I'm hitting a blocker with @Tobias's **Puzzle TF3**.

Suppose for simplicity that \\(\$ = I\\), so every \\(\mathcal{X}(x,y)\\) is either \\(0\\) (if \\(x \leq y\\)) or \\(\infty\\) (if \\(x \nleq y\\)).

Suppose further that we have \\(x, y, z \in X\\) with \\(x \leq z\\), \\(y \leq z\\), but \\(x \nleq y\\).

Then we have \\(\mathcal{X}(x, y) = \infty, \mathcal{X}(y, z) = 0\\) so \\(\mathcal{X}(x, y) + \mathcal{X}(y, z) = \infty \nleq 0 = \mathcal{X}(x, z)\\)

... which contravenes rule (b) for enriched categories.

**EDIT** to add: in fact it seems in general we have \\(\mathcal{X}(x, y) + \mathcal{X}(y, z) \geq \\mathcal{X}(x, z)\\) rather than \\(\leq\\)

**EDIT** to add: aaaargh just checked the Fong/Spivak book and of course the ordering *is* \\(\geq\\) not \\(\leq\\) on **Cost**. d'oh!

Suppose for simplicity that \\(\$ = I\\), so every \\(\mathcal{X}(x,y)\\) is either \\(0\\) (if \\(x \leq y\\)) or \\(\infty\\) (if \\(x \nleq y\\)).

Suppose further that we have \\(x, y, z \in X\\) with \\(x \leq z\\), \\(y \leq z\\), but \\(x \nleq y\\).

Then we have \\(\mathcal{X}(x, y) = \infty, \mathcal{X}(y, z) = 0\\) so \\(\mathcal{X}(x, y) + \mathcal{X}(y, z) = \infty \nleq 0 = \mathcal{X}(x, z)\\)

... which contravenes rule (b) for enriched categories.

**EDIT** to add: in fact it seems in general we have \\(\mathcal{X}(x, y) + \mathcal{X}(y, z) \geq \\mathcal{X}(x, z)\\) rather than \\(\leq\\)

**EDIT** to add: aaaargh just checked the Fong/Spivak book and of course the ordering *is* \\(\geq\\) not \\(\leq\\) on **Cost**. d'oh!