Sophie wrote:
> I found it helpful to write the ordering as \\( (a , b, c + n) \leq (a + 2n, b + n , c) \\), for all \\(a,b,c \in \mathbb{N}\\)

I agree. This is much nicer and probably also easier to understand. Thanks!

John wrote:
> I forget if someone answered my other question: which of these two relations define partial orders, as opposed to mere preorders?

**Puzzle 69/70** In the **irreversible case** there is no way to proceed along a reaction sequence and end up where you started. Thus, there are no equivalent elements and it forms a **partial order**.

In the **reversible case** one can also go backward along a reaction sequence, thus making all of the elements of a sequence equivalent. Thus, this is merely a **preorder**.

Side note: If we wanted to turn this preorder into a partial order we would have to identify all elements of any given reaction sequence. This new element would correspond to the general atomic composition of this reaction sequence. E.g \\( (2,1,0) \leftrightharpoons (0,0,1) \\) would become one point signifying 2 H atoms and 1 O atom (which can in any allowed molecular configuration).