Proof Condition (i) holds because every finite distributive lattice is a Heyting algebra and the structure of a Heyting algebra is uniquely determined by its order reduct. For Condition (ii), see, e.g., [Reference Köhler58].

Proof For $\mathsf {(b)ISL}$ and $\mathsf {PSL}$ the result is essentially [Reference Diego29, Corollary III.4.1] (see also [Reference Diego30, Reference Jones54]), while for $\mathsf {PDL}$ , see, e.g., [Reference Bergman3, Corollary 4.55] and [Reference Lee61].

Proof We begin by the case where ${\boldsymbol {A}} \in \mathsf {ISL}$ . Since every finitely generated subalgebra of ${\boldsymbol {A}}$ embeds into itself, in view of Proposition 2.9 there exist a family $\{ {\boldsymbol {A}}_i : i \in I \}$ of finitely generated subalgebras of ${\boldsymbol {A}}$ and an ultrafilter U on I with an embedding

$$\begin{align*}f \colon {\boldsymbol{A}} \to \prod_{i \in I}{\boldsymbol{A}}_i /U. \end{align*}$$

Since universal classes are closed under $\mathbb {S}$ and

, we have

In view of Proposition 2.8, each ${\boldsymbol {A}}_i$ is finite and, therefore, can be expanded to a Heyting algebra ${\boldsymbol {A}}_i^+$ by Proposition 2.5(ii). Clearly,

is a Heyting algebra, whose $\langle \land , \to \rangle $ -reduct ${\boldsymbol {B}}^-$ coincides with $\prod _{i \in I}{\boldsymbol {A}}_i / U$ . Since ${\boldsymbol {A}}$ embeds into ${\boldsymbol {B}}^-$ and ${\boldsymbol {B}}^- \in \mathbb {U}({\boldsymbol {A}})$ , we are done.

The same proof works for the case where ${\boldsymbol {A}}$ belongs to $\mathsf {bISL}$ or to $\mathsf {PDL}$ with the only difference that, when ${\boldsymbol {A}} \in \mathsf {PDL}$ , we apply Condition (i) of Proposition 2.5 instead of Condition (ii). Lastly, the case where ${\boldsymbol {A}}$ belongs to $\mathsf {HA}$ is straightforward, since can simply take .

It only remains to consider the case where ${\boldsymbol {A}} \in \mathsf {IL}$ . Let ${\boldsymbol {A}}^+$ be the expansion of ${\boldsymbol {A}}$ with a new constant $c_a$ for each of its elements a. Let also $0$ be another new constant. Then, consider the set of sentences

where $\mathsf {Th}({\boldsymbol {A}}^+)$ is the elementary theory of ${\boldsymbol {A}}^+$ and $0 \leqslant c_a$ is a shorthand for ${0 \land c_a \thickapprox 0}$ . Clearly, every finite part $\Gamma $ of $\Sigma $ is realizable in ${\boldsymbol {A}}^+$ , for if $c_{a_1}, \dots , c_{a_n}, 0$ are the new constants appearing in $\Gamma $ , we can interpret $0$ as $c_{a_1} \land \dots \land c_{a_n}$ in ${\boldsymbol {A}}^+$ . Therefore, we can apply the Compactness Theorem of first-order logic, obtaining that $\Sigma $ has a model $\boldsymbol {C}$ .

Let ${\boldsymbol {B}}^+$ be the subalgebra of $\boldsymbol {C}$ generated by $\{ 0 \} \cup \{ c_a : a \in A \}$ . As $\boldsymbol {C}$ is a model of $\mathsf {Th}({\boldsymbol {A}}^+)$ , the map $f \colon A \to C$ defined by the assignment is an elementary embedding of ${\boldsymbol {A}}$ into the $\langle \land , \lor , \to , 1 \rangle $ -reduct ${\boldsymbol {C}}^-$ of $\boldsymbol {C}$ . Consequently, ${\boldsymbol {A}}$ embeds also into the $\langle \land , \lor , \to , 1 \rangle $ -reduct ${\boldsymbol {B}}^-$ of ${\boldsymbol {B}}^+$ . Furthermore, as ${\boldsymbol {B}}^-$ embeds into the elementary extension ${\boldsymbol {C}}^-$ of ${\boldsymbol {A}}$ and the validity of universal sentences persists in elementary extensions and subalgebras, ${\boldsymbol {B}}^-$ satisfies all the universal sentences valid in ${\boldsymbol {A}}$ and, therefore, belongs to $\mathbb {U}({\boldsymbol {A}})$ .

To conclude the proof, it only remains to show that ${\boldsymbol {B}}^-$ is the $\langle \land , \lor , \to , 1 \rangle $ -reduct of a Heyting algebra ${\boldsymbol {B}}$ . Since ${\boldsymbol {A}} \in \mathsf {IL}$ and $\mathsf {IL}$ is a universal class, from ${\boldsymbol {B}}^-\in \mathbb {U}({\boldsymbol {A}})$ it follows that ${\boldsymbol {B}}^- \in \mathsf {IL}$ . Therefore, it suffices to prove that $0$ is the minimum element of ${\boldsymbol {B}}^-$ , as in this case we can let ${\boldsymbol {B}}$ be the expansion of ${\boldsymbol {B}}^-$ with $0$ .

To prove this, recall that ${\boldsymbol {B}}^+$ is the subalgebra of $\boldsymbol {C}$ generated by of $\{ 0 \} \cup \{ c_a : a \in A \}$ . Therefore, every element of ${\boldsymbol {B}}^-$ (equiv. of ${\boldsymbol {B}}^+$ ) has the form $\varphi ^{{\boldsymbol {C}}}(0, c_{a_1}, \dots , c_{a_n})$ for some $a_1, \dots , a_n \in A$ and $\langle \land , \lor , \to , 1 \rangle $ -term $\varphi (y, x_1, \dots , x_n)$ . We will prove by induction on the construction of $\varphi $ that

$$\begin{align*}0 \leqslant \varphi^{{\boldsymbol{C}}}(0, c_{a_1}, \dots, c_{a_n}) \end{align*}$$

for every $a_1, \dots , a_n \in A$ .

In the base case, $\varphi $ is either the constant $1$ or a variable. If it is the constant $1$ , then $\varphi ^{\boldsymbol {C}}(0, c_{a_1}, \dots , c_{a_n})$ is the maximum of ${\boldsymbol {C}}$ , whence the above display holds. Then we consider the case where $\varphi $ is a variable. Since $\varphi \in \{ y, x_1, \dots , x_n \}$ , we have $\varphi ^{\boldsymbol {C}}(0, c_{a_1}, \dots , c_{a_n}) \in \{ 0, c_{a_1}, \dots , c_{a_n}\}$ . If $\varphi ^{\boldsymbol {C}}(0, c_{a_1}, \dots , c_{a_n}) = 0$ , it is clear that the above display holds. Consider the case where $\varphi ^{\boldsymbol {C}}(0, c_{a_1}, \dots , c_{a_n}) = c_{a_i}$ for some $i \leqslant n$ . Since ${\boldsymbol {C}}$ is a model of the formula $0 \leqslant c_{a_i}$ (which belongs to $\Sigma $ ), we obtain $0 \leqslant c_{a_i} = \varphi ^{\boldsymbol {C}}(0, c_{a_1}, \dots , c_{a_n})$ as desired.

In the induction step, $\varphi $ is a complex formula. The case where $\varphi $ is of the form $\psi _1 \land \psi _2$ or $\psi _1 \lor \psi _2$ is straightforward. Accordingly, we detail only the case where $\varphi $ is of the form $\psi _1 \to \psi _2$ . By the inductive hypothesis, we have

$$\begin{align*}0 \leqslant \psi_2^{{\boldsymbol{C}}}(0, c_{a_1}, \dots, c_{a_n}) \end{align*}$$

and, therefore,

$$\begin{align*}0 \land \psi_1^{{\boldsymbol{C}}}(0, c_{a_1}, \dots, c_{a_n}) \leqslant \psi_2^{{\boldsymbol{C}}}(0, c_{a_1}, \dots, c_{a_n}). \end{align*}$$

Since ${\boldsymbol {C}}$ is an elementary extension of ${\boldsymbol {A}}$ , it satisfies Condition (4). Consequently, from the above display it follows

$$\begin{align*}0 \leqslant \psi_1^{{\boldsymbol{C}}}(0, c_{a_1}, \dots, c_{a_n}) \to^{{\boldsymbol{C}}} \psi_2^{{\boldsymbol{C}}}(0, c_{a_1}, \dots, c_{a_n}) = \varphi^{{\boldsymbol{C}}}(0, c_{a_1}, \dots, c_{a_n}). \end{align*}$$

Hence, we conclude that $0$ is the minimum of ${\boldsymbol {B}}^-$ as desired.

Proof Consider the poset $\mathbb {X}$ whose universe is

$$\begin{align*}\{ P : P\text{ is a filter of }{\boldsymbol{B}} \text{ such that }F \cup f[G] \subseteq P \text{ and }P \cap f[A \smallsetminus G] = \emptyset \} \end{align*}$$

and whose order is the inclusion relation. By assumption, $\mathbb {X}$ is nonempty because it contains the filter of ${\boldsymbol {B}}$ generated by $F \cup f[G]$ . Since $\mathbb {X}$ is closed under unions of chains, we can apply Zorn’s Lemma obtaining that $\mathbb {X}$ has a maximal element H.

Proof of the Claim

Suppose the contrary, with a view to contradiction. By assumption, G is a meet irreducible filter of ${\boldsymbol {A}}$ and, therefore, proper. Consequently, $A \smallsetminus G$ is nonempty, whence so is $f[A \smallsetminus G]$ . Since H is disjoint from $f[A \smallsetminus G]$ , we conclude that H is proper. Since H is not meet irreducible, this means that there are two filters $H_1, H_2$ of ${\boldsymbol {B}}$ other than H such that $H = H_1 \cap H_2$ . From the maximality of H in $\mathbb {X}$ it follows that neither $H_1$ nor $H_2$ is disjoint from $f[A \smallsetminus G]$ . Therefore, there are $a, b\in A \smallsetminus G$ such that

(5) $$ \begin{align} f(a) \in H_1 \, \, \text{ and } \, \, f(b) \in H_2. \end{align} $$

Now, from $a, b \in A \smallsetminus G$ it follows that G is properly contained in $G + {\uparrow }^{\boldsymbol {A}} a$ and $G + {\uparrow }^{\boldsymbol {A}} b$ . Since G a meet irreducible filter of ${\boldsymbol {A}}$ , this guarantees that

$$\begin{align*}G \subsetneq (G + {\uparrow}^{\boldsymbol{A}} a) \cap (G + {\uparrow}^{\boldsymbol{B}} a). \end{align*}$$

Accordingly, there are $c \in G$ and $d \in A \smallsetminus G$ such that

$$\begin{align*}a \land^{{\boldsymbol{A}}} c \leqslant d \, \, \text{ and } \, \, b \land^{{\boldsymbol{A}}} c \leqslant d. \end{align*}$$

Since f is a homomorphism, we obtain

(6) $$ \begin{align} f(a) \land^{\boldsymbol{B}} f(c) \leqslant f(d) \, \, \text{ and } \, \, f(b) \land^{\boldsymbol{B}} f(c) \leqslant f(d). \end{align} $$

Furthermore, from $c \in G$ and the assumption that $f[G] \subseteq H = H_1 \cap H_2$ it follows that $f(c) \in H_1 \cap H_2$ . Together with Conditions (5) and (6) and the fact that $H_1, H_2$ are filters of ${\boldsymbol {B}}$ , this implies that

$$\begin{align*}f(d) \in H_1 \cap H_2 = H, \end{align*}$$

a contradiction with the assumption that $d \in A \smallsetminus G$ and $H \cap f[A \smallsetminus G] = \emptyset $ . Hence, we conclude that H is meet irreducible.

By the claim, $H \in {\boldsymbol {B}}_\ast $ . Furthermore, since $H \in \mathbb {X}$ , we know that $F \subseteq H$ . Therefore, it only remains to prove that $G = f^{-1}[H]$ . The fact that $H \in \mathbb {X}$ guarantees that $f[G] \subseteq H$ and $H \cap f[A \smallsetminus G] = \emptyset $ . From $f[G] \subseteq H$ it follows $G \subseteq f^{-1}[f[G]] \subseteq f^{-1}[H]$ . To prove the other inclusion, consider $a \in f^{-1}[H]$ . Then $f(a) \in H$ . Since $H \cap f[A \smallsetminus G] = \emptyset $ and $a \in A$ , this implies that $a \in G$ as desired.

Proof The proof of the first part of the statement is straightforward. As for the second part, suppose that p is total and order preserving. The inclusion from left to right follows from the fact that the function ${\downarrow }^{\mathbb {X}}p^{-1}[-] \colon \mathscr {P}({Y}) \to \mathscr {P}({X})$ is order preserving. To prove the other inclusion, consider $x \in {\downarrow }^{\mathbb {X}} p^{-1}[Y \smallsetminus U] \cap {\downarrow }^{\mathbb {X}} p^{-1}[Y \smallsetminus V]$ . Then there are $u, v \in X$ such that $p(u) \in Y \smallsetminus U$ , $p(v) \in Y \smallsetminus V$ , and $x \leqslant ^{\mathbb {X}}u, v$ . Since p is a total function, $x \in \mathsf {dom}(p)$ . Furthermore, as p is order preserving and $x \leqslant ^{\mathbb {X}}u, v$ , we obtain $p(x) \leqslant ^{\mathbb {Y}}p(u), p(v)$ . Together with the assumption that U and V are upsets and that $p(u) \notin U$ and $p(v) \notin V$ , this yields $p(x) \in Y \smallsetminus (U \cup V)$ . Hence, we conclude that $x \in {\downarrow }^{\mathbb {X}} p^{-1}[Y \smallsetminus (U \cup V)]$ .

Proof Suppose, with a view to contradiction, that $f^{-1}[F] \subseteq G$ and that the filter of ${\boldsymbol {B}}$ generated by $F \cup f[G]$ is not disjoint from $f[A \smallsetminus G]$ . Then there exist $a_1, \dots , a_n \in G$ , $b_1, \dots , b_m \in F$ , and $c \in A \smallsetminus G$ such that

$$\begin{align*}f(a_1) \land^{\boldsymbol{B}} \dots \land^{\boldsymbol{B}} f(a_n) \land^{\boldsymbol{B}} b_1 \land^{\boldsymbol{B}} \dots \land^{\boldsymbol{B}} b_m \leqslant f(c). \end{align*}$$

Since G is maximal and proper, $G + {\uparrow }^{\boldsymbol {A}} c = A$ . Therefore, $c \land ^{\boldsymbol {A}} d = 0^{\boldsymbol {A}}$ for some $d \in G$ . By Condition (3), this yields $d \leqslant \lnot ^{\boldsymbol {A}} c$ and, since f is a homomorphism, $f(d) \leqslant \lnot ^{\boldsymbol {B}} f(c)$ .

As $d \in G$ , we may assume, without loss of generality, that $d \in \{ a_1, \dots , a_n \}$ . Therefore, from $f(d) \leqslant \lnot ^{\boldsymbol {B}} f(c)$ and the above display it follows

$$\begin{align*}f(a_1) \land^{\boldsymbol{B}} \dots \land^{\boldsymbol{B}} f(a_n) \land^{\boldsymbol{B}} b_1 \land^{\boldsymbol{B}} \dots \land^{\boldsymbol{B}} b_m \leqslant f(c) \land^{\boldsymbol{B}} \lnot^{\boldsymbol{B}} f(c) = 0^{\boldsymbol{B}}. \end{align*}$$

Since f is a homomorphism, we can apply Condition (3) obtaining

$$\begin{align*}b_1 \land^{\boldsymbol{B}} \dots \land^{\boldsymbol{B}} b_m \leqslant \lnot^{\boldsymbol{B}} f(a_1 \land^{\boldsymbol{A}} \dots \land^{\boldsymbol{A}} a_n) = f(\lnot^{\boldsymbol{A}}(a_1 \land^{\boldsymbol{A}} \dots \land^{\boldsymbol{A}} a_n)). \end{align*}$$

As F is a filter of ${\boldsymbol {B}}$ containing $b_1, \dots , b_m$ , the above display guarantees that $f(\lnot ^{\boldsymbol {A}}(a_1 \land ^{\boldsymbol {A}} \dots \land ^{\boldsymbol {A}} a_n)) \in F$ . As a consequence, $\lnot ^{\boldsymbol {A}}(a_1 \land ^{\boldsymbol {A}} \dots \land ^{\boldsymbol {A}} a_n) \in f^{-1}[F] \subseteq G$ . But together with the fact that $a_1, \dots , a_n \in G$ and that G is a filter of ${\boldsymbol {A}}$ , this implies

$$\begin{align*}0^{\boldsymbol{A}} = (a_1 \land^{\boldsymbol{A}} \dots \land^{\boldsymbol{A}} a_n) \land \lnot^{\boldsymbol{A}}(a_1 \land^{\boldsymbol{A}} \dots \land^{\boldsymbol{A}} a_n) \in G, \end{align*}$$

a contradiction with the assumption that G is proper.

Proof (i): The definition of $f_\ast $ guarantees that $f_\ast \colon {\boldsymbol {B}}_\ast \rightharpoonup {\boldsymbol {A}}_\ast $ is a well-defined partial order preserving map. Therefore, it suffices to prove that:

1. 1. ${\boldsymbol {B}}_\ast = {\downarrow }^{{\boldsymbol {B}}_\ast }\{ F \in {\boldsymbol {B}}_\ast : {\uparrow }^{{\boldsymbol {B}}_\ast }F \subseteq \mathsf {dom}(f_\ast )\}$ .

2. 2. For every $F \in \mathsf {dom}(f_\ast )$ and $G \in {\boldsymbol {A}}_\ast $ ,

   $$\begin{align*}\text{if }f_\ast(F) \subseteq G, \text{ there exists }H \in \mathsf{dom}(f_\ast)\text{ s.t. }F \subseteq H \text{ and }G \subseteq f_\ast(H). \end{align*}$$

Notice that inverse images under f of proper filters of ${\boldsymbol {B}}$ are proper filters of ${\boldsymbol {A}}$ , because f preserves binary meets and minimum elements. We will use this observation repeatedly.

To prove Condition (1), it suffices to establish the inclusion ${\boldsymbol {B}}_\ast \subseteq {\downarrow }^{{\boldsymbol {B}}_\ast }\{ F \in {\boldsymbol {B}}_\ast : {\uparrow }^{{\boldsymbol {B}}_\ast }F \subseteq \mathsf {dom}(f_\ast )\}$ as the other one is obvious. Accordingly, consider $F \in {\boldsymbol {B}}_\ast $ . Since F is a proper filter of ${\boldsymbol {B}}$ , the set $f^{-1}[F]$ is a proper filter of ${\boldsymbol {A}}$ . By Zorn’s Lemma, we can extend it to a maximal proper filter G of ${\boldsymbol {A}}$ . Being maximal and proper, G is meet irreducible and, therefore, it belongs to ${\boldsymbol {A}}_\ast $ . Furthermore, since $f^{-1}[F] \subseteq G$ , we can apply Lemma 3.9 obtaining that the filter of ${\boldsymbol {B}}$ generated by $F \cup f[G]$ is disjoint from $f[A \smallsetminus G]$ . By Lemma 3.6, there exists $H \in {\boldsymbol {B}}_\ast $ such that $F \subseteq H$ and $G = f^{-1}[H]$ . Thus, $H \in \mathsf {dom}(f_\ast )$ and $f_\ast (H) = G$ . To conclude the proof, it only remains to show that ${\uparrow }^{{\boldsymbol {B}}_\ast } H \subseteq \mathsf {dom}(f_\ast )$ . To this end, consider $H^+ \in {\uparrow }^{{\boldsymbol {B}}_\ast } H$ . Since $H^+$ is a proper filter of ${\boldsymbol {B}}$ , the set $f^{-1}[H^+]$ is a proper filter of ${\boldsymbol {A}}$ . Furthermore, $G = f^{-1}[H] \subseteq f^{-1}[H^+]$ . Since G is a maximal proper filter of ${\boldsymbol {A}}$ and $f^{-1}[H^+]$ is a proper filter of ${\boldsymbol {A}}$ , this implies $f^{-1}[H^+] = G \in {\boldsymbol {A}}_\ast $ . Hence, we conclude that $H^+ \in \mathsf {dom}(f_\ast )$ as desired.

Then we turn to prove Condition (2). Consider $F \in \mathsf {dom}(f_\ast )$ and $G \in {\boldsymbol {A}}_\ast $ such that $f_\ast (F) \subseteq G$ , that is, $f^{-1}[F] \subseteq G$ . We may assume, without loss of generality, that G is maximal and proper (otherwise we use Zorn’s Lemma to extend it to such a filter). Therefore, we can apply Lemma 3.9 obtaining that the filter of ${\boldsymbol {B}}$ generated by $F \cup f[G]$ is disjoint from $f[A \smallsetminus G]$ . By Lemma 3.6, there exists $H \in {\boldsymbol {B}}_\ast $ such that $F \subseteq H$ and $G = f^{-1}[H]$ . We conclude that $H \in \mathsf {dom}(f_\ast )$ and $G = f_\ast (H)$ .

(ii): To see that $\mathsf {Up}_{\mathsf {PSL}}(p) \colon \mathsf {Up}_{\mathsf {PSL}}(\mathbb {Y}) \to \mathsf {Up}_{\mathsf {PSL}}(\mathbb {X})$ preserves binary meets, recall from Lemma 3.8 that

$$\begin{align*}{\downarrow}^{\mathbb{X}} p^{-1}[Y \smallsetminus (U \cap V)] = {\downarrow}^{\mathbb{X}} p^{-1}[Y \smallsetminus U] \cup {\downarrow}^{\mathbb{X}} p^{-1}[Y \smallsetminus V], \end{align*}$$

for every $U,V \in \mathsf {Up}({\mathbb {Y}})$ . Therefore, we obtain

$$\begin{align*}X \smallsetminus {\downarrow}^{\mathbb{X}} p^{-1}[Y \smallsetminus (U \cap V)] = ( X \smallsetminus {\downarrow}^{\mathbb{X}} p^{-1}[Y \smallsetminus U]) \cap (X \smallsetminus {\downarrow}^{\mathbb{X}} p^{-1}[Y \smallsetminus V]), \end{align*}$$

that is, $\mathsf {Up}_{\mathsf {PSL}}(p)(U \cap V) = \mathsf {Up}_{\mathsf {PSL}}(p)(U) \cap \mathsf {Up}_{\mathsf {PSL}}(p)(V)$ , as desired.

We now detail the proof that $\mathsf {Up}_{\mathsf {PSL}}(p)$ preserves the operation $\lnot $ (this, in turn, guarantees that $\mathsf {Up}_{\mathsf {PSL}}(p)$ preserves also the constant $0$ , because the latter is term-definable as $x \land \lnot x$ ). Consider $U \in \mathsf {Up}_{\mathsf {PSL}}(\mathbb {Y})$ . We need to prove that

$$\begin{align*}\mathsf{Up}_{\mathsf{PSL}}(p) (\lnot^{\mathsf{Up}_{\mathsf{PSL}}(\mathbb{Y})} (U)) = \lnot^{\mathsf{Up}_{\mathsf{PSL}}(\mathbb{X})}\mathsf{Up}_{\mathsf{PSL}}(p)(U). \end{align*}$$

Using the definitions of $\mathsf {Up}_{\mathsf {PSL}}(p)$ and of the operation $\lnot $ in $\mathsf {Up}_{\mathsf {PSL}}(\mathbb {Y})$ and $\mathsf {Up}_{\mathsf {PSL}}(\mathbb {X})$ , this amounts to

$$\begin{align*}X \smallsetminus {\downarrow}^{\mathbb{X}}(p^{-1}[{\downarrow}^{\mathbb{Y}}U]) = X \smallsetminus {\downarrow}^{\mathbb{X}}(X \smallsetminus {\downarrow}^{\mathbb{X}}(p^{-1}[Y \smallsetminus U])). \end{align*}$$

Clearly, it suffices to show that the complements of the sets in the above display coincide, namely,

(7) $$ \begin{align} {\downarrow}^{\mathbb{X}}(p^{-1}[{\downarrow}^{\mathbb{Y}}U]) = {\downarrow}^{\mathbb{X}}(X \smallsetminus {\downarrow}^{\mathbb{X}}(p^{-1}[Y \smallsetminus U])). \end{align} $$

To prove the inclusion from left to right in Condition (7), consider $x \in {\downarrow }^{\mathbb {X}}(p^{-1}[{\downarrow }^{\mathbb {Y}}U])$ . Then there are $z \in X$ and $u \in U$ such that $z \in \mathsf {dom}(p)$ and

$$\begin{align*}x \leqslant^{\mathbb{X}} z \, \, \text{ and } \, \, p(z) \leqslant^{\mathbb{Y}} u. \end{align*}$$

Since $p(z) \leqslant ^{\mathbb {Y}} u$ and $p \colon \mathbb {X} \rightharpoonup \mathbb {Y}$ is a partial negative p-morphism, there exists $w \in \mathsf {dom}(p)$ such that $z \leqslant ^{\mathbb {X}} w$ and $u \leqslant ^{\mathbb {Y}} p(w)$ . Together with the above display, this yields $x \leqslant ^{\mathbb {X}} w$ . Therefore, to conclude that $x \in {\downarrow }^{\mathbb {X}}(X \smallsetminus {\downarrow }^{\mathbb {X}}(p^{-1}[Y \smallsetminus U]))$ , it suffices to show that $w \in X \smallsetminus {\downarrow }^{\mathbb {X}}(p^{-1}[Y \smallsetminus U])$ . Accordingly, consider $v \in \mathsf {dom}(p)$ such that $w \leqslant ^{\mathbb {X}} v$ . We need to show that $p(v) \in U$ . Since p is order preserving, from $w \leqslant ^{\mathbb {X}} v$ it follows $p(w) \leqslant ^{\mathbb {Y}} p(v)$ . Together with $u \leqslant ^{\mathbb {Y}} p(w)$ , this yields $u \leqslant ^{\mathbb {Y}}p(v)$ . Since U is an upset of $\mathbb {Y}$ and $u \in U$ , we conclude that $p(v) \in U$ as desired.

Lastly, we prove the inclusion from right to left in Condition (7). Consider $x \in {\downarrow }^{\mathbb {X}}(X \smallsetminus {\downarrow }^{\mathbb {X}}(p^{-1}[Y \smallsetminus U]))$ . Then there exists $z \in X$ such that $x \leqslant ^{\mathbb {X}} z$ and for every $w \in X$ ,

$$\begin{align*}\text{if }z \leqslant^{\mathbb{X}} w \text{ and }w \in \mathsf{dom}(p)\text{, then }p(w) \in U. \end{align*}$$

Since p is a partial negative p-morphism, $X = {\downarrow }^{\mathbb {X}}\mathsf {dom}(p)$ . Thus, there exists $w \in \mathsf {dom}(p)$ with $z \leqslant ^{\mathbb {X}} w$ . In view of the above display, we obtain $p(w) \in U$ . Since $x \leqslant ^{\mathbb {X}} z\leqslant ^{\mathbb {X}} w$ , this yields $x \in {\downarrow }^{\mathbb {X}}(p^{-1}[U]) \subseteq {\downarrow }^{\mathbb {X}}(p^{-1}[{\downarrow }^{\mathbb {Y}}U])$ .

Proof Suppose, with a view to contradiction, that $f^{-1}[F] \subseteq G$ and that the filter of ${\boldsymbol {B}}$ generated by $F \cup f[G]$ contains an element $c \in f[A \smallsetminus G]$ . Then there exist $a_1, \dots , a_n \in G$ and $b_1, \dots , b_m \in F$ such that

$$\begin{align*}f(a_1) \land^{\boldsymbol{B}} \dots \land^{\boldsymbol{B}} f(a_n) \land^{\boldsymbol{B}} b_1 \land^{\boldsymbol{B}} \dots \land^{\boldsymbol{B}} b_m \leqslant c. \end{align*}$$

Since f preserves binary meets, this yields

$$\begin{align*}f(a_1 \land^{\boldsymbol{A}} \dots \land^{\boldsymbol{A}} a_n) \land^{\boldsymbol{B}} b_1 \land^{\boldsymbol{B}} \dots \land^{\boldsymbol{B}} b_m \leqslant c. \end{align*}$$

Together $c \in f[A \smallsetminus G]$ , this implies that there exists $d \in A \smallsetminus G$ such that

$$\begin{align*}f(a_1 \land^{\boldsymbol{A}} \dots \land^{\boldsymbol{A}} a_n) \land^{\boldsymbol{B}} b_1 \land^{\boldsymbol{B}} \dots \land^{\boldsymbol{B}} b_m \leqslant f(d). \end{align*}$$

Applying Condition (4) and the fact that f preserves $\to $ to the above display, we obtain

$$\begin{align*}b_1 \land^{\boldsymbol{B}} \dots \land^{\boldsymbol{B}} b_m \leqslant f(a_1 \land^{\boldsymbol{A}} \dots \land^{\boldsymbol{A}} a_n) \to^{\boldsymbol{B}} f(d) = f((a_1 \land^{\boldsymbol{A}} \dots \land^{\boldsymbol{A}} a_n) \to^{\boldsymbol{A}} d). \end{align*}$$

Lastly, as F is a filter of ${\boldsymbol {B}}$ containing $b_1, \dots , b_m$ , the above display guarantees that $f((a_1 \land ^{\boldsymbol {A}} \dots \land ^{\boldsymbol {A}} a_n) \to ^{\boldsymbol {A}} d) \in F$ . As a consequence, $(a_1 \land ^{\boldsymbol {A}} \dots \land ^{\boldsymbol {A}} a_n) \to ^{\boldsymbol {A}} d \in f^{-1}[F] \subseteq G$ . Together with the fact that $a_1, \dots , a_n \in G$ and that G is a filter of ${\boldsymbol {A}}$ , this implies

$$\begin{align*}(a_1 \land^{\boldsymbol{A}} \dots \land^{\boldsymbol{A}} a_n) \land^{\boldsymbol{A}} ((a_1 \land^{\boldsymbol{A}} \dots \land^{\boldsymbol{A}} a_n) \to^{\boldsymbol{A}} d) \in G. \end{align*}$$

Since G is an upset and, by Condition (4), we have

$$\begin{align*}(a_1 \land^{\boldsymbol{A}} \dots \land^{\boldsymbol{A}} a_n) \land^{\boldsymbol{A}} ((a_1 \land^{\boldsymbol{A}} \dots \land^{\boldsymbol{A}} a_n) \to^{\boldsymbol{A}} d) \leqslant d, \end{align*}$$

this implies $d \in G$ , a contradiction with the assumption that $d \in A \smallsetminus G$ .

Proof (i): The definition of $f_\ast $ guarantees that $f_\ast \colon {\boldsymbol {B}}_\ast \rightharpoonup {\boldsymbol {A}}_\ast $ is a well-defined partial order preserving map. Therefore, it suffices to prove that for every $F \in \mathsf {dom}(f_\ast )$ and $G \in {\boldsymbol {A}}_\ast $ ,

$$\begin{align*}\text{if }f_\ast(F) \subseteq G, \text{ there exists }H \in \mathsf{dom}(f_\ast)\text{ s.t. }F \subseteq H \text{ and }G = f_\ast(H). \end{align*}$$

Accordingly, let $F \in \mathsf {dom}(f_\ast )$ and $G \in {\boldsymbol {A}}_\ast $ be such that $f_\ast (F) \subseteq G$ , that is, $f^{-1}[F] \subseteq G$ . By Lemma 3.11, the filter of ${\boldsymbol {B}}$ generated by $F \cup f[G]$ is disjoint from $f[A \smallsetminus G]$ . Therefore, we can apply Lemma 3.6 obtaining an $H \in {\boldsymbol {B}}_\ast $ that contains F such that $G = f^{-1}[H]$ . Since $G \in {\boldsymbol {A}}_\ast $ , we conclude that $H \in \mathsf {dom}(f_\ast )$ and $G = f_{\ast }(H)$ as desired.

(ii): The proof of this condition coincides with that of [Reference Bezhanishvili and Bezhanishvili4, Theorem 3.15] (although the respective statements are slightly different).

Proof The cases where $\mathsf {K}$ is $\mathsf {PSL}$ , $\mathsf {ISL}$ , or $\mathsf {bISL}$ follow from Propositions 3.10 and 3.12, while the case where $\mathsf {K} = \mathsf {HA}$ is well known (see Remark 3.4). Therefore it only remains to detail the cases of $\mathsf {PDL}$ and $\mathsf {IL}$ . We detail the case of $\mathsf {PDL}$ only, as that of $\mathsf {IL}$ is analogous.

To prove Condition (i), consider ${\boldsymbol {A}}, {\boldsymbol {B}} \in \mathsf {PDL}$ and let $f \colon {\boldsymbol {A}} \to {\boldsymbol {B}}$ be a homomorphism. Since f is a homomorphism of pseudocomplemented semilattices, $f_\ast \colon {\boldsymbol {B}}_\ast \rightharpoonup {\boldsymbol {A}}_\ast $ is a partial negative p-morphism by Proposition 3.10(i). To prove that $f_\ast $ is total, consider $F \in {\boldsymbol {B}}_\ast $ . Since ${\boldsymbol {B}}$ is a distributive lattice, F is a prime filter in view of Remark 3.3. As f preserves binary joins, $f^{-1}[F]$ is a prime filter of ${\boldsymbol {A}}$ , whence $f^{-1}[F] \in {\boldsymbol {A}}_\ast $ by Remark 3.3. Thus, we conclude that $F \in \mathsf {dom}(f_\ast )$ and, therefore, that $f_\ast $ is total. This shows that $f_\ast $ is a negative p-morphism.

To prove Condition (ii), let $p \colon \mathbb {X} \rightharpoonup \mathbb {Y}$ be a negative p-morphism. By Proposition 3.10(ii), $\mathsf {Up}_{\mathsf {PDL}}(p)$ is a homomorphism of pseudocomplemented semilattices. Therefore, it only remains to prove that it preserves binary joins. To this end, consider $U, V \in \mathsf {Up}(\mathbb {Y})$ . Since

$$ \begin{align*} \mathsf{Up}_{\mathsf{PDL}}(p)(U \lor^{\mathsf{Up}_{\mathsf{PDL}}(\mathbb{Y})} V)&= X \smallsetminus {\downarrow}^{\mathbb{X}} p^{-1}[Y \smallsetminus (U \cup V)]\\ \mathsf{Up}_{\mathsf{PDL}}(p)(U) \lor^{\mathsf{Up}_{\mathsf{PDL}}(\mathbb{X})} \mathsf{Up}_{\mathsf{PDL}}(p)(V)&= (X \smallsetminus {\downarrow}^{\mathbb{X}} p^{-1}[Y \kern1.3pt{\smallsetminus}\kern1.3pt U]) \cup (X \kern1.3pt{\smallsetminus}\kern1.3pt {\downarrow}^{\mathbb{X}} p^{-1}[Y \kern1.3pt{\smallsetminus}\kern1.3pt V]), \end{align*} $$

it suffices to show that

$$\begin{align*}(X \smallsetminus {\downarrow}^{\mathbb{X}} p^{-1}[Y \smallsetminus U]) \cup (X \smallsetminus {\downarrow}^{\mathbb{X}} p^{-1}[Y \smallsetminus V]) = X \smallsetminus {\downarrow}^{\mathbb{X}} p^{-1}[Y \smallsetminus (U \cup V)]. \end{align*}$$

As p is order preserving and total, we can apply the second part of Lemma 3.8 obtaining that the above display holds. As a consequence, $\mathsf {Up}_{\mathsf {PDL}}(p)$ is a homomorphism as desired.

It only remains to prove the last part of the statement, namely, that for every variety $\mathsf {K}$ among $\mathsf {PSL}, \mathsf {(b)ISL}, \mathsf {PDL}$ , $\mathsf {IL}$ , and $\mathsf {HA}$ , every homomorphism $f \colon {\boldsymbol {A}} \to {\boldsymbol {B}}$ in $\mathsf {K}$ , and every arrow $p \colon \mathbb {X} \rightharpoonup \mathbb {Y}$ in $\mathsf {K}^\partial $ , if f is injective (resp. p is surjective), then $f_\ast $ is surjective (resp. $\mathsf {Up}_{\mathsf {K}}(p)$ is injective).

Suppose first that $f \colon {\boldsymbol {A}} \to {\boldsymbol {B}}$ is injective and consider $G \in {\boldsymbol {A}}_\ast $ . Since G is proper, we can choose an element $a \in G$ and consider the filter of ${\boldsymbol {B}}$ . Since f is order reflecting, $f^{-1}[F] \subseteq {\uparrow }^{\boldsymbol {A}} a$ . Together with $a \in G$ and the fact that G is an upset, this implies $f^{-1}[F] \subseteq G$ . We will prove that the filter of ${\boldsymbol {B}}$ generated by $F \cup f[G]$ is disjoint from $f[A \smallsetminus G]$ . Suppose, on the contrary, that there exists some $b \in B$ in the filter of ${\boldsymbol {B}}$ generated by $F \cup f[G]$ and in $f[A \smallsetminus G]$ . Since $b \in f[A \smallsetminus G]$ there exists $c \in A \smallsetminus G$ such that $f(c) = b$ . Since $F = {\uparrow }^{\boldsymbol {B}} f(a)$ and b belongs to the filter of ${\boldsymbol {B}}$ generated by $F \cup f[G]$ , there are $a_1, \dots , a_n \in G$ such that

$$\begin{align*}f(a \land^{\boldsymbol{A}} a_1 \land^{\boldsymbol{A}} \dots \land^{\boldsymbol{A}} a_n )= f(a) \land^{\boldsymbol{B}} f(a_1) \land^{\boldsymbol{B}} \dots \land^{{\boldsymbol{B}}} f(a_n) \leqslant b = f(c). \end{align*}$$

As f is order reflecting, this implies $a \land ^{\boldsymbol {A}} a_1 \land ^{\boldsymbol {A}} \dots \land ^{\boldsymbol {A}} a_n \leqslant a$ . As G is a filter and $a, a_1, \dots , a_n \in G$ , we obtain that $c \in G$ , a contradiction with the assumption that $c \in A \smallsetminus G$ . In sum, $f^{-1}[F] \subseteq G$ and the filter of ${\boldsymbol {B}}$ generated by $F \cup f[G]$ is disjoint from $f[A \smallsetminus G]$ . Therefore, we can apply Lemma 3.6 obtaining an $H \in {\boldsymbol {B}}_\ast $ such that $G = f^{-1}[H]$ . Hence, $H \in \mathsf {dom}(f_\ast )$ and $f_\ast (H) = G$ and we conclude that $f_\ast $ is surjective.

Lastly, consider a surjective arrow $p \colon \mathbb {X} \rightharpoonup \mathbb {Y}$ in $\mathsf {K}^\partial $ and let $U, V$ be distinct upsets of $\mathbb {Y}$ . By symmetry, we may assume that there exists $y \in U \smallsetminus V$ . Since p is surjective, there exists $x \in \mathsf {dom}(p)$ such that $p(x) = y$ . Since p is order preserving, $p(x) \in U$ , and U is an upset, we have $x \notin {\downarrow }^{\mathbb {X}}p^{-1}[Y \smallsetminus U]$ , whence $x \in X \smallsetminus {\downarrow }^{\mathbb {X}}p^{-1}[Y \smallsetminus U] = \mathsf {Up}_{\mathsf {K}}(p)(U)$ . On the other hand, as $p(x) \in U \smallsetminus V \subseteq Y \smallsetminus V$ , we obtain $x \in {\downarrow }^{\mathbb {X}}p^{-1}[Y \smallsetminus V]$ , whence $x \notin X \smallsetminus {\downarrow }^{\mathbb {X}}p^{-1}[Y \smallsetminus V] = \mathsf {Up}_{\mathsf {K}}(p)(V)$ . Hence, we conclude that $\mathsf {Up}_{\mathsf {K}}(p)(U) \ne \mathsf {Up}_{\mathsf {K}}(p)(V)$ and, therefore, that $\mathsf {Up}_{\mathsf {K}}(p)$ is injective as desired.

Proof Suppose that ${\boldsymbol {A}} \vDash \Phi $ and consider $\vec {a} \in A$ . For every $i \leqslant n$ , we have

$$\begin{align*}\varphi_i(\vec{a}) \land 1 = \varphi_i(\vec{a}) \leqslant\varphi_1(\vec{a}) \lor \dots \lor \varphi_n(\vec{a}). \end{align*}$$

Since ${\boldsymbol {A}} \vDash \Phi $ , this implies $1 \leqslant \varphi _1(\vec {a}) \lor \dots \lor \varphi _n(\vec {a})$ . As $1$ is the maximum of ${\boldsymbol {A}}$ , we conclude that $\varphi _1(\vec {a}) \lor \dots \lor \varphi _n(\vec {a}) = 1$ as desired.

Conversely, suppose that ${\boldsymbol {A}} \vDash \varphi _1 \lor \dots \lor \varphi _n$ and consider $\vec {a}, b, c \in A$ such that $\varphi _i(\vec {a}) \land b \leqslant c$ for every $i \leqslant n$ . Using the distributive laws, we obtain

$$\begin{align*}(\varphi_1(\vec{a}) \lor \dots \lor \varphi_n(\vec{a})) \land b = (\varphi_1(\vec{a}) \land b) \lor \dots \lor(\varphi_n(\vec{a}) \land b) \leqslant c. \end{align*}$$

Since ${\boldsymbol {A}} \vDash \varphi _1 \lor \dots \lor \varphi _n$ , this yields $b = 1 \land b \leqslant c$ , whence ${\boldsymbol {A}} \vDash \Phi $ .

Modal Sahlqvist Theorem 4.9 [Reference Blackburn, de Rijke and Venema7, Theorems 3.54 and 5.91].

The following conditions hold for a modal Sahlqvist quasiequation $\Phi$

1. (i) Canonicity $:$ If a modal algebra ${\boldsymbol {A}}$ validates $\Phi $ , then also $\mathscr {P}_{\mathsf {M}}({{\boldsymbol {A}}_+})$ validates $\Phi $ .

2. (ii) Correspondence $:$ There is an effectively computable first-order sentence $\mathsf {mtr}(\Phi )$ Footnote ⁷ in the language of Kripke frames such that $\mathscr {P}_{\mathsf {M}}({\mathbb {X}}) \vDash \Phi $ iff $\mathbb {X} \vDash \mathsf {mtr}(\Phi )$ , for every Kripke frame $\mathbb {X}$ .

Proof (i): It is well known that

(8) $$ \begin{align} \mathsf{Up}(\mathbb{X}) \vDash \varphi \iff \mathscr{P}_{\mathsf{M}}({\mathbb{X}}) \vDash \varphi_g \end{align} $$

for every formula $\varphi $ of $\mathcal {L}$ and poset $\mathbb {X}$ (see, e.g., [Reference Chagrov and Zakharyaschev17, Corollary 3.82]). Then for every poset $\mathbb {X}$ and Sahlqvist quasiequation $\Phi = \varphi _1 \land y \leqslant z \, \& \cdots \& \, \varphi _n \land y \leqslant z \Longrightarrow y \leqslant z$ , we have

$$ \begin{align*} \mathsf{Up}(\mathbb{X}) \vDash \Phi \, \, &\Longleftrightarrow \, \, \mathsf{Up}(\mathbb{X}) \vDash \varphi_1 \lor \dots \lor \varphi_n\\ \, \, &\Longleftrightarrow \, \,\mathscr{P}_{\mathsf{M}}({\mathbb{X}}) \vDash (\varphi_1 \lor \dots \lor \varphi_n)_g \\ \, \,&\Longleftrightarrow \, \,\mathscr{P}_{\mathsf{M}}({\mathbb{X}}) \vDash \varphi_{1g} \lor \dots \lor \varphi_{ng}\\ \, \,&\Longleftrightarrow \, \,\mathscr{P}_{\mathsf{M}}({\mathbb{X}}) \vDash \Phi_g. \end{align*} $$

The equivalences above are justified as follows: the first and the last follow, respectively, from Corollary 4.7 and Proposition 4.6, the second holds by Condition (8), and the third by the definition of the Gödel–McKinsey–Tarski translation.

(ii): Let $\mathsf {f}({\boldsymbol {A}})$ be the subalgebra of $\mathscr {P}_{\mathsf {M}}({{\boldsymbol {A}}_\ast })$ generated by the sets of the form

for every $a \in A$ . In view of [Reference Maksimova and Rybakov62, Lemmas 3.1 and 3.2], for every formula $\varphi $ of $\mathcal {L}$ we have

$$\begin{align*}{\boldsymbol{A}} \vDash \varphi \, \,\Longleftrightarrow \, \,\mathsf{f}({\boldsymbol{A}})\vDash \varphi_g. \end{align*}$$

As in the proof of Condition (i), this implies that ${\boldsymbol {A}} \vDash \Phi $ iff $\mathsf {f}({\boldsymbol {A}}) \vDash \Phi _g$ , for every Sahlqvist quasiequation $\Phi $ . For a proof that ${\boldsymbol {A}}_\ast \cong \mathsf {f}({\boldsymbol {A}})_+$ , see, e.g., [Reference Esakia34, Construction 2.5.7 and Theorem 3.4.6(1)].

Intuitionistic Sahlqvist Theorem 4.12 [Reference Conradie, Palmigiano and Zhao22, Theorems 6.1 and 7.1].

The following conditions hold for a Sahlqvist quasiequation $\Phi :$

1. (i) Canonicity $:$ If a Heyting algebra ${\boldsymbol {A}}$ validates $\Phi $ , then also $\mathsf {Up}({\boldsymbol {A}}_\ast )$ validates $\Phi $ .

2. (ii) Correspondence $:$ There is an effectively computable first-order sentence $\mathsf {tr}(\Phi )$ in the language of posets such that $\mathsf {Up}(\mathbb {X}) \vDash \Phi $ iff $\mathbb {X} \vDash \mathsf {tr}(\Phi _g)$ , for every poset  $\mathbb {X}$ .

Proof (i): Suppose that ${\boldsymbol {A}} \vDash \Phi $ . In view of Proposition 4.11(ii), we have $\mathsf {f}({\boldsymbol {A}}) \vDash \Phi _g$ . As $\Phi _g$ is a modal Sahlqvist quasiequation by Lemma 4.10, we can apply the canonicity part of the Modal Sahlqvist Theorem obtaining $\mathscr {P}_{\mathsf {M}}({\mathsf {f}({\boldsymbol {A}})_+}) \vDash \Phi _g$ . Since ${\boldsymbol {A}}_\ast \cong \mathsf {f}({\boldsymbol {A}})_+$ by Proposition 4.11(ii), this amounts to $\mathscr {P}_{\mathsf {M}}({{\boldsymbol {A}}_\ast }) \vDash \Phi _g$ . Together with by Proposition 4.11(i), this implies that $\mathsf {Up}({\boldsymbol {A}}_\ast ) \vDash \Phi $ as desired.

(ii): From Proposition 4.11(i) it follows that $\mathsf {Up}(\mathbb {X}) \vDash \Phi $ iff $\mathscr {P}_{\mathsf {M}}({\mathbb {X}}) \vDash \Phi _g$ . Furthermore, as $\Phi _g$ is a modal Sahlqvist quasiequation by Lemma 4.10, we can apply the correspondence part of the Modal Sahlqvist Theorem obtaining that $\mathscr {P}_{\mathsf {M}}({\mathbb {X}}) \vDash \Phi _g$ iff $\mathbb {X} \vDash \mathsf {mtr}(\Phi _g)$ , where the first-order sentence $\mathsf {mtr}(\Phi _g)$ is effectively computable. Therefore, setting , we are done.

Proof It is well known that, in view of the poorness of the language $\mathcal {L}_\land $ , the class $\mathsf {K}$ of $\mathcal {L}_\land $ -subreducts of Heyting algebras is a minimal quasivariety.Footnote ⁹ This means that every quasiequation in $\mathcal {L}_\land $ is either true in $\mathsf {K}$ or false in all the nontrivial members of  $\mathsf {K}$ .

Suppose that $\Phi $ is valid in some ${\boldsymbol {A}} \in \mathsf {K}$ . If $\mathsf {K} \vDash \Phi $ , then $\Phi $ is also valid in the $\mathcal {L}_\land $ -reduct of the Heyting algebra $\mathsf {Up}({\boldsymbol {A}}_\ast )$ . This, in turn, implies that $\mathsf {Up}({\boldsymbol {A}}_\ast ) \vDash \Phi $ as desired. Then we consider the case where $\Phi $ is false in all the nontrivial members of $\mathsf {K}$ . In this case, the assumption that ${\boldsymbol {A}} \vDash \Phi $ forces ${\boldsymbol {A}}$ to be trivial. Therefore, ${\boldsymbol {A}}_\ast $ is the empty poset and the Heyting algebra $\mathsf {Up}({\boldsymbol {A}}_\ast )$ is trivial. As a consequence, $\mathsf {Up}({\boldsymbol {A}}_\ast )$ validates every quasiequation and, in particular, $\Phi $ .

Proof Consider a variety $\mathsf {K}$ among $\mathsf {(b)ISL}, \mathsf {PDL}$ , $\mathsf {IL}$ , and $\mathsf {HA}$ , a Sahlqvist quasiequation $\Phi $ in the language of $\mathsf {K}$ , and an algebra ${\boldsymbol {A}} \in \mathsf {K}$ such that ${\boldsymbol {A}} \vDash \Phi $ . By Theorem 2.7, ${\boldsymbol {A}}$ embeds into the appropriate reduct ${\boldsymbol {B}}^-$ of a Heyting algebra ${\boldsymbol {B}}$ such that ${\boldsymbol {B}}^- \in \mathbb {U}({\boldsymbol {A}})$ . Since $\Phi $ is a universal sentence valid in ${\boldsymbol {A}}$ , from ${\boldsymbol {B}}^- \in \mathbb {U}({\boldsymbol {A}})$ it follows ${\boldsymbol {B}}^- \vDash \Phi $ . As ${\boldsymbol {B}}^-$ is the reduct of ${\boldsymbol {B}}$ in the language of $\mathsf {K}$ , this guarantees that ${\boldsymbol {B}} \vDash \Phi $ .

Given that $\Phi $ is a Sahlqvist quasiequation, we can apply the canonicity part of the Intuitionistic Sahlqvist Theorem obtaining that $\mathsf {Up}({\boldsymbol {B}}_\ast ) \vDash \Phi $ . Since ${\boldsymbol {B}}_\ast = {\boldsymbol {B}}^-_\ast $ , the algebra $\mathsf {Up}_{\mathsf {K}}({\boldsymbol {B}}^{-}_\ast )$ is the reduct of $\mathsf {Up}({\boldsymbol {B}}_\ast )$ in the language of $\mathsf {K}$ . Consequently, from $\mathsf {Up}({\boldsymbol {B}}_\ast ) \vDash \Phi $ it follows that $\mathsf {Up}_{\mathsf {K}}({\boldsymbol {B}}_\ast ) \vDash \Phi $ .

Now, recall that there exists an embedding $f \colon {\boldsymbol {A}} \to {\boldsymbol {B}}^-$ . By Conditions (i) and (ii) of Proposition 3.5, the map $\mathsf {Up}_{\mathsf {K}}(f_\ast ) \colon \mathsf {Up}_{\mathsf {K}}({\boldsymbol {A}}_\ast ) \to \mathsf {Up}_{\mathsf {K}}({\boldsymbol {B}}^{-}_\ast )$ is a homomorphism between members of $\mathsf {K}$ . Furthermore, applying the last part of the same proposition to the assumption that f is injective, we obtain that $\mathsf {Up}_{\mathsf {K}}(f_\ast )$ is also injective, whence $\mathsf {Up}_{\mathsf {K}}({\boldsymbol {A}}_\ast ) \in \mathbb {I}\mathbb {S}(\mathsf {Up}_{\mathsf {K}}({\boldsymbol {B}}^{-}_\ast ))$ . Since the validity of universal sentences persists under the formation of subalgebras and isomorphic copies, from $\mathsf {Up}_{\mathsf {K}}({\boldsymbol {B}}^-_\ast ) \vDash \Phi $ it follows that $\mathsf {Up}_{\mathsf {K}}({\boldsymbol {A}}_\ast ) \vDash \Phi $ and, therefore, $\mathsf {Up}({\boldsymbol {A}}_\ast ) \vDash \Phi $ , thus concluding the proof.

Proof This is an immediate consequence of the fact that every element of a finite semilattice ${\boldsymbol {A}}$ is the join of a subset of $\mathsf {J}({\boldsymbol {A}})$ .

Proof Condition (i) follows from the fact that $\lnot $ is order reversing in $\mathsf {PSL}$ [Reference Frink38, Condition (9)], while $\land $ is order preserving in both arguments. For Condition (ii), see [Reference Frink38, Condition (19)].

Proof Notice that ${\boldsymbol {A}}^+$ coincides with the algebra $\mathsf {Up}_{\mathsf {PSL}}(\mathbb {X})$ , where $\mathbb {X}$ is the order dual of $\mathsf {J}({\boldsymbol {A}})$ . Since $\mathsf {Up}_{\mathsf {PSL}}(\mathbb {X})$ is a pseudocomplemented semilattice, we infer that so is ${\boldsymbol {A}}^+$ . Furthermore, the definition of ${\boldsymbol {A}}^+$ guarantees that it is a distributive lattice (whose join operation is $\cup $ ). Lastly, since ${\boldsymbol {A}}$ is finite, so is ${\boldsymbol {A}}^+$ . Therefore, ${\boldsymbol {A}}^+$ is a finite distributive pseudocomplemented lattice. By Proposition 2.5(i), we conclude that ${\boldsymbol {A}}^+$ is the $\langle \land , \lnot , 0, 1 \rangle $ -reduct of a Heyting algebra.

Then we turn to prove that $\epsilon _{\boldsymbol {A}} \colon {\boldsymbol {A}} \to {\boldsymbol {A}}^+$ is an embedding. Clearly, it is well defined and preserves $\land , 0$ , and $1$ . Furthermore, it is injective by Lemma 5.6(i). To prove that it also preserves $\lnot $ , consider $a \in A$ . We will prove that for every $b \in \mathsf {J}({\boldsymbol {A}})$ ,

$$ \begin{align*} b \in \lnot^{{\boldsymbol{A}}^+}\epsilon_{{\boldsymbol{A}}}(a) \, \,&\Longleftrightarrow \, \,\epsilon_{{\boldsymbol{A}}}(a) \cap {\downarrow}b = \emptyset \, \,\Longleftrightarrow \, \,c \nleqslant a \land^{\boldsymbol{A}} b, \text{ for every }c \in \mathsf{J}({\boldsymbol{A}})\\ \, \,& \Longleftrightarrow \, \, a \land^{\boldsymbol{A}} b = 0 \, \,\Longleftrightarrow \, \, b \leqslant \lnot^{\boldsymbol{A}} a \, \,\Longleftrightarrow\, \, b \in \epsilon_{{\boldsymbol{A}}}(\lnot^{{\boldsymbol{A}}} a). \end{align*} $$

The first of the above equivalences holds by the definition of $\lnot $ in ${\boldsymbol {A}}^+$ , the second and the last by the definition of $\epsilon _{\boldsymbol {A}}$ , the third by Lemma 5.6(ii), and the fourth by Condition (3). This shows that $\lnot ^{{\boldsymbol {A}}^+}\epsilon _{{\boldsymbol {A}}}(a) = \epsilon _{{\boldsymbol {A}}}(\lnot ^{{\boldsymbol {A}}} a)$ . Hence, we conclude that $\epsilon _{\boldsymbol {A}} \colon {\boldsymbol {A}} \to {\boldsymbol {A}}^+$ preserves $\lnot $ and, therefore, it is an embedding.

Claim 5.11. For every $D,V \in \mathsf {Dw}(\mathsf {J}({\boldsymbol {A}}))$ , we have

$$\begin{align*}\lnot^{{\boldsymbol{A}}^+}(D \cap V) = \lnot^{{\boldsymbol{A}}^+} \bigg(\epsilon_{{\boldsymbol{A}}}\bigg(\bigvee^{{\boldsymbol{A}}} D\bigg) \cap V\bigg). \end{align*}$$

Proof of the Claim

In order to prove the inclusion from right to left, observe that for every $a \in D$ we have $a \leqslant \bigvee ^{{\boldsymbol {A}}}D$ and, therefore, $a \in \epsilon _{{\boldsymbol {A}}}(\bigvee ^{{\boldsymbol {A}}} D)$ . Consequently, $D \subseteq \epsilon _{{\boldsymbol {A}}}(\bigvee ^{{\boldsymbol {A}}} D)$ . This, in turn, implies $D \cap V \subseteq \epsilon _{{\boldsymbol {A}}}(\bigvee ^{{\boldsymbol {A}}} D) \cap V$ . Bearing in mind that ${\boldsymbol {A}}^+ \in \mathsf {PSL}$ (Lemma 5.8), we can apply Lemma 5.6(i) obtaining that the operation $\lnot ^{{\boldsymbol {A}}^+}$ is order reversing. Thus, $\lnot ^{{\boldsymbol {A}}^+} (\epsilon _{{\boldsymbol {A}}}(\bigvee ^{{\boldsymbol {A}}} D) \cap V) \subseteq \lnot ^{{\boldsymbol {A}}^+}(D \cap V)$ as desired.