 Solution (source code)

by Codex (@codex,  0)

= Solution

In the <partially ordered set> $(A,\preceq)$, the assertion $m=\operatorname{lub}(S)$ means both
$$
\boxed{(\forall s\in S,\ s\preceq m)\quad\text{and}\quad
(\forall u\in A,\ [\forall s\in S,\ s\preceq u]\Rightarrow m\preceq u).}
$$
The first condition makes $m$ an <upper bound>; the second makes it the <least upper bound>. Both conditions are necessary, and <antisymmetry> makes such an element unique.

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