From 2d93d8d7683c417a5e9b2fd28d02cf415c664f27 Mon Sep 17 00:00:00 2001 From: Joshua Potter Date: Wed, 2 Aug 2023 11:02:33 -0600 Subject: [PATCH] Enderton. Add various lemmas/begin algebraic proofs. --- Bookshelf/Enderton/Set.tex | 365 +++++++++++++++++++++++++++++++++++-- 1 file changed, 345 insertions(+), 20 deletions(-) diff --git a/Bookshelf/Enderton/Set.tex b/Bookshelf/Enderton/Set.tex index 2fc343a..535891d 100644 --- a/Bookshelf/Enderton/Set.tex +++ b/Bookshelf/Enderton/Set.tex @@ -370,14 +370,14 @@ A \textbf{linear ordering} on $A$ (also called a \textbf{total ordering} on $A$) \end{definition} \section{\defined{Multiplication}}% -\hyperlabel{sec:multiplication} +\hyperlabel{ref:multiplication} For each $m \in \omega$, there exists (by the \nameref{sub:recursion-theorem-natural-numbers}) a unique \nameref{ref:function} $M_m \colon \omega \rightarrow \omega$ for which \begin{align*} M_m(0) & = 0, \\ - M_m(n^+) = M_m(n) + m. + M_m(n^+) & = M_m(n) + m. \end{align*} \textbf{Multiplication} ($\cdot$) is the \nameref{ref:binary-operation} on $\omega$ such that for any $m$ and $n$ in $\omega$, $$m \cdot n = M_m(n).$$ @@ -2840,11 +2840,11 @@ If not, then under what conditions does equality hold? \subsection{\verified{Lemma 3B}}% \hyperlabel{sub:lemma-3b} -\begin{theorem}[3B] +\begin{lemma}[3B] If $x \in C$ and $y \in C$, then $\pair{x, y} \in \powerset{\powerset{C}}$. -\end{theorem} +\end{lemma} \begin{proof} @@ -3058,7 +3058,12 @@ If not, then under what conditions does equality hold? \subsection{\verified{Lemma 1}}% \hyperlabel{sub:lemma-1} -For any one-to-one function $F$, $F^{-1}$ is also one-to-one. +\begin{lemma}[1] + + For any one-to-one function $F$, $F^{-1}$ is also one-to-one. + +\end{lemma} + \begin{proof} @@ -6502,24 +6507,344 @@ Show that $<_L$ is a linear ordering on $A \times B$. \end{proof} -\subsection{\sorry{Theorem 4K}} +\subsection{\pending{Lemma 2}}% +\label{sub:lemma-2} + +\begin{lemma}[2] + + For all $n \in \omega$, $A_0(n) = n$. + In other words, $$0 + n = n.$$ + +\end{lemma} + +\begin{proof} + + Let $S = \{n \in \omega \mid A_0(n) = n\}$. + 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:lemma-2-i} + + By definition of \nameref{ref:addition}, $A_0(0) = 0$. + Thus $0 \in S$. + + \paragraph{(ii)}% + \hyperlabel{par:lemma-2-ii} + + Suppose $n \in S$. + By definition of addition, $A_0(n^+) = A_0(n)^+$. + Since $n \in S$, $A_0(n) = n$ which in turn implies that $A_0(n)^+ = n^+$. + Thus $n^+ \in S$. + + \paragraph{(iii)}% + + By \nameref{par:lemma-2-i} and \nameref{par:lemma-2-ii}, $S$ is an + \nameref{ref:inductive-set}. + Hence \nameref{sub:theorem-4b} implies $S = \omega$. + Thus for all $n \in \omega$, $A_0(n) = n$. + +\end{proof} + +\subsection{\pending{Lemma 3}}% +\label{sub:lemma-3} + +\begin{lemma}[3] + + For all $m, n \in \omega$, $A_{m^+}(n) = A_m(n^+)$. + In other words, $$m^+ + n = m + n^+.$$ + +\end{lemma} + +\begin{proof} + + Let $m \in \omega$ and define + $$S = \{n \in \omega \mid A_{m^+}(n) = A_m(n^+)\}.$$ + 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)}% + \label{par:lemma-3-i} + + By definition of \nameref{ref:addition}, $A_{m^+}(0) = m^+$. + Likewise, $$A_m(0^+) = A_m(0)^+ = m^+.$$ + Thus $0 \in S$. + + \paragraph{(ii)}% + \label{par:lemma-3-ii} + + Suppose $n \in S$. + By definition of addition, $A_{m^+}(n^+) = \left[A_{m^+}(n)\right]^+$. + Since $n \in S$, $A_{m^+}(n) = A_m(n^+)$ which in turn implies that + $$\left[A_{m^+}(n)\right]^+ = \left[A_m(n^+)\right]^+ = A_m(n^{++}).$$ + Thus $n^+ \in S$. + + \paragraph{(iii)}% + + By \nameref{par:lemma-3-i} and \nameref{par:lemma-3-ii}, $S$ is inductive. + Hence \nameref{sub:theorem-4b} implies $S = \omega$. + Thus for all $n \in \omega$, $A_{m^+}(n) = A_m(n^+)$. + +\end{proof} + +\subsection{\pending{Lemma 4}}% +\hyperlabel{sub:lemma-4} + +\begin{lemma}[4] + + For all $n \in \omega$, $M_0(n) = 0$. + In other words, $$0 \cdot n = 0.$$ + +\end{lemma} + +\begin{proof} + + Define + \begin{equation} + \hyperlabel{sub:lemma-4-eq1} + S = \{n \in \omega \mid M_0(n) = 0\}. + \end{equation} + We prove that (i) $0 \in S$ and (ii) that if $n \in S$, then $n^+ \in S$. + Afterwards we show that (iii) our theorem holds. + + \paragraph{(i)}% + \hyperlabel{par:lemma-4-i} + + By definition of \nameref{ref:multiplication}, $M_0(0) = 0$. + Thus $0 \in S$. + + \paragraph{(ii)}% + \hyperlabel{par:lemma-4-ii} + + Suppose $n \in S$. + Then, by definition of \nameref{ref:multiplication}, + \begin{align*} + M_0(n^+) + & = M_0(n) + 0 \\ + & = 0 + 0 & \eqref{sub:lemma-4-eq1} \\ + & = 0. & \textref{ref:addition} + \end{align*} + Thus $n^+ \in S$. + + \paragraph{(iii)}% + + By \nameref{par:lemma-4-i} and \nameref{par:lemma-4-ii}, $S$ is an + \nameref{ref:inductive-set}. + Hence \nameref{sub:theorem-4b} implies $S = \omega$. + Thus for all $n \in \omega$, $M_0(n) = 0$. + +\end{proof} + +\subsection{\sorry{Lemma 5}}% +\hyperlabel{sub:lemma-5} + +\begin{lemma}[5] + + For all $m, n \in \omega$, $M_{m^+}(n) = M_m(n) + n$. + In other words, $$m^+ \cdot n = m \cdot n + n.$$ + +\end{lemma} + +\begin{proof} + + TODO + +\end{proof} + +\subsection{\pending{Theorem 4k (1)}}% +\label{sub:theorem-4k-1} + +\begin{theorem}[4K.1] + + Associative law for addition. + For $m, n, p \in \omega$, $$m + (n + p) = (m + n) + p.$$ + +\end{theorem} + +\begin{proof} + + Fix $n, p \in \omega$ and define + \begin{equation} + \hyperlabel{sub:theorem-4k-1-eq1} + S = \{m \in \omega \mid m + (n + p) = (m + n) + p\}. + \end{equation} + We show that (i) $0 \in S$ and (ii) if $m \in S$ then $m^+ \in S$. + Afterward we show that (iii) the associative law for addition holds. + + \paragraph{(i)}% + \hyperlabel{par:theorem-4k-1-i} + + By definition of \nameref{ref:addition}, + \begin{align*} + 0 + (n + p) + & = n + p & \textref{sub:lemma-2} \\ + & = (0 + n) + p. & \textref{sub:lemma-2} \\ + \end{align*} + Thus $0 \in S$. + + \paragraph{(ii)}% + \hyperlabel{par:theorem-4k-1-ii} + + Suppose $m \in S$. + By definition of \nameref{ref:addition}, + \begin{align*} + m^+ + (n + p) + & = m + (n + p)^+ & \textref{sub:lemma-3} \\ + & = (m + (n + p))^+ & \textref{sub:theorem-4i} \\ + & = ((m + n) + p)^+ & \eqref{sub:theorem-4k-1-eq1} \\ + & = (m + n) + p^+ & \textref{sub:theorem-4i} \\ + & = (m + n)^+ + p & \textref{sub:lemma-3} \\ + & = (m + n^+) + p & \textref{sub:theorem-4i} \\ + & = (m^+ + n) + p. & \textref{sub:lemma-3} + \end{align*} + Thus $m^+ \in S$. + + \paragraph{(iii)}% + \hyperlabel{par:theorem-4k-1-iii} + + By \nameref{par:theorem-4k-1-i} and \nameref{par:theorem-4k-1-ii}, $S$ is an + \nameref{ref:inductive-set}. + By \nameref{sub:theorem-4b}, $S = \omega$. + Thus for all $m, n, p \in \omega$, $m + (n + p) = (m + n) + p$. + +\end{proof} + +\subsection{\pending{Theorem 4k (2)}}% +\label{sub:theorem-4k-2} + +\begin{theorem}[4K.2] + + Commutative law for addition. + For $m, n \in \omega$, $$m + n = n + m.$$ + +\end{theorem} + +\begin{proof} + + Fix $n \in \omega$ and define + \begin{equation} + \hyperlabel{sub:theorem-4k-2-eq1} + S = \{m \in \omega \mid m + n = n + m\}. + \end{equation} + We show that (i) $0 \in S$ and (ii) if $m \in S$ then $m^+ \in S$. + Afterward we show that (iii) the commutative law for addition holds. + + \paragraph{(i)}% + \hyperlabel{par:theorem-4k-2-i} + + By definition of \nameref{ref:addition}, + \begin{align*} + 0 + n + & = n & \textref{sub:lemma-2} \\ + & = n + 0. \\ + \end{align*} + Thus $0 \in S$. + + \paragraph{(ii)}% + \hyperlabel{par:theorem-4k-2-ii} + + Suppose $m \in S$. + By definition of \nameref{ref:addition}, + \begin{align*} + m^+ + n + & = m + n^+ & \textref{sub:lemma-3} \\ + & = (m + n)^+ \\ + & = (n + m)^+ & \eqref{sub:theorem-4k-2-eq1} \\ + & = n + m^+. + \end{align*} + Thus $m^+ \in S$. + + \paragraph{(iii)}% + + By \nameref{par:theorem-4k-2-i} and \nameref{par:theorem-4k-2-ii}, $S$ + is an \nameref{ref:inductive-set}. + By \nameref{sub:theorem-4b}, $S = \omega$. + Thus for all $m, n \in \omega$, $m + n = n + m$. + +\end{proof} + +\subsection{\pending{Theorem 4k (3)}}% +\label{sub:theorem-4k-3} + +\begin{theorem}[4K.3] + + Distributive law. + For $m, n, p \in \omega$, $$m \cdot (n + p) = m \cdot n + m \cdot p.$$ + +\end{theorem} + +\begin{proof} + + Fix $n, p \in \omega$ and define + \begin{equation} + \hyperlabel{sub:theorem-4k-3-eq1} + S = \{m \in \omega m \cdot (n + p) = m \cdot n + m \cdot p\}. + \end{equation} + We show that (i) $0 \in S$ and (ii) if $m \in S$ then $m^+ \in S$. + Afterward we show that (iii) the distributive law holds. + + \paragraph{(i)}% + \hyperlabel{par:theorem-4k-3-i} + + By definition of \nameref{ref:multiplication} and \nameref{ref:addition}, + \begin{align*} + 0 \cdot (n + p) + & = 0 & \textref{sub:lemma-4} \\ + & = 0 + 0 & \textref{ref:addition} \\ + & = 0 \cdot n + 0 \cdot p. & \textref{sub:lemma-4} + \end{align*} + Thus $0 \in S$. + + \paragraph{(ii)}% + \hyperlabel{par:theorem-4k-3-ii} + + Suppose $m \in S$. + By definition of \nameref{ref:multiplication} and \nameref{ref:addition}, + \begin{align*} + m^+ \cdot (n + p) + & = m \cdot (n + p) + (n + p) & \textref{sub:lemma-5} \\ + & = m \cdot (n + p) + n + p & \textref{sub:theorem-4k-1} \\ + & = m \cdot n + m \cdot p + n + p + & \eqref{sub:theorem-4k-3-eq1} \\ + & = m \cdot n + n + m \cdot p + p & \textref{sub:theorem-4k-2} \\ + & = m^+ \cdot n + m^+ \cdot p. & \textref{sub:lemma-5} + \end{align*} + Thus $m^+ \in S$. + + \paragraph{(iii)}% + \hyperlabel{par:theorem-4k-3-iii} + + By \nameref{par:theorem-4k-3-i} and \nameref{par:theorem-4k-3-ii}, $S$ + is an \nameref{ref:inductive-set}. + By \nameref{sub:theorem-4b}, $S = \omega$. + Thus for all $m, n, p \in \omega$, + $m \cdot (n + p) = m \cdot n + m \cdot p$. + +\end{proof} + +\subsection{\sorry{Theorem 4k (4)}}% +\label{sub:theorem-4k-4} + +\begin{theorem}[4K.4] + + Associative law for multiplication. + For $m, n, p \in \omega$, $$m \cdot (n \cdot p) = (m \cdot n) \cdot p.$$ + +\end{theorem} + +\begin{proof} + + TODO + +\end{proof} + +\subsection{\sorry{Theorem 4K (5)}} \hyperlabel{sub:theorem-4k} -\begin{theorem}[4K] +\begin{theorem}[4K.5] - The following identities hold for all natural numbers. - \begin{enumerate} - \item Associative law for addition - $$m + (n + p) = (m + n) + p.$$ - \item Commutative law for addition - $$m + n = n + m.$$ - \item Distributive law - $$m \cdot (n + p) = m \cdot n + m \cdot p.$$ - \item Associative law for multiplication - $$m \cdot (n \cdot p) = (m \cdot n) \cdot p.$$ - \item Commutative law for multiplication - $$m \cdot n = n \cdot m.$$ - \end{enumerate} + Commutative law for multiplication. + For $m, n \in \omega$, $$m \cdot n = n \cdot m.$$ \end{theorem}