How to prove de Vries algebras morphisms are dense and full if their duals are into? Well, this is quite a short question, but I think it will require some explanations.
Let's say that a de Vries (or compingent) algebra is a Boolean algebra $B=(B,0,1, \wedge, \vee, \neg) $ with a binary relation $\prec$ such that for each $a,b,c,d \in B$ :
$0 \prec 0$,
$a \prec b \Rightarrow a \leq b$,
$ a \leq c \prec b \Rightarrow a \prec b$,
$a \leq b, c \leq d \Rightarrow a \wedge c \prec b \wedge d$,
$ a \prec b \Rightarrow \neg b \prec \neg a$
$ a \prec b \neq 0 \Rightarrow \exists c \in B \setminus \lbrace 0 \rbrace : a \prec c \prec b$.
Now, let's say that a homomorphism $h$ between compingent algebras $B$ and $C$ is a function $h : B \to C$ such that
$ h(0) = 0$
$h(a \wedge b) = h(a) \wedge h(b)$
$ a \prec b \Rightarrow \neg (h(\neg a)) \prec h(b)$.
Finally, a round (or compingent) filter $F$ in a de Vries algebra $B$ is a filter for Boolean algebra such that for every $b \in F$ there is $a \in F$ with $a \prec b$. It can be shown that a round filter $F$ is maximal if and only if for every $a, b \in B$, $a \prec b$ implies $b \in F$ or $\neg a \in F$.
And now, the question! Suppose $h : B \to C$ is a homomorphism. Then $m(h)$ defined as $$ m(h)(F) = \lbrace a \mid \exists b \in h^{-1}(F) : b \prec a \rbrace $$ is a function from the maximal round filters of $C$ to the maximal round filters of $B$.
What I want to prove is: if $m(h)$ is one-to-one then for every $c,d \in C$ such that $c \prec d $ there exist $a,b \in B$ such that $$ (1)\ a \prec \neg b,\ c \prec h(a) \text{ and } \neg d \prec h(b)\,.$$
According to de Vries, this implication should follow from the fact that if $m(h)$ is one-to-one, then for every maximal round filter $F$ of $C$ and every $d\in F$, there is $a,b \in B$ such that $$ (2) \ a \prec \neg b , \ a \in m(h)(F) \text{ and } \neg d \prec h(b).$$
While I have no problem proving that $m(h)$ being one-to-one implies the existence of elements $a,b$ for which $(2)$ holds, I can't prove that we can deduce $(1)$ from $(2)$ (this last step should be easy to show according to de Vries).
Thank you for any answer!
 A: The question is related to my father H. de Vries' 1962 Ph.D. Thesis "Compact Spaces and Compactifications. An Algebraic Approach". Prof. Georgi Dimov from Sofia University, Bulgaria kindly looked into your question and he provided me with the following answer, which I post here with his permission. I have done my best effort to reset his text in MathJax, and to proofread it carefully, but an occasional glitch might be present. The references are to the original thesis, https://www.illc.uva.nl/Research/Publications/Dissertations/HDS-23-Hendrik_de_Vries.text.pdf

Fact 0.1. Let $h : B_1 \to B_2$ be a homomorphism between compingent algebras and let $\mathfrak{m}(h) : \mathfrak{M}_{B_2} \to \mathfrak{M}_{B_1} $ be a homeomorphism into (i.e., an embedding). Then, for every $e,d \in B_2$ such that $e \ll d$, there exist $a,b \in B_1$ with $a \ll b^◦$, $e \ll h(a)$ and $d^◦ \ll h(b) $.
Proof. For simplicity, I will put $\omega_i \overset{\mathrm{df}}= \omega_{B_i} $, for $i = 1, 2$. Also, the closure of a set $M$ will be denoted by $\mathrm{cl}(M)$.
In the first part of the proof of Theorem 1.7.3, it is shown that for every $d \in B_2 $ and every $ \mathfrak{n} \in \omega_2(d)$, there exist $a,b \in B_1 $ such that $\mathfrak{n} \in \omega_2(h(a)) $ and $a \ll b^◦$, $ d^◦ \ll h(b)$.
Let now $e,d \in B_2$ and $e \ll d$. Then, by Lemma 1.3.5, $\mathrm{cl} (\omega_2(e)) \subseteq \omega_2(d)$. Set  $K \overset{\mathrm{df}} = \mathrm{cl}(\omega_2(e))$. Then, for every $\mathfrak{n} \in K$, by the first part of the proof of Theorem 1.7.3, there exist $ a_\mathfrak{n}, b_\mathfrak{n} \in B_1$ such that $\mathfrak{n} \in \omega_2(h(a_\mathfrak{n}))$ and $ a_\mathfrak{n} \ll b_\mathfrak{n}^◦, d^◦ \ll h(b_\mathfrak{n})$. Since
$K$ is compact, there exist $\mathfrak{n}_1, …, \mathfrak{n}_l$ such that $K \subseteq \bigcup \{ \omega_2(h(a_{\mathfrak{n}_i})) \, | \, i = 1, \ldots ,l\}$. Set $ a \overset{\mathrm{df}}=\bigvee \{a_{\mathfrak{n}_i} \, | \, i = 1, \ldots ,l \} $ and $b \overset{\mathrm{df}} = \bigwedge \{b_{\mathfrak{n}_i} \, | \, i = 1, \ldots,l\}$. Then $b^◦ = \bigvee 
\{b^◦_{\mathfrak{n}_i} \, | \, i = 1, \ldots ,l\}, a \ll b^◦,d^◦ \ll h(b) $ and $\omega_2(e) \subseteq \mathrm{cl}(\omega_2(e))=K \subseteq  \bigcup \{\omega_2(h(a_{\mathfrak{n}_i}))\, | \, i=1, \ldots ,l\}\subseteq 
\omega_2(h(a)) $. Thus, by Theorem 1.3.9(ii), $e \ll h(a)$.
