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).