**Puzzle 68/69 alternative** A simpler approach that doesn't use the inclusion ordering would be the following construction:

We start by fulfilling reflexivity by defining \$$(a,b,c) \le (a,b,c) \forall a,b,c\$$. Then we construct the ordering given by the Petri net as follows:

$$\text{As long as } a \ge 2n \text{ and } b \ge n, (a-2n,b-n,c+n) \le (a,b,c)$$

This gives us a set of 'reaction sequences' as well as discrete elements that can't react (sequences of length 0). E.g \$$(2,1,0) \rightarrow (0,0,1) \$$ and \$$(5,3,1) \rightarrow (3,2,2) \rightarrow (1,1,3) \$$ and \$$(0,1,0)\$$. These reaction sequences are linear (since we don't generate branches) and disjoint from one another (since each element has at most one direct predecessor). This construction forms a poset since reflexivity is fulfilled and transitivity is guaranteed in the linear sequences and is trivial on the discrete elements such as \$$(0,1,0)\$$. Furthermore there are no equivalent elements (loops) except when \$$(a,b,c) \le (a,b,c) \$$ .

It thus remains to be shown that we have a symmetric monoidal poset under addition of natural numbers with the unit \$$(0,0,0) \$$. Unit laws and associativity follow by the above proof in [#5]( https://forum.azimuthproject.org/discussion/comment/17960/#Comment_17960 ).

\$$x \le x' \textrm{ and } y \le y' \textrm{ imply } x \otimes y \le x' \otimes y' \$$ holds on the discrete elements since \$$(a,b,c) \le (a,b,c) \$$ and \$$(d,e,f) \le (d,e,f) \$$ do indeed imply \$$(a+d,b+e,c+f) \le (a+d,b+e,c+f) \$$. It also holds on elements in (the same) sequence(s) since as shown above:

> $$(a-2n,b-n,c+n) \le (a,b,c) \text{ and } (d-2m,e-m,f+m) \le (d,e,f) \text{ should imply } ((a-2n)+(d-2m),(b-n)+(e-m),(c+n)+(f+m)) \le (a+d,b+e,d+f)$$
>This is the case since:
>$$((a-2n)+(d-2m),(b-n)+(e-m),(c+n)+(f+m)) = ((a+d)-2(n+m),(b+e)+(n-m),(c+f)+(n+m)) \le (a+d,b+e,d+f)$$
>By taking \$$a'=a+d, b'=b+e, d'=d+f, n'=n+m\$$

This just takes you to a new sequence. Furthermore, it holds on combinations of a sequence and a discrete element since \$$(a-2n,b-n,c+n) \le (a,b,c)\$$ and \$$(d,e,f) \le (d,e,f)\$$ implies \$$(a-2n+d,b-n+e,c+n+f) \le (a+d,b+e,c+f)\$$. This also takes you to a different sequence.

In contrast to [#5]( https://forum.azimuthproject.org/discussion/comment/17960/#Comment_17960 ) we don't have to show that in general:
$$(a-2n,b-n,c+n) \le (a,b,c) \text{ and } (d,e,f) \le (d',e',f') \text{ should imply } (a-2n+d,b-n+e,c+n+f) \le (a+d',b+e',c+f')$$
since we have shown this for all the special cases that our construction creates.

Thus, provided the above arguments are correct we would have a symmetric monoidal poset.

I'm sure there is a more elegant way of showing that this and [#5]( https://forum.azimuthproject.org/discussion/comment/17960/#Comment_17960 ) work (provided they actually do). Looking forward to being corrected and/or refined.