \in S$.
Afterwards we show that (iii) our theorem holds.
\paragraph{(i)}%
\hyperlabel{par:exercise-4.17-i}
Consider $m^{n+0}$:
\begin{align*}
m^{n+0}
& = m^n & \textref{sub:theorem-4i} \\
& = m^n \cdot 1 & \textref{sub:right-multiplicative-identity} \\
& = m^n \cdot m^0. & \textref{ref:exponentiation}
\end{align*}
Thus $0 \in S$.
\paragraph{(ii)}%
\hyperlabel{par:exercise-4.17-ii}
Suppose $p \in S$.
Now consider $m^{n+p^+}$:
\begin{align*}
m^{n+p^+}
& = m^{(n + p)^+} & \textref{sub:theorem-4i} \\
& = E_m(n + p) \cdot m & \textref{ref:exponentiation} \\
& = m^n \cdot m^p \cdot m & \eqref{sub:exercise-4.17-eq1} \\
& = m^n \cdot (m^p \cdot m) & \textref{sub:theorem-4k-4} \\
& = m^n \cdot m^{p^+}. & \textref{ref:exponentiation}
\end{align*}
Thus $p^+ \in S$.
\paragraph{(iii)}%
By \nameref{par:exercise-4.17-i} and \nameref{par:exercise-4.17-ii},
$S \subseteq \omega$ is an \nameref{ref:inductive-set}.
By \nameref{sub:theorem-4b}, $S = \omega$.
Thus for all $m, n, p \in \omega$, it follows that
$m^{n+p} = m^n \cdot m^p$.
\end{proof} \end{proof}