Strict Ordering Preserved under Product with Cancellable Element
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Theorem
Let $\struct {S, \circ, \preceq}$ be an ordered semigroup.
Let $x, y, z \in S$ be such that:
- $(1): \quad z$ is cancellable for $\circ$
- $(2): \quad x \prec y$
Then:
- $x \circ z \prec y \circ z$
- $z \circ x \prec z \circ y$
Proof
Let $z$ be cancellable and $x \prec y$.
Then by the definition of ordered semigroup:
- $x \circ z \preceq y \circ z$
From the fact that $z$ is cancellable:
- $x \circ z = y \circ z \iff x = y$
Thus as $x \circ z \ne y \circ z$ it follows from Strictly Precedes is Strict Ordering that:
- $x \circ z \prec y \circ z$
Similarly, $z \circ x \prec z \circ y$ follows from $z \circ x \preceq z \circ y$.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 15$: Ordered Semigroups: Theorem $15.1: \ 1^\circ$