@[Matthew](https://forum.azimuthproject.org/profile/1818/Matthew%20Doty), you're right. There's probably some statement I could make about the free symmetric monoidal preorder generated by the desired reactions, but that's just kicking the can down the road. (See Section 25.2 of John's book, [Quantum Techniques for Stochastic Mechanics](https://arxiv.org/abs/1209.3632).) Let's see if I can recover this.

(**EDIT:** Also worth noting here that my thought process went badly off track when I said that Marius' and my rules were ultimately the same. Mine was _very_ wrong.)

First, consider that every complex \\((x, y, z)\\) can be written uniquely as \\((2x_q + x_r, y, z)\\) with \\(0 \le x_r < 2\\) -- in other words, dividing \\(x\\) by 2 and getting a quotient and remainder. (We'd only be dividing the others by 1, so we'll leave them alone.)

Now, we say that \\((2x_q + x_r, y, z) \le (2x'_q + x'_r, y', z')\\) iff the following statements hold for some \\(c \in \mathbb{N}\\):

$$c = x_q - x'_q$$

$$0 = x_r - x'_r$$

$$c = y - y'$$

$$c = z' - z$$

When \\(c = 0\\), we get the reflexive case. As \\(c\\) increases, we utilize more of the available resources.

(**EDIT:** Also worth noting here that my thought process went badly off track when I said that Marius' and my rules were ultimately the same. Mine was _very_ wrong.)

First, consider that every complex \\((x, y, z)\\) can be written uniquely as \\((2x_q + x_r, y, z)\\) with \\(0 \le x_r < 2\\) -- in other words, dividing \\(x\\) by 2 and getting a quotient and remainder. (We'd only be dividing the others by 1, so we'll leave them alone.)

Now, we say that \\((2x_q + x_r, y, z) \le (2x'_q + x'_r, y', z')\\) iff the following statements hold for some \\(c \in \mathbb{N}\\):

$$c = x_q - x'_q$$

$$0 = x_r - x'_r$$

$$c = y - y'$$

$$c = z' - z$$

When \\(c = 0\\), we get the reflexive case. As \\(c\\) increases, we utilize more of the available resources.