Enderton. Prove out addition/multiplication properties.

finite-set-exercises
Joshua Potter 2023-08-02 13:47:43 -06:00
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@ -3057,6 +3057,7 @@ If not, then under what conditions does equality hold?
\subsection{\verified{Lemma 1}}%
\hyperlabel{sub:lemma-1}
\hyperlabel{sub:one-to-one-inverse}
\begin{lemma}[1]
@ -3125,8 +3126,8 @@ If not, then under what conditions does equality hold?
{Enderton.Set.Chapter\_3.theorem\_3g\_ii}
Suppose $F$ is a one-to-one \nameref{ref:function}.
Then \nameref{sub:lemma-1} indicates $F^{-1}$ is a one-to-one function with
domain $\ran{F}$ and range $\dom{F}$.
Then \nameref{sub:one-to-one-inverse} indicates $F^{-1}$ is a one-to-one
function with domain $\ran{F}$ and range $\dom{F}$.
For all $x \in \dom{F}$, $\pair{x, F(x)} \in F$.
Then $\pair{F(x), x} \in F^{-1}$.
@ -3275,8 +3276,8 @@ If not, then under what conditions does equality hold?
F^{-1}(y) & \text{if } y \in \ran{F} \\
a & \text{otherwise}.
\end{cases}$$
$G$ is a function by virtue of \nameref{sub:lemma-1} and choice of mapping
for all values $y \not\in \ran{F}$.
$G$ is a function by virtue of \nameref{sub:one-to-one-inverse} and choice
of mapping for all values $y \not\in \ran{F}$.
Furthermore, for all $x \in A$, $F(x) \in \ran{F}$.
Thus $(G \circ F)(x) = G(F(x)) = F^{-1}(F(x)) = x$ by
\nameref{sub:theorem-3g}.
@ -6469,46 +6470,78 @@ Show that $<_L$ is a linear ordering on $A \times B$.
\section{Arithmetic}%
\hyperlabel{sec:arithmetic}
\subsection{\sorry{Theorem 4I}}
\subsection{\verified{Theorem 4I}}
\hyperlabel{sub:theorem-4i}
\begin{theorem}[4I]
For natural numbers $m$ and $n$,
\begin{align*}
m + 0 & = m, \\
m + n^+ & = (m + n)^+.
\end{align*}
\begin{align}
m + 0 & = m, \hyperlabel{sub:theorem-4i-eq1} \\
m + n^+ & = (m + n)^+. \hyperlabel{sub:theorem-4i-eq2}
\end{align}
\end{theorem}
\begin{proof}
TODO
\statementpadding
\lean*{Init/Data/Nat/Basic}{Nat.add\_zero}
\lean{Init/Prelude}{Nat.add}
\paragraph{\eqref{sub:theorem-4i-eq1}}%
Let $m$ be a \nameref{ref:natural-number}.
By definition of \nameref{ref:addition}, $m + 0 = A_m(0)$.
By definition of $A_m$, $A_m(0) = m$.
Thus $m + 0 = m$.
\paragraph{\eqref{sub:theorem-4i-eq2}}%
Let $m$ and $n$ be natural numbers.
By definition of \nameref{ref:addition},
$$m + n^+ = A_m(n^+) = A_m(n)^+ = (m + n)^+.$$
\end{proof}
\subsection{\sorry{Theorem 4J}}
\subsection{\verified{Theorem 4J}}
\hyperlabel{sub:theorem-4j}
\begin{theorem}[4J]
For natural numbers $m$ and $n$,
\begin{align*}
m \cdot 0 & = 0, \\
m \cdot n^+ & = m \cdot n + m.
\end{align*}
\begin{align}
m \cdot 0 & = 0, \hyperlabel{sub:theorem-4j-eq1} \\
m \cdot n^+ & = m \cdot n + m. \hyperlabel{sub:theorem-4j-eq2}
\end{align}
\end{theorem}
\begin{proof}
TODO
\statementpadding
\lean*{Init/Data/Nat/Basic}{Nat.mul\_zero}
\lean{Init/Prelude}{Nat.mul}
\paragraph{\eqref{sub:theorem-4j-eq1}}%
Let $m$ be a \nameref{ref:natural-number}.
By definition of \nameref{ref:multiplication}, $$m \cdot 0 = M_m(0) = 0.$$
\paragraph{\eqref{sub:theorem-4j-eq2}}%
Let $m$ and $n$ be natural numbers.
By definition of \nameref{ref:multiplication},
$$m \cdot n^+ = M_m(n^+) = M_m(n) + m = m \cdot n + m.$$
\end{proof}
\subsection{\pending{Lemma 2}}%
\label{sub:lemma-2}
\subsection{\verified{Left Additive Identity}}%
\hyperlabel{sub:left-additive-identity}
\begin{lemma}[2]
@ -6519,35 +6552,39 @@ Show that $<_L$ is a linear ordering on $A \times B$.
\begin{proof}
Let $S = \{n \in \omega \mid A_0(n) = n\}$.
\lean{Init/Data/Nat/Basic}{Nat.zero\_add}
Let $S = \{n \in \omega \mid 0 + n = n\}$.
We prove that (i) $0 \in S$ and (ii) if $n \in S$ then $n^+ \in S$.
Afterwards we show that (iii) our theorem holds.
\paragraph{(i)}%
\hyperlabel{par:lemma-2-i}
\hyperlabel{par:left-additive-identity-i}
By definition of \nameref{ref:addition}, $A_0(0) = 0$.
By \nameref{sub:theorem-4i}, $0 + 0 = 0$.
Thus $0 \in S$.
\paragraph{(ii)}%
\hyperlabel{par:lemma-2-ii}
\hyperlabel{par:left-additive-identity-ii}
Suppose $n \in S$.
By definition of addition, $A_0(n^+) = A_0(n)^+$.
Since $n \in S$, $A_0(n) = n$ which in turn implies that $A_0(n)^+ = n^+$.
By \nameref{sub:theorem-4i}, $0 + n^+ = (0 + n)^+$.
Since $n \in S$, $0 + n = n$ which in turn implies that $(0 + n)^+ = n^+$.
Thus $n^+ \in S$.
\paragraph{(iii)}%
By \nameref{par:lemma-2-i} and \nameref{par:lemma-2-ii}, $S$ is an
By \nameref{par:left-additive-identity-i} and
\nameref{par:left-additive-identity-ii}, $S$ is an
\nameref{ref:inductive-set}.
Hence \nameref{sub:theorem-4b} implies $S = \omega$.
Thus for all $n \in \omega$, $A_0(n) = n$.
Thus for all $n \in \omega$, $0 + n = n$.
\end{proof}
\subsection{\pending{Lemma 3}}%
\label{sub:lemma-3}
\subsection{\verified{Lemma 3}}%
\hyperlabel{sub:lemma-3}
\hyperlabel{sub:succ-add-eq-add-succ}
\begin{lemma}[3]
@ -6558,103 +6595,41 @@ Show that $<_L$ is a linear ordering on $A \times B$.
\begin{proof}
\lean{Std/Data/Nat/Lemmas}{Nat.succ\_add\_eq\_succ\_add}
Let $m \in \omega$ and define
$$S = \{n \in \omega \mid A_{m^+}(n) = A_m(n^+)\}.$$
$$S = \{n \in \omega \mid m^+ + n = m + n^+\}.$$
We prove that (i) $0 \in S$ and (ii) if $n \in S$ then $n^+ \in S$.
Afterwards we show that (iii) our theorem holds.
\paragraph{(i)}%
\label{par:lemma-3-i}
\hyperlabel{par:lemma-3-i}
By definition of \nameref{ref:addition}, $A_{m^+}(0) = m^+$.
Likewise, $$A_m(0^+) = A_m(0)^+ = m^+.$$
By \nameref{sub:theorem-4i}, $m^+ + 0 = m^+$.
Likewise, $m + 0^+ = (m + 0)^+ = m^+$.
Thus $0 \in S$.
\paragraph{(ii)}%
\label{par:lemma-3-ii}
\hyperlabel{par:lemma-3-ii}
Suppose $n \in S$.
By definition of addition, $A_{m^+}(n^+) = \left[A_{m^+}(n)\right]^+$.
Since $n \in S$, $A_{m^+}(n) = A_m(n^+)$ which in turn implies that
$$\left[A_{m^+}(n)\right]^+ = \left[A_m(n^+)\right]^+ = A_m(n^{++}).$$
By \nameref{sub:theorem-4i}, $m^+ + n^+ = (m^+ + n)^+$.
Since $n \in S$, $m^+ + n = m + n^+$.
Therefore $(m^+ + n)^+ = (m + n^+)^+ = m + n^{++}$.
Thus $n^+ \in S$.
\paragraph{(iii)}%
By \nameref{par:lemma-3-i} and \nameref{par:lemma-3-ii}, $S$ is inductive.
Hence \nameref{sub:theorem-4b} implies $S = \omega$.
Thus for all $n \in \omega$, $A_{m^+}(n) = A_m(n^+)$.
Thus for all $n \in \omega$, $m^+ + n = m + n^+$.
\end{proof}
\subsection{\pending{Lemma 4}}%
\hyperlabel{sub:lemma-4}
\begin{lemma}[4]
For all $n \in \omega$, $M_0(n) = 0$.
In other words, $$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.$$
\end{lemma}
\begin{proof}
TODO
\end{proof}
\subsection{\pending{Theorem 4k (1)}}%
\subsection{\verified{Theorem 4K-1}}%
\label{sub:theorem-4k-1}
\begin{theorem}[4K.1]
\begin{theorem}[4K-1]
Associative law for addition.
For $m, n, p \in \omega$, $$m + (n + p) = (m + n) + p.$$
@ -6663,6 +6638,8 @@ Show that $<_L$ is a linear ordering on $A \times B$.
\begin{proof}
\lean{Mathlib/Algebra/Group/Defs}{add\_assoc}
Fix $n, p \in \omega$ and define
\begin{equation}
\hyperlabel{sub:theorem-4k-1-eq1}
@ -6674,28 +6651,24 @@ Show that $<_L$ is a linear ordering on $A \times B$.
\paragraph{(i)}%
\hyperlabel{par:theorem-4k-1-i}
By definition of \nameref{ref:addition},
\begin{align*}
0 + (n + p)
& = n + p & \textref{sub:lemma-2} \\
& = (0 + n) + p. & \textref{sub:lemma-2} \\
\end{align*}
By \nameref{sub:left-additive-identity},
$$0 + (n + p) = n + p = (0 + n) + p.$$
Thus $0 \in S$.
\paragraph{(ii)}%
\hyperlabel{par:theorem-4k-1-ii}
Suppose $m \in S$.
By definition of \nameref{ref:addition},
Then
\begin{align*}
m^+ + (n + p)
& = m + (n + p)^+ & \textref{sub:lemma-3} \\
& = m + (n + p)^+ & \textref{sub:succ-add-eq-add-succ} \\
& = (m + (n + p))^+ & \textref{sub:theorem-4i} \\
& = ((m + n) + p)^+ & \eqref{sub:theorem-4k-1-eq1} \\
& = (m + n) + p^+ & \textref{sub:theorem-4i} \\
& = (m + n)^+ + p & \textref{sub:lemma-3} \\
& = (m + n)^+ + p & \textref{sub:succ-add-eq-add-succ} \\
& = (m + n^+) + p & \textref{sub:theorem-4i} \\
& = (m^+ + n) + p. & \textref{sub:lemma-3}
& = (m^+ + n) + p. & \textref{sub:succ-add-eq-add-succ}
\end{align*}
Thus $m^+ \in S$.
@ -6709,10 +6682,10 @@ Show that $<_L$ is a linear ordering on $A \times B$.
\end{proof}
\subsection{\pending{Theorem 4k (2)}}%
\subsection{\verified{Theorem 4K-2}}%
\label{sub:theorem-4k-2}
\begin{theorem}[4K.2]
\begin{theorem}[4K-2]
Commutative law for addition.
For $m, n \in \omega$, $$m + n = n + m.$$
@ -6721,6 +6694,8 @@ Show that $<_L$ is a linear ordering on $A \times B$.
\begin{proof}
\lean{Mathlib/Algebra/Group/Defs}{add\_comm}
Fix $n \in \omega$ and define
\begin{equation}
\hyperlabel{sub:theorem-4k-2-eq1}
@ -6732,25 +6707,21 @@ Show that $<_L$ is a linear ordering on $A \times B$.
\paragraph{(i)}%
\hyperlabel{par:theorem-4k-2-i}
By definition of \nameref{ref:addition},
\begin{align*}
0 + n
& = n & \textref{sub:lemma-2} \\
& = n + 0. \\
\end{align*}
By definition of \nameref{ref:addition} and
\nameref{sub:left-additive-identity}, $$0 + n = n = n + 0.$$
Thus $0 \in S$.
\paragraph{(ii)}%
\hyperlabel{par:theorem-4k-2-ii}
Suppose $m \in S$.
By definition of \nameref{ref:addition},
Then
\begin{align*}
m^+ + n
& = m + n^+ & \textref{sub:lemma-3} \\
& = (m + n)^+ \\
& = m + n^+ & \textref{sub:succ-add-eq-add-succ} \\
& = (m + n)^+ & \textref{sub:theorem-4i} \\
& = (n + m)^+ & \eqref{sub:theorem-4k-2-eq1} \\
& = n + m^+.
& = n + m^+. & \textref{sub:theorem-4i}
\end{align*}
Thus $m^+ \in S$.
@ -6763,10 +6734,116 @@ Show that $<_L$ is a linear ordering on $A \times B$.
\end{proof}
\subsection{\pending{Theorem 4k (3)}}%
\label{sub:theorem-4k-3}
\subsection{\verified{Zero Multiplicand}}%
\hyperlabel{sub:zero-multiplicand}
\begin{theorem}[4K.3]
\begin{lemma}[4]
For all $n \in \omega$, $M_0(n) = 0$.
In other words, $$0 \cdot n = 0.$$
\end{lemma}
\begin{proof}
\lean{Init/Data/Nat/Basic}{Nat.zero\_mul}
Define
\begin{equation}
\hyperlabel{sub:zero-multiplicand-eq1}
S = \{n \in \omega \mid 0 \cdot n = 0\}.
\end{equation}
We prove that (i) $0 \in S$ and (ii) that if $n \in S$, then $n^+ \in S$.
Afterwards we show that (iii) our theorem holds.
\paragraph{(i)}%
\hyperlabel{par:zero-multiplicand-i}
By \nameref{sub:theorem-4j}, $0 \cdot 0 = 0$.
Thus $0 \in S$.
\paragraph{(ii)}%
\hyperlabel{par:zero-multiplicand-ii}
Suppose $n \in S$.
Then
\begin{align*}
0 \cdot n^+
& = 0 \cdot n + 0 & \textref{sub:theorem-4j} \\
& = 0 + 0 & \eqref{sub:zero-multiplicand-eq1} \\
& = 0. & \textref{ref:addition}
\end{align*}
Thus $n^+ \in S$.
\paragraph{(iii)}%
By \nameref{par:zero-multiplicand-i} and \nameref{par:zero-multiplicand-ii},
$S$ is an \nameref{ref:inductive-set}.
Hence \nameref{sub:theorem-4b} implies $S = \omega$.
Thus for all $n \in \omega$, $0 \cdot n = 0$.
\end{proof}
\subsection{\verified{Successor Distribution}}%
\hyperlabel{sub:successor-distribution}
\begin{lemma}[5]
For all $m, n \in \omega$, $M_{m^+}(n) = M_m(n) + n$.
In other words, $$m^+ \cdot n = m \cdot n + n.$$
\end{lemma}
\begin{proof}
\lean{Init/Data/Nat/Basic}{Nat.succ\_mul}
Let $m \in \omega$ and define
\begin{equation}
\hyperlabel{sub:successor-distribution-eq1}
S = \{n \in \omega \mid m^+ \cdot n = m \cdot n + n\}.
\end{equation}
We prove that (i) $0 \in S$ and (ii) that if $n \in S$, then $n^+ \in S$.
Afterwards we show that (iii) our theorem holds.
\paragraph{(i)}%
\hyperlabel{par:successor-distribution-i}
By \nameref{sub:theorem-4j}, $m^+ \cdot 0 = 0$.
Likewise, by \nameref{sub:theorem-4i}, $m \cdot 0 + 0 = 0$.
Thus $0 \in S$.
\paragraph{(ii)}%
\hyperlabel{par:successor-distribution-ii}
Suppose $n \in S$.
Then
\begin{align*}
m^+ \cdot n^+
& = m^+ \cdot n + m^+ & \textref{sub:theorem-4j} \\
& = (m \cdot n + n) + m^+ & \eqref{sub:successor-distribution-eq1} \\
& = m \cdot n + (n + m^+) & \textref{sub:theorem-4k-1} \\
& = m \cdot n + (n^+ + m) & \textref{sub:succ-add-eq-add-succ} \\
& = m \cdot n + (m + n^+) & \textref{sub:theorem-4k-2} \\
& = (m \cdot n + m) + n^+ & \textref{sub:theorem-4k-1} \\
& = m \cdot n^+ + n^+. & \textref{sub:theorem-4j}
\end{align*}
Thus $n^+ \in S$.
\paragraph{(iii)}%
By \nameref{par:successor-distribution-i} and
\nameref{par:successor-distribution-ii}, $S$ is an
\nameref{ref:inductive-set}.
By \nameref{sub:theorem-4b}, $S = \omega$.
Thus for all $m, n \in \omega$, $m^+ \cdot n = m \cdot n + n$.
\end{proof}
\subsection{\verified{Theorem 4K-3}}
\hyperlabel{sub:theorem-4k-3}
\begin{theorem}[4K-3]
Distributive law.
For $m, n, p \in \omega$, $$m \cdot (n + p) = m \cdot n + m \cdot p.$$
@ -6775,10 +6852,12 @@ Show that $<_L$ is a linear ordering on $A \times B$.
\begin{proof}
\lean{Init/Data/Nat/Basic}{Nat.left\_distrib}
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\}.
S = \{m \in \omega \mid 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.
@ -6789,9 +6868,9 @@ Show that $<_L$ is a linear ordering on $A \times B$.
By definition of \nameref{ref:multiplication} and \nameref{ref:addition},
\begin{align*}
0 \cdot (n + p)
& = 0 & \textref{sub:lemma-4} \\
& = 0 & \textref{sub:zero-multiplicand} \\
& = 0 + 0 & \textref{ref:addition} \\
& = 0 \cdot n + 0 \cdot p. & \textref{sub:lemma-4}
& = 0 \cdot n + 0 \cdot p. & \textref{sub:zero-multiplicand}
\end{align*}
Thus $0 \in S$.
@ -6802,17 +6881,17 @@ Show that $<_L$ is a linear ordering on $A \times B$.
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:successor-distribution} \\
& = 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}
& = m^+ \cdot n + m^+ \cdot p. & \textref{sub:successor-distribution}
\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}.
@ -6822,10 +6901,174 @@ Show that $<_L$ is a linear ordering on $A \times B$.
\end{proof}
\subsection{\sorry{Theorem 4k (4)}}%
\label{sub:theorem-4k-4}
\subsection{\verified{Successor Identity}}%
\hyperlabel{sub:successor-identity}
\begin{theorem}[4K.4]
\begin{lemma}[6]
For all $m \in \omega$, $A_m(1) = m^+$.
In other words, $$m + 1 = m^+.$$
\end{lemma}
\begin{proof}
\lean{Init/Prelude}{Nat.succ}
Let
\begin{equation}
\hyperlabel{sub:successor-identity-eq1}
S = \{m \in \omega \mid m + 1 = m^+\}.
\end{equation}
We prove that (i) $0 \in S$ and (ii) if $m \in S$, then $m^+ \in S$.
Afterwards we show that (iii) our theorem holds.
\paragraph{(i)}%
\hyperlabel{par:successor-identity-i}
By \nameref{sub:left-additive-identity}, $0 + 1 = 1$.
By definition of the \nameref{ref:successor},
$0^+ = \emptyset \cup \{\emptyset\} = 1$.
Thus $0 \in S$.
\paragraph{(ii)}%
\hyperlabel{par:successor-identity-ii}
Let $m \in S$.
Then
\begin{align*}
m^+ + 1
& = m + 1^+ & \textref{sub:succ-add-eq-add-succ} \\
& = (m + 1)^+ & \textref{sub:theorem-4i} \\
& = (m^+)^+. & \eqref{sub:successor-identity-eq1}
\end{align*}
Thus $m^+ \in S$.
\paragraph{(iii)}%
By \nameref{par:successor-identity-i} and
\nameref{par:successor-identity-ii}, $S$ is an
\nameref{ref:inductive-set}.
Hence \nameref{sub:theorem-4b} implies $S = \omega$.
Thus for all $m \in \omega$, $m + 1 = m^+$.
\end{proof}
\subsection{\verified{Right Multiplicative Identity}}%
\hyperlabel{sub:right-multiplicative-identity}
\begin{lemma}[7]
For all $m \in \omega$, $M_m(1) = m$.
In other words, $$m \cdot 1 = m.$$
\end{lemma}
\begin{proof}
\lean{Init/Data/Nat/Basic}{mul\_one}
Let
\begin{equation}
\hyperlabel{sub:right-multiplicative-identity-eq1}
S = \{m \in \omega \mid m \cdot 1 = m\}.
\end{equation}
We prove that (i) $0 \in S$ and (ii) if $m \in S$, then $m^+ \in S$.
Afterwards we show that (iii) our theorem holds.
\paragraph{(i)}%
\hyperlabel{par:right-multiplicative-identity-i}
By \nameref{sub:zero-multiplicand}, $0 \cdot 1 = 0$.
Thus $0 \in S$.
\paragraph{(ii)}%
\hyperlabel{par:right-multiplicative-identity-ii}
Suppose $m \in S$.
Then
\begin{align*}
m^+ \cdot 1
& = m \cdot 1 + 1 & \textref{sub:successor-distribution} \\
& = m + 1 & \eqref{sub:right-multiplicative-identity-eq1} \\
& = m^+. & \textref{sub:successor-identity}
\end{align*}
Thus $m^+ \in S$.
\paragraph{(iii)}%
By \nameref{par:right-multiplicative-identity-i} and
\nameref{par:right-multiplicative-identity-ii}, $S$
is an \nameref{ref:inductive-set}.
By \nameref{sub:theorem-4b}, $S = \omega$.
Thus for all $m \in \omega$, $m \cdot 1 = m$.
\end{proof}
\subsection{\verified{Theorem 4K-5}}
\hyperlabel{sub:theorem-4k-5}
\begin{theorem}[4K-5]
Commutative law for multiplication.
For $m, n \in \omega$, $$m \cdot n = n \cdot m.$$
\end{theorem}
\begin{note}
We prove commutativity before associativity, though Enderton orders these
two properties in the opposite direction.
\end{note}
\begin{proof}
\lean{Mathlib/Algebra/Group/Defs}{mul\_comm}
Fix $n \in \omega$ and define
\begin{equation}
\hyperlabel{sub:theorem-4k-5-eq1}
S = \{m \in \omega \mid m \cdot n = n \cdot m\}.
\end{equation}
We show that (i) $0 \in S$ and (ii) if $m \in S$ then $m^+ \in S$.
Afterward we show that (iii) the commutative law for multiplication holds.
\paragraph{(i)}%
\hyperlabel{par:theorem-4k-5-i}
By \nameref{sub:theorem-4j} and \nameref{sub:zero-multiplicand},
$$0 \cdot n = 0 = n \cdot 0.$$
Thus $0 \in S$.
\paragraph{(ii)}%
\hyperlabel{par:theorem-4k-5-ii}
Suppose $m \in S$.
Then
\begin{align*}
m^+ \cdot n
& = m \cdot n + n & \textref{sub:successor-distribution} \\
& = n \cdot m + n & \eqref{sub:theorem-4k-5-eq1} \\
& = n \cdot m + n \cdot 1
& \textref{sub:right-multiplicative-identity} \\
& = n \cdot (m + 1) & \textref{sub:theorem-4k-3} \\
& = n \cdot m^+. & \textref{sub:successor-identity}
\end{align*}
Thus $m^+ \in S$.
\paragraph{(iii)}%
\hyperlabel{par:theorem-4k-5-iii}
By \nameref{par:theorem-4k-5-i} and \nameref{par:theorem-4k-5-ii}, $S$
is an \nameref{ref:inductive-set}.
By \nameref{sub:theorem-4b}, $S = \omega$.
Thus for all $m, n \in \omega$, $m \cdot n = n \cdot m$.
\end{proof}
\subsection{\verified{Theorem 4K-4}}%
\hyperlabel{sub:theorem-4k-4}
\begin{theorem}[4K-4]
Associative law for multiplication.
For $m, n, p \in \omega$, $$m \cdot (n \cdot p) = (m \cdot n) \cdot p.$$
@ -6834,23 +7077,46 @@ Show that $<_L$ is a linear ordering on $A \times B$.
\begin{proof}
TODO
\lean{Mathlib/Algebra/Group/Defs}{mul\_assoc}
\end{proof}
Fix $m, n \in \omega$ and define
\begin{equation}
\hyperlabel{sub:theorem-4k-4-eq1}
S = \{p \in \omega \mid m \cdot (n \cdot p) = (m \cdot n) \cdot p\}.
\end{equation}
We show that (i) $0 \in S$ and (ii) if $p \in S$ then $p^+ \in S$.
Afterward we show that (iii) the associative law for multiplication holds.
\subsection{\sorry{Theorem 4K (5)}}
\hyperlabel{sub:theorem-4k}
\paragraph{(i)}%
\hyperlabel{par:theorem-4k-4-i}
\begin{theorem}[4K.5]
By \nameref{sub:theorem-4j},
$$m \cdot (n \cdot 0) = 0 = (m \cdot n) \cdot 0.$$
Thus $0 \in S$.
Commutative law for multiplication.
For $m, n \in \omega$, $$m \cdot n = n \cdot m.$$
\paragraph{(ii)}%
\hyperlabel{par:theorem-4k-4-ii}
\end{theorem}
Suppose $p \in S$.
Then
\begin{align*}
m \cdot (n \cdot p^+)
& = m \cdot (n \cdot p + n) & \textref{sub:theorem-4j} \\
& = m \cdot (n \cdot p) + m \cdot n & \textref{sub:theorem-4k-3} \\
& = (m \cdot n) \cdot p + m \cdot n & \eqref{sub:theorem-4k-4-eq1} \\
& = p \cdot (m \cdot n) + m \cdot n & \textref{sub:theorem-4k-5} \\
& = p^+ \cdot (m \cdot n) & \textref{sub:successor-distribution} \\
& = (m \cdot n) \cdot p^+ & \textref{sub:theorem-4k-5}
\end{align*}
Thus $p^+ \in S$.
\begin{proof}
\paragraph{(iii)}%
TODO
By \nameref{par:theorem-4k-4-i} and \nameref{par:theorem-4k-4-ii}, $S$
is an \nameref{ref:inductive-set}.
By \nameref{sub:theorem-4b}, $S = \omega$.
Thus for all $m, n, p \in \omega$,
$m \cdot (n \cdot p) = (m \cdot n) \cdot p$.
\end{proof}
@ -7249,7 +7515,7 @@ Show that each natural number is either even or odd, but never both.
\subsection{\sorry{Exercise 4.15}}%
\hyperlabel{sub:exercise-4.15}
Complete the proof of part (1) of \nameref{sub:theorem-4k}.
Complete the proof of \nameref{sub:theorem-4k-1}.
\begin{proof}
@ -7260,7 +7526,7 @@ Complete the proof of part (1) of \nameref{sub:theorem-4k}.
\subsection{\sorry{Exercise 4.16}}%
\hyperlabel{sub:exercise-4.16}
Complete the proof of part (5) of \nameref{sub:theorem-4k}.
Complete the proof of \nameref{sub:theorem-4k-5}.
\begin{proof}