The Lattice NExtS41 as Composed of Replicas of NExtInt, and Beyond

Abstract

We study the lattice of all normal consistent extensions of the modal logic $\mathbf {S4}$, $\mathrm{{NExt}}\mathbf {S4}$, from a structural point of view. We show that a pattern isomorphic to the lattice of intermediate logics is present as a sublattice in $\mathrm{{NExt}}\mathbf {S4}$ in many, in fact, in infinitely many places, and this pattern itself is isomorphic to a quotient lattice of $\mathrm{{NExt}}\mathbf {S4}$. We also designate three “dark spots” of $\mathrm{{NExt}}\mathbf {S4}$, three sublattices of it, where, although we can characterize the logics belonging to each of these sublattices, their structural picture is invisible at all.

Dedicated to Vadim A. Yankov (Jankov)

7.9 Appendix

The terminology, notions and facts used in this section are standard. The main references here are Rasiowa and Sikorski (1970), Rasiowa (1974) and Chagrov and Zakharyaschev (1997). Although we avoid using the term Kripke frame, this concept underlies our exposition below.

Lat $\mathfrak {A}$ be a Heyting algebra and $\textsf {S}_{\mathfrak {A}}$ be the set of all prime filters of $\mathfrak {A}$. The map $h:a\mapsto \mathbf {Set}{x\in \textsf {S}_{\mathfrak {A}}}{a\in x}$, where $a\in |\mathfrak {A}|$, is called Stone embedding, because it is an embedding of $\mathfrak {A}$ into $\text {H}(\textsf {S}_{\mathfrak {A}})$, where the last algebra is a Heyting algebra associated with the partially ordered set $\langle \textsf {S}_{\mathfrak {A}},\subseteq \rangle$.

Let us fix $x^{*}\in \textsf {S}_{\mathfrak {A}}$ and let $x_0$ and $x_1$ be two new distinct elements that do not belong to $\textsf {S}_{\mathfrak {A}}$. We define

$$ \textsf {S}_{\mathfrak {A}}^{*}:=(\textsf {S}_{\mathfrak {A}}\setminus \lbrace x^{*}\rbrace )\cup \lbrace x_0,x_1\rbrace ; $$

and, then, define the following relation R on $\textsf {S}_{\mathfrak {A}}^{*}$ as follows:

$$ xR^{*}y{\mathop {\Longleftrightarrow }\limits ^{\text {df}}} {\left\{ \begin{array}{ll} \begin{array}{l} x,y\in \textsf {S}_{\mathfrak {A}}\setminus \lbrace x^{*}\rbrace \text { and } x\subseteq y ;\\ x\in \lbrace x_0,x_1\rbrace \text { and } y\in \lbrace x_0,x_1\rbrace ;\\ x\in \textsf {S}_{\mathfrak {A}}\setminus \lbrace x^{*}\rbrace , y\in \lbrace x_0,x_1\rbrace , \text { and } x\subseteq x^{*} ;\\ y\in \textsf {S}_{\mathfrak {A}}\setminus \lbrace x^{*}\rbrace , x\in \lbrace x_0,x_1\rbrace , \text { and } x^{*}\subseteq y . \end{array} \end{array}\right. } $$

We note that the relation R is transitive and reflexive on $\textsf {S}_{\mathfrak {A}}^{*}$.

Next we define: for any $X\subseteq \textsf {S}_{\mathfrak {A}}$,

$$\begin{aligned} X^{\oplus }:={\left\{ \begin{array}{ll} \begin{array}{cl} X, &{}\text {if } x^{*}\notin X ,\\ (X\setminus \lbrace x^{*}\rbrace )\cup \lbrace x_1,x_2\rbrace , &{}\text {if } x^{*}\in X ; \end{array} \end{array}\right. } \end{aligned}$$

(7.25)

and for any $X\subseteq \textsf {S}_{\mathfrak {A}}^{*}$,

$$\begin{aligned} X^{\ominus }:={\left\{ \begin{array}{ll} \begin{array}{cl} X, &{}\text {if } X\cap \lbrace x_1,x_2\rbrace =\emptyset ,\\ X\setminus \lbrace x_1,x_2\rbrace ,&{}\text {if } X\cap \lbrace x_1,x_2\rbrace =\lbrace x_i\rbrace , \text {where } i=1 \text { or } i=2 ,\\ (X\setminus \lbrace x_1,x_2\rbrace )\cup \lbrace x^{*}\rbrace , &{}\text {if } X\cap \lbrace x_1,x_2\rbrace \ne \emptyset . \end{array} \end{array}\right. } \end{aligned}$$

(7.26)

We note that for any $X\subseteq \textsf {S}_{\mathfrak {A}}^{*}$,

$$\begin{aligned} \forall y\in \textsf {S}_{\mathfrak {A}}\setminus \lbrace x^{*}\rbrace .~y\in X^{\ominus }\Longleftrightarrow y\in X. \end{aligned}$$

(7.27)

Also, for any $X\subseteq \textsf {S}_{\mathfrak {A}}$,

$$\begin{aligned} X\setminus \lbrace x^{*}\rbrace =X^{\oplus }\setminus \lbrace x_1,x_2\rbrace ; \end{aligned}$$

(7.28)

and for any $X\subseteq \textsf {S}_{\mathfrak {A}}^{*}$,

$$\begin{aligned} X\setminus \lbrace x_1,x_2\rbrace =X^{\ominus }\setminus \lbrace x^{*}\rbrace . \end{aligned}$$

(7.29)

It is obvious that if $X\subseteq \textsf {S}_{\mathfrak {A}}$, then $X^{\oplus }\subseteq \textsf {S}_{\mathfrak {A}}^{*}$; and if $X\subseteq \textsf {S}_{\mathfrak {A}}^{*}$, then $X^{\ominus }\subseteq \textsf {S}_{\mathfrak {A}}$. The last observation can be refined as follows.

Lemma 7.6

For any $X\in \mathcal{U}(\textsf {S}_{\mathfrak {A}})$, $X^{\oplus }\in \mathcal{U}(\textsf {S}_{\mathfrak {A}}^{*})$; and for any $X\in \mathcal{U}(\textsf {S}_{\mathfrak {A}}^{*})$, $X^{\ominus }\in \mathcal{U}(\textsf {S}_{\mathfrak {A}})$.

Proof

Let $X\in \mathcal{U}(\textsf {S}_{\mathfrak {A}})$. If $x^{*}\notin X$, then $X=X^{\oplus }$. Let us take $x\in X$ and $y\in \textsf {S}_{\mathfrak {A}}^{*}$ with xRy. For contradiction assume that $y=x_i$. Then $x\subseteq x^{*}$ and hence $x^{*}\in X$. This implies that $y\in \textsf {S}_{\mathfrak {A}}$ and $x\subseteq y$. Hence $y\in X$.

Now we suppose that $x^{*}\in X$. Then $\lbrace x_1,x_2\rbrace \subseteq X^{\oplus }$. Let us take $x\in X^{\oplus }$ and $y\in \textsf {S}_{\mathfrak {A}}^{*}$ with xRy. We have to consider the following cases.

Case: $y\in \lbrace x_1,x_2\rbrace$. Then, obviously, $y\in X^{\oplus }$.

Case: $x\in X\setminus \lbrace x_1,x_2\rbrace$ and $y\notin \lbrace x_1,x_2\rbrace$. Then $x\subseteq y$ and, by premise, $y\in X$. Since $y\ne x^{*}$, $y\in X^{\oplus }$.

Case: $x\in \lbrace x_1,x_2\rbrace$ and $y\notin \lbrace x_1,x_2\rbrace$. Then $x^{*}\subseteq y$ and, by supposition, $y\in X$. And since $y\ne x^{*}$, $y\in X^{\oplus }$.

The other part of the statement can be proven in a similar way.

The last observation can be extended as follows.

Proposition 7.13

Restricted to the upward sets of $\textsf {S}_{\mathfrak {A}}$ and, respectively, to that of $\textsf {S}_{\mathfrak {A}}^{*}$, the maps $g:X\mapsto X^{\oplus }$ and $g^{-1}:X\mapsto X^{\ominus }$ are mutually inverse isomorphisms between $\langle \mathcal{U}(\textsf {S}_{\mathfrak {A}}),\subseteq \rangle$ and $\langle \mathcal{U}(\textsf {S}_{\mathfrak {A}}^{*}),\subseteq \rangle$, and hence between $\mathrm{H}({\textsf {S}_{\mathfrak {A}}})$ and $\mathrm{H}({\textsf {S}_{\mathfrak {A}}^{*}})$.

Proof

It should be obvious that for any X and Y of $\mathcal{U}(\textsf {S}_{\mathfrak {A}})$,

$$ X\subseteq Y\Longrightarrow X^{\oplus }\subseteq Y^{\oplus }, $$

and for any X and Y of $\mathcal{U}(\textsf {S}_{\mathfrak {A}}^{*})$,

$$ X\subseteq Y\Longrightarrow X^{\ominus }\subseteq Y^{\ominus }. $$

Also, for any $X\in \mathcal{U}(\textsf {S}_{\mathfrak {A}})$ and any $Y\in \mathcal{U}(\textsf {S}_{\mathfrak {A}}^{*})$,

$$ X^{\oplus \,\ominus }=X~~\text {and}~~Y=Y^{\ominus \,\oplus }. $$

Lemma 7.7

For any $X\cup Y\subseteq \textsf {S}_{\mathfrak {A}}^{*}$, if $X\cap \lbrace x_1,x_2\rbrace =\lbrace x_i\rbrace \ne \lbrace x_j\rbrace =Y\cap \lbrace x_1,x_2\rbrace$, then

$$\begin{aligned} (X\cup Y)^{\ominus }=X^{\ominus }\cup Y^{\ominus } \end{aligned}$$

(7.30)

Proof

We consider the following cases.

Case: $X\cap \lbrace x_1,x_2\rbrace =\lbrace x_1,x_2\rbrace$ or $Y\cap \lbrace x_1,x_2\rbrace =\lbrace x_1,x_2\rbrace$. Then $(X\cup Y)\cap \lbrace x_1,x_2\rbrace =\lbrace x_1,x_2\rbrace$. This yields:

• to obtain $X^{\ominus }$, $x^{*}$ replaces $\lbrace x_1,x_2\rbrace$ in X;

• or to obtain $Y^{\ominus }$, $x^{*}$ replaces $\lbrace x_1,x_2\rbrace$ in Y;

• and to obtain $(X\cup Y)^{\ominus }$, $x^{*}$ replaces $\lbrace x_1,x_2\rbrace$ in $X\cup Y$.

Thus (7.30) holds.

Case: $X\cap \lbrace x_1,x_2\rbrace \subset \lbrace x_1,x_2\rbrace$ and $Y\cap \lbrace x_1,x_2\rbrace \subset \lbrace x_1,x_2\rbrace$. Taking into account the premise, this leads to the following three cases:

• $X\cap \lbrace x_1,x_2\rbrace =\lbrace x_i\rbrace$ and $Y\cap \lbrace x_1,x_2\rbrace =\emptyset$;

• $X\cap \lbrace x_1,x_2\rbrace =\emptyset$ and $Y\cap \lbrace x_1,x_2\rbrace =\lbrace x_i\rbrace$; and

• $X\cap \lbrace x_1,x_2\rbrace =\emptyset$ and $Y\cap \lbrace x_1,x_2\rbrace =\emptyset$.

In all these cases, (7.30) is true.

Lemma 7.8

For any $X\subseteq \textsf {S}_{\mathfrak {A}}^{*}$, .

Proof

We consider two cases.

Case: $X\cap \lbrace x_1,x_2\rbrace =\emptyset$. Then .  This also implies that, although ,  $x^{*}\notin X$ and hence .

Next, we observe that if $x\ne x^{*}$, then

$$ (\forall y\in \textsf {S}_{\mathfrak {A}}^{*}.~xR y\Longrightarrow y\in X)\Longleftrightarrow (\forall y\in \textsf {S}_{\mathfrak {A}}.~x\subseteq y\Longrightarrow y\in X). \qquad \qquad \qquad {(\text {L}~7.8{\textendash }*)} $$

Indeed, suppose the left-hand implication holds and $x\subseteq y$ with $y\in \textsf {S}_{\mathfrak {A}}$. It is clear that $y\notin \lbrace x_1,x_2\rbrace$ and xRy. This implies that $y\in X$. Now we assume that the right-hand implication holds and xRy with $y\in \textsf {S}_{\mathfrak {A}}^{*}$. If it were that $y\in \lbrace x_1,x_2\rbrace$, then, we would have that $x\subseteq x^{*}$ and, by the right-hand implication, $x^{*}\in X$. Thus, $y\notin \lbrace x_1,x_2\rbrace$. This means that $y\in \textsf {S}_{\mathfrak {A}}$ and hence, by the right-hand implication, $y\in X$.

We continue considering the first case as follows.

Case: $X\cap \lbrace x_1,x_2\rbrace =\lbrace x_i\rbrace$. Without loss of generality, we count that $X\cap \lbrace x_1,x_2\rbrace =\lbrace x_1\rbrace$. Hence, $X^{\ominus }\subseteq \textsf {S}_{\mathfrak {A}}$ and $x^{*}\notin X^{\ominus }$. This implies that $x^{*}\notin \square X^{\ominus }$. On the other hand, .  Hence  and and, therefore, .

Now, assume that $x\ne x^{*}$. We obtain:

Case: $X\cap \lbrace x_1,x_2\rbrace =\lbrace x_1,x_2\rbrace$. Then $x^{*}\in X^{\ominus }$. It should be clear that

Now assume that $x\ne x^{*}$. First, we show that

$$ (\forall y\in \textsf {S}_{\mathfrak {A}}^{*}.~xR y\Longrightarrow y\in X)\Longleftrightarrow (\forall y\in \textsf {S}_{\mathfrak {A}}.~x\subseteq y\Longrightarrow y\in X^{\ominus }). \quad \quad {(\text {L}~7.8{\textendash }**)} $$

Indeed, suppose the left-hand side is valid and $x\subseteq y$ with $y\in \textsf {S}_{\mathfrak {A}}$. Assume first that $y\ne x^{*}$. Then, in virtue of the premise, $y\in X$ and, applying (7.27), we conclude that $y\in X^{\ominus }$. If $y=x^{*}$, then, by definition (7.26), $x^{*}\in X^{\ominus }$.

Next, assume that the right-hand side of (L 7.8–$**$) is true. Let xRy, where $y\in \textsf {S}_{\mathfrak {A}}^{*}$. If $y\in \lbrace x_1,x_2\rbrace$, then, by premise, $y\in X$. Now assume that $y\notin \lbrace x_1,x_2\rbrace$, that is $y\in \textsf {S}_{\mathfrak {A}}\setminus \lbrace x^{*}$. This implies that $y\in X^{\ominus }$ and, in virtue of (7.27), $y\in X$.

Now, we obtain for $x\ne x^{*}$:

Lemma 7.9

For any $a\in \mathfrak {A}$, $(\textsf {S}_{\mathfrak {A}}^{*}\setminus h(a)^{\oplus })^{\ominus }=\textsf {S}_{\mathfrak {A}}\setminus h(a)$.

Proof

We consider the following two cases.

Case: $x^{*}\in h(a)$. Then $\textsf {S}_{\mathfrak {A}}^{*}\setminus h(a)^{\oplus }=\textsf {S}_{\mathfrak {A}}\setminus h(a)$. The conclusion is obvious.

Case: $x^{*}\notin h(a)$. Then $h(a)^{\oplus }=h(a)$ and $\lbrace x_1,x_2\rbrace \subseteq (\textsf {S}_{\mathfrak {A}}^{*}\setminus h(a)^{\oplus })$. The conclusion is obvious again.

Proposition 7.14

Let $X\subseteq \textsf {S}_{\mathfrak {A}}^{*}$ and the following conditions be satisfied:

$$ \begin{array}{cl} (a) &{}(\textsf {S}_{\mathfrak {A}}^{*}\setminus X)^{\ominus }=\textsf {S}_{\mathfrak {A}}\setminus h(a^{*}),~\text {for some } a^{*}\in \mathfrak {A},\\ (b) &{}X^{\ominus }=h(b^{*}),~\text {for some } b^{*}\in \mathfrak {A}. \end{array} $$

Then, if $\mathfrak {B}$ is the subalgebra of $\text {B}(\textsf {S}_{\mathfrak {A}}^{*})$ generated by the set $\mathbf {Set}{h(a)^{\oplus }}{a\in \mathfrak {A}}\cup \lbrace X\rbrace$, then $\mathfrak {B}^{\circ }=g(h(\mathfrak {A}))$.

Proof

Generating the algebra $\mathfrak {B}$, if we use only Boolean operations, according to Rasiowa and Sikorski (1970), Theorem II.2.1, we get meets of terms of the following three categories:

(7.31)

If we apply the operation  to those meets and, then, distribute ,  applying it to each of the terms listed above (which is possible, for $\text {B}(\textsf {S}_{\mathfrak {A}}^{*})$ is an $\mathbf {S4}$-algebra), we obtain meets of terms of the following three forms:

We aim to show that these terms belong to $g(h(\mathfrak {A}))$. This conclusion will be reached, if we show that the $g^{-1}$-images of these terms belong to $h(\mathfrak {A})$. Thus, according to Proposition 7.13, we have to consider the following terms:

According to Lemma 7.8, we will focus on the terms:

$$ \begin{array}{rl} \bullet &{}\square ((\textsf {S}_{\mathfrak {A}}^{*}\setminus h(a)^{\oplus })\cup h(b)^{\oplus })^{\ominus }, ~\text {for some } a,b\in \mathfrak {A},\\ \bullet &{}\square ((\textsf {S}_{\mathfrak {A}}^{*}\setminus X)\cup h(b)^{\oplus })^{\ominus }, ~\text {for some } b\in \mathfrak {A},\\ \bullet &{}\square ((\textsf {S}_{\mathfrak {A}}^{*}\setminus h(a)^{\oplus })\cup X)^{\ominus }, ~\text {for some } a\in \mathfrak {A}. \end{array} $$

We note that the premise of Lemma 7.7 is satisfied, when we apply the operation $X\mapsto X^{\ominus }$ to the terms (7.31). Thus, applying successively Lemmas 7.7, 7.9 and Proposition 7.13, we obtain that the above terms are equal, respectively, to the following terms:

$$ \begin{array}{cl} \bullet &{}\square ((\textsf {S}_{\mathfrak {A}}\setminus h(a))\cup h(b))=h(a)\boldsymbol{\rightarrow }h(b),\\ \bullet &{}\square ((\textsf {S}_{\mathfrak {A}}^{*}\setminus X)^{\ominus }\cup h(b))= h(a^{*})\boldsymbol{\rightarrow }h(b)~\text {[by the first premise]},\\ \bullet &{}\square ((\textsf {S}_{\mathfrak {A}}\setminus h(a))\cup X^{\ominus })=h(a)\boldsymbol{\rightarrow }h(b^{*})~\text {[by the second premise]}. \end{array} $$

This completes the proof.

Proposition 7.15

Let $\mathfrak {A}$ be a finitely generated Heyting algebra. Then there is a prime filter $x^{*}\in \textsf {S}_{\mathfrak {A}}$ and a set $X\subseteq \textsf {S}_{\mathfrak {A}}^{*}$ such that the conditions (a) and (b) of Proposition 7.14 are satisfied; in addition, the valuation $v:p\mapsto X$ refutes the formulas $\square (\square (p\rightarrow \square p)\rightarrow p)\rightarrow p$ and $\square \Diamond p\rightarrow \Diamond \square p$ in the algebra $\mathfrak {B}$ of Proposition 7.14.

Proof

According to Kuznetsov (1973), the algebra $\mathfrak {A}$ is atomic.²¹ Let $a^{*}$ be an atom of $\mathfrak {A}$. Then we define

$$ x^{*}:=[a^{*}). $$

We note that, since $a^{*}$ is an atom, $h(a^{*})=\lbrace x^{*}\rbrace$.

Remembering that new elements $x_1,x_2$ replace $x^{*}$ in $\textsf {S}_{\mathfrak {A}}$ to generate $\textsf {S}_{\mathfrak {A}}^{*}$, we define:

$$ X:=\lbrace x_1\rbrace . $$

It should be clear that $\textsf {S}_{\mathfrak {A}}^{*}\setminus X=(\textsf {S}_{\mathfrak {A}}\setminus \lbrace x^{*}\rbrace )\cup \lbrace x_2\rbrace$. Therefore, $(\textsf {S}_{\mathfrak {A}}^{*}\setminus X)^{\ominus }=\textsf {S}_{\mathfrak {A}}\setminus h(a^{*})$; that is the condition (a) of Proposition 7.14 is satisfied.

On the other hand, $X^{\ominus }=\emptyset =h(\mathbf {0})$; that is the condition (b) of Proposition 7.14 is also fulfilled.

In Table 7.1 we show a refutation of the formula $\square (\square (p\rightarrow \square p)\rightarrow p)\rightarrow p$ in the algebra $\mathfrak {B}$, where A denotes a subformula of this formula. We aim to prove that $v(\square (\square (p\rightarrow \square p)\rightarrow p)\rightarrow p)\ne \textsf {S}_{\mathfrak {A}}^{*}$.

Table 7.1 Refutation of $\square (\square (p\rightarrow \square p)\rightarrow p)\rightarrow p$

In the next table we show a refutation of the formula $\square \Diamond p\rightarrow \Diamond \square p$ in $\mathfrak {B}$. We aim to prove that $v(\square \Diamond p\rightarrow \Diamond \square p)\ne \textsf {S}_{\mathfrak {A}}^{*}$ (Table 7.2).

Table 7.2 Refutation of $\square \Diamond p\rightarrow \Diamond \square p$