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+\section{Monomial Orders}
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+
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+A Gröbner basis always pertains to a particular order on monomials. Let us
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+therefore introduce the most fundamental ones.
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+
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+Before we actually define a monomial order, let us start with a concise
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+discussion about binary relations so that it is convenient to prove that certain
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+orders are in fact monomial orders.
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+
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+\begin{df}
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+ Let $S$ be a non-empty set. A \textbf{binary relation} on $S$ is a subset $r$
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+ of $S \times S$. The relation $\Delta\!\left( S \right) = \set*{\left( a, a \right)~|~a
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+ \in S}$ is the \textbf{diagonal} of $S$.
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+\end{df}
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+
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+We will use only binary relations in our work and so we will refer to them
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+simply as relations. In order to simplify the notation, we will also employ
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+infix notation to denote that two elements are in a relation, i.e., if $r$ is a
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+binary relation on $S$ and $a, b \in S$, then $a~r~b$ will mean $\left( a, b
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+\right) \in r$.
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+
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+\begin{df}
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+ \sloppy Let $r$ and $s$ be relations on $S$. The relation $r^{-1} =
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+ \set*{\left( a, b \right) | \left( b, a \right) \in r}$ is the \textbf{inverse}
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+ of $r$. The \textbf{strict part} of $r$ is the relation $r_s = r \setminus r^{-1}$,
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+ and \[s \circ r = \set*{\left( a, c \right)~|\text{ there is } b \in S \text{ such
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+ that } \left( a, b \right) \in r \text{ and } \left( b, c \right) \in s}\]
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+ is the \textbf{product} of $r$ and $s$.
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+\end{df}
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+
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+\begin{df}
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+ Let $r$ be a relation on $S$. Then $r$ is
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+ \begin{enumerate}[label=(\roman*)]
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+ % \item \textbf{reflexive} if $\Delta\!\left( S \right) \subseteq r$,
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+ \item \textbf{transitive} if $r \circ r \subseteq r$,
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+ \item \textbf{antisymmetric} if $r \cap r^{-1} \subseteq \Delta\!\left( S \right)$,
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+ \item \textbf{connex} if $r \cup r^{-1} = S \times S$,
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+ % \item a \textbf{partial order on} $S$ if $r$ is reflexive, transitive and
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+ % antisymmetric,
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+ \item a \textbf{linear order on} $S$ if $r$ is transitive, antisymmetric and
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+ connex.
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+ \end{enumerate}
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+\end{df}
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+
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+\begin{df}
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+ Let $r$ be a relation on $S$ with strict part $r_s$ and let $R \subseteq S$. An
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+ element $a \in R$ is \textbf{minimal} if there is no $b \in R$ such that
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+ $b~r_s~a$. A \textbf{strictly descending} (or \textbf{strictly decreasing})
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+ \textbf{sequence} in $S$ is an infinite sequence of elements $a_n \in S$ such
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+ that $a_{n+1}~r_s~a_n$ for all $n \in \mathbb{N}$. The relation $r$ is
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+ \textbf{noetherian} if every non-empty subset $R$ of $S$ has a minimal
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+ element. The relation $r$ is a \textbf{well-order} on $S$ if it is a
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+ noetherian linear order on $S$.
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+\end{df}
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+
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+A natural way to think about the strict part of a relation is to consider the
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+natural order on $\mathbb{N}$, which is a linear order, where for each $m, n \in
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+\mathbb{N}$; $m > n$ means $m \ge n$ and $m \ne n$. The symbol $>$ denotes the
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+strict part of the relation $\ge$. We will also denote our orders on monomials by
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+$\succeq$, the inverse will be $\preceq$ and the strict parts will be denoted
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+$\succ$ and $\prec$.
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+
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+We will denote by $\mathcal{M}\!\left( x_1, \ldots, x_n \right)$, or simply
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+$\mathcal{M}$, the set of all monomials in the indeterminates $x_1, \ldots, x_n$. It
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+turns out that $\mathcal{M}$ forms an Abelian monoid under natural
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+multiplication where we add corresponding exponents of the indeterminates. The
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+multiplicative identity is the monomial 1. Note that we can associate any
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+monomial $x^\alpha \in \mathcal{M}\!\left( x_1, \ldots, x_n \right)$ with its $n$-tuple of
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+exponents $\alpha = \left(\alpha_1, \ldots, \alpha_n\right) \in \mathbb{N}_0^n$ in a one-to-one
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+fashion. Thus, we can use the sets $\mathcal{M}$ and $\mathbb{N}_0^n$
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+interchangeably.
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+
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+\begin{lm}
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+ \label{order_lm}
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+ A linear order $\ge$ on $S$ is a well-order if and only if there is no
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+ strictly descending sequence in $S$.
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+\end{lm}
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+
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+\begin{p}
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+ Let us turn the lemma into its contrapositive form: $\ge$ is not a well-order if
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+ and only if there is a strictly descending sequence in $S$; and prove this
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+ version of the lemma.
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+
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+ \begin{itemize}
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+ \item[$\implies$] Suppose $\ge$ is not a well-order. Then there is a non-empty
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+ subset $R \subseteq S$ that has no minimal element. We can choose $a \in R$ and
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+ since $a$ is not the minimal element, we can choose again $b \in R$ such that
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+ $a > b$, which leads to a strictly descending sequence.
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+
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+ \item[$\impliedby$] Suppose there is a strictly descending sequence in $S$.
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+ The elements of such a sequence form a non-empty subset $R$ of $S$ that has
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+ no minimal element. Hence, $\ge$ is not a well-order. \qedhere
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+ \end{itemize}
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+\end{p}
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+
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+\begin{df}
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+ \label{monomial_order_df}
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+ A \textbf{monomial order} $\succeq$ is a well-order on $\mathcal{M}$, which
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+ satisfies the \textbf{property of respecting multiplication}: if $m_1 \succeq m_2$,
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+ then $n \cdot m_1 \succeq n \cdot m_2$ for all $m_1, m_2, n \in \mathcal{M}$.
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+\end{df}
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+
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+The purpose of the property of respecting multiplication is that the relative
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+ordering of monomials in a polynomial does not change when we multiply the
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+polynomial by a monomial. Such behavior is necessary for the division algorithm
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+described in the next section.
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+
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+\begin{df}[Lexicographic order]
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+ Let $x^\alpha, x^\beta \in \mathcal{M}\!\left( x_1, \ldots, x_n \right)$ be monomials. We say
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+ $x^\alpha \succeq_{lex} x^\beta$ if $\alpha = \beta$ or if there is $1 \le i \le n$ such that $\alpha_j = \beta_j$
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+ for $1 \le j < i$ and $\alpha_i > \beta_i$.
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+\end{df}
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+
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+Note that $\succ_{lex}$ compares the exponent $n$-tuples $\alpha, \beta \in \mathbb{N}_0^n$
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+so that $x^\alpha \succ_{lex} x^\beta$ if the left-most non-zero component of the difference
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+$\alpha - \beta \in \mathbb{N}_0^n$ is positive.
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+
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+\begin{re}
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+ \label{lex_re}
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+ Also note that the lexicographic order depends on how the underlying
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+ indeterminates $x_1, x_2, \ldots, x_n$ are ordered. In general, there are $n!$ ways
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+ to order $n$ indeterminates and each of these orders has its respective
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+ lexicographic order. We will only assume the standard order where $x_1 > x_2 >
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+ \cdots > x_n$, or the alphabetical order where $x > y > z$.
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+\end{re}
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+
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+\begin{e}
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+ \label{lex_e}
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+ ~\begin{enumerate}[label=(\roman*)]
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+ \item Let $xy^2z^3$ and $xy^3$ be monomials in $\mathcal{M}\!\left( x, y, z
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+ \right)$. Then $xy^3 \succ_{lex} xy^2z^3$ since there is $i = 2$ and $j = 1$
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+ such that $\alpha_j = \beta_j$ and $\alpha_i > \beta_i$, where $\alpha = (1, 3, 0)$ and $\beta = (1, 2,
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+ 3)$. Also, the left-most non-zero component of the difference $\beta - \alpha =
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+ \left( 0, 1, -3 \right)$ is positive.
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+ \item Let $x, y, z$ be monomials in $\mathcal{M}\!\left( x, y, z \right)$.
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+ Then considering remark \ref{lex_re} and example (i), we get $x \succ_{lex} y
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+ \succ_{lex} z$.
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+ \item In the lexicographic order, note that a monomial that contains the most
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+ significant indeterminate (as regards the underlying order) is greater than
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+ any other monomial that does not contain such an indeterminate. For example,
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+ if $x$ and $y^3z^2$ are monomials in $\mathcal{M}\!\left( x, y, z \right)$,
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+ then $x \succ_{lex} y^3z^2$. The reasoning is the same as in (i) and (ii).
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+ \end{enumerate}
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+\end{e}
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+
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+The intuitive outlook on the lexicographic order is that it looks for the most
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+significant indeterminate that appears in one of the monomials and then gives
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+preference to the monomial in which this indeterminate has greater power.
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+
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+\begin{pr}
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+ \label{lex_pr}
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+ The lexicographic order $\succeq_{lex}$ on $\mathcal{M}$ is a monomial order.
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+\end{pr}
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+
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+\begin{p}
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+ Following the definition of the lexicographic order and the fact that the
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+ regular numerical order on $\mathbb{N}_0$ is a linear order, it is
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+ straightforward to show that for any monomials $x^{\alpha}, x^\beta, x^\gamma \in
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+ \mathcal{M}\!\left( x_1, \ldots, x_n \right)$ and $\alpha, \beta, \gamma \in \mathbb{N}_0^n$,
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+ the following conditions hold:
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+ \begin{description}
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+ \item[(transitivity)] if $x^\alpha \succeq_{lex} x^\beta$ and $x^\beta \succeq_{lex} x^\gamma$, then $x^\alpha
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+ \succeq_{lex} x^\gamma$;
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+ \item[(antisymmetry)] if $x^\alpha \succeq_{lex} x^\beta$ and $x^\alpha \preceq_{lex} x^\beta$, then
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+ $x^\alpha = x^\beta$; and
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+ \item[(connexity)] either $x^\alpha \succeq_{lex} x^\beta$ or $x^\alpha \preceq_{lex} x^\beta$.
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+ \end{description}
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+ These properties show that $\succeq_{lex}$ is a linear order on $\mathcal{M}$.
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+
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+ Let us prove the property of respecting multiplication explicitly. If $x^\alpha
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+ \succeq_{lex} x^\beta$, then either $\alpha = \beta$, or there is $1 \le i \le n$ such that $\alpha_i -
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+ \beta_i > 0$ with $\alpha_j = \beta_j$ for $1 \le j < i$. Also, $x^\alpha \cdot x^\gamma = x^{\alpha + \gamma}$ and
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+ $x^\beta \cdot x^\gamma = x^{\beta + \gamma}$. Comparing the results gives us $\left( \alpha + \gamma \right)
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+ - \left( \beta + \gamma \right) = \alpha - \beta$ and we see that $\alpha_i - \beta_i > 0$ with $\alpha_j =
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+ \beta_j$ for $1 \le j < i$ again; or if $\alpha = \beta$, then $\left( \alpha + \gamma \right) = \left(
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+ \beta + \gamma \right)$. This shows that also $x^{\alpha+\gamma} \succeq_{lex} x^{\beta+\gamma}$.
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+
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+ The last part to prove is to show that $\succeq_{lex}$ is also noetherian, i.e a
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+ well-order. We will prove this by the following contradiction:
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+
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+ By lemma \ref{order_lm}, if $\succeq_{lex}$ is not a well-order, then there is a
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+ strictly decreasing sequence \[\mathlarger{x^{\alpha_{(1)}} \succ_{lex} x^{\alpha_{(2)}}
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+ \succ_{lex} \cdots}\] of elements in $\mathcal{M}\!\left( x_1, \ldots, x_n \right)$,
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+ where each $\alpha_{(i)} = \left( \alpha_1, \ldots, \alpha_n \right) \in \mathbb{N}_0^n$. By the
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+ definition of $\succeq_{lex}$, we also know that there exists a $j$ such that all
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+ the first components of the $n$-tuples $\alpha_{(k)}$ with $k \ge j$ are equal.
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+ Continuing further, there is an $l \ge j$ such that all the second components of
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+ the $n$-tuples $\alpha_{(m)}$ with $m \ge l$ are all equal. We see that there must be
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+ a $p \ge l$, for which the whole $n$-tuples $\alpha_{(p)} = \alpha_{(p+1)} = \cdots$ are all
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+ equal. This means that the sequence is not strictly decreasing, which
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+ contradicts the lemma.
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+\end{p}
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+
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+\begin{df}[Reverse Colexicographic Order]
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+ \sloppy Let $x^\alpha, x^\beta \in \mathcal{M}\!\left( x_1, \ldots, x_n \right)$ be monomials.
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+ We say $x^\alpha \succeq_{rclex} x^\beta$ if $\alpha = \beta$ or if there is $1 \le i \le n$ such that
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+ $\alpha_j = \beta_j$ for $i < j \le n$ and $\alpha_i < \beta_i$.
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+\end{df}
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+
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+Observe that $\succ_{rclex}$ compares the exponent $n$-tuples $\alpha, \beta \in
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+\mathbb{N}_0^n$ so that $x^\alpha \succ_{rclex} x^\beta$ if the right-most non-zero component
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+of the difference $\alpha - \beta \in \mathbb{N}_0^n$ is negative. Remark \ref{lex_re} also
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+applies.
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+
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+\begin{e}
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+ \label{rclex_e}
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+ ~\begin{enumerate}[label=(\roman*)]
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+ \item Let $xy^2z^3$ and $xy^3$ be monomials in $\mathcal{M}\!\left( x, y, z
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+ \right)$. Then $xy^3 \succ_{rclex} xy^2z^3$ as well as in example \ref{lex_e}
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+ (i), but for a different reason. There is $i = 3$ such that $\alpha_i < \beta_i$,
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+ where $\alpha = (1, 3, 0)$ and $\beta = (1, 2, 3)$. Also, the right-most non-zero
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+ component of the difference $\beta - \alpha = \left( 0, 1, -3 \right)$ is negative.
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+ \item The lexicographic order coincides with the reverse colexicographic order
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+ for monomials in one and two indeterminates. These orders may differ for
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+ monomials in three and more variables, as shown by the following example:
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+ let $xz$ and $y^2$ be monomials in $\mathcal{M}\!\left( x, y, z \right)$.
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+ Then $xz \succ_{lex} y^2$, as explained in example \ref{lex_e} (i), but $y^2
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+ \succ_{rclex} xz$, as explained in example (i).
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+ \end{enumerate}
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+\end{e}
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+
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+The intuitive outlook on the reverse colexicographic order is that it looks for
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+the least significant indeterminate that appears in one of the monomials and
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+then gives preference to the monomial in which this indeterminate has lesser
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+power. It can be thought of as a double reversal of the lexicographic order ---
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+we first reverse the underlying order of the indeterminates and then their
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+powers.
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+
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+Equivalently to the lexicographic order, it is straightforward to show that the
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+reverse colexicographic order is a linear order as well. However, it is not a
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+well-order since it is possible to define the following strictly decreasing
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+sequence \[x_1x_2 \succ_{rclex} x_1x_2^2 \succ_{rclex} x_1x_2^3 \succ_{rclex} \cdots\] of
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+monomials in $\mathcal{M}\!\left( x_1, x_2 \right)$. In this sequence, let $x^\alpha
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+= x^{(1,n)}$ and $x^\beta = x^{(1,n+1)}$ for $n \in \mathbb{N}_{>0}$. We see that it
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+is always the case that $x^\alpha \succ_{rclex} x^\beta$ since $\alpha_1 = \beta_1$ and $\alpha_2 < \beta_2$,
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+and we get a strictly decreasing sequence. Hence, by lemma \ref{order_lm},
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+$\succeq_{rclex}$ is not a well-order and by definition \ref{monomial_order_df},
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+$\succeq_{rclex}$ cannot be a monomial order either. For this reason, we will not use
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+it to order monomials on its own, but we will use it as a ``sub-order'' in the
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+definition of the next order, which will be a monomial order.
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+
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+Examples \ref{lex_e} and \ref{rclex_e} show that the lexicographic and reverse
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+colexicographic orders do not take into consideration the total degree of
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+monomials. Later in our work, we will see that in certain cases, it is desirable
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+to order the monomials in a polynomial according to their total degree. Let us
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+therefore introduce the following order, which allows for the total degree.
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+
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+\begin{df}[Graded Reverse Lexicographic Order]
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+ \sloppy Let $x^\alpha, x^\beta \in \mathcal{M}\!\left( x_1, \ldots, x_n \right)$ be monomials.
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+ We say $x^\alpha \succeq_{grlex} x^\beta$ if $\lvert x^\alpha \rvert > \lvert x^\beta \rvert$, or
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+ $\lvert x^\alpha \rvert = \lvert x^\beta \rvert$ and $x^\alpha \succeq_{rclex} x^\beta$.
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+\end{df}
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+
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+Notice that despite its name, the graded reverse lexicographic order actually
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+makes use of the reverse colexicographic order. There is a general consensus on
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+such a name, so we will follow it.
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+
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+\begin{e}
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+ ~\begin{enumerate}[label=(\roman*)]
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+ \item Let $x, y^2, xz \in \mathcal{M}\!\left( x, y \right)$ be monomials. Then
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+ $y^2 \succeq_{grlex} x$ since $\lvert y^2 \rvert = 2 > \lvert x \rvert = 1$; and
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+ $y^2 \succeq_{grlex} xz$ since $\lvert xz \rvert = \lvert y^2 \rvert$ and $y^2
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+ \succeq_{rclex} xz$.
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+ \item Let $x, y, z \in \mathcal{M}\!\left( x, y z \right)$ be monomials. Then $x
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+ \succeq_{grlex} y \succeq_{grlex} z$ since $\lvert x \rvert = \lvert y \rvert = \lvert z
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+ \rvert$ and $x \succeq_{rclex} y \succeq_{rclex} z$.
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+ \end{enumerate}
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+\end{e}
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+
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+\begin{pr}
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+ The graded reverse lexicographic order $\succeq_{grlex}$ on $\mathcal{M}$ is a
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+ monomial order.
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+\end{pr}
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+
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+\begin{p}
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+ Since $\succeq_{grlex}$ first uses the usual well-order order on the total degree of
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+ monomials $\lvert x^\alpha \rvert \in \mathbb{N}_0$ and when $\lvert x^\alpha \rvert =
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+ \lvert x^\beta \rvert$, it decides ties using the reverse colexicographic order
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+ (which is a linear order), grlex is also linear.
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+
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+ It is also straightforward to show that $\succeq_{grlex}$ is a well-order since we
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+ consider only the strict part $\succ_{grlex}$, which is solely the well-order on
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+ $\lvert x^\alpha \rvert \in \mathbb{N}_0$.
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+
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+ In order to show that the property of respecting multiplication holds,
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+ consider the monomials $x^\alpha, x^\beta, x^\gamma \in \mathcal{M}\!\left( x_1, \ldots, x_n
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+ \right)$ with the $n$-tuples $\alpha, \beta, \gamma \in \mathbb{N}_0^n$. Also, $x^\alpha \cdot x^\gamma =
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+ x^{\alpha+\gamma}$ and $x^\beta \cdot x^\gamma = x^{\alpha+\gamma}$. Assume $x^\alpha \succeq_{grles} x^\beta$. If $\lvert x^\alpha
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+ \rvert > \lvert x^\beta \rvert$, then $x^{\alpha+\gamma} \succ_{grlex} x^{\beta+\gamma}$ since $\lvert
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+ x^{\alpha+\gamma} \rvert = \lvert x^\alpha \rvert + \lvert x^\gamma \rvert > \lvert x^\beta \rvert +
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+ \lvert x^\gamma \rvert = \lvert x^{\beta+\gamma} \rvert$. Also, if $\lvert x^\alpha \rvert =
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+ \lvert x^\beta \rvert$, we get $\lvert x^{\alpha+\gamma} \rvert = \lvert x^{\beta+\gamma} \rvert$ by
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+ the same argument as above and we use the reverse colexicographic order. So if
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+ $\lvert x^\alpha\rvert = \lvert x^\beta \rvert$, then $x^\alpha \succeq_{rclex} x^\beta$ (since we
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+ have assumed that $x^\alpha \succeq_{grlex} x^\beta$\big), which means that either $\alpha = \beta$,
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+ or there is $1 \le i \le n$ such that $\alpha_i - \beta_i < 0$ with $\alpha_j = \beta_j$ for $i < j
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+ \le n$. As in the proof of proposition \ref{lex_pr}, comparing the results gives
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+ us $\left( \alpha + \gamma \right) - \left( \beta + \gamma \right) = \alpha - \beta$ and we see that $\alpha_i
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+ - \beta_i < 0$ with $\alpha_j = \beta_j$ for $i < j \le n$ again; or if $\alpha = \beta$, then $\left(
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+ \alpha + \gamma \right) = \left( \beta + \gamma \right)$. This shows that $x^{\alpha+\gamma} \succeq_{grlex}
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+ x^{\beta+\gamma}$ and completes the proof.
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+\end{p}
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