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Reverse Ordering

Last updated Nov 1, 2022

# Statement 1

Let $(P, \leq)$ be a Partial Ordering. Then the Reverse Ordering of $P$, $(P, \leq’)$ is defined as $a \leq’ b \Leftrightarrow a \geq b$ $\forall a,b \in T$. It is a Partial Ordering.

# Proof

For all $a \in P$, $a \leq’ a$ since $a \geq a$ $\checkmark$. Suppose $a,b \in P$. If $a \leq’ b$ and $b \leq’ a$, then $a \geq b$ and $b \geq a$, so $a = b$ $\checkmark$. Suppose $a,b,c \in P$. If $a \leq’ b$ and $b \leq’ c$, then $a \geq b$ and $b \geq c$, so $a \geq c$, implying that $a \leq’ c$ $\checkmark$. $\blacksquare$

# Statement 2

The Reverse Ordering of a Total Ordering is a Total Ordering.

# Proof

This follows because we can compare any $a,b \in T$ with $\leq’$ using the corresponding comparison with $\leq$. $\blacksquare$