If $[0,a] \cong [0,b]$ and $a \wedge b = 0$, does it follow that $[0,a'] \cong [0,b']$? Let $\mathbf{B} = \langle B, \wedge, \vee, ', 0, 1 \rangle$ be a Boolean Algebra.
Then $[0,a] = \{ x \in B : 0 \leq x \leq a \}$, and this can be made into a Boolean Algebra $\mathbf{B}_a = \langle [0,a], \wedge, \vee, ^*, 0, a \rangle$, where $x^* = x' \wedge a$.
Suppose $\mathbf{B}_a \cong \mathbf{B}_b$, for some $a,b \in B$ such that $a \wedge b = 0$.
Does it follow that $\mathbf{B}_{a'} \cong \mathbf{B}_{b'}$?
I would like to have an isomorphism $\psi:\mathbf{B}_{a'} \to \mathbf{B}_{b'}$ explicitly defined, and I suppose that if this is true, it can be made such that $\psi(b) = a$ (notice that $b \leq a'$ follows from $a \wedge b = 0$), so that it would extend the inverse of the given isomorphism $\varphi:\mathbf{B}_{a} \to \mathbf{B}_{b}$.
Notice that it is not true if we drop the hypothesis that $a \wedge b = 0$.
For example, if $\mathbf{B}$ is the powerset of $\mathbb{N}$, define $a = \mathbb{N} \setminus \{1\}$ and $b = \mathbb{N}$.
Then $\mathbf{B}_a \cong \mathbf{B}_b = \mathbf{B}$; however, $B_a = \{ \varnothing, \{1\} \}$, and $B_b = \{\varnothing\}$, so they can't give isomorphic Boolean Algebras.
 A: Let us start with two auxiliary results.


*

*If $\mathbf{A}$ and $\mathbf{B}$ are Boolean algebras and $f : A \to B$ is a map, then $f$ is a homomorphism if and only if it is a bounded lattice homomorphism between the bounded lattice reducts of the algebras.

*Let $\mathbf{L}$ and $\mathbf{K}$ be lattices and $f:L\to K$ a bijection. Then $f$ is an isomorphism if and only if both $f$ and $f^{-1}$ are isotone if and only if $f$ preserves joins (or equivalently, meets).


Proof: 


*

*For $a \in A$, we have 
$$f(a) \vee f(a') = f(a \vee a') = f(1) = 1,$$
$$f(a) \wedge f(a') = f(a \wedge a') = f(0) = 0,$$
and so $f(a') = (f(a))'$ whence $f$ preserves $'$ and is a Boolean Algebra homomorphism.

*For the first equivalence, see, for example, Burris and Sankappanavar, Ch.I, Theorem 2.3. Now suppose that $f$ preserves joins, and let $a, b \in L$ such that $f(a) \leq f(b)$. Then,
$$f(a \vee b) = f(a) \vee f(b) = f(b) \Longrightarrow a \vee b = b \Longrightarrow a \leq b,$$
so that $f^{-1}$ is isotone. It is clear that if $f$ preserves joins, then $f$ is isotone. Hence $f$ is an isomorphism.


It follows from 1 and 2 that if a bijection between Boolean Algebras preserves joins, then it is an isomorphism.

Now since $\mathbf{B}_a$ and $\mathbf{B}_b$ are isomorphic, let $\alpha:\mathbf{B}_a \to \mathbf{B}_b$ be an isomorphism and $\beta = \alpha^{-1}$.
Define $\varphi: \mathbf{B}_{a'} \to \mathbf{B}_{b'}$ by
$$x \mapsto \beta(x \wedge b) \vee (x \wedge b'),$$
and $\psi: \mathbf{B}_{b'} \to \mathbf{B}_{a'}$ by
$$x \mapsto \alpha(x \wedge a) \vee (x \wedge a').$$
For $x \in [0,b']$ we have 
\begin{align}
\varphi( \psi(x) )
&= \varphi( \alpha(x \wedge a) \vee (x \wedge a') )\\
&= \varphi( \alpha(x \wedge a) ) \vee \varphi( x \wedge a' )\\
&= \beta( \alpha(x \wedge a) \wedge b ) \vee ( \alpha(x \wedge a) \wedge b' ) \vee \beta( x \wedge a' \wedge b ) \vee (x \wedge a' \wedge b)\\
&= (\beta( \alpha(x \wedge a) ) \wedge \beta(b)) \vee (\alpha(x \wedge a) \wedge b') \vee 0 \vee (x \wedge a' \wedge b')\tag{$x \wedge b = 0$}\\
&= ((x \wedge a) \wedge a) \vee (\alpha(x \wedge a) \wedge b') \vee (x \wedge a')\tag{$x \leq b'$}\\
&= (x \wedge a) \vee (x \wedge a') \vee ( \alpha(x \wedge a) \wedge b' )\\
&= (x \wedge (a \vee a')) \vee (\alpha(x \wedge a) \wedge b' )\\
&= x \vee (\alpha(x \wedge a) \wedge b' )\\
&= x,
\end{align}
where $\alpha(x \wedge a) \wedge b' = 0$ because $\alpha(x \wedge a) \leq b$.
Analogously, if $x \leq a'$ then $\psi(\varphi(x)) = x$, and $\varphi, \psi$ are bijections.  
Let us see that $\varphi$ preserves joins:
\begin{align}
\varphi(x \vee y)
&= \beta( (x \vee y) \wedge b ) \vee ( (x \vee y) \wedge b' )\\
&= \beta( (x \wedge b) \vee (y \wedge b) ) \vee ( (x \wedge b') \vee (y \wedge b') )\\
&= \beta(x\wedge b) \vee \beta(y \wedge b) \vee (x \wedge b') \vee (y \wedge b')\\
&= \varphi(x) \vee \varphi(y).
\end{align}
If also follows that $\varphi(b) = a$ and $\varphi(0) = 0$, and therefore, $\varphi$ is an isomorphism.
