Completion unique up to isomorphism I have a question concerning having a unique completion (up to isomorphism), which I managed to solve partially, but still need some help for finishing.
Namely, 
Let $\mathbb P$ = $(P, ≤, . . .)$ be a partial order.  Given $A ⊆ P$, we
say that
$a =$ sup $A$
in case 
$a ∈ P$ is the least upper bound of $A$ in $P$, that is
• b ≤ a for every b ∈ A and
• a ≤ c whenever $c ∈ P$ is such that $b ≤ c$,  for every $b ∈ A$.
For a Boolean algebra $B$ and some $A ⊆ B$, sup $A$ may or may not exist in $B$.
We say that $B$ is complete in case sup $A$ exists in $B$ for every $A ⊆ B$. 
If $P$ is any
separative partial order, we say that $B$ is a completion of $P$ in case $P$ is a dense
subset of $B$ \ {$0$} and $B$ is a complete Boolean algebra.
Now my question is, how to show that if $B$ and $C$ are both completions of a given separative partial
order $P$, that then $B$ and $C$ are isomorphic.
Okay, it is easy to send $0$ to $0$, unit to unit and put the identity on restriction to $P$ (since $P$ is a subset), but what to do with the rest of elements? 

Every element $a\in B$ is the supremum of the elements in $P$ below it, and by completeness, there is also such an element in $C$, and this gives the isomorphism. Every element in the completion is the join of a subset (in fact, an antichain) of the partial order. 
