At least Anindya understood the questions well enough to correctly answer them!

The point of the questions was this: after we think about it a minute, the definition of feasibility relation says that

$\text{if } (x,y) \le (x',y') \text{ in } X^{\text{op}} \times Y, \text{ then }\Phi(x,y) \text{ implies } \Phi(x',y') .$

But the theorem claims that this one condition is equivalent to _two_ conditions:

1. if \$$x' \le x \$$ in \$$X\$$, then \$$\Phi(x,y) \$$ implies \$$\Phi(x',y)\$$ for all \$$y \in Y\$$,

2. if \$$y \le y'\$$ in \$$Y\$$, then \$$\Phi(x,y) \$$ implies \$$\Phi(x,y')\$$ for all \$$x \in X\$$.

So, somehow we must be able to take a _single_ inequality in \$$X^{\text{op}} \times Y\$$ and break it into _two_ inequalities, one in \$$X\$$ and one in \$$Y\$$.

The proof of the theorem explains how... but it might not be clear to everyone. So I wanted you to figure it out:

**Puzzle 169.** The one inequality \$$(x,y) \le (x',y')\$$ in \$$X^{\text{op}} \times Y\$$ is equivalent to two inequalities, one in \$$X\$$ and one in
\$$Y\$$. What are they?

But preorders are just a special case of categories, and this fact is just a special case of a fact about categories:

**Puzzle 170.** Any morphism in a product of categories, \$$\mathcal{C} \times \mathcal{D}\$$, is a composite of two morphisms, one built from a morphism in \$$\mathcal{C}\$$ and one built from a morphism in \$$\mathcal{D}\$$. What are they?

Both puzzles have _two correct answers!_ Why?

This stuff isn't very deep, but it's good to know when you're working with products of categories, or preorders.