Jonathan

I'm not sure if everything I did here is legal but if you look at the progression below you can literally see associativity spring up from reflexivity and monoidal preorder laws:

![associativity proof](http://aether.co.kr/images/associativity_proof.svg)

So you start with \\(x \otimes y \otimes z \\). Then apply reflexivity to x and y to add in the "reactions." Then you use monoidal preorder law to parallel contract the preorder. Then you apply reflexivity and monoidal preorder law again to y and z leaving you with the associativity identity. You can do this starting with y and z instead of x and y and the two diagrams will be the same since we haven't done anything to the three wires. The parentheses are kind of a way to give vertical ordering.

This proof is kind of hard to see how "reactions" come into play when using regular symbols but you can still do it since you can do it with pictures. Start with :

$$x \leq x$$

$$y \leq y$$

$$z \leq z$$

Next add the first two and then last.

$$(x \otimes y) \otimes z \leq (x \otimes y) \otimes z$$

Then add the last two and then the first.

$$x \otimes (y \otimes z) \leq x \otimes (y \otimes z)$$

Then notice that adding all three equals x+y+z since that is what we started with. So:

$$(x \otimes y) \otimes z=x \otimes y \otimes z=x \otimes (y \otimes z)$$

The way I drew it in comment 23 is just shorthand for the last step in the progression since you can always write \\(x \otimes y\\) as two wires or one wire. But going with the flow of the all the pictures, I think this last picture portrays it the best.

I'm not sure if everything I did here is legal but if you look at the progression below you can literally see associativity spring up from reflexivity and monoidal preorder laws:

![associativity proof](http://aether.co.kr/images/associativity_proof.svg)

So you start with \\(x \otimes y \otimes z \\). Then apply reflexivity to x and y to add in the "reactions." Then you use monoidal preorder law to parallel contract the preorder. Then you apply reflexivity and monoidal preorder law again to y and z leaving you with the associativity identity. You can do this starting with y and z instead of x and y and the two diagrams will be the same since we haven't done anything to the three wires. The parentheses are kind of a way to give vertical ordering.

This proof is kind of hard to see how "reactions" come into play when using regular symbols but you can still do it since you can do it with pictures. Start with :

$$x \leq x$$

$$y \leq y$$

$$z \leq z$$

Next add the first two and then last.

$$(x \otimes y) \otimes z \leq (x \otimes y) \otimes z$$

Then add the last two and then the first.

$$x \otimes (y \otimes z) \leq x \otimes (y \otimes z)$$

Then notice that adding all three equals x+y+z since that is what we started with. So:

$$(x \otimes y) \otimes z=x \otimes y \otimes z=x \otimes (y \otimes z)$$

The way I drew it in comment 23 is just shorthand for the last step in the progression since you can always write \\(x \otimes y\\) as two wires or one wire. But going with the flow of the all the pictures, I think this last picture portrays it the best.