Preordering of Products under Operation Compatible with Preordering

Theorem
Let $\struct {S, \circ}$ be an algebraic structure.

Let $\precsim$ be a preordering on $S$.

Then $\precsim$ is compatible with $\circ$


 * $\forall x_1, x_2, y_1, y_2 \in S: x_1 \precsim x_2 \land y_1 \precsim y_2 \implies \paren {x_1 \circ y_1} \precsim \paren {x_2 \circ y_2}$

Proof
By definition, $\precsim$ is compatible with $\circ$ :

Sufficient Condition
Let $\precsim$ be compatible with $\circ$.

Then for all $x_1, x_2, y_1, y_2 \in S$:

As $\precsim$ is a preordering it is by definition transitive.

Thus it follows that:
 * $x_1 \precsim x_2 \land y_1 \precsim y_2 \implies \paren {x_1 \circ y_1} \precsim \paren {x_2 \circ y_2}$

Necessary Condition
Let $\precsim$ fulfil the property that:
 * $\forall x_1, x_2, y_1, y_2 \in S: x_1 \precsim x_2 \land y_1 \precsim y_2 \implies \paren {x_1 \circ x_2} \precsim \paren {y_1 \circ y_2}$

As $\precsim$ is a preordering it is by definition reflexive.

That is:
 * $\forall z \in S: z \precsim z$

Make the substitutions:

It follows that:
 * $\forall x, y, z \in S: x \precsim y \implies \paren {x \circ z} \precsim \paren {y \circ z}$

Similarly, make the substitutions:

It follows that:
 * $\forall x, y, z \in S: x \precsim y \implies \paren {z \circ x} \precsim \paren {z \circ y}$

So for all $x, y, z \in S$, both conditions are fulfilled:

for $\precsim$ to be compatible with $\circ$.

The result follows.