Proposition 2.20 implies that
$$ \textbf{Cost}^{op} = ( [ 0, \infty ], \le, 0, + ) $$
is a perfectly good symmetric monoidal preorder.

Using \\( 0 \\) for the monoidal unit, and \\( + \\) as the monoidal product seem fine.

Thinking of cost as a negative value \\( [ -\infty, 0 ] \\) gives an alternate \\(op\\) mapping that I will call \\(neg\\).

An alternate **Proposition : Preorder Negation**
Suppose \\( \mathcal{X} = (X, \le) \\) is a preorder and \\( \mathcal{X}^{neg} = (X^{neg}, \le) \\) is an opposite.
That is to say, all the arrows are reversed but the meaning of order function is retained.
If \\( (X, ≤, I, \otimes) \\) is a symmetric monoidal preorder then so is its negation, \\( (X^{neg}, \le, I, \otimes) \\).
As an example, consider the following.
$$ \textbf{Cost}_{neg} = ( [ -\infty, 0 ], \ge, 0, + ) $$
![Does this generalize?](https://docs.google.com/drawings/d/e/2PACX-1vSaZS_iF5odXxpSMFNiPtH58VMEAgYuLuXV5JMT4dOwc01ZplZW1rU0oZ95wLFTlhxXDU8l_nqPR2V4/pub?w=327&h=204)

$$ \textbf{X} = \textbf{Cost}^{op} = ( [ 0, \infty ], \le, 0, + ) $$

$$ \textbf{X}^{op} = \textbf{Cost} = ( [ 0, \infty ], \ge, 0, + ) $$

Using the negative real numbers augmented with \\( - \infty \\).

$$ \textbf{X}^{neg} = \textbf{Cost}^{neg^{op}} = ( [ - \infty, 0 ], \ge, 0, + ) $$

$$ \textbf{X}^{neg^{op}} = \textbf{Cost}^{neg} = ( [ - \infty, 0 ], \le, 0, + ) $$

Does the \\(neg\\) already have a name?