[Sophie wrote](https://forum.azimuthproject.org/discussion/comment/18107/#Comment_18107):
> > in other words, we have \$$(x, y) \le (x', y')\$$ iff \$$x \le x' \land y \le y'\$$
>
> I assume you used \$$\land\$$ to mean "and" and not "join". Is that correct?

I can definitely see how that's ambiguous! I did mean "and". (Though technically, "and" is a join on some lattice of propositions, so it's not an _incorrect_ reading, as long as the right parentheses are inferred.)

(EDIT: Also, lest I fall back into bad habits, I think \$$\wedge\$$ is the symbol for “meet”, the greatest lower bound. I do find it more natural sometimes to think of “true” as the informationless bottom element though.)

> Second I am still unclear about what ordering you are getting on \$$\mathbb R ^2 \$$ from product. You wrote:
> > this lifts to the product straightforwardly
>
> Any chance you could make that explicit?

Sure thing. Generally when working with products, things "lift" componentwise. So we say \$$(x, y) \le (x', y')\$$ iff both \$$x \le x'\$$ and \$$y \le y'\$$. The lifting I'm referring to in the quote is that the fact that \$$+\$$ respects the order on \$$R\$$ lifts in the “usual” sense to how the component-wise \$$+\$$ respects the order on \$$R^2\$$. Explicitly, if \$$(x, y) \le (x', y')\$$ and we have some \$$(r, s) \in \mathbb{R}^2\$$, we know that \$$(x, y) + (r, s) \le (x', y') + (r, s)\$$ because \$$x + r \le x' + r\$$ and \$$y + s \le y' + s\$$ individually.