Anindya - right! There are always two ways to "factorize" an inequality in a product of preorders, and as Keith noted, this is a special case of a more general fact: there are always two ways to factorize a morphism in a product of categories.

Namely, as you said a while back, any morphism in a product of categories \$$\mathcal{C} \times \mathcal{D}\$$ is of the form

$(f,g) : (x,y) \to (x',y')$

and we always have

$(f,g) = (f,1_{y'}) \circ (1_x, g)$

but also

$(f,g) = (1_{x'}, g) \circ (f, 1_y)$

It's nice to think of this using a commutative square:

$\begin{matrix} & & (f, 1_y) & & \\\\ & (x,y) & \rightarrow & (x',y) &\\\\ (1_x, g) & \downarrow & & \downarrow & (1_{x'},g) \\\\ & (x,y') & \rightarrow & (x',y') &\\\\ & & (f,1_{y'}) & & \\\\ \end{matrix}$

The morphism \$$(f,g)\$$ goes down the diagonal of this square, from upper left to lower right - but this typesetting system is too wimpy for me to draw it here!