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