How to prove that Supremum preorder coincides with Hoare preorder?
I’m going to use normal notation, as the square symbols are a pain to typeset.

Proposition: Let $L$ be a complete lattice, and $B$ its set of completely join-irreducible elements. If every element of $L$ is a join of a subset of $B$, then the following are equivalent:

$\bigvee X\le\bigvee Y\implies \forall x\in X\,\exists y\in Y\,x\le y$ for all $X,Y\subseteq B$.

All $x\in B$ are completely join-prime, i.e., $x\le\bigvee Y\implies\exists y\in Y\,x\le y$ for all $Y\subseteq L$.

$L$ satisfies the infinitary distributive law $x\land\bigvee_{i\in I}y_i=\bigvee_{i\in I}(x\land y_i)$.



Note that condition 3 is not equivalent to its dual, and it is weaker than full complete distributivity.
Proof:
1 → 2: By the assumption on $L$, $\bigvee Y=\bigvee\{u\in B:\exists y\in Y\,u\le y\}$. Thus, if $x\le\bigvee Y$, then $\bigvee\{x\}\le\bigvee\{u\in B:\exists y\in Y\,u\le y\}$, hence by 1, there exists $u\in B$ and $y\in Y$ such that $x\le u\le y$.
2 → 1: For any $x\in X$, $x\le\bigvee X\le\bigvee Y$, hence $x\le y$ for some $y\in Y$ as $x$ is completely join-prime.
3 → 2: If $x$ is completely join-irreducible and $x\le\bigvee Y$, then $x=\bigvee\{x\land y:y\in Y\}$ by 3, hence $x=x\land y$ (i.e., $x\le y$) for some $y\in Y$ by complete irreducibility.
2 → 3: $x\land\bigvee_{i\in I}y_i\ge\bigvee_{i\in I}(x\land y_i)$ always holds. For the other inequality, since the left-hand side is a join of a subset of $B$ by assumption, it suffices to show that
$$u\le x\land\bigvee_{i\in I}y_i\implies u\le\bigvee_{i\in I}(x\land y_i)$$
for all $u\in B$. The premise implies $u\le x$, and since $u$ is completely join-prime by 2, there exists $i\in I$ such that $u\le y_i$. Then $u\le x\land y_i\le\bigvee_{i\in I}(x\land y_i)$.