Enderton. Prove out addition/multiplication properties.
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@ -3057,6 +3057,7 @@ 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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\hyperlabel{sub:one-to-one-inverse}
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\begin{lemma}[1]
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@ -3125,8 +3126,8 @@ If not, then under what conditions does equality hold?
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{Enderton.Set.Chapter\_3.theorem\_3g\_ii}
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Suppose $F$ is a one-to-one \nameref{ref:function}.
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Then \nameref{sub:lemma-1} indicates $F^{-1}$ is a one-to-one function with
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domain $\ran{F}$ and range $\dom{F}$.
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Then \nameref{sub:one-to-one-inverse} indicates $F^{-1}$ is a one-to-one
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function with domain $\ran{F}$ and range $\dom{F}$.
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For all $x \in \dom{F}$, $\pair{x, F(x)} \in F$.
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Then $\pair{F(x), x} \in F^{-1}$.
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@ -3275,8 +3276,8 @@ If not, then under what conditions does equality hold?
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F^{-1}(y) & \text{if } y \in \ran{F} \\
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a & \text{otherwise}.
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\end{cases}$$
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$G$ is a function by virtue of \nameref{sub:lemma-1} and choice of mapping
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for all values $y \not\in \ran{F}$.
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$G$ is a function by virtue of \nameref{sub:one-to-one-inverse} and choice
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of mapping for all values $y \not\in \ran{F}$.
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Furthermore, for all $x \in A$, $F(x) \in \ran{F}$.
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Thus $(G \circ F)(x) = G(F(x)) = F^{-1}(F(x)) = x$ by
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\nameref{sub:theorem-3g}.
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@ -6469,46 +6470,78 @@ Show that $<_L$ is a linear ordering on $A \times B$.
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\section{Arithmetic}%
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\hyperlabel{sec:arithmetic}
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\subsection{\sorry{Theorem 4I}}
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\subsection{\verified{Theorem 4I}}
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\hyperlabel{sub:theorem-4i}
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\begin{theorem}[4I]
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For natural numbers $m$ and $n$,
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\begin{align*}
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m + 0 & = m, \\
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m + n^+ & = (m + n)^+.
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\end{align*}
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\begin{align}
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m + 0 & = m, \hyperlabel{sub:theorem-4i-eq1} \\
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m + n^+ & = (m + n)^+. \hyperlabel{sub:theorem-4i-eq2}
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\end{align}
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\end{theorem}
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\begin{proof}
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TODO
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\statementpadding
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\lean*{Init/Data/Nat/Basic}{Nat.add\_zero}
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\lean{Init/Prelude}{Nat.add}
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\paragraph{\eqref{sub:theorem-4i-eq1}}%
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Let $m$ be a \nameref{ref:natural-number}.
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By definition of \nameref{ref:addition}, $m + 0 = A_m(0)$.
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By definition of $A_m$, $A_m(0) = m$.
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Thus $m + 0 = m$.
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\paragraph{\eqref{sub:theorem-4i-eq2}}%
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Let $m$ and $n$ be natural numbers.
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By definition of \nameref{ref:addition},
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$$m + n^+ = A_m(n^+) = A_m(n)^+ = (m + n)^+.$$
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\end{proof}
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\subsection{\sorry{Theorem 4J}}
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\subsection{\verified{Theorem 4J}}
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\hyperlabel{sub:theorem-4j}
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\begin{theorem}[4J]
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For natural numbers $m$ and $n$,
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\begin{align*}
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m \cdot 0 & = 0, \\
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m \cdot n^+ & = m \cdot n + m.
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\end{align*}
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\begin{align}
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m \cdot 0 & = 0, \hyperlabel{sub:theorem-4j-eq1} \\
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m \cdot n^+ & = m \cdot n + m. \hyperlabel{sub:theorem-4j-eq2}
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\end{align}
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\end{theorem}
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\begin{proof}
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TODO
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\statementpadding
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\lean*{Init/Data/Nat/Basic}{Nat.mul\_zero}
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\lean{Init/Prelude}{Nat.mul}
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\paragraph{\eqref{sub:theorem-4j-eq1}}%
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Let $m$ be a \nameref{ref:natural-number}.
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By definition of \nameref{ref:multiplication}, $$m \cdot 0 = M_m(0) = 0.$$
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\paragraph{\eqref{sub:theorem-4j-eq2}}%
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Let $m$ and $n$ be natural numbers.
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By definition of \nameref{ref:multiplication},
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$$m \cdot n^+ = M_m(n^+) = M_m(n) + m = m \cdot n + m.$$
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\end{proof}
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\subsection{\pending{Lemma 2}}%
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\label{sub:lemma-2}
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\subsection{\verified{Left Additive Identity}}%
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\hyperlabel{sub:left-additive-identity}
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\begin{lemma}[2]
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@ -6519,35 +6552,39 @@ Show that $<_L$ is a linear ordering on $A \times B$.
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\begin{proof}
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Let $S = \{n \in \omega \mid A_0(n) = n\}$.
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\lean{Init/Data/Nat/Basic}{Nat.zero\_add}
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Let $S = \{n \in \omega \mid 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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\hyperlabel{par:left-additive-identity-i}
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By definition of \nameref{ref:addition}, $A_0(0) = 0$.
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By \nameref{sub:theorem-4i}, $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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\hyperlabel{par:left-additive-identity-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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By \nameref{sub:theorem-4i}, $0 + n^+ = (0 + n)^+$.
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Since $n \in S$, $0 + n = n$ which in turn implies that $(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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By \nameref{par:left-additive-identity-i} and
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\nameref{par:left-additive-identity-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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Thus for all $n \in \omega$, $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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\subsection{\verified{Lemma 3}}%
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\hyperlabel{sub:lemma-3}
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\hyperlabel{sub:succ-add-eq-add-succ}
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\begin{lemma}[3]
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@ -6558,103 +6595,41 @@ Show that $<_L$ is a linear ordering on $A \times B$.
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\begin{proof}
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\lean{Std/Data/Nat/Lemmas}{Nat.succ\_add\_eq\_succ\_add}
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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^+)\}.$$
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$$S = \{n \in \omega \mid m^+ + n = m + n^+\}.$$
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We prove that (i) $0 \in S$ and (ii) if $n \in S$ then $n^+ \in S$.
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Afterwards we show that (iii) our theorem holds.
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\paragraph{(i)}%
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\label{par:lemma-3-i}
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\hyperlabel{par:lemma-3-i}
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By definition of \nameref{ref:addition}, $A_{m^+}(0) = m^+$.
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Likewise, $$A_m(0^+) = A_m(0)^+ = m^+.$$
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By \nameref{sub:theorem-4i}, $m^+ + 0 = m^+$.
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Likewise, $m + 0^+ = (m + 0)^+ = m^+$.
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Thus $0 \in S$.
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\paragraph{(ii)}%
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\label{par:lemma-3-ii}
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\hyperlabel{par:lemma-3-ii}
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Suppose $n \in S$.
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By definition of addition, $A_{m^+}(n^+) = \left[A_{m^+}(n)\right]^+$.
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Since $n \in S$, $A_{m^+}(n) = A_m(n^+)$ which in turn implies that
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$$\left[A_{m^+}(n)\right]^+ = \left[A_m(n^+)\right]^+ = A_m(n^{++}).$$
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By \nameref{sub:theorem-4i}, $m^+ + n^+ = (m^+ + n)^+$.
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Since $n \in S$, $m^+ + n = m + n^+$.
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Therefore $(m^+ + n)^+ = (m + n^+)^+ = m + n^{++}$.
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Thus $n^+ \in S$.
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\paragraph{(iii)}%
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By \nameref{par:lemma-3-i} and \nameref{par:lemma-3-ii}, $S$ is inductive.
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Hence \nameref{sub:theorem-4b} implies $S = \omega$.
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Thus for all $n \in \omega$, $A_{m^+}(n) = A_m(n^+)$.
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Thus for all $n \in \omega$, $m^+ + n = m + n^+$.
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\end{proof}
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\subsection{\pending{Lemma 4}}%
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\hyperlabel{sub:lemma-4}
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\begin{lemma}[4]
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For all $n \in \omega$, $M_0(n) = 0$.
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In other words, $$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.$$
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\end{lemma}
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\pending{Theorem 4k (1)}}%
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\subsection{\verified{Theorem 4K-1}}%
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\label{sub:theorem-4k-1}
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\begin{theorem}[4K.1]
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\begin{theorem}[4K-1]
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Associative law for addition.
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For $m, n, p \in \omega$, $$m + (n + p) = (m + n) + p.$$
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@ -6663,6 +6638,8 @@ Show that $<_L$ is a linear ordering on $A \times B$.
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\begin{proof}
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\lean{Mathlib/Algebra/Group/Defs}{add\_assoc}
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Fix $n, p \in \omega$ and define
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\begin{equation}
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\hyperlabel{sub:theorem-4k-1-eq1}
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@ -6674,28 +6651,24 @@ Show that $<_L$ is a linear ordering on $A \times B$.
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\paragraph{(i)}%
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\hyperlabel{par:theorem-4k-1-i}
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By definition of \nameref{ref:addition},
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\begin{align*}
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0 + (n + p)
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& = n + p & \textref{sub:lemma-2} \\
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& = (0 + n) + p. & \textref{sub:lemma-2} \\
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\end{align*}
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By \nameref{sub:left-additive-identity},
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$$0 + (n + p) = n + p = (0 + n) + p.$$
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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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Then
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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:succ-add-eq-add-succ} \\
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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:succ-add-eq-add-succ} \\
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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:succ-add-eq-add-succ}
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\end{align*}
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Thus $m^+ \in S$.
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@ -6709,10 +6682,10 @@ Show that $<_L$ is a linear ordering on $A \times B$.
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\end{proof}
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\subsection{\pending{Theorem 4k (2)}}%
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\subsection{\verified{Theorem 4K-2}}%
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\label{sub:theorem-4k-2}
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\begin{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.$$
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@ -6721,6 +6694,8 @@ Show that $<_L$ is a linear ordering on $A \times B$.
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\begin{proof}
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\lean{Mathlib/Algebra/Group/Defs}{add\_comm}
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Fix $n \in \omega$ and define
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\begin{equation}
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\hyperlabel{sub:theorem-4k-2-eq1}
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@ -6732,25 +6707,21 @@ Show that $<_L$ is a linear ordering on $A \times B$.
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\paragraph{(i)}%
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\hyperlabel{par:theorem-4k-2-i}
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By definition of \nameref{ref:addition},
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\begin{align*}
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0 + n
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& = n & \textref{sub:lemma-2} \\
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& = n + 0. \\
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\end{align*}
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By definition of \nameref{ref:addition} and
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\nameref{sub:left-additive-identity}, $$0 + n = n = n + 0.$$
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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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Then
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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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& = m + n^+ & \textref{sub:succ-add-eq-add-succ} \\
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& = (m + n)^+ & \textref{sub:theorem-4i} \\
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& = (n + m)^+ & \eqref{sub:theorem-4k-2-eq1} \\
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& = n + m^+.
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& = n + m^+. & \textref{sub:theorem-4i}
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\end{align*}
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Thus $m^+ \in S$.
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@ -6763,10 +6734,116 @@ Show that $<_L$ is a linear ordering on $A \times B$.
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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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\subsection{\verified{Zero Multiplicand}}%
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\hyperlabel{sub:zero-multiplicand}
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\begin{theorem}[4K.3]
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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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\lean{Init/Data/Nat/Basic}{Nat.zero\_mul}
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Define
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\begin{equation}
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\hyperlabel{sub:zero-multiplicand-eq1}
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S = \{n \in \omega \mid 0 \cdot 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:zero-multiplicand-i}
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By \nameref{sub:theorem-4j}, $0 \cdot 0 = 0$.
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Thus $0 \in S$.
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\paragraph{(ii)}%
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\hyperlabel{par:zero-multiplicand-ii}
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Suppose $n \in S$.
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Then
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\begin{align*}
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0 \cdot n^+
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& = 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}
|
||||
|
||||
|
|
|
|||
Loading…
Reference in New Issue