**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.

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.