TITLE: Relationship between Boolean algebras isomorphism and partially ordered sets isomorphism
QUESTION [2 upvotes]: Is the following proposition correct?

Two Boolean algebras $B_1$ and $B_2$ are isomorphic (i.e., there exists a bijection $f$ from $B_1$ to $B_2$ under which the operations $+,\ast, '$ are preserved) if and only if they are isomorphic as ordered sets (i.e., there exists a bijection $f$ from $B_1$ to $B_2$ under which the ordering is preserved).

REPLY [2 votes]: Yes, it is true.
Suppose $f: \mathbf{B}_1 \to \mathbf{B}_2$ is an order-isomorphism between the underlying posets.
Then $$a \leq b \Leftrightarrow f(a) \leq f(b).$$
Using this equivalence, it follows immediately that $f(0) = 0$ and $f(1)=1$.
Also, since $a \wedge b \leq a$, we have $f(a \wedge b) \leq f(a)$; also $f(a \wedge b) \leq f(b)$, whence $f(a \wedge b) \leq f(a) \wedge f(b)$.
Now if $d \leq f(a) \wedge f(b)$, then, taking $c$ such that $f(c) = d$, we have $c \leq a, b$ because $f(c) = d \leq f(a), f(b)$. 
So $c \leq a \wedge b$ whence $d \leq f(a \wedge b)$.
$$\therefore f(a) \wedge f(b) = f(a \wedge b).$$
Analogously $f(a) \vee f(b) = f(a \vee b)$.
Now 
$$0 = f(0) = f(a \wedge a') = f(a) \wedge f(a'),$$
and analogously, $1 = f(a) \vee f(a')$, and therefore $f(a') = f(a)'$.
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