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?