Is it correct that if \\( \otimes \\) is commutative, we only have to prove that:

\[ \forall (x, x', y), x \le x' \textrm{ implies } x \otimes y \le x' \otimes y \]

By using different variables and then using commutativity, we get:

\[ \forall (y, y', x'), y \le y' \textrm{ implies } y \otimes x' \le y' \otimes x' \textrm{ implies } x' \otimes y \le x' \otimes y' \]

And then chaining the two and using transitivity:

\[ \forall (x, x', y, y'), x \le x' \textrm{ and } y \le y' \textrm{ implies } x \otimes y \le x' \otimes y \le x' \otimes y' \]

If so, this helps a lot with visualizing **Puzzle 71**. For all \\( x, x', y \in \mathbb{C} \\) the values used in the equation (i.e. \\(x, x', x+y, x'+y)\\) form a parallelogram, and \\( x \le x' \textrm{ implies } x + y \le x' + y \\). This is to say that opposite edges of any parallelogram must have the same "orientation" if we imagine edges pointing from the smaller to the larger value.

Anyways, using the bullet-proof method of imagining parallelograms sliding around on the surface of a 3D plot, I think a superset of Matthew's set of solutions is

\[ x \preceq_{p,F} y \iff F(p \cdot x) \leq F(p \cdot y) \]

for each point \\(p \in \mathbb{C}\\) and monotone function \\(F : \mathbb{R} \to \mathbb{R} \\). I think this is \\(2^{2^{\aleph_0}}\\) solutions.