Enderton (set). Add cardinal arithmetic theorems and exercise prompts.
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@ -77,6 +77,19 @@
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\lean{Mathlib/Data/Set/Prod}{Set.prod}
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\section{\defined{Cardinal Arithmetic}}%
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\hyperlabel{sec:cardinal-arithmetic}
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Let $\kappa$ and $\lambda$ be any cardinal numbers.
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\begin{enumerate}[(a)]
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\item $\kappa + \lambda = \card{(K \cup L)}$, where $K$ and $L$ are any
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disjoint sets of cardinality $\kappa$ and $\lambda$, respectively.
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\item $\kappa \cdot \lambda = \card{(K \times L)}$, where $K$ and $L$ are
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any sets of cardinality $\kappa$ and $\lambda$, respectively.
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\item $\kappa^\lambda = \card{^L{K}}$, where $K$ and $L$ are any sets of
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cardinality $\kappa$ and $\lambda$, respectively.
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\end{enumerate}
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\section{\defined{Compatible}}%
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\hyperlabel{ref:compatible}
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@ -8811,11 +8824,73 @@
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Hence $S'$ is a finite set.
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\end{proof}
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\subsection{\sorry{Theorem 6H}}%
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\hyperlabel{sub:theorem-6h}
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Assume that $\equinumerous{K_1}{K_2}$ and $\equinumerous{L_1}{L_2}$.
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\begin{enumerate}[(a)]
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\item If $K_1 \cap L_1 = K_2 \cap L_2 = \emptyset$, then
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$\equinumerous{K_1 \cup L_1}{K_2 \cup L_2}$.
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\item $\equinumerous{K_1 \times L_1}{K_2 \times L_2}$.
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\item $\equinumerous{^{(L_1)}{K_1}}{^{(L_2)}{K_2}}$.
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\end{enumerate}
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\sorry{Theorem 6I}}%
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\hyperlabel{sub:theorem-6i}
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For any cardinal numbers $\kappa$, $\lambda$, and $\mu$:
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\begin{enumerate}
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\item $\kappa + \lambda = \lambda + \kappa$ and
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$\kappa \cdot \lambda = \lambda \cdot \kappa$.
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\item $\kappa + (\lambda + \mu) = (\kappa + \lambda) + \mu$ and
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$\kappa \cdot (\lambda \cdot \mu) = (\kappa \cdot \lambda) \cdot \mu$.
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\item $\kappa \cdot (\lambda + \mu) =
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\kappa \cdot \lambda + \kappa \cdot \mu$.
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\item $\kappa^{\lambda + \mu} = \kappa^\lambda \cdot \kappa^\mu$.
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\item $(\kappa \cdot \lambda)^\mu = \kappa^\mu \cdot \lambda^\mu$.
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\item $(\kappa^\lambda)^\mu = \kappa^{\lambda \cdot \mu}$.
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\end{enumerate}
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\sorry{Theorem 6J}}%
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\hyperlabel{sub:theorem-6j}
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Let $m$ and $n$ be finite cardinals.
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Then
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\begin{align*}
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m + n & = m +_\omega n, \\
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m \cdot n = m \cdot_\omega n, \\
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m^n = m^n,
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\end{align*}
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where on the right side we use the operations of $\omega$ defined via
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recursion and on the left side we use the operations of cardinal arithmetic.
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\sorry{Corollary 6K}}%
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\hyperlabel{sub:corollary-6k}
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If $A$ and $B$ are finite, then $A \cup B$, $A \times B$, and $^B{A}$ are also
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finite.
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\begin{proof}
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TODO
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\end{proof}
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\section{Exercises 6}%
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\hyperlabel{sec:exercises-6}
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\subsection{\unverified{Exercise 6.1}}%
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\hyperlabel{sub:exercise-6-1}
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\hyperlabel{sub:exercise-6.1}
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Show that the equation $$f(m, n) = 2^m(2n + 1) - 1$$ defines a one-one-one
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correspondence between $\omega \times \omega$ and $\omega$.
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@ -8894,7 +8969,7 @@
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\end{proof}
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\subsection{\unverified{Exercise 6.2}}%
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\hyperlabel{sub:exercise-6-2}
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\hyperlabel{sub:exercise-6.2}
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Show that in Fig. 32 we have:
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\begin{align*}
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@ -8933,7 +9008,7 @@
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\end{proof}
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\subsection{\unverified{Exercise 6.3}}%
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\hyperlabel{sub:exercise-6-3}
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\hyperlabel{sub:exercise-6.3}
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Find a one-to-one correspondence between the open unit interval $\ioo{0}{1}$
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and $\mathbb{R}$ that takes rationals to rationals and irrationals to
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@ -9087,7 +9162,7 @@
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\end{proof}
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\subsection{\sorry{Exercise 6.4}}%
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\hyperlabel{sub:exercise-6-4}
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\hyperlabel{sub:exercise-6.4}
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Construct a one-to-one correspondence between the closed unit interval
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$$\icc{0}{1} = \{x \in \mathbb{R} \mid 0 \leq x \leq 1\}$$
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@ -9098,7 +9173,7 @@
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\end{proof}
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\subsection{\verified{Exercise 6.5}}%
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\hyperlabel{sub:exercise-6-5}
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\hyperlabel{sub:exercise-6.5}
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Prove \nameref{sub:theorem-6a}.
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@ -9107,7 +9182,7 @@
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\end{proof}
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\subsection{\sorry{Exercise 6.6}}%
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\hyperlabel{sub:exercise-6-6}
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\hyperlabel{sub:exercise-6.6}
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Let $\kappa$ be a nonzero cardinal number.
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Show there does not exist a set to which every set of cardinality $\kappa$
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@ -9118,7 +9193,7 @@
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\end{proof}
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\subsection{\sorry{Exercise 6.7}}%
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\hyperlabel{sub:exercise-6-7}
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\hyperlabel{sub:exercise-6.7}
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Assume that $A$ is finite and $f \colon A \rightarrow A$.
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Show that $f$ is one-to-one iff $\ran{f} = A$.
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@ -9128,7 +9203,7 @@
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\end{proof}
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\subsection{\sorry{Exercise 6.8}}%
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\hyperlabel{sub:exercise-6-8}
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\hyperlabel{sub:exercise-6.8}
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Prove that the union of two finite sets is finite, without any use of
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arithmetic.
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@ -9138,7 +9213,7 @@
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\end{proof}
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\subsection{\sorry{Exercise 6.9}}%
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\hyperlabel{sub:exercise-6-9}
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\hyperlabel{sub:exercise-6.9}
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Prove that the Cartesian product of two finite sets is finite, without any use
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of arithmetic.
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@ -9147,4 +9222,60 @@
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TODO
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\end{proof}
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\subsection{\sorry{Exercise 6.10}}%
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\hyperlabel{sub:exercise-6.10}
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Prove part 4 of \nameref{sub:theorem-6i}.
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\sorry{Exercise 6.11}}%
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\hyperlabel{sub:exercise-6.11}
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Prove part 5 of \nameref{sub:theorem-6i}.
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\sorry{Exercise 6.12}}%
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\hyperlabel{sub:exercise-6.12}
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The proof to \nameref{sub:theorem-6i} involves eight instances of showing two
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sets to be equinumerous.
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(The eight are listed in the proof of the theorem as statements numbered 1-6.)
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In which of these eight cases does equality actually hold?
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\sorry{Exercise 6.13}}%
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\hyperlabel{sub:exercise-6.13}
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Show that a finite union of finite sets is finite.
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That is, show that if $B$ is a finite set whose members are themselves finite
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sets, then $\bigcup{B}$ is finite.
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\sorry{Exercise 6.14}}%
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\hyperlabel{sub:exercise-6.14}
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Define a \textit{permutation} of $K$ to be any one-to-one function from $K$
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onto $K$.
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We can the define the factorial operation on cardinal numbers by the equation
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$$\kappa! = \card{\{f \mid f \text{ is a permutation of } K\}},$$
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where $K$ is any set of cardinality $\kappa$.
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Show that $\kappa!$ is well defined, i.e. the value of $\kappa!$ is
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independent of just which set $K$ is chosen.
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\begin{proof}
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TODO
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\end{proof}
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\end{document}
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@ -166,6 +166,7 @@
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% ========================================
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\newcommand{\abs}[1]{\left|#1\right|}
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\newcommand{\card}[1]{\mathop{\text{card}}{#1}}
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\newcommand{\ceil}[1]{\left\lceil#1\right\rceil}
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\newcommand{\ctuple}[2]{\left< #1, \cdots, #2 \right>}
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\newcommand{\dom}[1]{\textop{dom}{#1}}
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