In the process of writing this comment, I realized that I don't like my answer as an answer to Puzzzle 5. It is a separation into three subsets, but one of these subsets does not feel that enlightening to me. The answer by @JaredSummers feels much nicer: "Because for elements x and y there are exactly three possibilities: x

However, I decided to post this comment as an observation:

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> "Puzzle 5. Why is this property called "trichotomy"?"

Because it partitions any set S with a preorder into three disjunct sets X, Y and Z, where members of X only appear on the left-hand-side of an binary relation and members of Z only appear on the right-hand-side (and the set Z holds everything else). Intuitively, X can be thought of the set of sources, Z as set of sinks and Y as set of in-betweens and other stuff. Formally:

$$\forall x \in X . (\exists y \in (S \setminus X) . x \leq y) \land (\exists! y \in (S \setminus X) . y \leq x)$$

$$\forall z \in Z . (\exists y \in (S \setminus Z) . y \leq z) \land (\exists! y \in (S \setminus Z) . z \leq y)$$

$$Y = S \setminus X \cup Z$$

$$S = X \cup Y \cup Z $$

$$ X \cap Y \cap Z = \emptyset$$

This is feels like extracting the set of lower and upper bound out of an preorder. The trichotomy would then be "Any member of the preorder is either an lower bound (exclusively on the left-hand-side of the order relation), an upper bound or neither."

However, I decided to post this comment as an observation:

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> "Puzzle 5. Why is this property called "trichotomy"?"

Because it partitions any set S with a preorder into three disjunct sets X, Y and Z, where members of X only appear on the left-hand-side of an binary relation and members of Z only appear on the right-hand-side (and the set Z holds everything else). Intuitively, X can be thought of the set of sources, Z as set of sinks and Y as set of in-betweens and other stuff. Formally:

$$\forall x \in X . (\exists y \in (S \setminus X) . x \leq y) \land (\exists! y \in (S \setminus X) . y \leq x)$$

$$\forall z \in Z . (\exists y \in (S \setminus Z) . y \leq z) \land (\exists! y \in (S \setminus Z) . z \leq y)$$

$$Y = S \setminus X \cup Z$$

$$S = X \cup Y \cup Z $$

$$ X \cap Y \cap Z = \emptyset$$

This is feels like extracting the set of lower and upper bound out of an preorder. The trichotomy would then be "Any member of the preorder is either an lower bound (exclusively on the left-hand-side of the order relation), an upper bound or neither."