If a poset is Dedekind-complete, is its opposite necessarily Dedekind-complete?

Question

Call a poset $X$ Dedekind-complete iff for all non-empty sets $A \subseteq X$, the following are equivalent.

1. $A$ has an upper bound
2. $A$ has a least upper bound.

Supposing $X = (X, \leq)$ is Dedekind-complete, is $X^\mathrm{op} = (X,\geq)$ necessarily Dedekind complete?

Answer 1

Yes, the infimum can be characterized as the supremum of the lower bounds, and conversely.