The typical textbook definition of the Dedekind-MacNeille completion is as follows where $X^u$ denotes the set of all upper bounds of a set $X$ and $X^l$ the set of all lower bounds of a set $X$:
Given a poset $(A,\leq)$ then the completion $(L,\subseteq)$ is defined as
$$L = \{ D \in 2^A \mid (D^u)^l = D \}$$
with the subset inclusion ordering.
I was wondering if the completion as follows is the same
$$L' = \{ D \in 2^A \mid (D^l)^u = D \}$$
That means does $L = L'$ hold and is the ordering the same on $L$ and $L'$? My gut feeling says yes, but so far I was not able to show that.
$\langle ^{u}, ^l \rangle$ is a Galois connection in the powerset of $A$.
It follows that if $X \subseteq Y$ then $Y^{u} \subseteq X^{u}$ and $Y^l \subseteq X^l$.
Since $X^{ulu} = X^{u}$ and $X^{lul} = X^l$, the operators $^{u}$ and $^l$ turn the structure upside down and your lattices $L$ and $L'$ are dually isomorphic.