“That Will Do”: Logics of Deontic Necessity and Sufficiency

Abstract

We study a logic for deontic necessity and sufficiency (often interpreted as obligation, resp. strong permission), as originally proposed in van Benthem (Bull Sect Log 8(1):36–41, 1979). Building on earlier work in modal logic, we provide a sound and complete axiomatization for it, consider some standard extensions, and study other important properties. After that, we compare this logic to the logic of “obligation as weakest permission” from Anglberger et al. (Rev Symb Log 8(4):807–827, 2015).

Appendices

Appendix 1: Completeness for $\mathbf{DNS}$: Copy-and-Merge

1.1 The Need for Copies

Completeness boils down to the following claim: for every consistent set $\Delta \subseteq {\mathcal {W}}$, there is a $\mathbf {DNS}$-model M and a world w such that $M,w\models \varphi$ for all $\varphi \in \Delta$. Below, we prove a slightly stronger claim: there is a $\mathbf {DNS}$-model M such that for all consistent $\Delta \subseteq {\mathcal {W}}$, there is a w such that $M,w\models \varphi$ for all $\varphi \in \Delta$. This model M will be called $M^+$.

We first define the canonical model $M^c= \langle W^c, R_\square ^c, R_\mathsf{N}^c, v^c\rangle$ in the standard way, viz. as follows:

• 1. $W^c$ is the set of all subsets of ${\mathcal {W}}$ that are maximally consistent w.r.t. $\mathbf{DNS}$

• 2. $R^c_\square = \{( \Delta , \Theta )\mid \{\psi \mid \square \psi \in \Delta \}\subseteq \Theta \}$

• 3. $R^c_\mathsf{N}= \{( \Delta , \Theta )\mid \{\psi \mid \mathsf{N}\psi \in \Delta \}\subseteq \Theta \}$

• 4. $v^c(\psi ) = \{\Delta \mid \psi \in \Delta \}$ for all $\psi \in \mathcal {S}$

It can now be established by standard means that $M^c$ is an $\mathbf{DNS}$-model, and for all formulas $\varphi \in \mathcal {W}_{\overline{\mathsf{S}}}$ and all $\Delta \in W^c$, $M^c,\Delta \models \varphi$ iff $\varphi \in \Delta$. This is usually called the truth lemma.⁴² However, this lemma does not hold for the entire language. Consider a model M that consists of only two worlds, $w_0$ and $w_1$, such that all $\psi \in \mathcal {S}$ are true in both worlds, $R_\mathsf{N}$ is the identity relation, and $R_\square$ is the total relation. Let $\Theta _\star = \{\varphi \mid M,w_0\models \varphi \}$. Note that, since $M,w_1 \models p$ and $w_1\in R_\square (w_0) - R_\mathsf{N}(w_0)$, $M,w_0 \not \models {\mathsf{S}} p$ and hence ${\mathsf{S}} p\not \in \Theta _\star$.

By standard means we can show that $\Theta _\star \in W^c$. Moreover, $R_\square ^c(\Theta _\star ) = \{\Theta _\star \}$. That is, $\square \varphi \in \Theta _\star$ iff $M,w_0\models \square \varphi$ iff [by the construction and symmetry of M] $M,w_0\models \varphi$ iff $\varphi \in \Theta _\star$.⁴³ By the same reasoning, $R_\mathsf{N}^c(\Theta _\star ) = \{\Theta _\star \}$. But then $M^c, \Theta _\star \models {\mathsf{S}} p$, whereas ${\mathsf{S}} p\not \in \Theta _\star$.

In other words, if we simply construct the canonical model in the standard way, then we lack a “witness” for the formula $\lnot {\mathsf{S}}p$, i.e. a deliberative alternative in which p holds, but which is not deontically acceptable from the viewpoint of $\Theta _\star$.

To solve this problem, we make two disjoint copies $M^1$ and $M^2$ of $M^c$, take their union, and make some “smart” connections between the points in both. The idea is that whenever some world in $M^1$ needs a witness for a formula of the form $\lnot \mathsf{S}\psi$, we take it from $M^2$ (and vice versa). This can however be done in many ways. We will first consider arbitrary ways to copy-and-merge $M^c$ and establish a sufficient condition for the truth lemma to hold in such constructions. After that, we will consider three specific, concrete ways to copy-and-merge $M^c$. Whereas the first is aimed at preserving as many properties of $R_\mathsf{N}$ as possible, the second and third are aimed at preserving as many properties of $R_\square$ as possible.

First some more notation. In the remainder of this appendix, we use i, j, k to range over $\{1,2\}$. Let $W^i = \{\langle \Delta ,i\rangle \mid \Delta \in W^c\}$. So the members of $W^1$ and $W^2$ are not sets of formulas, but indexed sets of formulas. In order to enhance readibility, we will write $\Delta ^i$ to refer to $\langle \Delta ,i\rangle$. Also, let $W^+ = W^1\cup W^2$. Note that $W^+ = \{\Delta ^1,\Delta ^2\mid \Delta \in W^c\}$.

We define four accessibility relations on $W^+$ and a valuation over $W^+$:

• $R_\square ^i = \{(\Delta ^i,\Theta ^i)\mid (\Delta ,\Theta )\in R_\square ^c\}$

• $R^\cup _\square = R^1_\square \cup R^2_\square$

• $R_\mathsf{N}^i = \{(\Delta ^i,\Theta ^i)\mid (\Delta ,\Theta )\in R_\mathsf{N}^c\}$

• $R^\cup _\mathsf{N}= R^1_\mathsf{N}\cup R^2_\mathsf{N}$

• $v^i(\varphi ) = \{\Delta ^i \mid \Delta \in v^c(\varphi )\}$ for all $\varphi \in {\mathcal {S}}$

where ${\mathcal {X}}\subseteq W^+\times W^+$, let $\overline{i}({\mathcal {X}}) = \{(\Delta ,\Theta ) \mid (\Delta ^i,\Theta ^j)\in {\mathcal {X}}\}$.

1.2 A Sufficient Condition for the Truth Lemma

Definition 8

A $\mathbf {DNS}$-model $M^+ = \langle W^+,R^+_\square , R^+_\mathsf{N}, v^+\rangle$ is a smart copy-merge of $M^c$ iff each of the following hold:

• 1. for all $\varphi \in {\mathcal {S}}$, $v(\varphi )= v^1(\varphi )\cup v^2(\varphi )$

• 2. $\overline{i}(R^+_\square ) = R^c_\square$

• 3. $\overline{i}(R^+_\mathsf{N}) = R^c_\mathsf{N}$

• 4. $\overline{i}(R^+_\square \setminus R^+_\mathsf{N}) = \{(\Delta ,\Theta )\in R_\square ^c \mid \{\lnot \sigma \mid \mathsf{S}\sigma \in \Delta \}\subseteq \Theta \}$

Theorem 9

If $M^+$ is a smart copy-merge of $M^c$, then for all $i\in \{1,2\}$, $\Delta \in W^c$, and $\psi \in {\mathcal {W}}$: $M^+,\Delta ^i \models \psi$ iff $\psi \in \Delta$.

Proof

Suppose that the antecedent holds. We prove the consequent by an induction on the complexity of $\psi$. The base case ($\psi \in \mathcal {S}$) is immediate in view of Definition 8.1. For the induction step, the connectives are routine and hence safely left to the reader. This leaves us with the three modal operators:

Case 1: $\psi = \square \tau$. We have: $M^+,\Delta ^i\models \square \tau$ iff for all $\Theta ^j\in R^+_\square (\Delta ^i)$, $M^+,\Theta ^j\models \tau$ iff [by the induction hypothesis] for all $\Theta ^j\in R^+_\square (\Delta _i)$, $\tau \in \Theta$ iff [by Definition 8.2] ($\dagger$) for all $\Theta \in R^c_\square (\Delta )$, $\tau \in \Theta$. Suppose now that ($\dagger$) holds. Hence by item (ii) of the construction of $M^c$, every maximally consistent set $\Theta \supseteq \{\varphi \mid \square \varphi \in \Delta \}$ contains $\tau$. By a standard proof (relying on the compactness of $\mathbf {DNS}$ and the axioms and rules for $\square$), we can derive that $\square \tau \in \Delta$.

For the other direction, suppose that $\square \tau \in \Delta$. Hence, every maximal consistent set $\Theta \supseteq \{\varphi \mid \square \varphi \in \Delta \}$ contains $\tau$. By item (ii) of the construction of $M^c$, ($\dagger$) holds.

Case 2: $\psi = \mathsf{N}\tau$. Analogous to the preceding case: replace $\square$ with $\mathsf{N}$, (ii) with (iii) and Definition 8.2 with Definition 8.3.

Case 3: $\psi = \mathsf{S}\tau$. ($\Rightarrow$) Suppose that $M^+,\Delta ^i \models \mathsf{S}\tau$. Hence, for all $\Theta ^j\in R^+_\square (\Delta ^i)$ such that $M^+,\Theta ^j\models \tau$, $\Theta ^j\in R^+_\mathsf{N}(\Delta ^i)$. By Definition 8.4 there is no $\Theta \in R_\square (\Delta )$ such that $\tau \in \Theta$ and $\{\lnot \sigma \mid \mathsf{S}\sigma \in \Delta \}\subseteq \Theta$. Hence, there is no maximal consistent extension of $\{\xi \mid \square \xi \in \Delta \} \cup \{\tau \} \cup \{\lnot \sigma \mid \mathsf{S}\sigma \in \Delta \}$. By the compactness of $\mathbf{DNS}$ and $\mathbf {CL}$-properties, this means that there are $\xi _1, \ldots , \xi _n$ and $\sigma _1, \ldots , \sigma _n$ such that each of the following hold:

• (a) $\square \xi _1, \ldots , \square \xi _n \in \Delta$

• (b) $\{\xi _1, \ldots , \xi _n\} \cup \{\tau \} \vdash _{\mathbf{DNS}} \sigma _1\vee \cdots \vee \sigma _n$

• (c) $\mathsf{S}\sigma _1, \ldots , \mathsf{S}\sigma _n \in \Delta$

By Observation 2.7 and (c),

$$ \mathsf{S}(\sigma _1\vee \cdots \vee \sigma _n)\in \Delta $$

(1)

By (Nec) and (a),

$$ \square ((\xi _1\wedge \cdots \wedge \xi _n \wedge \tau )\supset (\sigma _1\vee \cdots \vee \sigma _n)) \in \Delta $$

(2)

By (a) and (2),

$$ \square (\tau \supset (\sigma _1\vee \cdots \vee \sigma _n)) \in \Delta $$

(3)

By axiom (OR), (1) and (3), $\mathsf{S}\tau \in \Delta$.

($\Leftarrow$) Suppose that $\mathsf{S}\tau \in \Delta$. Let $\Theta ^j\in R^+_\square (\Delta ^i)$ be arbitrary such that $M^+,\Theta ^j \models \tau$—we need to prove that $\Theta ^j\in R^+_\mathsf{N}(\Delta ^i)$. By the induction hypothesis, $\tau \in \Theta$. Hence, $\lnot \tau \not \in \Theta$. It follows that $\{\lnot \sigma \mid \mathsf{S}\sigma \in \Delta \}\not \subseteq \Theta$. By Definition 8.4, $(\Delta ,\Theta )\not \in \overline{i}(R^+_\square \setminus R^+_\mathsf{N})$, and hence $(\Delta ^i,\Theta ^j)\not \in R^+_\square \setminus R^+_\mathsf{N}$. It follows that $(\Delta ^i,\Theta ^j)\in R^+_\mathsf{N}$ and hence $\Theta ^j\in R^+_\mathsf{N}(\Delta ^i)$. $\square$

1.3 Copy-and-Merge Version 1

As promised, we will now define three concrete ways to copy-and-merge $M^c$ in a smart way. For the first, put

• (a) $R^+_\square = R^\cup _\square \cup \{(\Delta ^i,\Theta ^j) \mid (\Delta ,\Theta )\in R^c_\square \text{ and } \{\lnot \sigma \mid \mathsf{S}\sigma \in \Delta \}\subseteq \Theta \}$

• (b) $R^+_\mathsf{N}= R^\cup _\mathsf{N}$

To see that $M^+$ is an $\mathbf{DNS}$-model, it suffices to check that $R^+_\mathsf{N}\subseteq R^+_\square$. So suppose that $(\Delta ^i,\Theta ^j)\in R_\mathsf{N}^+$. By (b), $(\Delta ,\Theta )\in R_\mathsf{N}^c$ and $i = j$. Let $\tau$ be arbitrary such that $\square \tau \in \Delta$. By axiom (B) and (MP), also $\mathsf{N}\tau \in \Delta$. Hence, by (iii), $\tau \in \Theta$. So we have shown that $\{\tau \mid \square \tau \in \Delta \}\subseteq \Theta$. Hence, by (ii), $(\Delta ,\Theta )\in R_\square ^c$. Finally, by (b) and since $i = j$, $(\Delta ^i,\Theta ^j)\in R_\square ^+$.

So it remains to check that $M^+$ is a smart copy-merge of $M^c$. But this is obvious in view of the way we defined $R^+_\square$ and $R^+_\mathsf{N}$. So by Theorem 9 we get the truth lemma for $M^+$, and hence we obtain a model for every $\mathbf {DNS}$-consistent set $\Gamma \subseteq {\mathcal {W}}$.

1.4 Copy-and-Merge Version 2

For the second version of copy-and-merge, we put

• (a′) $R^+_\square = \{(\Delta ^i,\Theta ^j) \mid (\Delta ,\Theta ) \in R^c_\square \}$

• (b′) $R^+_\mathsf{N}= R^\cup _\mathsf{N}\cup \{(\Delta ^i,\Theta ^j)\in R^+_\square \mid \{\lnot \sigma \mid \mathsf{S}\sigma \in \Delta \}\not \subseteq \Theta \}$

The main difference with the previous construction is that here, we connect every $\Delta ^i$ with $\Theta ^j$ whenever $(\Delta ,\Theta )\in R^+_\square$. This allows us to preserve a great number of properties of $R^c_\square$. However, to make sure that formulas of the form $\mathsf{S}\psi$ are respected, we need to enlarge the relation $R_\mathsf{N}$ in such a way that whenever some $\tau$ is sufficient at $\Delta ^i$, then every $R_\square$-accessible world at which $\tau$ holds is also $R_\mathsf{N}$-accessible.

Checking that $M^+$ is a $\mathbf{DNS}$-model is again easy in view of the (B)-axiom and the construction. Also, in view of (a’) and (b’) we can easily verify that $M^+$ is a smart copy-merge of $M^c$. So we can again infer by Theorem 9 that the truth lemma holds for $M^+$, which finishes the proof of completeness.

1.5 Copy-and-Merge Version 3

The third and last copy-merge operation is defined as follows:

• (a″) $R^+_\square = \{(\Delta ^i,\Theta ^j) \mid (\Delta ,\Theta ) \in R^c_\square \}$

• (b″) $R^+_\mathsf{N}= \{(\Delta ^i,\Theta ^j) \mid (\Delta ,\Theta )\in R^c_\mathsf{N} \text{ and } j = 1 \text{ or } \{\lnot \sigma \mid \mathsf{S}\sigma \in \Delta \}\not \subseteq \Theta \}$

We leave it to the reader to check that once more, every $M^+$ obtained in this way is a smart copy-merge of $M^c$ and is a $\mathbf {DNS}$-model. Note that, in contrast to the two previous constructions, this one is asymmetric w.r.t. the indices 1 and 2. This will turn out instrumental in proving that transitivity and uniformity of $R_\mathsf{N}$ transfers to $R_\mathsf{N}^+$ (see Appendices 2.2 and 2.3 respectively).

Appendix 2: Extensions of $\mathbf{DNS}$

1.1 Proof of Theorems 2 and 3

To prove the left-right direction of Theorems 2 and 3, it suffices to check that the validity of axioms is preserved in the richer setting of $\mathbf {DNS}$-models—we safely leave this to the reader. For the other direction, we need to apply the two techniques of copy-merge from Sect. 1 to the stronger logics obtained by adding the axioms. In the remainder, we first illustrate how this works for some cases covered by Theorem 2; the reasoning is completely analogous for the other cases and Theorem 3.

Let $\mathbf {\mathbf {DNS}.4_\mathsf{N}}$ be the logic obtained by adding the axiom $(4_\mathsf{N})$ to $\mathbf {DNS}$. We need to prove that this is the logic of $\mathbf {DNS}$-models in which $R_\mathsf{N}$ is transitive. The proof proceeds in two steps: first, we define the model $M^c$ as in Sect. 1. We prove that, due to this definition and the axiom $(4_\mathsf{N})$, $R^c_\mathsf{N}$ is transitive. This is done in the standard way as for the logic $\mathbf {K4}$: suppose that $(\Delta ,\Theta ),(\Theta ,\Lambda )\in R^c_\mathsf{N}$. By the construction of $M^c$, $\{\varphi \mid \mathsf{N}\varphi \in \Delta \}\subseteq \Theta$ and $\{\psi \mid \mathsf{N}\psi \in \Theta \subseteq \Lambda$. Let $\mathsf{N}\varphi \in \Delta$. Hence by $(4_\mathsf{N})$, $\mathsf{N}\mathsf{N}\varphi \in \Delta$. It follows that $\mathsf{N}\varphi \in \Theta$ and hence $\varphi \in \Lambda$. So we have that $\{\varphi \mid \mathsf{N}\varphi \in \Delta \}\subseteq \Lambda$ and hence by the construction of $M^c$, $(\Delta ,\Lambda )\in R^c_\mathsf{N}$.

For the second step, let $M^+$ be defined according to the first method of copy-merge—see Sect. 1. Then we have: $R^+_\mathsf{N}= R^1_\mathsf{N}\cup R^2_\mathsf{N}$. It follows immediately that also $R^+_\mathsf{N}$ is transitive. Hence, we obtain a canonical model for $\mathbf {\mathbf {DNS}.4}$ in which $R^+_\mathsf{N}$ is transitive.

The same reasoning can be applied to the other conditions on $R_\mathsf{N}$: we first prove that the construction of $M^c$ and the axioms ensure that $R^c_\mathsf{N}$ satisfy them, and next we observe that these conditions are preserved in $R^+_\mathsf{N}$ in view of its definition.

We now consider the conditions on $R_\square$ that are covered by Theorem 2. Again, we prove these in the same basic steps: first prove that they hold for $R^c_\square$, and second show that this transfers to $R^+_\square$. Since the reasoning in the second step is slightly more intricate here, we summarize the main points here:

($\hbox {CD}_\square$):

Suppose that $R_\square ^c$ is serial. Let $\Delta ^i\in W^+$ be arbitrary. Hence, $\Delta \in W^c$ and hence by seriality, there is a $\Theta$ such that $(\Delta ,\Theta )\in R_\square ^c$. But then by the construction, $(\Delta ^i,\Theta ^i)\in R_\square ^+$. Hence, $R_\square ^+$ is serial.

($\hbox {CT}_\square$):

Suppose that $R_\square ^c$ is reflexive. Let $\Delta ^i\in W^+$. Hence, $\Delta \in W^c$ and hence by reflexivity, $(\Delta ,\Delta )\in R_\square ^c$. By the construction, $(\Delta ^i,\Delta ^i)\in R_\square ^+$. Hence, $R_\square ^+$ is reflexive.

($\hbox {CM}_\square$):

Suppose that $R_\square ^c$ is shift reflexive and that $(\Delta ^i,\Theta ^j)\in R_\square ^+$. By the construction, $(\Delta ,\Theta )\in R_\square ^c$. Hence, by shift reflexivity, $(\Theta ,\Theta )\in R_\square ^c$. But then, by the construction, $(\Theta ^j,\Theta ^j)\in R_\square ^+$ and we are done.

To prove Theorem 3, we use the second construction of copy-and-merge—see Sect. 1. Otherwise, the reasoning is the same: first prove that $R^c_\mathsf{N}$ ($R^c_\square$) satisfy the respective conditions, and next show how those conditions are transferred to $R^+_\mathsf{N}$ ($R^+_\square$).

1.2 Transitivity of $R_\mathsf{N}$

We now prove that the logic of frames with $R_\square$ euclidian and $R_\mathsf{N}$ transitive is completely axiomatized by $\mathbf {DNS}$ plus the following axioms:

($5_\square$):

$\Diamond \varphi \supset \square \Diamond \varphi$

($4_\mathsf{N}$):

$\mathsf{N}\varphi \supset \mathsf{N}\mathsf{N}\varphi$

($\hbox {Trans}_\mathsf{S}$):

$\lnot \mathsf{N}\lnot \mathsf{S}\varphi \supset \mathsf{S}\varphi$

The proof of soundness is again safely left to the reader. For completeness, we use the third type of copy-and-merge (see Appendix 1). This gives us the $\mathbf {DNS}$-model $M^+$ which is a smart copy-merge of $M^c$ for the logic $\mathbf {\mathbf {DNS}.5_\square .4_\mathsf{N}.Trans_\mathsf{S}}$. To show that $M^+$ satisfies the right conditions, we again proceed by two steps: first, show that $R^c_\square$ is euclidian and that $R^c_\mathsf{N}$ is transitive. This is done in the standard way, relying on (respectively) the axioms $(5_\square )$ and $(4_\mathsf{N})$. Second, show that (1) $R^+_\square$ is euclidian and (2) $R^+_\mathsf{N}$ is transitive.

Proving (1) is straightforward in view of (a″) and (b″) in the definition of $M^+$. For the proof that $R^+_\mathsf{N}$ is transitive, suppose that $(\Delta ^i,\Theta ^j),(\Theta ^j,\Lambda ^k)\in R^+_\mathsf{N}$. Hence $(\Delta ,\Theta ),(\Theta ,\Lambda )\in R^c_\mathsf{N}$ and hence, since $R^c_\mathsf{N}$ is transitive, ($\dagger$) $(\Delta ,\Lambda )\in R^c_\mathsf{N}$. We distinguish the following cases:

• $k = 1$. By (b″) and ($\dagger$), we know at once that $(\Delta ^i, \Lambda ^k)\in R^+_\mathsf{N}$.

• $k = 2$. By (b″), there is a $\psi$ such that $\mathsf{S}\psi \in \Theta$ and $\psi \in \Lambda$. Hence by the construction of $M^c$, $\lnot \mathsf{N}\lnot \mathsf{S}\psi \in \Delta$ and hence, by ($\hbox {Trans}_\mathsf{S}$), $\mathsf{S}\psi \in \Delta$. But then also $(\Delta ^i,\Lambda ^k)\in R^+_\mathsf{N}$.

1.3 Completeness for ${\mathbf{DNS}^\mathsf{u}.\mathbf{U}}$ and ${\mathbf{DNS}^\mathsf{u}.\mathbf{U.D}_\mathsf{N}}$

To prove strong completeness for $\mathbf {\mathbf {DNS}^\mathsf{u}.U}$, we again use the third copy-and-merge technique (see Sect. 1). Starting from a canonical model $M^c$ for $\mathbf {\mathbf {DNS}^\mathsf{u}.U}$. This gives us a new model $M^+$. It can be easily checked that $R^+_\square$ is an equivalence relation.

Note that by a standard argument (relying on the axiom (U)), we can show that ($\star$) if $(\Delta ,\Theta )\in R^c_\mathsf{N}$, then for all $\Lambda \in R^c_\square (\Delta )$, $(\Lambda ,\Theta )\in R^c_\mathsf{N}$.

Suppose now that $(\Delta ^i,\Theta ^j)\in R^+_\mathsf{N}$ and let $\Lambda ^k\in R^+_\square (\Delta ^i)$ be arbitrary. It follows that $(\Delta ,\Theta )\in R^c_\mathsf{N}$ and $(\Delta ,\Lambda )\in R^c_\square$. By ($\star$), $(\Lambda ,\Theta )\in R^c_\mathsf{N}$. If $j = 1$, then $(\Lambda ^k,\Theta ^j)\in R^+_\mathsf{N}$ by the construction. If $j \ne 1$, this means that there is a $\sigma$ such that $\mathsf{S}\sigma \in \Delta$ and $\sigma \in \Theta$. Hence in view of the axiom (U), $\square \mathsf{S}\sigma \in \Delta$ and hence $\mathsf{S}\sigma \in \Lambda$. But then by the construction, $(\Lambda ^k,\Theta ^j) \in R^+_\mathsf{N}$. So we have shown that

($\star \star$):

if $(\Delta ^i,\Theta ^j)\in R^+_\mathsf{N}$, then for all $\Lambda ^k\in R^+_\square (\Delta ^i)$, $(\Lambda ^k,\Theta ^j)\in R^+_\mathsf{N}$.

Next, consider an arbitrary maximal $\mathbf {\mathbf {DNS}^\mathsf{u}.U}$-consistent set $\Xi$. We know that $\Xi \in W$ and hence $\Xi ^1 \in W^+$. Consider now the generated submodel $M^\Xi$ of $M^+$, which consists of the following four elements:

• 1. $W^\Xi = \{\Delta ^i \mid (\Xi ^1,\Delta ^i)\in R^+_\square \}$

• 2. $R^\Xi _\square = R^+_\square \cap (W^\Xi \times W^\Xi ) = W^\Xi \times W^\Xi$

• 3. $R^\Xi _\mathsf{N}= R^+_\mathsf{N}\cap (W^\Xi \times W^\Xi )$

• 4. $v^\Xi (\psi ) = v^+(\psi ) \cap W^\Xi$ for all $\psi \in {\mathcal {S}}$

It can be easily observed that $M^+$ and $M^\Xi$ are pointwise equivalent for all $\Delta ^i\in W^\Xi$, and that $R^\Xi _\square = W^\Xi \times W^\Xi$. Moreover, in view of ($\star \star$), we can derive that $M^\Xi$ is uniform. Hence, we have obtained a $\mathbf {\mathbf {DNS}^\mathsf{u}.U}$-model which verifies all the members of $\Xi$ in at least one world.

The completeness proof for $\mathbf {\mathbf {DNS}^\mathsf{u}.U.D_\mathsf{N}}$ is entirely analogous; one just needs to observe that $R^c_\mathsf{N}$ is serial and hence so is $R^+_\mathsf{N}$ and $R^\Xi _\mathsf{N}$.

Appendix 3: Finite Model Property

For the proof of the finite model property, we combine the standard technique of filtration with the copy-and-merge variants that we used for strong completeness of $\mathbf {DNS}$ and its extensions. Below, we give the outline of the proof for all three types of construction. We start from the supposition that $\not \!\!\Vdash _{\mathbf{DNS}} \varphi$. Hence, there is an $\mathbf{DNS}$-model $M = \langle W, R_\square , R_\mathsf{N},v\rangle$ and $t\in W$ such that $M,t \not \models \varphi$. We then construct a finite model $M^f$ from M, such that also $M^f$ falsifies $\varphi$ at some state.

The three different variants defined below can be used to establish the finite model property for three different groups of extensions of $\mathbf {DNS}$, which in turn correspond to three groups of frame conditions. The three groups of frame conditions are:

• 1. all frame conditions in Table 1

• 2. all frame conditions in Table 2

• 3. uniformity and seriality for $R_\mathsf{N}$.

It then suffices to show that whenever M satisfies one or more conditions within one of these groups, so does $M^f$, when constructed according to the corresponding variant of the copy-and-merge technique. We safely leave this part of the proof to the reader.

Some notation: let $\Sigma$ be the set of all subformulas of $\varphi$.⁴⁴ For all $w\in W$, let $|w| = \{v\in W\mid \text{ for } \text{ all } \psi \in \Sigma : M,w\models \psi \,{ iff }\,M,v\models \psi \}$.

1.1 Filtration Plus Copy-and-Merge Version 1

Let $M^f = \langle W^f,R_\square ^f, R_\mathsf{N}^f,v^f\rangle$, where

• 1. $W^f = \{|w|^1, |w|^2\mid w\in W\}$

• 2. $R_\square ^f = \{( |w|^i,|v|^i)\mid (w,v)\in R_\square \} \cup \{(|w|^i,|v|^j) \mid (w,v)\in R_\square \text{ and } \text{ there } \text{ is } \text{ no } \psi \in \Sigma : M,w\models \mathsf{S}\psi , M,v\models \psi \}$

• 3. $R_\mathsf{N}^f = \{(|w|^i, |v|^i)\mid (w,v)\in R_\mathsf{N}\}$

• 4. For all $\psi \in \Sigma$, $v^f(\psi ) = \{|w|^i, |w|^j\mid M,w\models \psi \}$

• 5. For all $\psi \in \mathcal {S}-\Sigma$, $v^f(\psi ) = W^f$.

Since $\Sigma$ is finite, $W^f$ is also finite (it contains at most $2\times 2^{|\Sigma |}$ nodes). To see that $M^f$ is a $\mathbf{DNS}$-model, suppose that $(|w|^i,|v|^j)\in R_\mathsf{N}^f$. By (ii), $i = j$ and there are $w'\in |w|, v'\in |v|$ with $(w',v')\in R_\mathsf{N}$. Hence, since M is a $\mathbf{DNS}$-model, $(w',v')\in R_\square ^f$. So by (iii) and since $i = j$, $(|w|^i,|v|^j)\in R_\square ^f$. We now prove the following crucial lemma:

Lemma 1

Where $i\in \{1,2\}$, $\psi \in \Sigma$ and $w\in W$: $M,w\models \psi$ iff $M^f, |w|^i \models \psi$.

Proof

We proceed by an induction on the complexity of $\psi$. The base case ($\psi$ is a propositional variable) and the induction step for the classical connectives are safely left to the reader. It remains to prove the induction step for the three modal operators:

Case 1 $\psi = \square \tau$. ($\Rightarrow$) Suppose that $M,w\not \models \square \tau$. So there is a $v\in R_\square (w)$ such that $M,v\not \models \tau$. By the definition of $R_\square ^f$, $|v|^i\in R_\square (|w|^i)$ and by the induction hypothesis, $M^f,|v|^i \not \models \tau$. It follows that $M^f,|w|^i\not \models \square \tau$.

($\Leftarrow$) Suppose that $M^f,|w|^i\not \models \square \tau$. Hence, there is a $|v|^j\in R_\square ^f(|w|^i)$ such that $M^f, |v|^j\not \models \tau$. By the symmetry of the construction, $M^f, |v|^i\not \models \tau$. By the induction hypothesis, $M,v\not \models \tau$. By the definition of $R_\square ^f$, there are $v'\in |v|$ and $w'\in |w|$ such that $v'\in R_\square (w')$. Since $\tau \in \Sigma$, $M,v'\not \models \tau$ and hence $M,w'\not \models \square \tau$. Since $\square \tau \in \Sigma$, $M,w\not \models \square \tau$.

Case 2 $\psi = \mathsf{N}\tau$. Analogous to case 1; just replace every occurrence of $\square$ (also in subscripts) with $\mathsf{N}$.

Case 3 $\psi = \mathsf{S}\tau$. ($\Rightarrow$) Suppose that $M,w\models \mathsf{S}\tau$. Let $|v|^j\in R_\square ^f(|w|^i)$ be arbitrary such that $M^f,|v|^j\models \tau$—we need to prove that $|v|^j\in R_\mathsf{N}^f(|w|^i)$. Note that by the induction hypothesis, $M,v\models \tau$ and hence by the construction $i = j$. By the definition of $R_\square ^f$, there is a $v'\in |v|$, $w'\in |w|$ such that $v'\in R_\square (w')$. Since $w'\in |w|$, $M,w'\models \mathsf{S}\tau$. It follows that $v'\in R_\mathsf{N}(w')$ and hence $(|w'|^i, |v'|^i) = (|w|^i,|v|^i) = (|w|^i,|v|^j)\in R_\mathsf{N}^f$.

($\Leftarrow$) Suppose that $M,w\not \models \mathsf{S}\tau$. Hence, there is a $v\in R_\square (w)\setminus R_\mathsf{N}(w)$ such that $M,v\models \tau$. It follows that, for no $\tau '\in {\mathcal {W}}$, $M,w\models \mathsf{S}\tau '$ and $M,v\models \tau '$. Let $i\ne j$. By the induction hypothesis and the construction of $R_\square ^f$ and $R_\mathsf{N}^f$, $M^f, |v|^j\models \tau$ and $|v|^j \in R_\square ^f(|w|^i) \setminus R_\mathsf{N}^f(|w|^i)$. Hence, $M^f,|w|^i\not \models \mathsf{S}\tau$. $\square$

1.2 Filtration Plus Copy-and-Merge Version 2

Let $M^f = \langle W^f,R_\square ^f, R_\mathsf{N}^f,v^f\rangle$, where

• 1. $W^f = \{|w|^1, |w|^2\mid w\in W\}$

• 2. $R_\square ^f = \{( |w|^i,|v|^j)\mid (w,v)\in R_\square \}$

• 3. $R_\mathsf{N}^f = \{(|w|^i, |v|^j)\mid (w,v)\in R_\mathsf{N} \text{ and } i=j \text{ or } \text{ there } \text{ is } \text{ a } \psi \in \Sigma : M,w\models \mathsf{S}\psi , M,v\models \psi \}$

• 4. For all $\psi \in \Sigma$, $v^f(\psi ) = \{|w|^i, |w|^j\mid M,w\models \psi \}$

• 5. For all $\psi \in \mathcal {S}-\Sigma$, $v^f(\psi ) = W^f$.

Again, it is easily verified that $M^f$ is a $\mathbf {DNS}$-model. So we are left proving:

Lemma 2

Where $i\in \{1,2\}$, $\psi \in \Sigma$ and $w\in W$: $M,w\models \psi$ iff $M^f, |w|^i \models \psi$.

Proof

Analogous to the proof for Lemma 1, except for the case where $\psi = \mathsf{S}\tau$:

($\Rightarrow$) Suppose that $M,w\models \mathsf{S}\tau$. Let $|v|^j\in R_\square ^f(|w|^i)$ be arbitrary such that $M^f,|v|^j\models \tau$—we need to prove that $|v|^j\in R_\mathsf{N}^f(|w|^i)$. Note that by the induction hypothesis, $M,u\models \tau$ for all $u\in |v|$. Also, by the definition of $R_\square ^f$, there is a $v'\in |v|$, $w'\in |w|$ such that $v'\in R_\square (w')$. Since $w'\in |w|$, $M,w'\models \mathsf{S}\tau$. It follows that ($\dagger$) $v'\in R_\mathsf{N}(w')$. We now distinguish two cases:

• 3.1 $j = i$. By ($\dagger$) and the definition of $R_\mathsf{N}^f$, $|v|^j\in R_\mathsf{N}(|w|^i)$.

• 3.2 $j\ne i$. By ($\dagger$) it suffices to note that, for all $v'\in |v|$ and all $w'\in |w|$, $M,v'\models \tau$ and $M,w'\models \mathsf{S}\tau$.

($\Leftarrow$) Suppose that $M,w\not \models \mathsf{S}\tau$. Hence, there is a $v\in R_\square (w)\setminus R_\mathsf{N}(w)$ such that $M,v\models \tau$. By the induction hypothesis and the construction of $R_\square ^f$ and $R_\mathsf{N}^f$, $M^f, |v|^j\models \tau$ and $|v|^j \in R_\square ^f(|w|^i) \setminus R_\mathsf{N}^f(|w|^i)$. Hence, $M^f,|w|^i\not \models \mathsf{S}\tau$. $\square$

1.3 Filtration Plus Copy-and-Merge Version 3

Let $M^f = \langle W^f,R_\square ^f, R_\mathsf{N}^f,v^f\rangle$, where

• 1. $W^f = \{|w|^1, |w|^2\mid w\in W\}$

• 2. $R_\square ^f = \{( |w|^i,|v|^j)\mid (w,v)\in R_\square \}$

• 3. $R_\mathsf{N}^f = \{(|w|^i, |v|^j)\mid (w,v)\in R_\mathsf{N} \text{ and } j=1 \text{ or } \text{ there } \text{ is } \text{ a } \psi \in \Sigma : M,w\models \mathsf{S}\psi , M,v\models \psi \}$

• 4. For all $\psi \in \Sigma$, $v^f(\psi ) = \{|w|^i, |w|^j\mid M,w\models \psi \}$

• 5. For all $\psi \in \mathcal {S}-\Sigma$, $v^f(\psi ) = W^f$.

We leave the proof of the following to the reader (it proceeds wholly analogously to the proof of Lemma 2):

Lemma 3

Where $i\in \{1,2\}$, $\psi \in \Sigma$ and $w\in W$: $M,w\models \psi$ iff $M^f, |w|^i \models \psi$.

Appendix 4: Proof of Theorem 6

We will first prove the theorem for the base case where $\mathbf {\mathbf {DNS}^+} = \mathbf {DNS}$. So suppose that $\mathsf{S}$ does not occur in $\Gamma$ or in $\varphi _1,\ldots , \varphi _n$. We need to prove:

$$ \Gamma\;\Vdash _{\mathbf{DNS}} \mathsf{S}\varphi _1\vee \cdots \vee \mathsf{S}\varphi _n\,{ iff }\,\Gamma\;\Vdash _{\mathbf{DNS}} \square \lnot \varphi _1\vee \cdots \vee \square \lnot \varphi _n $$

The right to left direction is easy, in view of the fact that $\square \lnot \varphi \supset \mathsf{S}\varphi$ is a theorem in $\mathbf{DNS}$. For the other direction, we will need some more work.

Suppose that $\Gamma \not \!\!\Vdash _{\mathbf{DNS}} \square \lnot \varphi _1\vee \cdots \vee \square \lnot \varphi _n$. Let $M = \langle W,R_\square , R_\mathsf{N},v\rangle$ and $w_0\in W$ be such that $M,w_0 \models \psi$ for all $\psi \in \Gamma$ and $M,w_0\models \lnot \square \lnot \varphi _1, \ldots ,\lnot \square \lnot \varphi _n$. We construct $M' = \langle W',R_\square ',R_\mathsf{N}',v'\rangle$ as follows:

• 1. $W' = \{w^1,w^2\mid w\in W\}$

• 2. $R_\square ' = \{(w^i,v^j)\mid i,j\in \{1,2\}, (w,v)\in R_\square \}$

• 3. $R_\mathsf{N}' = \{(w^i,v^i)\mid (w,v)\in R_\mathsf{N}\}$

• 4. $v'(\psi ) = \{w^1,w^2\mid w\in v(\psi )\}$ for all $\psi \in \mathcal {S}$

We need to show that (a) for all $\psi$ that do not contain $\mathsf{S}$ and all $w\in W$, $M,w\models \psi$ iff $M',w^i\models \psi$ and (b) for all $\psi \in {\mathcal {W}}_{\overline{\mathsf{S}}}$ and $w^i\in W'$ such that $M',w^i \models \lnot \square \lnot \psi$, $M',w^i\not \models \mathsf{S}\psi$. By (a) and the supposition, $M',w_0^1\models \psi$ for all $\psi \in \Gamma$. By (b), $M',w_0^1\not \models \mathsf{S}\varphi _1 \vee \cdots \vee \mathsf{S}\varphi _n$. Hence, $\Gamma \not \!\!\Vdash _{\mathbf{DNS}} \mathsf{S}\varphi _1 \vee \cdots \vee \mathsf{S}\varphi _n$.

Ad (a) This is shown by an induction on the complexity of $\psi$. The base case ($\psi \in \mathcal {S}$) is trivial in view of (iv). So is the inductive step for the classical connectives. For $\psi = \square \tau$, it suffices to observe that $v\in R_\square (w)$ iff $v^1,v^2\in R_\square '(w^i)$, and hence, by the induction hypothesis, $M,w\models \square \tau$ iff $M',w^i\models \square \tau$. For $\psi = \mathsf{N}\tau$, it suffices to observe that $v\in R_\mathsf{N}(w)$ iff $v^i\in R_\mathsf{N}(w^i)$ and hence, by the induction hypothesis, $M,w\models \mathsf{N}\tau$ iff $M',w^i\models \mathsf{N}\tau$.

Ad (b) Suppose that $M',w^i \models \lnot \square \lnot \psi$ where $\psi$ contains no occurrences of $\mathsf{S}$. Hence, there is a $v^j\in R'_\square (w^i)$ such that $M',v^j\models \psi$. Let $k \ne i$. By the construction, $v^k \in R_\square '(w^i) \setminus R_\mathsf{N}'(w^i)$. By (a), $M, v \models \psi$ and hence $M',v^k\models \psi$. Hence, $M,w^i\not \models \mathsf{S}\psi$.

The generalization of this proof to obtain a proof for Theorem 6 is straightforward: we just need to note that whenever a condition on $R_\mathsf{N}$ ($R_\square$) is preserved by copy (preserved by copy-merge), and whenever it holds for the model M of $\Gamma \cup \{\lnot \square \lnot \varphi _1, \ldots , \lnot \square \lnot \varphi _n\}$, then it will also hold for the model $M'$ we constructed. But this is just what Definitions 4 and 5 tell us.

Note that in the above construction, $R_\mathsf{N}$ is not uniform. Hence, proving the same theorem for $\mathbf {\mathbf {DNS}^\mathsf{u}.U}$ and $\mathbf {\mathbf {DNS}^\mathsf{u}.U.D_\mathsf{N}}$ requires a slightly different construction. Here we define $M' = \langle W',R_\square ',R_\mathsf{N}',v'\rangle$ as follows:

• 1. $W' = \{w^1,w^2\mid w\in W\}$

• 2. $R_\square ' = \{(w^i,v^j)\mid i,j\in \{1,2\}, (w,v)\in R_\square \}$ $(= W' \times W')$

• 3. $R_\mathsf{N}' = \{(w^i,v^1)\mid (w,v)\in R_\mathsf{N}\}$

• 4. $v'(\psi ) = \{w^1,w^2\mid w\in v(\psi )\}$ for all $\psi \in \mathcal {S}$

Note that this construction is not symmetric. We can now prove each of (a) and (b) as before, with two minor differences. For (a), the case $\psi = \mathsf{N}\tau$, one needs to observe that $v\in R_\mathsf{N}(w)$ iff $v^1\in R_\mathsf{N}(w^i)$. For (b), we need to take $k = 2$ instead of $k\ne i$.

Appendix 5: Reducing $\mathbf{DNS}$ to $\mathbf {K_d}$

The left-right direction of the reduction theorem is straightforward: it suffices to check that for every $\mathbf{DNS}$-axiom $\varphi$, $t(\varphi )$ is a $\mathbf {K}$-axiom. We safely leave this to the reader.

For the other direction, a more elaborate argument is needed. Suppose that $\Gamma \not \!\!\Vdash _{\mathbf{DNS}} \varphi$. Hence, there is an $\mathbf{DNS}$-model $M = \langle W,R_\square ,R_\mathsf{N},v\rangle$ and $w_0\in W$ such that $M,w_0\models \Gamma$ and $M,w_0\not \models \varphi$. The proof now proceeds in two steps.

First, we unravel the model M around the node $w_0$, obtaining a model $M' = \langle W',R'_\square , R_\mathsf{N}', v'\rangle$, where

• 1. $W' = \{\langle w_0,\ldots ,w_n\rangle \mid (w_0,w_1), \ldots , (w_{n-1},w_n)\in R_\square \}$

• 2. $R_\square ' = \{(\langle w_0,\ldots ,w_i\rangle , \langle w_0,\ldots ,w_i,w_{i+1}\rangle ) \mid (w_i,w_{i+1})\in R_\square \}$

• 3. $R_\mathsf{N}' = \{(\langle w_0,\ldots ,w_i\rangle , \langle w_0,\ldots ,w_i,w_{i+1}\rangle ) \mid (w_i,w_{i+1})\in R_\mathsf{N}\}$

• 4. $v'(\psi ) = \{\langle w_0, \ldots , w_n\rangle \mid w_n\in v(\psi )$

By standard means, we can prove that (a.1) for all $w_n\in W$ and all $\psi$, $M',\langle w_0, \ldots , w_n\rangle \models \psi$ iff $M,w_n\models \psi$; and (a.2) for all $x\in W'$, there is at most one $y\in W'$ with $x\in R_\square (y)$.⁴⁵

In the second step of the proof, we transform $M'$ into a $\mathbf {K_d}$-model $M'' = \langle W'',R_\square '',v\rangle$, as follows: $W'' = W'$, $R_\square '' = R_\square '$, for all $\psi \in \mathcal {S}$, $v''(\psi ) = v'(\psi )$, and $v(d) = \{x \in W' \mid \text{ there } \text{ is } \text{ a } y\in W': x \in R_\mathsf{N}'(y)\}$. By induction on the complexity of $\psi$, we can show that for all $\psi$ and all $x\in W'$, $M',x \models \psi$ iff $M'',x\models t(\psi )$. In view of (a.1), $M'',\langle w_0\rangle$ verifies all members of $t(\Gamma \cup \{\lnot \varphi \})$ whence we are done.

The unraveling is a necessary step in this proof—even if for the standard proof of the Andersonian reduction as spelled out e.g. in Åqvist (2002), this is not needed. The reason for this complication is that the present theorem concerns the entire language of $\mathbf{DNS}$, which includes the operator $\square$. As a result, we cannot just define the accessibility relation of our $\mathbf {K_d}$-model ad libitum: we have to take it over from the original $\mathbf{DNS}$-model. This in turn makes it impossible to simply make d true at all worlds w that are deontically acceptable from some world. Instead, we first make sure by the unraveling that whenever a world w is deontically acceptable for a world, then it is deontically acceptable for exactly one such world. Only then do we apply the usual trick, viz. making d true in those worlds that are acceptable for their (unique) predecessor.

As promised, we also briefly outline the proof for the reduction of $\mathbf {\mathbf {DNS}^\mathsf{u}.U}$ to $\mathbf {S5_d}$. The translation proceeds in the same way as before. We leave it to the reader to check that, under this translation, all $\mathbf {\mathbf {DNS}^\mathsf{u}.U}$-axioms are valid in $\mathbf {S5_d}$.

For the other direction, suppose that $\Gamma\not\Vdash_{\mathbf{DNS^{\sf u}.U}} \varphi$. Let $M = \langle W,R_\square ,R_\mathsf{N},v\rangle$ be an $\mathbf {\mathbf {DNS}^\mathsf{u}.U}$-model and $w\in W$, such that $M,w\models \psi$ for all $\psi \in \Gamma$ and $M,w\not \models \varphi$. Construct the $\mathbf {S5_d}$-model $M' = \langle W,R_\square , v'\rangle$, putting $v'(\tau ) = v(\tau )$ for all $\tau \in {\mathcal {S}}$ and $v'(\mathsf{d}) = \{w'\in W \mid R_\mathsf{N}(w,w')\}$. Then prove by induction that, for every $\tau$ and every $u\in W$, $M,u\models \tau$ iff $M',u \models t(\tau )$. The only difficult cases are $\tau = \mathsf{N}\psi$ and $\tau = \mathsf{S}\psi$; for these, we rely on the fact that $R_\mathsf{N}(u) = R_\mathsf{N}(w)$ for all $u\in W$.

For the reduction of $\mathbf {\mathbf {DNS}^\mathsf{u}.U.D_\mathsf{N}}$ to $\mathbf {S5_d} + \{\Diamond \mathsf{d}\}$, we can run the same argument. Observe that, since $R_\mathsf{N}$ is serial in M, $v'(\mathsf{d}) \ne \emptyset$.

Appendix 6: No Interpolation for $\mathbf {DNS.T_\square }$

As we argued in the main text, $\mathsf{S}p\wedge p\, \Vdash _{\mathbf {DNS.T_\square }} \mathsf{N}q \supset q$. Assume now that there is an interpolant for $\langle \mathsf{S}p\wedge p, \mathsf{N}q\supset q\rangle$—let us call it $\varphi$. Note that $\varphi$ contains no propositional variables. We prove two lemmas about every such $\varphi$:

Lemma 4

For every $\mathbf {DNS}$ -frame $F = \langle W,R_\square ,R_{\mathsf{N}}\rangle$, for all $w\in W$, and for all valuations $v,v': {\mathcal {S}} \rightarrow \wp (W)$: $\langle F,v\rangle ,w \models \varphi$ iff $\langle F,v'\rangle , w\models \varphi$.

Proof

By an induction on the complexity of $\varphi$. Note that the base case is $\varphi = \bot$. For the induction step, we simply rely on the induction hypothesis and the semantic clauses for the various connectives and modal operators. $\square$

Lemma 5

Where $F = \langle W,R_\square ,R_\mathsf{N}\rangle$, $w\in W$, and $v: {\mathcal {S}}\rightarrow \wp (W)$: $\langle F,v\rangle ,w \models \varphi$ iff $w\in R_{\mathsf{N}}(w)$.

Proof

($\Rightarrow$) Suppose that $w\not \in R_{\mathsf{N}}(w)$. Let $v': {\mathcal {S}}\rightarrow \wp (W)$ be such that $v'(q) = R_{\mathsf{N}}(w)$. It follows that $\langle F,v'\rangle ,w \models \mathsf{N}q \wedge \lnot q$ and hence $\langle F,v'\rangle ,w \models \lnot (\mathsf{N}q\supset q)$. Since $\varphi$ is an interpolant for $\langle \mathsf{S}p\wedge p, \mathsf{N}q\supset q\rangle$, this means that $\langle F,v'\rangle ,w \not \models \varphi$. Hence by Lemma 4, $\langle F,v\rangle ,w \not \models \varphi$.

($\Leftarrow$) Suppose that $w\in R_\mathsf{N}(w)$. Let $v': {\mathcal {S}}\rightarrow \wp (W)$ be such that $v'(p) = w$. It follows that $\langle F,v'\rangle ,w \models \mathsf{S}p \wedge p$. Since $\varphi$ is an interpolant for $\langle \mathsf{S}p\wedge p, \mathsf{N}q\supset q\rangle$, this means that $\langle F,v'\rangle ,w \models \varphi$. By Lemma 4, $\langle F,v\rangle ,w\models \varphi$. $\square$

Consider now two models $M^1 = \langle W^1,R^1_\square ,R^1_\mathsf{N},v^1\rangle$ and $M^2= \langle W^2,R^2_\square ,R^2_\mathsf{N},v^2\rangle$, which are defined as follows:

• 1. where $i\in \{1,2\}$: $W^i = \{w,w'\}$ and $R_\square ^i = W^i \times W^i$

• 2. where $i\in \{1,2\}$ and $\varphi \in {\mathcal {S}}$: $v^i(\varphi ) = W^i$

• 3. $R^1_\mathsf{N}= \{(w,w),(w',w')\}$

• 4. $R^2_\mathsf{N}= \{(w,w'),(w',w)\}$

Note that these models only differ in one respect, viz. the relation $R_\mathsf{N}$. In $M^1$ this relation is reflexive; in $M^2$ it is not. We can now prove the following (by induction on the complexity of $\psi$):

Lemma 6

For all $\psi$ and $u\in \{w,w'\}$: $M^1,u \models \psi$ iff $M^2,u\models \psi$.

But this means that, in particular, $M^1,w \models \varphi$ iff $M^2,w\models \varphi$. Since $R^1_\mathsf{N}$ is reflexive, we can derive by Lemma 5 that $M^1,w\models \varphi$. But then also $M^2,w\models \varphi$, which contradicts the fact that $R^2_\mathsf{N}$ is not reflexive and Lemma 5.

The proof for the failure of interpolation for $\mathbf {DNS.SR_\square }$ proceeds in a similar way; the only real difference concerns Lemma 5. The idea here is that $\varphi$ expresses exactly that there is a $w'\in R_\square (w)$ such that $w'\in R_{\mathsf{N}}(w')$. In the proof of the lemma, we define $v'(q) = \bigcup _{w'\in R_\square (w)} R_{\mathsf{N}}(w')$ for the left to right direction, and $v'(p) = \{w'\}$ for the right to left direction. The construction of $M^1$ and $M^2$ can just as well be used for this case, since $R^1_{\mathsf{N}}$ is shift reflexive and $R^2_{\mathsf{N}}$ is not.

Appendix 7: Proof of Theorem 8

We will first give the outline of our proof for the base logic $\mathbf {DNS_{O}}$, which proceeds along the same lines as the soundness and completeness proof for $\mathbf {DNS}$. One direction (soundness of $\mathbf {DNS_{O}}$ with respect to $\mathbf {DNS}$) is easy and has been discussed in the main text. For the other direction (completeness w.r.t. the $O/P/\square$-fragment of $\mathbf {DNS}$), we rely on the soundness and completeness of $\mathbf {DNS}$, so that it suffices to show that every $\mathbf {DNS_{O}}$-consistent set $\Gamma$ is satisfiable at a point in a $\mathbf {DNS}$-model. In order to arrive there, we use once more the copy-and-merge technique, adapting it to present needs.

Let $M= \langle W,R_\square ,R_{\mathsf{N}},v\rangle$, where

• 1. W is the set of all maximal consistent (w.r.t. $\mathbf {DNS_{O}}$) subsets of ${\mathcal {W}}_{O}$.

• 2. $(\Delta ,\Theta )\in R_\square$ iff $\{\varphi \mid \square \varphi \in \Delta \}\subseteq \Theta$

• 3. $(\Delta ,\Theta ) \in R_{\mathsf{N}}$ iff $(\Delta ,\Theta )\in R_\square$ and there is a $\varphi$ such that $P\varphi \in \Delta$, $\varphi \in \Theta$

• 4. $v(p) = \{\Delta \in W\mid p\in \Delta \}$ for all $p\in {\mathcal {S}}$

Let $M = \langle W^+,R_\square ^+,R_{\mathsf{N}}^+,v^+\rangle$ where

• 1′. $W^+ = \{\Delta ^1,\Delta ^2\mid \Delta \in W\}$

• 2′. $R_\square ^+ = \{(\Delta ^i,\Theta ^j)\mid (\Delta ,\Theta )\in R_\square , i,j\in \{1,2\}\}$

• 3′. $R_{\mathsf{N}}^+ = \{(\Delta ^i,\Theta ^j)\mid (\Delta ,\Theta )\in R_{\mathsf{N}} \mid i=j \text{ or } \{\varphi \mid O\varphi \in \Delta \}\subseteq \Theta \}$

• 4′. $v^+(p) = \{\Delta ^1,\Delta ^2\mid \Delta \in v(p)\}$ for all $p\in {\mathcal {S}}$

The first step in the proof is to check that $M^+$ is a $\mathbf {DNS}$-model. It suffices to check that $R^+_{\mathsf{N}} \subseteq R^+_\square$, which is immediate in view of the construction.

Next, we need to establish the following version of the truth lemma:

Lemma 7

For all $\varphi \in {\mathcal {W}}_{O}$ and where $i\in \{1,2\}$: $M^+, \Delta ^i\models \varphi$ iff $\varphi \in \Delta ^i$.

Proof

The base case and the induction step for the connectives and $\square$ are a matter of routine—we safely leave this to the reader. So we are left with two cases:

Case 1: $\varphi = P\psi = \mathsf{S}\psi$. ($\Rightarrow$) Here, we can follow exactly the same reasoning as in the proof of Theorem 9, left-right direction of Case 3. ($\Leftarrow$) Suppose that $\mathsf{S}\varphi \in \Delta$. Let $\Theta ^j\in R^+_\square (\Delta ^i)$ be such that $M^+,\Theta ^j \models \varphi$. By the induction hypothesis, $\varphi \in \Theta$. Hence by items (iii) and (iii’) of the construction, $\Theta ^j\in R^+_\mathsf{N}(\Delta ^i)$. Hence, $M^+,\Delta ^i\models \mathsf{S}\varphi$.

Case 2: $\varphi = O\psi = \mathsf{N}\psi \wedge \mathsf{S}\psi$. ($\Rightarrow$) Suppose that $O\psi \not \in \Delta$. If $P\psi \not \in \Delta$, then we can infer at once (relying on case 1 of the present proof) that $M^+,\Delta ^i\not \models P\psi$ and hence $M^+,\Delta ^i\not \models O\psi$. So suppose moreover that $P\psi \in \Delta$. It follows that $\square \psi \not \in \Delta$—otherwise, we can use (EQ$_P$), (Taut-Perm) and (EQ$_O$) to derive that $O\psi \in \Delta$, contradicting our initial supposition.⁴⁶ Hence, $\Diamond \lnot \psi \in \Delta$.

We now distinguish two cases:

• (a) there is no $\tau$ such that $O\tau \in \Delta$. Let $\Theta \in R_\square (\Delta )$ be such that $\psi \not \in \Theta$.⁴⁷ By item (iii’) of the construction and $\Theta ^i \in R^+_\mathsf{N}(\Delta ^i)$. Hence, $M^+,\Delta ^i\not \models \mathsf{N}\psi$ and hence $M^+,\Delta ^i\not \models O\psi$.

• (b) there is a $\tau$ such that $O\tau \in \Delta$. By axiom (Weakest-Perm), $\square (\psi \supset \tau )\in \Delta$, but by (EQ$_O$) and since $O\psi \not \in \Delta$, $\square (\psi \equiv \tau )\not \in \Delta$. Hence, $\Diamond (\tau \wedge \lnot \psi ) \in \Delta$. We can infer that there is a $\Theta \in R_\square (\Delta )$ such that $\tau \in \Theta$, $\psi \not \in \Theta$. Note that for all $\tau '$ such that $O\tau '\in \Delta$, also $\square (\tau '\equiv \tau )\in \Delta$, and hence $\tau '\in \Theta$. In view of item (iii’) of the construction, $\Theta ^i \in R_\mathsf{N}^+(\Delta ^i)$. Hence, $M^+,\Delta ^i\not \models \mathsf{N}\psi$ and hence $M^+,\Delta ^i\not \models O\psi$.

($\Leftarrow$) Suppose that $O\psi \in \Delta$. Hence, $\mathsf{N}\psi , \mathsf{S}\psi \in \Delta$ and hence by case 1 of the present proof, $M^+,\Delta ^i \models \mathsf{S}\psi$. Also, by axiom (Weakest-Perm), for all $\tau$, $S\tau \supset \square (\tau \supset \varphi ) \in \Delta$. Assume now that $M^+\not \models \mathsf{N}\psi$. Hence by the induction hypothesis, there is a $\Theta ^j\in R^+_\mathsf{N}(\Delta ^i)$ such that $\psi \not \in \Theta$. In view of item (iii’) of the construction, this can only mean two things:

• (c) There is a $\tau$ such that $\mathsf{S}\tau \in \Delta$, $\tau \in \Theta$. It follows that $\square (\tau \supset \psi )\in \Delta$ and hence $\tau \supset \psi \in \Theta$ and hence $\psi \in \Theta$—contradiction.

• (d) For all $\tau$ such that $O\tau \in \Delta$, $\tau \in \Theta$. But then $\psi \in \Theta$—contradiction again.

So we have shown that $M^+,\Delta ^i\models \mathsf{N}\psi$ and hence $M^+,\Delta ^i\models O\psi$. $\square$

The extension of the above proof to cover the additional frame conditions, resp. axioms, is straightforward. For completeness, it suffices to check that whenever the axioms are added, the resulting canonical model will be one that satisfies the associated frame conditions. For soundness, it suffices to check that the axioms are valid whenever the conditions are in place.